SCE3337 Quantum Mechanics III (Quantum Mechanics and Quantum Optics) Teaching Team

SCE3337 Quantum Mechanics III 1 SCE3337 Quantum Mechanics III (Quantum Mechanics and Quantum Optics) Teaching Team Dr Robert Sang Robert Sang can be found in Science II Level Room.15, non-personal contact is possible via email at the following address: R.Sang@sct.gu.edu.au or via the telephone on 3875 3848. Lecture notes/problem Sheets can be found on the web page: http://www.sct.gu.edu.au/~sctsang It is your responsibility to ensure that you are up-to-date with them and should be downloaded prior to each lecture. 1. Introduction This course has been divided into two parts; The first part deals with the introduction of operators and techniques that are applied on a regular basis in dealing with problems that require the use of Quantum Mechanics these include: The application of New Notations (Bra/Ket) Operator Techniques Perturbation Theory Selection Rules for Single Photon Dipole Allowed Transitions The second half of the course deals with Quantum Optics which includes The interactions of light with matter Optical Processes such as Absorption, Spontaneous/Stimulated Emission Lasers The two level atom treatment of Strong Interactions between laser light and atoms Course Structure, Tutorials and Examinations There will be eight lectures that deal with the development of the tools of Quantum Mechanics which will also involve three tutorials. There will be a mid-semester 1 hour examination which has a weighting of 5% of the total marks for the course. Following the exam will be a further nine lectures in quantum optics which will also have two tutorials. There will be a final examination at the conclusion of the course which will cover all aspects of the course. Useful Texts Modern Physics and Quantum Mechanics- Anderson Quantum Mechanics- Cohen-Tannoudi, Diu, Laloë The Quantum Theory of Light- Loudon Optical Resonance and Two Level Atoms-Allen and Eberly

SCE3337 Quantum Mechanics III. Basics of One Particle Wave Function Space The quantum state of a particle is defined at any given instant by a wavefunction ψ(r,t). Recall from QMII that the wave function of a particle ψ(r,t): ψ(r,t) d 3 r represents the probability of finding, at time t, the particle in volume d 3 r =dxdydz about the point r. Therefore the total probability of finding the particle over all space is equal to 1, mathematically this is expressed as: (r,t) d 3 r = 1 where the integration extends over all space. The above integral must converge (the sum of the probabilities must yield 1) and this type of integral is called a square integrable function. Mathematicians call this set of functions L. In QM this set of functions is to wide in scope since ψ(r,t) has an actual physical meaning therefore we can only keep functions ψ(r,t) which are everywhere defined (we can t have particles disappearing!), continuous and differentiable. It is also possible to define wavefunctions that have a bounded domain, this allows us to be certain that a particle can be found in a finite region of space. For example we can define a space inside the laboratory in which our particle can be found. We shall call this group of functions that satisfy our conditions the F set of wavefunctions which is a subset of the L set. This statistical interpretation of Quantum Mechanics is due to Born, Heisenberg and Bohr..1 Structure of the Wave Function Space F a) F as a vector Space It is easy to show that F satisfies all the criteria of a vector space, for example if ψ 1 (r) and ψ (r) exists in F then the sum of the two vectors should yield another vector that belongs to the vector space F: (r) = 1 1 (r)+ (r) where λ 1 and λ are two arbitrary complex numbers. Now in order to show that ψ(r) belongs to the wavefunction space F then we need to demonstrate that it is square integrable. Squaring ψ(r) gives (r) = 1 (r) + (r) + 1 * (r)* (r)+ 1 * (r) (r)* The last two terms of this expression have the same modulus which has the upper limit when ψ 1 (r)= ψ (r) then the final terms equal 1 (r) + (r) ] ψ(r) is therefore equal to or smaller than the maximum function whose integral must converges since ψ 1 (r) and ψ (r) are square integrable then ψ(r) must also be square integrable and hence exists in the vector space F.

SCE3337 Quantum Mechanics III 3 b) The Scalar Product The scalar product is defined as (, ) = *(r) (r) d 3 r The integral always converges if ϕ(r) and ψ(r) belong to F. The scalar product has the following properties ( )=( )* ( 1 + )= 1 )+ ) ( 1 + )= 1 * )+ * ) The scalar product is said to be linear in the second relationship and anti-linear with respect to the third relationship. If (ϕ,ψ)= then the two functions are said to be orthogonal. (, ) = (r,t) d 3 r This number is always real and positive and can only equal zero if ψ(r) =. norm of ψ(r). (ψ,ψ) is the c) Linear Operators An operator is something that transforms a wavefunction to a new wavefunction. If A is an operator it is said to be linear if it produces the following transformation: (r)'=a (r) A (r) = A 1 (r)+ (r)] = A 1 (r)+ (r) Examples of Linear Operators The X operator; multiplies a function by x: Xψ(x,y,z)=xψ(x,y,z) The differential operator D x which differentiates wrt x: D x ψ(x,y,z) = ψ(x,y,z) x

SCE3337 Quantum Mechanics III 4 d) Products of Operators Let A and B be two linear operators. The product AB is defined as (AB) (r) = A[B (r)] We say that B operates on ψ(r) which produces a new function ϕ(r)=bψ(r), A then operates on the new function ϕ(r). In general AB BA i.e. they do not commute. The commutator of the operators A and B is written as [A,B] and has the following definition [ A, B] = AB- BA If [A,B] = then the operators are said to commute. Example: [X, D x ], we shall use the arbitrary function ψ(r). [X, D x ] ψ(r) = XD x ψ(r) - D x X ψ(r) = x ψ(r) x x [xψ(r)] = x ψ(r) {( x x x)ψ(r) + x x ψ(r)] = -1. ψ(r) Hence [X, D x ] = -1. Since the commutator is non-zero X does not commute with D x. Exercise 1: Evaluate the commutation [D x, X ] where the operator X =x. Is it equal to [X, D x ]?

SCE3337 Quantum Mechanics III 5 e) Discrete Orthonormal Bases in F Consider a countable set of functions in F which are labelled by a discrete index i (i=1,,3,..,n,..): u 1 (r), u (r),... u i (r),.. which all exist in F. The set of functions {u i (r)} are said to be orthonormal if ( u i (r), u (r)) = u i (r)* u (r)d 3 r = i where δ i is the Kronecker delta function which takes the following values δ i = 1 when i= δ i = when i This set of functions is said to constitute a basis if every function (r) that exists in F can be expanded in one and only one way in terms of u i (r): (r) = c i u i (r) i The components of the wave function in the u i (r) basis my be found by multiplying both sides of this equation by u (r) and integrating over all space. Proof: (u (r), (r)) = (u (r), c i u i (r)) i = c i (u (r),u i (r)) i = c i u (r)* u i (r)d 3 r (applying orthonormality) i = c i i i = c c = (u (r), (r)) = u i (r)* (r)d 3 r The coefficient c of the wave function ψ(r) on u (r) is equal to the scalar product of wave function ψ(r) with u (r). Once the basis set {u i (r)} is chosen, it is completely equivalent to express the wave function ψ(r) in terms of the c i coefficients with respect to their basis functions. The set of numbers c is said to represent (r) in the {u i (r)} basis.

SCE3337 Quantum Mechanics III 6 f) Expression for the Scalar Product in Terms of the Components Let ϕ(r) and ψ(r) be two wave functions which have the following expansions in the basis {u i (r)}: (r) = b i u i (r) i (r) = c u (r) The scalar product can be calculated such that Therefore (, ) = ( b i u i (r), c u (r)) i = b i * c (u i (r),u (r)) i, = b i * c i i, = b i * c i i (, ) = c i i The scalar product of two wave functions can be expressed in terms of the components of the functions in the basis {u i (r)}.

SCE3337 Quantum Mechanics III 7 3. Dirac Notation and the Postulates of Quantum Mechanics We now look at using a new notation which is simpler to write down that the notation that has been used until present (saves us having to write all the messy integrals in QM). This new notation is called Dirac Notation as it was first introduced by Dirac. We will assume that each quantum state can be represented by a state vector which belongs to an abstract space we will call E which is called the state space of the particle. E is a subspace of Hilbert Space. Postulate 1: The quantum state of any physical system is characterised by a state vector, belonging to a space E which is that state space of the system. 3.1 Dirac Notation Any element or vector of E space is called ket therefore by postulate 1 we can define any wavefunction which describes a state α by a ket: ψ α α> This is a ket vector The complex conugate defines the bra vector ψ α * <α This is a bra vector For every ket there is a bra and vice versa. We will now consider some more of the postulates of QM using this notation. Postulate : For every physical observable the is an associated operator Q such that Q α> = q α α> This is know as an eigenvalue equation where q α is called the eigenvalue which in general will be complex and α> is known as an eigenfunction. Note that operators act on kets from the left and bras from the right: <α Q = q α *<α Hilbert Space you may recall from second year is an infinite dimensional complex space. A Hilbert space is a vector space in which there is a well defined scalar product and a norm in terms of this scalar product and the vector space is complete and all function defined within this space must be square integrable.

SCE3337 Quantum Mechanics III 8 Postulate 3: The set of eigenfunctions { i>} associated with an operator Q form a complete orthogonal set of wavefunctions and satisfy the following eigenvalue equation; Q i> = q i i> Recall the orthogonality condition for normalised wavefunctions from lecture 1 ψ i * ψ dτ = δ i In bra/ket notation this is expressed as ψ i * ψ dτ = <i > = δ i Another way to think of this is that there is no overlap of the wavefunctions in all space hence the dot product must yield zero. Therefore no eigenfunction depends on any either eigenfunction for its definition and as such is independent. Postulate 4: The complete orthonormal set of eigenfunctions { >} can be used to represent or expand any general wavefunction of the same Hilbert subspace: ψ> = N c =1 > and <ψ = c i * < i N i =1 The expansion coefficients are in general complex. Note that N does not have to be finite as there can also be an infinite number i.e. N of eigenfunctions to provide a full and complete description of an interaction process (eg: electron-atom collisions) although it is often that only a few of these eigenfunctions are necessary for a accurate calculation. It should also be noted that the expansion may require an integration over continuous states, as in the case where the interaction takes place in continuum. This case can occur for an electron-atom collision in which the atom is ionised after the interaction and hence the wavefunction describing the system has to take into account that the incident, scattered and ionised electrons are free of a constraining potential of the atom. For a general normalised wavefunction we may write: <ψ ψ> = 1 = c i *c < i >= c i *c i, i, = c i *c i = c i = 1 i i δ i As we would expect for a normalised wavefunction the sum of the coefficients must be unity since the probability must be conserved.

SCE3337 Quantum Mechanics III 9 Postulate 5: The expectation of an operator is described in this notation by <Q> = < Q >. <ψ Q ψ> = * Q d 3 r * d 3 r = ψ *Qψd 3 r for normalised wavefunctions The expectation value of an operator corresponds to the average of that operator over all space and time. For example if the operator was X, the position operator, then the expectation value of this operator corresponds to the average value of position of a particle over all space and time. Now from postulate 4 we can rewrite ψ> as ψ> = c > Q ψ> = Q c > = c Q > using postulate reveals Q > = q > = c q > <ψ Q ψ> = ( c i * < i ) c q > i = c i *c q < i > = i = c i i q i i c i *c q δ i This equation yields a weighted average of the possible eigenvalues of Q. The probability of a single measurement yielding q i is therefore given by c i. The probability of finding a system in the state i> is c i. Postulate 6: The adoint of an operator Q is Q relationship: and is defined by the following <a Q b> = <b Q a>* An operator is said to be Hermitian if Q = Q All Hermitian operators have real eigenvalues All Measurable Observables (which are real quantities) have Hermitian Operators

SCE3337 Quantum Mechanics III 1 Proof: Given that Q i> = q i i> and that Q = Q then we need to show that q i are real then <i Q i>= <i q i i> = q i <i i> = q i From the definition of the adoint <i Q i> = <i Q i>* = <i q i i>* = (q i <i i>)* = q i * Now the operator Q has been defined to be Hermitian i.e. <i Q i> = <i Q i> Recall that for a complex number q that q = Re(q) + Im(q) and q* = Re(q) - Im(q) Therefore equating the two outcomes above reveals that the only way that q i * = q i is if they have no imaginary terms so q i must be real. 3. The Proection Operator Suppose that we have a complete orthonormal set of wavefunctions { a>}. We defined the Proection Operator # such that P a = a><a For an arbitrary wavefunction ψ> as given by postulate 4 may be written as N ψ> = c i i > i =1 Operating the proection operator on the wavefunction yields P a ψ> = c i a >< a i >= i = c a a> i c i a > δ ai Thus the proection operator proects a wavefunction onto a particular basis function. Summing over all proection operators in the basis { a>} gives P a ψ >= c a a > a a The term c a a > is ust the wavefunction written in the { a>} basis set hence a # Note the difference between the proection operator given by a><a and the integral that we introduced earlier <a a>.

SCE3337 Quantum Mechanics III 11 P a ψ >= I ψ > a where I is the Identity Operator which maps a wavefunction back onto itself. The complete sum of proector operators therefore yields P a = I or a >< a a a = I

SCE3337: QMIII 1 Lecture 3.3 Matrix Representation of Operators We will now look at an alternative representation of operators involved in quantum mechanics which was developed by Heisenberg. Assume that we have an operator Q that we allow to act on the wavefunction ψ a > ψ b > = Q ψ a > This is a typical quantum mechanical process where we say that Q acts on ψ a > to produce a new wavefunction ψ b >. We now apply postulate 4 and let ψ b > be represented by a superposition of basis functions { k>} such that ψ b > = b k k > k ψ a > = a > The connection between the a and b coefficients can be defined since ψ b > = b k k > = Q a > = a Q > k We now multiply the expression by the bra-vector <i which yields for the LHS: <i ψ b > = b k < i k > = b k δ ik k k = b i For the RHS we obtain <i ψ b > = < i Q a > = a < i Q > = a Q i Equating the LHS and RHS reveals b i = a Q i Q i is called the matrix element of the operator Q and is given by Q i = <i Q > It is easy to see how this process defined by the expression ψ b > = Q ψ a > can be represent by the following matrix equation

SCE3337: QMIII b 1 Q 11 Q 1 Q 13... Q 1n a 1 b Q 1 Q Q 3... Q n a b 3 Q 31 Q 3 Q 33... Q 3n a 3. =.......................... b n Q n1 Q n Q n3... Q nn a n Wavefunction kets are represented by column vectors whereas their complex conugate (bras) are represented by row vectors (b 1 *, b *, b 3 *,..., b n *). 3.4 Matrix Inversions A very useful property of the matrix representation is a technique that allows matrix inversions: If ψ b > = Q ψ a > then ψ a > = Q -1 ψ b > where Q -1 is the inverse of the matrix Q. This expression is valid provided that the determinate of the matrix Q, i.e. Q is non-singular. Recall from first year maths that Q Q -1 = I where I is the Identity Matrix (in QM we call it the Identity Operator since I ψ b > = ψ b >). This matrix is given by: 1... 1... 1... I =........................ 1 If Q is non singular a very simple technique that can be used to find the inverse is the cofactor technique: * Q -1 = 1 Q C T * It is also possible to use the Gaussian elimination or row reduction techniques to find the inverse.

SCE3337: QMIII 3 C T is the transpose 1 of the cofactor matrix which is a matrix of cofactors, the elements of which are cofactors of the original matrix Q. The individual elements are called cofactors c i. To find the inverse, one simply finds the cofactor matrix, then take the transpose (exchange rows and columns), and then multiply each matrix element by 1/ Q The cofactor matrix is defined as C i = (-1) i+ q i q i is the minor of the element Q i. It is a scalar value given by the (N-1) x (N-1) determinant remaining in the original matrix when the ith row and th column are struck out. Example: Consider the general matrix A below: A = a b 1 b c a The cofactors of the matrix A are First Row: (1) 1 1+1 c a = a, (1)1+ b a 1 3 =, (1)1+ b c = b Second Row: (1) b +1 c a a b = bc, (1)+ b a = a + b, (1) a + 3 b c = ac Third Row: (1) b 3+1 1 = b, (1)3+ a b =, (1)3+3 a 1 = a The cofactor matrix is therefore C = a b bc a + b ac b a Now to calculate the inverse matrix A -1 we need the transpose of the matrix: C T = a bc b a + b b ac a 1 To find the transpose ust swap the rows with the columns, ie C i T = C i

SCE3337: QMIII 4 The determinate of the matrix A is A = +(a) 1 c a - () b a + (-b) 1 b c = a +b The inverse matrix is therefore: A -1 = 1 a + b a bc b a + b b ac a Exercise: Confirm that this indeed is the inverse of A 3.5 The Matrix Form of the Eigenvalue Equation Consider the eigenvalue equation below Q ψ> = λ ψ> Expanding ψ> with respect to the complete orthonormal basis set of functions { >} yields N ψ> = c > Substitution into the eigenvalue equation yields N c Q > = λ c > Multiplication by the bra vector <i gives N N c < i Q > =λ c <i > N c Q i = λ c δ i N N c (Q i λδ i ) = N The only non-trivial solution of this expression (c ) is when the determinate Q i -λδ i = This is know as the secular equation.

SCE3337: QMIII 5 Q i -λδ i = Q 11 λ Q 1 Q 13. Q 1 Q λ Q 3. Q 31 Q 3 Q 33 λ..... = The determinant yields a polynomial of order N where N is the dimension of the matrix. The roots of the polynomial are the eigenvalues of the matrix Q. Note that if Q was a diagonal matrix then the secular equation would be Q i -λδ i = Q 11 λ. Q λ. Q 33 λ..... = = (Q 11 -λ)(q -λ)(q 33 -λ)...(q NN -λ) = Hence the eigenvalues of a diagonal matrix are ust equal to the diagonal elements Q 11, Q,... Q NN. It can be shown that there is a transformation, called a unitary transformation, such that any Hermitian matrix may be written in diagonal form. This is a very useful property when Q is very large as it enables on to find the eigenvalues of a large system very simply using a computer program. Example: We now solve an example using the secular equation Consider the following matrix representation of an operator Q which is given by Q = 1 4 Find the eigenvalues and hence the eigenvectors for this operator. Step 1: Write down the eigenvalue equation Q a> = λ a> The matrix representation of this eigenvalue equation is 3 c (Q i ) = i =1 In matrix notation the secular equation is given by

SCE3337: QMIII 6 1 λ λ 4 λ = Step : Solve the determinate to find the eigenvalues (1 λ) λ 4 λ () λ +( ) 4 λ = (1-λ){- λ (4- λ)} - + 4 λ = (1- λ){ - 4λ + λ } + 4 λ = -4 λ + λ + 4 λ - λ 3 + 4 λ = λ{ 5λ - λ } = λ{ λ ( 5- λ ) } = This equation is satisfied when λ =,,5 and hence these are the eigenvalues λ 1 =, λ = λ 3 =5. Step 3: We now determine the three eigenfunctions which must be orthogonal. Going back to the eigenvalue equation and substituting in the value for λ 1 : 3 c (Q i ) = c i (Q i. ) = c i Q i = =1 3 =1 3 =1 1 c 1 c = 4 c 3 Multiplication of this expression gives c 1 - c 3 = = -c 1 +4c 3 = The middle equation specifies that c is arbitrary since it is undefined. Accordingly we require a value for c and for simplicity we choose c =. Also we find that c 1 = c 3. We choose c 3 = 1 and hence c 1 =.

SCE3337: QMIII 7 Hence the eigen vector for the eigen value λ 1 is: a 1 > = c 3 1 At this point c 3 is arbitrary but since it is usual to normalise the eigenfunctions we can solve for this scalar using the normalisation condition: c 1 + c + c3 =.c3 + c3 =1 c 3 = 1 5 Therefore the normalised eigenfunction is a 1 > = 1 5 1 Now for the second eigenvalue λ = we would obtain the same result from our secular equation ie c would be arbitrary and c 1 = c 3. But you recall that the solutions for each eigenvalue must have eigenvectors that are orthogonal. Hence a > must be orthogonal to a 1 >. This is simple to evaluate since. a > = c 1 c c 3 c 3 = c c 3 (this uses the fact that c 1 = c 3 ) Now the orthogonality condition specifies that <a 1 a > = <a 1 = 1 T = [ 1] Thus c 3 <a 1 a > = [ 1] c = c 3 4 c 3 + c 3 = c 3 = (note that. c = does not define c ) Therefore the orthogonal vector in its general form is

SCE3337: QMIII 8 a > = c For the normalised eigenfunction c can be determined from the normalisation condition. Thus c 1 + c + c3 = c = 1 c = 1 a > = 1 Now we determine the final eigenfunction for the eigenvalue λ 3 =5. Applying the identical procedure as before 1 5 5 4 5 c 1 c c 3 = 4 5 1 c 1 c c 3 = This yields Thus -4c 1 - c 3 = -5 c = -c 1 - c 3 = c 1 = 1 c 3 and -5 c = c = The final eigenfunction in its general form is a 3 > = c 3 1 Applying normalisation gives c 1 + c + c3 = c3 + c3 =1 c3 = 1 5

SCE3337: QMIII 9 a 3 > = 1 5 1 Exercise: Check that the eigenvector a 3 > is orthogonal to a 1 > and a >

SCE3337: QMIII 1 Lecture 3 4. The Simple Harmonic Oscillator Equilibrium Position x= x m F=-kx Consider the simple harmonic oscillator in the figure above. When the mass m is pulled back and released it undergoes Simple Harmonic Motion (SHM) which is expressed by the equation: m d x dt + kx = k is known as the stiffness of the spring or the spring constant for the restoring force shown in the figure. This equation is well known and can be rewritten in the following form d x dt = x k where = is the resonant frequency of the SHO. The equation of motion has the m solution (this is very simple ust try trial solution e qt and sub into equation above); x = x cos( t) The simple harmonic oscillator plays a large role in many areas of both classical and quantum physics. It provides a model for systems diverse as Atoms Molecules Electromagnetic Radiation Atom Traps The classical equations of motion for a SHO are given by T = p x m (Kinetic Energy) V = 1 kx (Potential Energy) Recall that the Hamiltonian is H = T + V = p x m + 1 kx

SCE3337: QMIII We now use the principle of correspondence to determine the quantum mechanical Hamiltonian for the SHO. By direct analogy we exchange the linear momentum operator p x by p x ih x Thus H = h m x + 1 m x Let > be an eigenfunction of the operator H. Recall that the Hamiltonian operating on a wavefunction yields energy eigenvalues and we now week solutions to the eigenvalue equation: H > =E > Substitution of our quantum mechanical expression for the Hamiltonian gives ( h m x + 1 m x ) >= E > h m x > +(1 m x E) >= For convenience we multiply through by h which yields h m x > +( m h x E ) >= h This equation may be solved by a power technique developed by Dirac which is quite often used in solving equations involving non-commuting operators. It also forms the basis of much advanced theoretical work in quantum mechanics. Let = m h then the equation above can be recast: 1 x >+ x >= E > h we now make the substitutions q = x q = x and p = i x p = 1 x then the SE equation can be re-written as (p + q ) >= > where is a dimensionless energy term given by = E h

SCE3337: QMIII 3 4.1 The Quantum Mechanical Properties of p and q It is instructive to consider the properties of the operators p and q. The commutator of p and q is equal to -i. Proof: Let > be an eigenfunction of the operators p and q then [p,q] > = pq > - qp > Therefore [p,q] = -i = ( i x )( x) > ( x)( i x ) > = i{( x x ) > +x > }+ ix > x x = i > 4. Annihilation and Creation Operators Since p and q do not commute we can not simply factor the expression p + q into (p+iq)(p-iq) in the eigenvalue equation. Instead consider the following linear forms: (q+ip)(q-ip) = q + ipq iqp + p = q + i(pq qp) + p = q + i[ p,q] + p = q + p + i( i) = q + p +1 (q-ip) (q+ip) = p + iqp ipq + q = p + i(qp pq) + q = p + i[q, p] + q = q + p + i(i) = q + p 1 The addition of these two expression yields; (q + ip)(q - ip)+(q -ip)(q + ip) =( p + q ) p +q = 1 q +ip { ( )( q ip) + ( q ip) ( q +ip) } We have now transformed or operator form of the eigenvalue equation into an equation that contains linear terms of the operators p and q. This effectively transforms a second order differential equation into two consecutive first order differential equations. We now introduce two new operators called the raising and lowering operators as a = 1 (q + ip) The lowering operator which is commonly called the annihilation operator a = 1 (q ip) The raising operator also called the creation operator

SCE3337: QMIII 4 The Hamiltonian expression can now be written in terms of these operators: p + q = { 1 (q + ip) 1 (q ip) + 1 (q ip) 1 (q +ip)} p + q = {aa + a a} Hence the eigenvalue equation is given by (aa + a a) >= > We can further rewrite the operators p and q in terms of the operators a and a : a + a = 1 (q + ip) + 1 (q ip) = q q = 1 ( a + a ) And a a = 1 (q + ip) 1 (q ip) = ip p = i ( a ) a The commutation relation of the raising and lowering operators [a,a ] is [ a,a ] = a a a a = { 1 (q + ip) 1 (q ip) 1 (q ip) 1 (q +ip)} = 1 {(q + p +1) (q + p 1)} [ a,a ] = a a a a =1 a a = 1+ a a Important result! a a = a a -1 It is also useful to note that using the commutation relation we can re-write the Hamiltonan in terms of these operators as H = aa + a a = 1 + a a = aa 1 The multiplication of the two operators a a is called the number operator n, the reason for this will become more clear later.

SCE3337: QMIII 5 4.3 Eigenvectors and Eigenvalues of a a and a a : Using the results derived in the last page we can now evaluate our eigenvalue equation. H > = ( p + q ) > = > (aa + a a) >= > we now substitute for a a by using our commutation relation a a = 1 + a a (1+ a a + a a) >= > a a >= > 1 > a a >= ( 1) > a a >= n ˆ >= ( 1) > Hence the number operator ˆ n has the same eigenfunctions as the Hamiltonian. The eigenvalue of the number operator ˆ n = a a is given by ( 1) Consider a further operation in which we operate with a a: a a >= ( 1) > (aa 1) >= ( 1) > aa >= ( 1) >+1 > aa >= ( +1) > Once again we have shown that the operator aa has the same eigenfunctions as the Hamiltonian. The eigenvalue of aa is given by ( +1)

SCE3337: QMIII 6 4.4 Eigenvectors and Eigenvalues of the Annihilation and Creation Operators a and a : We now consider the application of the raising and lowering operators to the Hamiltonian eigenvalue equation. Let n> be an eigenstate of the Hamiltonian which has the eigenvalue n. As a consequence the eigenstate n> is also and eigenstate of the operators a a and aa as proved earlier. The eigenvalue equations is then H n> = n n> (aa + a a) n >= n n > (1+ a a) n >= n n > we now act on this equation with a. (a + a a a) n >= ε n a n > we want to factor out a the reason will become apparent further along, to accomplish this we use the commutation relation and replace a a with aa 1 (a + a { aa 1}) n >= n a n > (1+ a a )a n>= n a n > (1+ a a)a n >= ( n a + a ) n> (1+ a a)[a n >] = ( n + )[a n >] We now define n+1> = a n>, then the above expression is reduced to (1+ a a) n+1>= ( n + ) n+ 1> (aa + a a) n + 1 >= ( n + ) n +1 > H n + 1 >= n+1 n +1 >= ( n + ) n +1> This is exactly the same form of the energy eigenvalue equation given above. This shows that given a state n> of dimensionless energy n there exists another state which has two units of energy more than this state. The upper state has been defined as n+1>. One can repeat this procedure which yields a ladder of levels increase to infinity. From this we can conclude that The operator a INCREASES the energy and the eigenstate of the system

SCE3337: QMIII 7 We now consider the action of the operator a using the same technique: ah n> = n a n> a(aa + a a) n >= n a n > a(1+ a a) n >= n a n > (a + aa a) n >= n a n > (a + (1+ a a)a) n >= n a n > (a + a + a aa) n >= n a n > {(1+ a a)a + a} n >= n a n > (1+ a a)[a n >] = ( n )[a n >] We now define a n> = n-1> H n-1> = n-1 n-1> = ( n -) n-1> This is the same as the eigenvalue equation. This equation demonstrates that given a state n> of dimensionless energy n there exists a state which has units of lower energy than this state. The lower state is defined as n-1>. The operator a DECREASES the energy and lowers the eigenstate of the system Note that the operators a and a do not have eigenfunctions defined by n> unlike a a and aa which had the same eigenfunctions as H.

SCE3337: QMIII 1 Lecture 4 4.5 Higher Order Eigenfunctions of H Recall from section 4.3 aa n >= ( n +1) n > and (1) a a n >= ( 1) n n > () Summing equations (1) and () yields: ( aa + a a) n >= ( +1) n H n >= ( +1) n + ( n 1) + ( n 1) n > n > H n >= n n > (3) Which is what we expect i.e. we recover the original eigenvalue equation. If now operate with a on equation (1) we can determine a higher order eigenfunction of H: a a[a n >] = ( +1) n [a n >] (4) Recall from last lecture a n = n + 1, hence a a n +1 >= ( +1) n n + 1> Replacing a a with aa -1 (commutation relation) in equation (4) reveals (aa 1)[a n >] = ( +1) n [a n >] aa [a n >] 1[a n >] = ( +1) n [a n >] aa [a n >] = ( + 1) n [a n >] +1[a n >] aa [a n >] = ( + 3) n [a n >] (5) We now sum equations (4) and (5) giving

SCE3337: QMIII ( aa + a a)[a n >] = ( + 3) n + ( + 1) n [a n >] H[ a n > ] = ( n + ) [ a n > ] (6) Operating with a on equation (5) gives a aa [a n >] = ( + 3) n a [a n >] a a(a ) n >= ( + 3) n (a ) n > a a[(a ) n >] = ( + 3) n [(a ) n >] (7) a a n + >= ( n + 3) n + > We now replace a a in equation (7) by aa -1 : ( aa 1)[(a ) n >] = ( + 3) n [(a ) n >] aa [(a ) n >] 1[(a ) n >] = ( + 3) n [(a ) n >] aa [(a ) n >] = ( + 5) n [(a ) n >] (8) Summing equations (7) and (8) gives; ( a a + aa )[(a ) n >] = ( +3) n + ( n + 5) [(a ) n >] Therefore H[(a ) n >] = ( n + 4)[(a ) n >] (9) If we do this procedure m times then equations (3), (6) and (9) allow us by induction to deduce the general result: H(a ) m n >= (p + q )(a ) m n >= { n + m}[(a ) m n >] = { n + m} n + m > It is easy to see why a is called the raising operator as each successive application a higher energy is obtained.

SCE3337: QMIII 3 One can show that successive operations by a decreases the energy and as such it is called the lowering operator. One can show by induction the general result: H(a) m n >= ( p + q )(a) m n >= { n m}[(a) m n >] = { n m} n m > 4.6 Summary of Wavefunctions, Operators and Eigenvalues for the Simple Harmonic Oscillator: Eigenvalue of Wavefunction Eigenvalue of aa a a n> + 1 n 1 n a n>= n+1> + 3 n + 1 n (a ) m n>= n+m> + m + 1 n + m 1 n a n>= n-1> 1 n 3 n (a) m n>= n-m> n (m 1) n (m +1) Eigenvalue of H n + n +m n - n -m n Energy E n E + h n E + m n h E - h n E - m n h 4.7 Lower Energy Boundary Conditions and State Normalisation: It is clear that physically we must have a lower limit when applying the lowering operator since we have postulated that the energy of the simple harmonic oscillator can not be negative. Thus there exist a lowest positive energy state which we define to be the ground state. This means that the measurement of energy, or the expectation value of this energy after a measurement has been obtained must be non-negative. The expectation value is given by: Hence for < n > = H = p + q =<n n n > = <n p +q n> = <n p n> + <n q n> < n > <n p n> + <n q n> Now we define the wavefunction n> = > as the ground state (the state with lowest energy). Clearly acting on this wavefunction with the lowering operator must result in a condition of no energy (otherwise the state > would not be the lowest state). Hence a > = If we now act on this state with the creation operator gives

SCE3337: QMIII 4 a a > =. a = (1) but recall that a a n >= 1 n n > Hence for the ground state a a >= 1 > () So assuming that > exists, for equation () to be consistent with equation (1) requires that a a >= =1 1 >= Recall from last lecture that n was dimensionless energy term given by n = E n h putting n = : = E h 1 = E h E = 1 h This says that the lowest energy state is non-zero which is strikingly different to the classical SHO which has a lowest energy state equal to zero. This lowest energy quantum mechanical state is defined as the Zero Point Energy

SCE3337: QMIII 5 Using the result for the ground state energy plus the results summarised in the table (section 4.6) it is possible to find a general form for the energy eigenvalues. E = 1 h E 1 = h + 1 h = 1 + 1 h E = h + 1 h = + 1 h Therefore by induction E n = n + 1 h where n=,1,,... Energy En +1 = n + 3 h En = n + 1 h En 1 = n 1 h a a E 1 = 3 h E = 1 h E= This enables us to diagrammatically represent the energy states of the quantum SHO: Note that the dimensionless energy term is given by n = E n h = h n + 1 h = n+ 1 The meaning of the operator a a as the number operator ˆ n is now clear since: a a n >= ˆ n n >= 1 n n > Substitution of the value of n from above gives n +1 1 n ˆ n >= n >= n n > Thus ˆ n n >= n n > i.e. n is the nth energy state of the harmonic oscillator.

SCE3337: QMIII 6 4.8 Normalisation of Eigenfunctions It should be noted that the derivations that we have used so far for the wavefunctions of quantum mechanical SHO namely n+1> = a n> have not been normalised. To achieve this we require that <n-1 n-1> = <n n> = 1 As a result we need to introduce a numerical factor into the expressions which have been derived above to account for the normalisation. Therefore we should write the expressions as a n = B n 1 n 1 a n = B n+1 n +1 where B m are coefficients that are yet to be determined. As an example we will determine the normalised ground state wavefunction. Recall a = 1 (q + ip) Operating on the ground state gives a = 1 (q + ip) (1) Also recall that q = x and p = i x then q x = x = q sub ( x) in for p gives p = i x = i q = i q Therefore substituting back into (1) yields a = 1 (q + q ) = (q + q ) = q = q

SCE3337: QMIII 7 = q q ln = q + c Thus = C e q where C is an arbitrary integration constant which can be determined by normalising the wavefunction: < >= 1 * < >= dq = C e q dq where the integral 1 = C e q dq = Yielding for the normalisation constant: C = 1 1 Hence the normalised ground state wavefunction is = 1 1 q e Problem Sheet will be used to determine the normalisation coefficients for an arbitrary wavefunction n>.

SCE3337: QMIII 1 Lecture 5 5. Approximation Techniques in Quantum Mechanics In a quantum mechanical treatment of the physical world the determination of physical processes is governed by the Schrödinger equation. The equation may be time independent or time dependant: H (r) >= E (r) > (Time Independent S.E.) H (r,t ) >= ih t (r,t) > (Time Dependent S.E.) We have already solve the Schrödinger equation for the time independent case for the simple harmonic oscillator. In principle, the physical system is described by either of these equations depending whether we are interested in behaviour or a system that is time dependent or time independent. As it turns out real nature is not simple and there are few exactly solvable problems. Some examples of exact solutions are: The harmonic oscillator bounds states of a particle in a square box The hydrogen atom Actually even in reality the hydrogen atom is not exactly solvable even though the wavefunction can be written down exactly for this system. Small perturbations such as spin orbit effects have to be considered to allow for real experimental observations. If a system consists of more than one particle then interactions between the particles also has to be taken into consideration. As an example the He atom has two electrons that not only interact with the ionic core but also interact with each other. It is impossible to solve such a three body problem exactly. Early quantum physicists did not however completely give up on solving these problems they made allowances for difficult problems by introducing approximation techniques.

SCE3337: QMIII 5.1 Time Independent Perturbation Theory The basic idea of time independent perturbation theory is as follows. Suppose we have a system which has a Hamiltonian H and we apply a small perturbation, h, to the system such that the system Hamiltonian is: H = H + h Here H has a much greater influence over the system than h does. We also assume that the Schrödinger equation can be solved exactly: H i >= E i i > i> are the associated eigenkets of the Hamiltonian H and E i are the corresponding eigenvalues. The eigenkets form a complete orthonormal set as we have shown in previous lectures. Then any ket vector can be written as a linear superposition of the eigenkets i> with < i >= i and n >= a ni i > i For convenience we rewrite the system Hamiltonian as H = H + h where is a free parameter defined in the interval 1. This allows us to turn the perturbation on and off. We now want to solve the following Schrödinger equation: H i' >= E i' i' > Where i'> are the eigenkets of the perturbed system and as such are not the same eigenkets as i> and as such the eigenvalues E i E i'. We further assume that the sets { i'> } and { i> } are non-degenerate (i.e. their eigenvalues are unique). This point will be important in the following discussion as it allows us to get around the problem of a division by zero that will come up later. Perturbation theory covering degenerate states will be covered later in the course. We will also say in the limit that as lim i' > i > and lim E i' E i We now let the new eigenkets and eigenvalues be represented by the following power series: i' >= i >+ i 1 > + i > +...+ n i n >+.. E i' = E i + E i1 + E i +...+ n E in +... Thus i'>= i> and E =E i i if =.

SCE3337: QMIII 3 It is assumed that successive terms of this power series gets smaller and as a result the series converges. Substitution of the new eigenkets and eigenvalues into H i' >= E i' i' > gives; (H + h) i' >= E i' i' > Substitution of the power series for i'> and E i' yield { } { } (H + h) i > + i 1 > + i > +...+ n i n > +.. ( ) i > + i 1 >+ i >+...+ n i n > +.. = E i + E i1 + E i +...+ n E in +... { } { } = (H + h) i > + i 1 > + i > +...+ n i n > +.. ( ) i >+ i 1 > + i > +...+ n i n > +.. E i + E i1 + E i +...+ n E in +... Grouping terms in powers of : Order 1st Order 64 47 444 8 644 444 4 744 444 448 { H i > E i i > } + H i 1 > +h i > E i i 1 > E i1 i> + { } { } H i >+h i 1 > E i i> E i1 i 1 > E i i > 144 444 444 4 444 444 444 3 + 3 {...}+...= nd Order For the above equation to be valid each of the terms in brackets must separately equate to zero. The first and second terms give us the th order and 1st order terms respectively: H i >= E i i > Zero Order H i 1 > +h i >= E i i 1 > +E i1 i > First Order H i > +h i 1 >= E i i > +E i1 i 1 >+E i i > Second Order E i1 is defined as the first order energy correction.

SCE3337: QMIII 4 5. First Order Time Independent Perturbation Theory 5.a First Order Energy Correction The first order equation can be solved by noting that the eigenket i 1 > can be expressed as a linear superposition of the unperturbed eigenkets i>: i 1 >= a 1 > Substitution into the first order term gives H a 1 > +h i >= E i a 1 > + E i 1 i > We now multiply through by the bra vector <k : a 1 < k H >+ <k h i >= E i a 1 recall that H >= E > thus < k > +E i1 < k i > E a 1 < k >+ < k h i >= E i a 1 k + E i1 ki evaluating the summations gives E k a 1k + < k h i >= E i a 1k + E i1 ki (1) now for k = i we have E i a 1i + < i h i >= E i a 1i + E i1 < i h i >= E i1 The first order energy correction term is therefore given by the matrix element <i h i> taken between the unperturbed eigenkets i>.

SCE3337: QMIII 5 5.b First Order Eigenvector Correction: We can now evaluate what the perturbed ket vector i'> to first order will be. The first order correction to the eigenket i> is given by i' >= i >+ i 1 > Recall that i 1 > can be written as a linear superposition of basis states: i 1 >= a 1 > substitution back into the first order term yields i' >= i >+ a 1 > () The coefficients a 1 are evaluated from equation (1) derived in the previous section, hence for the case when k i (i.e. non-degenerate case) then: 67 =8 E k a 1k + < k h i >= E i a 1k + E i1 ki < k h i >= E i a 1k E k a 1k < k h i >= a 1k ( E i E k ) a 1k = < k h i > ( E i E k ) k is a dummy variable hence we can call it anything so make it. Substitution of this expression into equation() gives; i' >= i >+ < h i > > i E i E ( ) We now set =1 thus giving us the corrected eigenket to first order: < h i> i' >= i >+ > i E i E ( ) Notice that each eigenket > which has a non-zero matrix element with i> that is < h i > will contribute to the new eigenket i'>. The perturbation produces an interference # or mixing of the original eigenkets. States that are close together will have a large amount of mixing. We can see this mathematically as the denominator in the above expression will get large. As the states 1 become separated the mixing will become weaker since E i E ( ). # It is valid to call this effect an interference as both the amplitudes and the phases of the matrix elements play a role.

SCE3337: QMIII 6 5.3 Second Order Time Independent Perturbation Theory The extension to second order is relatively straight forward by following the same procedure as for the first order corrections. 5.3a Second Order Energy Correction We start with the terms in our perturbation expansion: H i > +h i 1 >= E i i > +E i1 i 1 >+E i i > We now expand i 1 > and i >: i 1 >= a 1 > and i >= a m m > m Substituting into the above equation gives H a m m > m +h a 1 > = E i a m m > m + E i i >+E i1 a 1 > We now multiply by the bra vector <k : a m < k H m > + a 1 < k h >= E i a m < k m > +E i1 a 1 < k > +E i < k i > m a m E m km + a 1 < k h >= E i a + E m km i 1 a 1 k + E i ki m m a k E k + a 1 < k h >= E i a k + E i1 a 1k + E i ki a k E k E i k h > E i1 a 1 k = E i ki ( ) + a 1 < Now for the case when k=i we get E i = a 1 < i h > E i1 a 1i m Recall that the first energy correction term is given by E i1 =< i h i > thus E i = a 1 < i h > a 1i < i h i > This can be rewritten as E i = a 1 < i h > +a 1i < i h i > a 1i <i h i> i = a 1 < i h > i

SCE3337: QMIII 7 Recall that a 1k = < k h i > ( E i E k ) E i = i < h i >< i h > E i E ( ). Upon substitution into the above equation yields: Therefore the second order correction to the energy is E i = i < i h > ( E i E ) The second order correction to the eigenket is messy to derive (see problem sheet 3) but straight forward using the same procedure as the first order correction and is given by Therefore i' >= i >+ i 1 > + i > we let =1 then, i' >= i >+ i 1 > + i > Expanding i 1 and i in basis sets yields i' >= i >+ a 1 > + a m m > m < h i> i' >= i >+ > + i E i E ( ) m i < m h >< h i > < i h i >< m h i > i ( E i E m )( E i E ) ( E i E m ) m > Exercise: Derive this expression for the corrected eigenvector to second order. Beyond second order, perturbation theory is seldom used as it becomes very messy quickly.

SCE3337: QMIII 8 5.4 The Zeeman Effect: An Application of Time Independent Perturbation Theory You may recall from second year that if we apply an external uniform magnetic field B to an atom with a magnetic moment then it will experience a perturbation. The magnetic moment is given by: J = g J B h J If the magnetic field is weak (<1 4 Gauss) the total angular momentum J=L+S is a good quantum number the perturbation to the system Hamiltonian is therefore given by h = J B = g J B h J B The first order energy shift is then E i1 =< Jm J h Jm J > If B is taken in the Z direction then J B = J Z B, hence the energy shift is given by E i1 = g J B h B < Jm J J Z Jm J > J Z is an operator of the eigenvector Jm J > with eigenvalue m J h : E i1 = g J B h Bm J h < Jm J Jm J > E i1 = g J B Bm J As an example applied to a real atom; consider the 6 1 S 6 3 P 1 transition in Hg. If we have no perturbing field then we get the following energy level structure: m = -1 m = J J m = 1 J 3 6 P 1 E i 53.7nm Radiation 1 6 S No Perturbing Field

SCE3337: QMIII 9 If we now turn on a weak magnetic field we can calculate the perturbation of the energy levels of the 6 1 S and 6 3 P 1 states due to the field. To do this we need to determine the Landè g factor which is given by g J = 1 + J (J +1) + S(S +1) L(L +1) J(J +1) For the 6 1 S state g J =1 and for the 6 3 P 1 state g J = 3. Now in the ground state the only m J value is hence this level is not perturbed since the perturbed energy is proportional to m J. In the case of the 6 3 P 1 state the degenerate energy levels will be split into three nondegenerate states by the amount E i1 = 3 BB Thus the perturbed energy levels of the 6 3 P 1 state will be E i' = E i + E i1 This is represented in the energy level diagram below: m J = 1 m J = -1 E i1 m J = E i1 3 6 P 1 E i 53.7nm Radiation 1 6 S B-field ON It should be noted that we have used the non-degenerate perturbation theory even though in the 6 3 P 1 state the energy levels were degenerate in m J. In this special case it is ok to do so as the operator J Z has a definite value whether there is a perturbation or not as a result there is little mixing of these states.

SCE3337: QMIII 1 Lecture 6 6. Perturbation Theory for Degenerate States In previous lectures we had assumed that there was no degeneracy among the perturbed states. However in practice it is often encountered that there are several states that have the same energy, that is they are degenerate. The effect of a perturbation is usually to lift the degeneracy (this was seen in the Zeeman effect example last lecture) in the first order correction. There are problems associated with the application of time independent perturbation theory if only part of the degeneracy is lifted in the first order correction. This is seen when one tries to apply second order corrections to degenerate states, recalling the equation to second order: < k h i > +a 1k E k = E i1 ki + E i a 1k Now if E k =E i and ki = then < k h i >= Therefore the second order energy correction term yields, < i h > E i = = i E i E ( ) This is of course undefined which means that there is a procedural problem with this technique. As such we need to develop a method that can handle this by removing this problem. Consider two states i> and k> that are nearly degenerate, with all other states well removed from these states. We also suppose that the matrix element <k h i>. The first order correction to the ket i> is given by < h i> i' >= i >+ > i E i E ( ) running the summation over reveals that when =k that this term dominates since the denominator E i -E k gets small and as such that term becomes large therefore: i' > i >+ < k h i> ( E i E k ) k > A similar expression can be found for the perturbed state k>: k' > k > + < i h k > ( E k E i ) i > These two expressions show that for states with nearly degenerate energy levels, the perturbed wavefunction for these levels is to a good approximation a mixture of the two unperturbed levels.

SCE3337: QMIII In the case of degenerate energy levels i> and k> we would expect that we could write them as; i' >= C ii i >+C ik k > k' >= C ki i >+C kk k > For a number of degenerate states we could write n' >= C n > We now substitute in to the eigenvalue equation (H + h) n' >= E n ' n' > : (H + h) C n ' >= E n C n > ' We now write the perturbed energy term as E n unperturbed energy plus the correction term; = E n + which is ust the sum of the (H + h) C n >= E n + > ( ) C n C n H > + C n C n E > + C n h >= E n C n > + C n > h >= E n C n > + C n > Since the states are degenerate E =E n then C n h >= C n > Multiplying by the bra vector <k reveals C n < k h >= C n < k > C n < k h >= C n C n < k h > k [ ] = k This equation has a solution when the determinant < k h > k = cf: The matrix form of the eigenvalue equation

SCE3337: QMIII 3 Example Consider two degenerate states 1> and > h 11 h 1 h h 1 = ( h 11 )( h ) h 1 h 1 ( h 11 + h ) + h 11 h h 1 h 1 Solving for yields = 1 h + h 11 Exercise: Show this ( ) ± ( h 11 h ) 4h 1 h 1 Therefore this system with two degenerate levels, the correction term takes on two values. As an example let h 11 = h = A and h 1 = h 1 = B, then from the above equation = A ± B. We can determine the coefficients C n by reconsidering the equation [ ] = then C n < k h > k A B B A C 1 = using the solution =A+B we get C B B B C 1 B = C BC 1 + BC = BC 1 BC = C 1 = C 1 >= C 1 1 From normalisation C i i = 1 thus C 1 C For C 1 = C then ( C 1 C )= C 1 +C =1 C 1 + C 1 = 1 C 1 = 1

SCE3337: QMIII 4 Therefore we obtain for the wavefunction >: >= 1 ( 1 > + > ) Called a symmetric wavefunction Similarly we find for =A-B the anti-symmetric wavefunction. >= 1 ( 1 > > ) 6.1 Example Two: The DC Stark Shift of the n= level of Hydrogen As an example of one of the uses of degenerate perturbation theory we will calculate the electric field Stark splitting of the n= level of hydrogen by a constant DC electric field E which we will assume points in the Z direction. The n= level has four degenerate levels which all have the form nlm = R nl (r)θ lm ( )Φ m ( ) : = 1 4 a 3 1 = 1 4 a 3 1±1 = 1 4 a 3 r exp r a a r exp r cos a a r exp r sin e ±i a a where (x,y,z) = (rsin cos,rsin sin,rcos ) transform into the usual spherical coordinates. The perturbation Hamiltonian is given by the dot product between the electric dipole of the atom and the electric field vector E: h = E D D is the dipole moment of the atom and is given by D=qd, where q is the charge (in our case q=-e) and d = zˆ k = r cos k ˆ. Hence our perturbation operator h is given by h = E ˆ k er cos ˆ k = eer cos We now need to solve the determinant < k h > k = where < k h >=< l' m l' eer cos lm l > = ee < l' m l ' rcos lm l > = ee < l' m l ' z lm l >

SCE3337: QMIII 5 We can now apply parity (i.e. is <k h i> an even or odd function of the spatial coordinate?) arguments to eliminate some terms in the perturbation Hamiltonian, since each wavefunction lm l > has a definite parity. For example the matrix element < l' m l' z lm l > must have odd parity since changing sign of the coordinate makes z The diagonal matrix elements are integrals over products of either: even function x odd function x even function = odd function or eg: z x z x z 4 =z 7 (odd) odd function x odd function x odd function = odd function eg: z x z x z=z 3 (odd) -z. Recall that the definite integral over symmetric limits of an odd function = (c.f. a a xdx = ) therefore the diagonal matrix elements must be equal to zero since the integral < lm l z lm l > is over symmetric limits. As a consequence only the off-diagonal elements are non zero. The parity of the wavefunction is defined by the factor (-1) l. Therefore is follows that for non-zero matrix elements: l and l' can't be both even or both odd Furthermore m l =m l' for a non-zero integral. This is seen when considering the form of the wavefunction above, where the m component is explicit in the function Φ m ( ) = e im. The matrix elements take the form: < l' m l' r cos lm l >= R * nl (r)r 3 * R nl' dr Θ lm l ( )sin Θ l' ml' d r { } e i(m l m l' ) =...drd =...drd { } cos(m l m l' ) d d =1 m l = m l' = m l m l' So the integral over the wavefunction = unless m l =m l'. Using these results the only non-zero matrix elements are: e im l e im l' d 6 7 =,always 44 444 8 14 4 44 3 i sin(m l m l ' ) d < h 1> = <1 h > = ee < r cos 1 >. The integral is given by < rcos 1 >= 3a Exercise: See problem sheet 3.

SCE3337: QMIII 6 We notice that there can be no first order correction since the diagonal terms are zero and as such the degeneracy persists. We therefore have a degenerate perturbation so we need to solve < k h > k = : > 1 > 11 > 1 1> > 1> 11> 1-1> 3a ee 3a ee The solution to the determinant is given by ( [ 3a ee] ) = Exercise: Show this There are four solutions to this equation =±3a ee, 1, 3,4 =. Therefore the degeneracy is partially lifted by the electric field. We now evaluate the perturbed wavefunctions which require solutions to the equation: 3a ee 3a ee Substituting in ε=+3a ee we get C 1 C C 3 C 4 = 3a ee 3a ee 3a ee 3a ee 3a ee 3a ee C 1 C C 3 C 4 = 1 1 1 1 3a ee 1 1 C 1 C C 3 C 4 = 1 1 1 1 1 1 C 1 + C = C 1 + C = C 3 = C 4 = C 1 C C 3 C 4 = multiplying out gives Hence C 1 =-C. Applying normalisation gives

SCE3337: QMIII 7 C 1 + C = 1 so C 1 = 1 and C = 1 Therefore the perturbed wavefunction with energy eigenvalue =+3a ee is 1 >= 1 > 1 1 > Similarly one can show using the same procedure that for =-3a ee >= 1 > + 1 1 > The perturbation has therefore coherently mixes these two states. For the eigenvalues =, the matrix element <11 h 1-1>= from the previous page. Recall the expansion for two nearly degenerate states i> and k> we had i' > i >+ < k h i> ( E i E k ) k > Thus if the matrix element is zero clearly we can have no mixing between the two states and the states are unperturbed since i' > i >. The eigenvalues and eigenvectors for the n= Stark shifted states of hydrogen are given by =+3a ee 1 >= 1 > 1 1 > =-3a ee >= 1 > + 1 1 > = >= 11 > 3 = 4 >= 1 1 >

SCE3337: QMIII 8 Energy E > 1> 11> 1-1> Before Perturbation Energy Apply perturbating electric field E + 3a ee E E 3a ee > 1> >= 1 1 > 1 1 > 3a ee 3a ee >= 1 > + 1 1 > With perturbing electric field

SCE3337: QMIII 1 Lecture 7 7. Time Dependent Perturbation Theory In lectures so far we have covered perturbations to the Hamiltonian that are time independent. We now develop an approximation technique to allow us to handle perturbations that have a time dependence. We write the time dependent perturbation Hamiltonian as H(t)=H +h(t) Where H is the time independent unperturbed Hamiltonian and h(t) is a perturbation which is time dependent. We also make the assumption that h(t) << H The goal is to find a solution to the time dependent Schrödinger equation: H Ψ' >= ih t Ψ' > The unperturbed eigenfunctions are given as before: H i >= E i i > This equation leads to stationary states (time independent) which have eigenvalues given by E i. It should be noted that this equation is formally a solution of the Schrödinger equation when the potential is time independent. It is more formally correct when considering the time dependent phenomena to consider the wavefunction i > which is associated with these stationary states. This wave function i > can be evaluated by considering the full Schrödinger equation, which evaluates the wavefunctions and their time dependence. We begin with: H Ψ i (r,t) >= ih t Ψ i (r,t) > where the Hamiltonian is given by Thus H = h m r + V(r) h m r Ψ i (r,t) >+V(r) Ψ i (r,t) >= ih t Ψ (r,t) > i

SCE3337: QMIII We now try separation of variables by splitting the wavefunction into time independent and time dependent part such that Ψ i (r,t) >= i (r) > i (t) > Substitution into the Schrödinger equation gives h m r (r) > (t) > +V(r) (r) > (t) >= ih i i i i t (r) > (t) > i i h m (t) > i r (r) > +V(r) (r) > (t) >= (r) > ih i i i i t (t) > i we now multiply both sides of the equation by 1 (r) > which gives i i (t) > 1 (r) > h i m r (r) >+V(r) (r) > i i = 1 (t)>ih i t (t)> i Since both sides are only dependent on either equation are equal to constants: (r) > or i i (t) > both sides of the 1 (r) > h i m r (r) >+V(r) (r) > i i =C = 1 (t)>ih i t (t) > i Consider the LHS of the equation h m r (r) > +V(r) (r) >= C (r) > i i i H i (r) >= C i (r) > We have recovered the time-independent SE and is ust the usual time independent eigenvalue equation, hence C = E i

SCE3337: QMIII 3 Considering the LHS of the separated Schrödinger equation 1 (t) > ih i t i (t) >= C 1 i (t) > ih t i (t) >= E i t i i (t) >= h E i (t ) > i d (t) > i i = i (t) > h E dt i Integrating reveals d i (t) > i (t) > = i h E i dt ln ( (t ) > ) = i i h E t + C' i i (t) >= Ae i h E i t i = e h E i t The constant A is found by the normalisation condition. The wavefunction for the stationary states (time independent potential) is therefore given by i Ψ i (r,t) >= (r) > (t) >= (r) > e i i i Consider the wavefunctions of the full Hamiltonian H(t). Since Ψ i > form a complete orthonormal set thus we can expand the wavefunctions as h E i t Ψ'(r,t) >= a n (t) Ψ n (r,t) > n where the coefficients a n (t) are time dependent.

SCE3337: QMIII 4 Substitution into the Schrödinger equation gives H Ψ'(r,t) >= ih t Ψ'(r,t)> n n a n (t)( H + h(t) ) Ψ n (r,t) >= ih a t n (t) Ψ n (r,t) > n a n (t)( H + h(t) ) Ψ n (r,t) >= ih a n (t) Ψ n (r,t) > +ih a n (t) t Ψ (r,t) > n n n a n (t) H Ψ (r,t) > ih Ψ (r,t) > n t n n = ih a n (t) Ψ n (r,t) > a n (t)h(t) Ψ n (r,t)> 144 444 44 444 43 n n This term = for the unperturbed wavefunction since H Ψ n (r,t )>=ih t Ψ n (r,t )> a n (t)h(t) Ψ n (r,t) > ih n n a (t) Ψ n n (r,t) >= We now multiply by the bra vector < Ψ k (r,t) : a n (t) <Ψ k (r,t) h(t) Ψ n (r,t)> ih a n (t) < Ψ k (r,t) Ψ n (r,t) >= n We now separate the spatial and time dependence in the wavefunction so that the bra and ket vectors above are defined as < Ψ k (r,t) = e i E k t Ψ n (r,t) >= e i h E n t < ψk (r) h ψn (r) > Substitution into the equation above reveals a n (t) < k (r) h(t) (r) > e i n n a n (t) < k (r) h(t) (r) > e i n n We now let E k E n = h kn therefore n [ E k E n ]t h [ E k E n ]t h [ E i k E n ]t ih h a n (t)e = kn n = ih a k (t) a n (t)h kn (t)e i kn t n ( ) = ih t a k (t) where h kn (t) =< k (r) h(t) n (r) >

SCE3337: QMIII 5 This is a set of coupled first order differential equations, one for each a k (t) which determine the a n (t) coefficients. These equations are not able to be solved exactly since all of the a n (t) coefficients are related to only the derivative of the kth coefficient. Notice that if the perturbation is zero then a k (t) =,therefore a k (t) must be a constant. This suggests that provided the perturbation is small, the coefficients change slowly. As a first approximation we will assume that the coefficients a n (t) on the LHS of the equation are constant. Suppose at t=, the system is in some state Ψ (r,t = ) > this requires that a ()=1 and a n ()= for n. This can be interpreted as the system is totally in the state at the start. At some time t later we will have ( ih a t (t) = a n (t)h n (t)e i n t ) n After t=, since the perturbation is weak a (t) 1 still and all other coefficients a n (t) hence only this term will contribute to the sum therefore the right hand side of the equation reduces the sum to RHS= a (t) { h (t)e ( i t) =1 h (t) Therefore ih t a (t) = h (t) Integrating gives t a i (t) = h h (t) t d a dt (t)dt = i h t h (t) dt a (t) a () = i h h (t) dt a (t) = a () i h h (t) dt a (t) = 1 i h h (t) dt t t t

SCE3337: QMIII 6 The coefficients other than a (t) are given by ih d dt a (t) = a k n(t)h kn (t)e i n ( ) kn t But since all a n (t) except a (t) then only one term in the sum of the RHS significantly contributes ih d dt a (t) = a (t) k { h ( i kt ) k (t)e 1 d dt a k (t) i h h k (t)e i k t ( ) Integrating this expression: t d a dt k (t)dt i h h (t) ( k e i k t ) dt t a k (t) a 1 k () 3 i h h (t) ( k e i k t ) dt t a k (t) = i h h (t) ( k e i k t ) dt t

SCE3337: QMIII 7 7.1 Example 1: Constant Perturbation As an example we will consider a step function perturbation which is turned on at t= and remains constant thereafter. The matrix elements of the perturbation are therefore constants and can be taken outside the integral: a (t) = 1 i h h t The other coefficients are a k (t) = i h h ( k e i k t ) dt t a k (t) = i h h i k k t ( e i k t ) a k (t) = i h h i ( k e i k t ) 1 k a k (t) = h k ( e i kt ) 1 h k The probability of finding the system in the state Ψ k > with the system starting at time t= in state Ψ > is a k (t). Then for k we find a k (t) = = = = h k h k h k h k h k h k h k h k e ( i k t ) 1 ( e i k t ) 1 ( e i k t ) i e k t e ( i k t ) 1 e ( i k t ) 1 ( ) ( i e k t ) ( i e k t ) +1 { } ( e i k t ) ( i + e k t ) *

SCE3337: QMIII 8 = h k h k [ { cos( t k )}] now cos( A) = 1 sin ( A) cos t k = h k h k ( ) = 1 sin kt 1 sin kt sin kt a k (t) = h k h k Thus the probability of finding the system in state Ψ k > oscillates at an angular frequency which corresponds to the transition frequency as shown in the figure below: 1.8.6.4. x -6-4 - 4 6

SCE3337: QMIII 1 Lecture 8 7. The Harmonic Perturbation and Fermi's First Golden Rule We now look at one of the more important time dependant perturbations which is an oscillatory perturbation. These types of perturbations are important as nearly all matterelectromagnetic radiation interactions can be thought of as an oscillatory perturbations produced by an electric or a magnetic field. We have already seen the effect of DC electric and magnetic fields we now consider the effect of AC fields. Consider a time dependent perturbed Hamiltonian given by: h(t) = V cos( t) This type of perturbation could be the due to the interaction of a monochromatic electromagnetic wave from a laser with a atom. The exact details of the interaction are contained in the V term. If there were more than one frequency involved, the perturbation would be a Fourier series over all of the frequencies present. The above Hamiltonian may be written as h(t) = V cos( t) = V ei t + e i t Substituting into our expression for the amplitudes a k (t) gives t a k (t) = i h h (t) ( k e i k t ) dt a k (t) = iv k h a k (t) = iv k h a k (t) = iv k h a k (t) = iv k h t ( e i t + e i t ( ) e i k t ) dt t e i ( k + )t + e i k e i ( k + )t i k + ( )t dt i ( k )t ( ) + e i( k ) e i ( k + )t 1 i k + i( k )t t 1 ( ) + e i( k ) We notice that if k then the term with the negative sign dominates all the other terms. This process corresponds to absorption of a photon from the perturbing field with energy h k exciting the system from the lower state Ψ > to the higher energy state Ψ k >, the transition has the frequency:

SCE3337: QMIII E k E h = k = The other term dominates when k, this case corresponds to stimulated emission whereby the incident photon from the perturbing field stimulates the atom to make a downward transition. One can see that if we are exactly on resonance i.e. k = or k = then we get a singularity and as such becomes undefined. This concept will be discussed further into the course when a finite lifetime for an excited state is introduced. In short, the terms in the denominator have a complex term added which stops it going zero. Clearly for there to be an upward or downward transition = k, this means that the photon of the field must be approximately equal to the transition frequency. Consider the case for near resonant absorption In this case the second term in the brackets dominates such that a k (t) V k h e i ( k )t ( k ) 1 a k (t) = V k h 1 ( ) e i ( )t = V k h ei t e i t e i ( ) t = V k i h ei t e i t e i i t = iv k h t t = iv k h tei e i t t t sin sin t t

SCE3337: QMIII 3 Therefore the probability of being in the state Ψ k > is: a k (t) = V sin t k 4h t t One can show that a similar expression can be found for the stimulated emission term. You sin x may recognise that the factor in the brackets has the form x which is identical to the intensity profile of the single slit diffraction pattern as shown below: 1.8.6.4. -8-6 -4-4 6 8 x For all practical purposes, the transition probability is zero unless t Multiplying through by h gives: h h t h h Et h Note that h is Planck's constant. This sets an upper limit on the energy difference multiplied by the time t which has elapsed since the field was turned on: Et h This is ust the uncertainty principle! In effect it says that the longer that the perturbation is on, the more nearly ω =ω k.

SCE3337: QMIII 4 This is an interesting result, since it says that for near resonant absorption or emission the probability of the process is proportional to the time squared, however experimentally it is observed that the probability is directly proportional to the time. The reason for this difference is due to the fact that even under ideal conditions there is an intrinsic width to the energy levels for an excited atom (due to the uncertainty principle) which causes the probability to be proportional to t and not t. It is assumed that even for a perfectly monochromatic source of excitation with a single frequency such that = k means that will have a range of values. Then to get the probability of getting a transition from state Ψ > to Ψ k > is given by the integration over (large range): W(t) = V sin t k 4h t t d( ) = Taking t to be fixed we let X = t, hence d( ) = dx and substituting into the above t equation gives ( ) W(t) = V k 4h t sin X t X dx 1 X 4 44 43 V W(t) = k h t This result is known as Fermi's First Golden Rule. End of Part A of the Course

SCE3337: QMIII 1 8. Selection Rules Lecture 9 The selection rules allow us to determine which optical transitions are allowed and which are forbidden. They can be determined by considering the matrix element of the electric dipole operator. For optical transitions, the electric field dominates so we may write V = E D where D is dipole operator and is given by D=-er. Thus the matrix element is V k = ee < Ψ k (r,t) r Ψ (r,t) > The radial matrix element can be expanded in terms of spherical polar coordinates, since the relation between the Cartesian coordinates and the spherical polar coordinates are defined by z θ r y x φ x = rsin cos y = rsin sin z = rcos Taking ust the x component we have < k (r) x (r) >= * k (r) rsin cos (r)r sin drd d

SCE3337: QMIII Where we have separated the time dependence of the wavefunctions. We now separate the wavefunction as done previously: Then (r) = R n l (r)θ l m ( )Φ m ( ) < k (r) x (r) >= R * n k l k r 3 * R n l dr Θ lk m k sin * Θ l m d Φ mk cos Φ m d 8.1 The (m) Selection Rule Evaluating the integral in gives: * Φ mk cos Φ m d = e im k cos e im d = 1 e i ( m m k ) [ e i + e i ]d = 1 e i ( m m k +1 ) + e i ( m m k 1 ) d ( ) = or ( m m k 1) =. The This integral is equal to zero unless either m m k + 1 matrix element < k (r,t) x (r,t) > is therefore zero unless m m k = m = ±1. We find that we get the same result if we try to find the matrix element of the y component. For the z component the matrix element < k (r,t) z (r,t) > for the dependent part of the integral is Φ * Φ mk d = e im k e im m d = e i ( m m k ) d 1 4 443 = if m m k Therefore the only non-zero matrix element for the z component is for m=. Thus the selection rules for the magnetic proection quantum numbers for dipole allowed transitions are: m =,±1

SCE3337: QMIII 3 8. Spin Selection Rule Consider the spin of the electron. The total wavefunction may be written as the product of the spatial and spin terms: = nlml s The electric dipole operator does not act on the spin wavefunction (it does with magnetic fields, but very weakly with time varying electromagnetic waves) so we can write: Therefore < k (r) s k er (r) s >= e * k (r) * s k r (r) s d * = e s k s d * k (r)r (r)d = e * k (r)r (r)d s k s s = 8.3 J and L Selection Rules We can use the parity of the wavefunction to determine the L selection rule. Consider the effect of changing the sign of the x,y,z coordinates of the one electron atom wavefunctions. In spherical polar coordinates changing the sign is equivalent to: Thus r r,, + Under Parity Transformation ψ nlml (r,θ,φ) ψ nlm l (r,π θ,π+ φ) =( 1) l ψ nlml (r,θ,φ) The parity of the wavefunction is determined on whether l is even or odd. This result is applicable for all bound or unbound eigenfunctions for any potential that is spherical in form. Now consider the matrix element < k (r) er (r) >= e * k (r)r (r)d We have shown previously that the integration or r which is an odd function requires that the wavefunctions < k (r) and (r) > must have different parity so as to yield a symmetric definite integral of an even function (see the section on the DC Stark shift in hydrogen). If < k (r) has even parity then l k must also be even. If (r) > has odd parity then l must be odd. Clearly then l k l = l= ±1,±3,±5 etc and l = ±,±4,±6... must be excluded. Thus l can only change by odd integer values.

SCE3337: QMIII 4 To establish the possible range of values of l, conservation of angular momentum of the atom before and after it emits a photon must be considered. It is found that if an atom is in an excited state with quantum numbers =1 and m =1 and it decays to a ground state with =, m = then if the photon which is emitted travels along the z direction it is Left Hand Circularly (LHC) Polarised and carries away one unit of angular momentum h. Clearly, since the atom started off with one unit of angular momentum h, the atom after emitting the photon has zero angular momentum since the photon has carried away one unit of the angular momentum. Consider the case when the atom starts of in the =1 and m =-1 states and decays via a photon to the =, m = then if the photon which is emitted travels along the z direction it is Right Hand Circularly (RHC) polarised and carries away one unit of angular momentum - h. The maximum amount of angular momentum that a photon can carry is h, so this means that the total change in angular momentum is =±1. We can see that photons can have two angular momentum states, it is also possible to form a superposition state, the wavefunction of which is given by: 1 LHC + 1 RHC This corresponds to linearly polarised light. As such an atom emitting such a photon must exhibit no change in angular momentum, thus = Hence the selection rules for the total angular momentum are =±1, Clearly if =±1, then only l = ±1 are allowed for single photon emission since s=. Thus the selection rule for l is: l = ±1 Also note that = transferred. = transitions are not possible since no angular momentum will be The single photon dipole allowed selection rules are summarised as follows: =±1, l = ±1 m l = ±1, s = You should memorise these selection rules!

QMIII: SCE3337 1 9. The Einstein A and B coefficients Lecture 1 The basic model of the interaction process between electromagnetic radiation and atoms can treated by a model first introduced by Albert Einstein. This theory was phenomenonolgical in nature and makes no explicit use of quantum mechanics except that the energy levels of the atoms are assumed to be quantised and it is convenient to regard the electromagnetic field as photons. Suppose that we have N identical atoms in a gas and each atom has only two energy levels E 1 and E (we will label the states 1> and >) with E 1 < E as shown below: - hω N E g B 1 W A 1 > Spontaneous Emission B 1 W Absorbtion 1> N 1 E 1 g 1 Stimulated Emission We also have the transition frequency defined by h = E E 1 which is the difference in energy between the two energy levels. We also assume that the two energy levels have N 1 and N numbers of atoms each with degeneracies of g 1 and g respectively. There are three basic radiative processes that can occur: Absorption Suppose that the energy density of the radiation passing through the gas is W( ) where W( ) is the average energy density, that is the energy per unit volume per unit bandwidth. An atom in the lower energy state 1> can make an up going transition to the higher energy state > by absorbing a photon of energy h = E E 1. We assume in this case that the transition of this type is proportional to the energy density with a constant of proportionality of B 1. The upward transition probability is B 1 W. Stimulated Emission If an atom is in the state > and another photon that is resonant with the transition passes this atom, the presence of this photon can induce the atom to emit an identical type of photon which will then take the atom back to state 1>. This rate is proportional to the energy density of the incident radiation, with a constant of proportionality of B 1. Thus the stimulated emission probability from state > to 1> is B 1 W.

QMIII: SCE3337 Spontaneous Emission Consider an atom in the higher energy state >. There is a finite probability given by A 1 that the atom will pass from this state to the state 1>. In doing so it will emit a photon which has a random direction and polarisation with energy hω (hence the term spontaneous emission). Actually this process was a maor problem for quantum theory, the explanation of which had to wait until the advent of Quantum Electrodynamics (QED). This process can be explained as follows: the states > and 1> are eigenstates of the Hamilitonian and for an isolated system, once it is in this energy eigenstate, if it is unperturbed should remain there forever. QED relies on the basis that even when there are no photons around, the electromagnetic field still has a zero point energy (recall your lectures on the SHO!) and it is this field which induces the atom to decay. The so called vacuum fluctuations are an infinite virtual supply of zero point energy photons which cover all of the frequency range of the EM spectrum, which induces the atom to emit photons (like stimulated emission). You might actually think this is a crazy idea since that means the vacuum field has an infinite energy and this still remains one of the unsolved problems of QED. But there have been experiments such as the Casmir Effect which shows conclusively that this zero point energy exists. 9.1 Rate Equations We now consider the influence of these three processes on the energy level populations, N 1 and N. The total number of atoms is given by N which is the sum of the populations: N = N 1 + N The rate of change of population of the two energy levels is: dn 1 dt = dn dt Now the rate of change of the population in 1> is: dn 1 dt =(rate at which atoms enter the lower state) - (rate at which they leave) = (Stimulated Emission Rate + Spontaneous emission rate) - (Absorption Rate) = N B 1 W + N A 1 N 1 B 1 W This rate equation holds for the general case of electromagnetic radiation interacting with these two level atoms. Now consider the special case of thermal equilibrium. In this case, the population levels are constant, thus our rate equation becomes: Hence N B 1 W + N A 1 N 1 B 1 W = N B 1 W + N 1 B 1 W = N A 1

QMIII: SCE3337 3 ( N 1 B 1 N B 1 )W = N A 1 W = A 1 N 1 B N 1 B 1 For thermal equilibrium with no external radiation field on the gas, the relative number of atoms in various energy states is given by the ratio of the Maxwell-Boltzman distributions for each level: N 1 N = g 1 e g e E 1 kt E kt = g 1 e g h kt Substitution into the above equation reveals: W = g 1 e g A 1 h kt B1 B 1 This can be rearranged to give W = g 1 B 1 e g A 1 1 h kt B 1 A 1 Now this expression for the energy density must be consistent with the Planck's law for the radiative energy distribution of a body in thermal equilibrium which is given by : Wd = h 3 d c 3 h kt e 1 For single W = c 3 h 3 1 e h kt 1 Equating the our derived expression for the energy density with the Planck relation gives: 1 h g 1 B 1 kt B e 1 g A 1 A 1 = c 3 h 3 1 e h kt 1 Hence matching coefficients yields B 1 A 1 = c 3 h B = c 3 3 1 h 3 A 1

QMIII: SCE3337 4 Also But B 1 = g 1 g B 1 A 1 = c 3 h 3 c 3 h A A = h 3 3 1 1 c 3 B 1 Substitution into our expression yields B 1 B 1 = g g 1 As shown the three Einstein coefficients are inter-related. It can be seen that without introducing the stimulated emission process, consistency between the Planck formula and the Einstein expression could not be achieved. Furthermore it must be stressed that although the relationship between the Einstein coefficients have been derived for the thermal equilibrium case, they hold generally since the coefficients are independent of the magnitude of the energy density or the temperature. It should also be noted that for thermal equilibrium, the radiative energy density W is distributed isotropically in space. This is of course not so with a light beam. The relationships do remain valid in systems such as a gas or a fluid in which the atoms or molecules have random orientations so that the interaction within the gas as a whole is isotropic. However in solids, the constituent atoms or molecules may be locked in a common orientation. In this case, the bulk material may have quite anisotropic optical properties. One point of interest is the ratio of power emitted in the spontaneous emission process compared to the stimulated process: A 1 B 1 W = Ratio of spontaneous to stimulated emission We need to look in detail at the form of the energy density W which is dependent on the frequency of the radiation. Recall for the case of thermal equilibrium that the energy density is given by Planck's Law for a frequency interval of + d is: Wd = h 3 d c 3 h kt e 1 For room temperature at approximately 3K then the ratio frequency is around 6x1 1 Hz with h k B T 1, the corresponding 5µm which is in the infrared part of the spectrum.

QMIII: SCE3337 5 h For frequencies where << 1 (that is for longer wavelengths such as microwave, k B T radiowave etc) then the energy density is large since the exponent in the denominator is close to 1 as such A 1 << B 1 W This means that for frequencies less than 6x1 1 Hz, the thermally induced spontaneous rate is much less than the stimulates emission process. In the case where frequencies are larger than 6x1 1 Hz i.e. h k B T >> 1 then the exponential in the term for the energy density dominates and as such the energy density is small thus A 1 >> B 1 W As such for frequencies that are greater than 6x1 1 Hz, the thermally induced spontaneous emission rate is much larger than the stimulated process. 9. The Quantum Theory of the Einstein B Coefficient So far in this section of the course, we have treated the Einstein A and B coefficients as parameters that are experimentally determined. But, in fact, the absorption and emission of the radiation can be calculated by using Time Dependent Perturbation Theory. Consider now an idealised atom consisting of two energy levels 1> and > with transition frequency ω 1. We now apply a harmonic perturbation on this system with radiation that has frequency ω and is near resonant with the transition frequency as shown below. > E - hω ω 1 1> E 1 The transition probability is given by Fermi's first Golden Rule (see section 7. of the lecture notes): V W(t) = k h t

QMIII: SCE3337 6 Recall that in the harmonic perturbation case the perturbation had the form h = V cos t. This term more accurately should be written as h = Vcos( t + k r) which means that there is a spatial dependence to the perturbation (without this term we have ust assumed that the perturbation is at the origin). This perturbation Hamiltonian describes the interaction between an atom and the electric field of the light. Consider the electromagnetic wave to be polarised along the x direction and propagates along the z direction. The atom consists of a nucleus and is surrounded by n electrons, over the dimensions of the atom, which is in the order of 1-1 m. For optical frequencies >1 15 Hz then k r =kz <<1. Thus the spatial variations of the electric and magnetic fields are virtually constant over the atomic dimension and hence the electrons in the atom experience the same electromagnetic fields. This approximation is known as the Dipole Approximation. The total electric dipole moment of the atom is given by: D = n =1 er The interaction Hamiltonian is given by the following h = E D = E Dcos t (we have dropped the spatially dependent term) Thus the matrix elements are V k = < k E D >= E < k D > with E parallel to the x-direction. Substituting this back into Fermi's First Golden Rule gives the transition probability of going from state 1> to state > is <1 D W 1 (t) = E x > t h Now the energy density of an electromagnetic wave is given by W = 1 E Hence the probability of transition can be related to the energy density such that; W 1 (t) = h <1 D x > Wt It follows therefore if we divide this equation by t we will get the transition rate: W 1 (t)/ t = h <1 D x > W

QMIII: SCE3337 7 We can equate this to the induced transition rate ( B 1 W ): B 1 W = h <1 D x > W Thus the Einstein B coefficient is B 1 = h <1 D x > for a single atomic system. What about a gas or molecules or atoms? In this case all of the electric dipoles will be orientated in different directions at any particular time in space in random directions. Given that θ is the angle between the dipole and the E field then < D x >=< D > cos we need the average value of <1 D x > <1 D x > = <1 D > cos Thus we need the average value of cos is given by integrated over the solid angle of a sphere which cos = 1 3 Exercise: Show this Hence the Einstein B coefficient is B 1 = 3 <1 D > h The B 1 coefficient can be found by the relationship derived earlier: B 1 = g B B 1 g 1 = g 1 B 1 g 1 B 1 = g 1 <1 D > g 3 h

QMIII: SCE3337 8 This approach does not yield the Einstein A coefficient directly but we can get this from the relationship between the A and B coefficients derived using the rate equation model: A 1 = h 3 c 3 B 1 Therefore upon substitution into the relationship for the A coefficient reveals; A 1 = g 1 g 3 3 hc 3 < 1 D >

QMIII: SCE3337 1 1. Optical Excitation of Atoms Lecture 11 Suppose that we consider atoms contained in a thin slice of gas perpendicular to an incident light beam, so that the intensity of the incident light changes by a negligible amount as shown below: Let the beam be switched on at time t= and all atoms are in their ground state. We want to find the number of atoms in their excited states at some time t later. Recall the rate equation for a two level system (lecture 1) was given by dn 1 dt = dn dt = N A + ( N N 1 )BW If the total population is N then N = N 1 + N N 1 = N N Substitution into the above differential equation reveals dn dt dn dt = N A + ( N N)BW = N ( A + BW) NBW This differential equation can be solved by separating the variables. let N dn A + BW ( ) NBW = dt c 1 = A + BW c = NBW

QMIII: SCE3337 Integrating yields N dn = dt N c 1 c t recall the standard integral dx = 1 ln ax + b ax + b a ( ) Hence 1 ln( N c c 1 c ) 1 N t = [ t] Therefore ln( N c 1 c ) ln( c ) = c 1 t ln N c c 1 = c c 1 t N c c 1 = e c 1 t c N = c c 1 ( ) ( 1 e c 1t ) { } Substitution for c and c 1 gives N as { } NBW N = A + BW 1 ( e A+ BW t ) Graphically the solution looks like:.8 Excited State Population Vs Time.7.6 N Population.5.4.3..1 4 6 8 1 time

QMIII: SCE3337 3 It is interesting to look at this population for different ranges of the exponential in this expression. For A + BW ( )t << 1. In this case we can Taylor expand the exponential: ( A+ BWt ) e 1 ( A + BW )t + H.O.T. Substituting back into our expression to N above gives N = N = NBW A + BW { 1 [ 1 ( A + BW)t] } NBW ( A + BW A + BW )t N = NBWt Therefore initially the population dependence of N is initially linear as is seen in the graphical solution above. We can determine the long time dependence of the population in the excited state quite easily as the exponential term becomes very small and as such negligible: N NBW A + BW ( ) We could also obtain this solution for long times by solving the d.e. for N at steady state. Recall that in the visible spectrum and for an ordinary light source (such as a fluorescent lamp) that the spontaneous emission rate is greater that then stimulated rate i.e. A >> BW as such from the above equation we deduce that N N << 1 This means that most of the atoms are in the ground state (N 1 ). If we used a much more powerful light source such as a laser beam then BW >> A (stimulated emission processes are much greater that the spontaneous processes) using this condition on our steady state solution gives N = NBW BW = N The interpretation of this expression is that for powerful excitation, the limiting population in the excited state is half of the total population, so no matter how many more extra photons we put into the system the population in the excited state can not exceed 5% of the total population. The effect is called Saturation and can be seen in the graph above. It is simple to explain this effect as in the high light intensity case the simulated emission effects balance the absorption effects (where we have assumed that spontaneous emission is negligible in high intensity fields).

QMIII: SCE3337 4 If we now turned the incident light beam off, the excited atoms will return to their ground state via spontaneous emission of photons. Then our rate equation for the excited population becomes with W set to zero dn = N dt ( A + BW) NBW = N A dn = N dt A This equation is separable and easily solved with the boundary condition that at time t= there are N steady state atoms in the excited state then N dn = Adt N Therefore N t N = N e At A plot of this function is shown below:.6 N Population Decay Curve.5.4 N Population.3..1 4 6 8 1 time The radiation emitted in this process is called fluorescent radiation and the measurement of this provides another experimental technique of measuring the Einstein A coefficient. The reciprocal of A gives the average lifetime for the state r : r = 1 A

QMIII: SCE3337 5 1.1 Simple Optical Processes You have already encountered in your electromagnetism courses, the macroscopic theory of absorption and dispersion of electromagnetic radiation passing through a medium. In this lecture we will establish the connection between the Einstein A and B coefficients and this macroscopic theory. This is important as we need to be able to make the quantum theory converge with the classical macroscopic theory on a large scale. Let us first re-visit some of the aspects of the macroscopic theory. Consider the picture below: Incident Light GAS Transmitted Light We have a light beam incident on a gas and this medium we consider as a dielectric material where the polarisability is proportional to the applied electric field: P = E where P is the polarisability of the medium and χ is the susceptibility. For an isotropic dielectric, the refractive index is related to the susceptibility by n = 1 + For a plane, monochromatic wave passing through a medium, the electric field vector is given by E(z,t) = E e i This can be rewritten as ( t nkz) E(z,t) = E e i( t kz ) e i( n 1 )kz = E (z,t)e i n 1 ( )kz The effect of the medium on the electromagnetic wave is given by the second exponential term. For small susceptibilities i.e. <<1 then n = ( 1+ ) 1 1 1 + (this is found by Taylor expanding the function about the point = ) is in general complex so it may be separated into its real and imaginary components such that = + i

QMIII: SCE3337 6 Substitution into the expression for the electric field gives E(z,t) = E (z,t)e i Refractive 67 8 Index kz e Absorption 67 8 kz The first exponential is related to the refraction of the wave (Snell's Law) as it enters the medium and the second term is related to the absorption of the radiation. The electromagnetic energy density is I = 1 c E(z,t) Thus the intensity of the wave after passing through the medium is then I = 1 c E (z,t) e kz This is known as Beer's Law. 1. Connection between the Einstein A and B coefficients and the susceptibility Consider a gas of two level atoms once again. We know that if we shine near resonant light on the medium that three process can occur which will effect the transmission intensity of the incident light as depicted above.

QMIII: SCE3337 7 1) Absorption will reduce the transmission intensity the rate of which is proportional to BW ) The stimulated emission process can be thought of a producing a carbon copy on the incident photon which travels in the same direction with identical polarisation etc and adds in phase to the incident light beam. The rate of this process is proportional to BW. 3) The spontaneous emission process is due to the vacuum electromagnetic field which has all frequencies, polarisations etc stimulating the excited atom to emit a photon and as such the atom emits into a 4π steradian solid angle thus reducing the intensity of the transmitted beam. This rate of emission is proportional to A. Suppose that now that we have our atoms in steady state and for simplicity we assume the medium has identical degeneracies: g 1 =g then our rate equation for the ground state population is dn 1 dt = N BW + N A N 1 BW = Thus N A = ( N 1 N )BW This tells us that the rate at which photons are scattered out of the beam due to spontaneous emission is balanced by the difference due to absorption and stimulated emission processes. Up to this point we have assumed that the frequencies of photons absorb/emitted by the atoms have been a single frequency but in reality the emitted photons have a spread of frequencies over which the atoms can absorb of emit photons. The spread of frequencies under these gas slice conditions is mainly due to the Doppler effect (atoms moving at different velocities will see the photons with different frequencies) as well as collisions of the atoms in the gas. To account for this frequency spread we introduce a function F( ) which is the fraction of atoms with frequency : F( ) d =1 Therefore over some range of frequencies say + d, the net rate at which light energy is absorbed by N atoms is ( N 1 N )BWh F( )d

QMIII: SCE3337 8 Consider a slice of gas medium shown below: The energy contained in the slice of the medium of cross sectional area a and length dz in the frequency range of + d is Energy in Medium = a[ I(z)d I(z + dz)d ] Recall that the small change in a function F is given by F = i F dx i x i Hence I(z + dz) = I(z) + I z dz The expression for the energy in the medium becomes: Energy in Medium = a I(z)d I(z) + I z dz d = a I z dzd That is the energy in the medium is equal to the change in energy of the incident light beam over the volume V. Recall the rate of energy transferred to a medium by the rate equation approach was ( N 1 N )BWh F( )d N 1 is the number of atoms in the ground state and N the number atoms in the excited state. In a slice of gas which has the volume V;

QMIII: SCE3337 9 Number of ground state atoms = N 1 adz V Number of excited state atoms = N adz V Therefore our expression becomes ( ) adz Rate of energy change in volume V = N 1 N V BWh F( )d Equating this to the other expression for the energy in the medium gives Therefore a I z dzd = ( N N 1 ) adz BWh F( )d V I z = - N 1 N ( ) Bh V F( )W From the rate equation steady state expression Then Hence N = NBW A + BW ( ) ( N 1 N ) = N N = NA A+ BW ( ) I z = - NA A+ BW ( ) Bh V F( )W We also note that W = I, thus our expression becomes c I z = - NA A+ B I c Bh V F( ) I c Rearranging this expression to separated the variables yields 1 + BI Ac di I = - NBh Vc F( )dz

QMIII: SCE3337 1 There are two interesting cases that we can investigate. Case 1: Suppose that the rate of spontaneous emission>>absorption rate or the stimulated emission rate then the quantity i.e. the intensity of the light is weak then, BI Ac = B W << 1 and as such is negligible compared to the 1 term on the RHS of A the equation. Therefore di I =- NBh Vc F( )dz We now make the substitution K = NBh Vc F( ) Hence Integrating Thus di I =-Kdz I di = -Kdz I I z I(z) = I e Kz This is ust an exponential fall off in intensity and provided that the atom density and the spectral profile are known we can then obtain the absorption coefficient B. Essentially we have recovered Beer's Law. K( ) = N V Bh c F( ) Rearranging this expression BF( ) = 1 c h K( ) where is the density: = N V Integrating this expression over all frequencies gives B F( )d = c h 1 K( )d

QMIII: SCE3337 11 Recall for a normalised spectral profile that F( ) d =1 Hence B = c h 1 K( )d Usually the frequency spread is small and the 1 term can be factored out as for the Einstein B coefficient 1 yielding B = c h K( )d Case : If the absorption or the stimulated emission rate >> spontaneous decay rate (eg: strong laser excitation), thus the term Therefore BI Ac >> 1 BI di Ac I =- Bh c F( )dz di =- Ah c F( )dz Integrating Thus I I di = - Ah c I(z) = I Ah z F( ) dz c F( ) z We get the interesting result that the intensity falls off linearly with distance rather than exponentially. We also notice that the intensity is proportional to the Einstein A coefficient unlike our low intensity case which had a rapid exponential decay dependence on the B coefficient.

SCE3337: QMII 1 11.1 Population Inversion Lecture 1 The Laser Consider our two level atom once again being irradiated with near resonant light as shown below: The usual three processes will occur: absorption, spontaneous emission and stimulated emission. If the number of excited atoms, N is greater than N 1, then instead of absorption of the incident light beam, amplification will take place. It can be thought of as if the gas had a negative absorption coefficient. The condition in which N >N 1 is known as population inversion. This is impossible for a two level atom as we have seen in previous lectures 11. The Three Level System We now consider a method of creating a population inversion via the use of three energy levels. Consider a system that consists of three non-degenerate energy levels, level > is a ground state and states 1> and > are excited states as shown below: We have a light beam with energy density W p which is used to excite population from the > to the > transition. A light beam used in this manner is called a pump beam. The

SCE3337: QMII 1 fraction of the total number of atoms pumped up is usually small and in the order of 1. 6 An atom in level > can emit light by making a downward transition by either > 1> or > > and an atom in level 1> can decay to the ground state ( >). We will soon see via rate equation that it is possible to achieve a population inversion between levels > and 1> i.e. N >N 1. We have some radiation with a frequency ω that corresponds to the transition frequency between levels > and 1> that is incident on the system i.e. ( = E E 1) h This radiation has energy density W and hence will induce transitions from 1> > (absorption) to as well as > 1> stimulated emission. Now since there is a population inversion the number of simulated photons is greater than the absorbed photons and hence the incident radiation will be amplified. We assume that only the three levels in our diagram participate hence the total population is N = N 1 + N + N The three rate equations that describe the change in population of the three levels are given by dn dt = N B W p N B W p N A N B 1 W + N 1 B 1 W N A 1 = ( N N )B W p ( N N 1 )B 1 W N A N A 1 (1) dn 1 dt = N B 1 W N 1 B 1 W + N A 1 N 1 A 1 = ( N N 1 )B 1 W + N A 1 N 1 A 1 () dn dt = N B W p + N B W p + N A + N 1 A 1 = ( N N )B W p + N A + N 1 A 1 (3) These equations must balance so that the sum of the rates is equal to zero exercise: Show that this is true In the steady state condition all the equation above are set to zero (i.e. the population in all levels stay the same). Under this condition, equations may be found for the level populations N, N 1, N in terms of N, W and W p. The terms for this are rather lengthy and

SCE3337: QMII 3 the main features of the steady state solution may be appreciated more easily by the following procedure. Let us define a pumping rate r which describes the population in the N level due to the pump: rn = N B W p N B W p = ( N N )B W p rn is the net rate at which atoms are excited into the state > by the pump. Consider the steady state condition for the second rate equation () then, = ( N N 1 )B 1 W + N A 1 N 1 A 1 = N ( A 1 + B 1 W) N 1 ( A 1 + B 1 W) N ( A 1 + B 1 W ) = N 1 ( A 1 + B 1 W) (4) Now from the third rate equation we have under steady state conditions From above this term = Nr 64 47 44 8 = ( N N )B W p + N A + N 1 A 1 = N A + N 1 A 1 Nr Nr = N A + N 1 A 1 (5) From equation (4) we can show that N N 1 ( ) ( ) = A + B W 1 1 A 1 + B 1 W Thus in order to get a population inversion between > and 1> we must have ( A 1 + B 1 W) A 1 + B 1 W ( ) > 1 which will occur for A 1 > A 1 This tells us that atoms that are excited to state > must decay relatively slowly into state 1> from which they drop rapidly into the ground state. We can solve the above equations to give N 1 = N = A 1 A 1 N( A 1 + B 1 W)r A 1 + B 1 W ( ) + A ( A 1 + B 1 W) N( A 1 + B 1 W )r A 1 + B 1 W ( ) + A ( A 1 + B 1 W)

SCE3337: QMII 4 We have now established a method of obtaining a population inversion which is capable of amplifying the radiation density W as it passes through the gas. It is then possible to substitute in the various Einstein A and B coefficients for different transitions and determine whether a particular set of transitions is suitable for lasing action and determining the population. 11.3 Lasing Action It is all very well to get a population inversion in a gas, it is another problem to obtain the so called lasing action. In essence, a laser consists of some medium, in which population inversion occurs as well as an enclosed cavity (sometimes it is also referred to as a resonator) with reflecting walls. A generic simple laser is shown below: Suppose that we have created a population inversion in a gas medium which has an energy level structure similar to the three level system we considered above. The population inversion may be created by a flashlamp or another laser. A spontaneously emitted photon with a frequency that matches the transition energy will also stimulate the emission of other identical photons. Surrounding the medium by two mirrors allows us to redirect photons in one dimension back and forth an in doing so stimulate more photons. Only photons travelling at right angles to the mirrors will be significantly amplified and of course will be useful to the laser beam. As the photons travel back and forth through the medium, the number of photons identical with the first photon will build up provided that the amplification or the gain of the medium is greater than the cavity losses due to imperfect mirrors. The point at which the gain equals the losses is called threshold. One of the mirrors in the laser cavity is made to be slightly transparent and allows a fraction of the photons to escape, these escaped photons constitute the laser beam. The mirrors enforce boundary conditions (i.e. the electric field must vanish at the surface of the mirrors) on the photons propagating within the cavity and hence only standing waves will be formed in this cavity (remember that the photons are also electromagnetic waves). As such the cavity will only support wavelengths with n = L where L is the length of the laser cavity. The first four frequency modes are shown below:

SCE3337: QMII 5 Recall that c = f f = c From the previous page Thus = L n f n = nc L = nf 1 where f 1 is the fundamental mode. This means that the possible output frequencies are spaced by c. For a 1 meter laser the frequency spacing is f=15mhz. L

SCE3337: QMII 6 11.4 Longi tudinal Modes c L The figure above displays the output from the laser which consists of a number of discrete lines within a Gaussian envelope of the Doppler broadened line (essentially this broadening is due to the motion of the atoms, since we have some thermal distribution of atoms (i.e. different speeds) then they atoms will see the stimulated photons with different frequencies hence the Gaussian envelope). The discrete lines are called the longitudinal modes of the laser and such operation is multimode behaviour. It is possible to run in the so called single mode by filtering out undesired frequencies by using frequency selecting devices within the laser cavity such as an etalon. 11.5 Transverse Modes A laser also behaves like a waveguide and as such each longitudinal mode also has a transverse mode structure (TEM). Essentially this comes from the fact that our waves are three dimensional. The transverse profile of the electric field within the laser resonator can be found by using Huygen's principle. It is possible to show that the transverse profile of a spherical wave is Gaussian in nature which is what we expect to see from the output of the laser. Many laser resonators use spherical mirrors rather than planar mirrors as it is simpler to align these systems. These system of mirrors also support higher order Gaussian beam modes. If we consider our laser longitudinal modes in the z direction then the TEM mode structure is denoted as TEM mn where m and n define the number of nodes (ie the number of intensity points) in the x and y directions respectively. Some of the lower order modes are shown below: TEM TEM 1 TEM 11 TEM

SCE3337: QMII 7 11.6 Excitation Mechanisms To Create Population Inversion Optical Excitation One of the obvious techniques that we can use to create our population inversion is to use a high intensity light source on our medium. One such source is known as a flash lamp which produces a high intensity burst of incoherent light. The ruby laser is one such laser that works on this principle and is surrounded by a flashlamp as shown below The ruby (Al O 3 ) is doped with chromium atoms which creates Cr 3+ ions in the ruby (it is the chromium atoms in ruby that give them the characteristic redness). The presence of the ruby with the Cr 3+ ions modifies the ions energy level structure as shown below

SCE3337: QMII 8 The flashlamp provides the pump beam with radiation at approximately 55nm in which the ground state Cr 3+ ions absorb the radiation and are promoted to the 4 F energy level. This state decays almost entirely to the E state via rapid non-radiative decay. The E state is a metastable state with a long lifetime in the order of 5msec and as such it is possible to obtain a population inversion between the ground state and this metastable state provide that the pump is strong. The transition from this state to the ground state is the lasing transition. One of the problems associated with this laser as it requires a great deal of pump energy and hence a low efficiency. This however was the first optical laser which was created by Maiman. It was interesting to see that Maiman publish this paper in 196 in Nature # in a communication that was only 3 words long! Other optical excitation is also possible such as the use of another laser such as in the case of a dye laser which use another laser for their pump. Another possibility is a diode laser array which produces intense pump beams for high laser power such as those used in Nd:YAG lasers. These types of lasers are also solid state and the host crystal in this case is Y 3 AL 5 O 1 (called YAG, an acronym for yttrium aluminium garnet) in which Y 3+ ions are replaced (doped) with Nd 3+ ions (Neodymium). It is also possible to use flash lamps for population inversion in this medium. Transfer of Excitation This technique is used in many gas lasers such as the helium-neon laser (He-Ne). Essentially this laser consists of a gas of He and Ne which has a high voltage place between two electrodes which creates a gas discharge. Electrons from the gas discharge inelastically collide with ground state He atoms exciting them to the 3 S metastable state. The figure below shows the relevant energy levels for He and Ne. Thermalising collisions occur between the excited He and the ground state Ne which can promote a transfer of energy from the excited He atoms to the ground state neon atoms leaving the Ne atoms in the s state which has comparable energy to the 3 S state of He (naturally the He atoms go back to the ground state if they lose their energy). The s state actually consists of 5 different energy levels separated by.15ev. A population inversion exists between the s states and the lower energy p states which consists of 1 closely spaced energy levels. The lifetime of the p states is about half as long as the S states and it is possible to get lasing action on thirty of the allowed s p transitions. The excited p Ne atoms can then decay via photon allowed transitions to lower energy levels and finally to the ground state for repumping. # T.H. Maiman, Nature 187, 493 (196).

SCE3337: QMII 9