Lecture #13 1. Incorporating a vector potential into the Hamiltonian 2. Spin postulates 3. Description of spin states 4. Identical particles in

Transcription

1 Lecture #3. Incorporating a vector potential into the Hamiltonian. Spin postulates 3. Description of spin states 4. Identical particles in classical and QM 5. Exchange degeneracy - the fundamental problem 6. Symmetrization postulate 7. Pauli s exclusion principle 8. Constructing many electron wavefunctions the Slater determinant 9. The Helium atom in the independent electron approximation 0. The fine structure of the He atom singlets and triplets.

2 Introduction Until now we have treated the electron as a point particle with three degrees of freedom associated with it s position described by x,y,z coordinates. Therefore the ψ xyz,,. wavefunction describing the electron depended on the three coordinates ( ) Using the postulates which were mentioned in the beginning of the course we were able to study the physical properties of a number of systems, of particular interest was the Hydrogen atom which allowed us to reproduce the emission spectra of this atom to a high degree of accuracy. While this approach works well for non-relativistic particles (v<<c) a more accurate theory is needed when analyzing systems where relativistic effects are important. The Dirac equation provides the basis for the incorporation of relativistic effects into quantum mechanics. In this course we will not study this equation but will to a certain extent include it s consequences in the form of postulates. One of it s more important consequences was the prediction of the existence of an additional intrinsic property of particles called spin (on equal footing to mass and charge!). When a spinless particle of mass m e is placed in a system with a vector and scalar potential the Hamiltonian is of the following form: Hˆ = ( Pˆ qaˆ ( Rˆ) ) + Vˆ( Rˆ) m e in particular the effects of a magnetic field B are incorporated via a vector potential A related to the magnetic field by: B = A The simplest and commonly referred to case is that of a uniform magnetic field which can be shown to have a vector potential equal to A= r B In this case B is a constant and therefore all operators commute with it which allow us to simplify the form of the Hamiltonian: Hˆ = ( Pˆ qaˆ ( Rˆ) ) + Vˆ( Rˆ) m e ( ) ( ) q ( ( ) ( ) ) ( ) ˆ q ˆ ˆ ˆ ˆ ˆ = P + B P R R P B + RB Rˆ B + Vˆ Rˆ me 4 The expression was written as to bring out the explicit dependence on the angular momentum: Lˆ = Rˆ Pˆ One can then rewrite the Hamiltonian as: H ˆ = H ˆ + H ˆ + H ˆ 0

3 Where, ˆ ˆ P H ˆ 0 = + V me ˆ µ B H ˆ = L B paramagnetic ħ qb Hˆ ˆ = R diamagnetic 8me 4 Joule qħ µ B = m e Tesla is the Bohr magneton, and in a coordinate system where the z component is in the direction of the magnetic field Rˆ = Xˆ + Yˆ. Order of magnitude analysis: E Hz >> E 0 Hz >> E 0 Hz h h h The physical interpretation of the second term is the coupling of the magnetic field and the magnetic moment M associated with revolution of the electron in its orbit and can thus be seen as an orbital magnetic moment. Hˆ ˆ = M B this term is called the paramagnetic coupling term. The eigenvalues associated with this term are therefore directly related to the eigenvalues of the L ˆz mµ Bwhere m is an integer. B Metals in Groups A (Alkali s) and B (Cu, Ag, Au) are well represented by an electron configuration where the valence electron is in an outer s shell and all the other electrons are in orbitals of lower principle quantum number: For example: Cu s s p 3s 3p 3d 4s What happen to an atom of silver placed in a magnetic field? What should the magnetic moment of the atom be due to the motion of the electron the s electron should not contribute anything because it has an orbital angular momentum of 0 yet it was observed (Stern and Gerlach) that when you pass a beam of silver atoms through a magnetic field you get a splitting of the atoms! The description of an electron by a wavefunction which contains only the spatial coordinates is incomplete: We thus introduce additional degrees of freedom or variables which describe the particle s spin state. The spin of a particle is an intrinsic property which cannot be changed (similar to mass and charge). 3

4 The spin intrinsic physical property is described by the spin operator: S ˆ = S ˆ, S ˆ, S ˆ ( x y z) In order to completely describe the particle one cannot solely use orbital variables spin variables are needed. These variables satisfy the following conditions: () The spin variable is an angular momentum: Sx, Sy = iħ Sz () The spin operators act on a new space which is called spin space in this space: ˆ S, S ˆ z are a CSCO. This space is thus spanned by eigenvectors common to these two operators ˆ Su = ħ s s+ u sm ( ) Su ˆ z sm = ħmusm As we saw in the case of the orbital angular momentum s must be an integer or half integer and m takes on values, s s+, s+,..., s + s (3) A given particle type is described by a unique value of s the particle is said to have a spin s. The spin state space is always of dimension s+ and all spin states are eigenvectors of Ŝ with the same eigenvalue s. (4) Unlike the orbital angular momentum where l can take on arbitrary values for a given particle the spin s is a property of the particle and does not change. (5) The state space of a particle is a tensor product of the orbital and spin spaces. All spin operators commute with all orbital operators. ψ r = φ r χ ( ) ( ) nlmms nlm ms ( ) ( ) = ( ) ( ) ( ) = ( ) sm in a slightly different notation, s a s s b s (6) The electron is a spin/ particle. (7) The particle interacts with a magnetic field via the spin angular momentum and the associated spin magnetic moment: ˆ µ M B ˆ s = S ħ Examples of other spin properties of particles proton, neutron = 0.5, photons =. 4

5 Identical Particles Definition: Two particles are said to be identical if all of their intrinsic properties (mass, charge, spin) are exactly the same. No experiment can distinguish one of them from the other. An important consequence follows from this definition: when a physical system contains two particles there is no change in its properties if the roles of the two particles are exchanged. Classically all particles can be distinguished because all trajectories are deterministic particles can be labeled by procedures which do not effect the behavior of the particles. In QM it is impossible to distinguish between two interacting identical particles resulting from the finite extent of the wavefunctions. Example on the board: collision of two particles two distinct wavefunctions can be used to describe the final outcome. The fundamental problem exchange degeneracy The problem will be illustrated using a system of spin ½ particles: A] The system: spin ½ particles with one having a component of the spin in the z direction equal to +ħ / and the other with ħ /. B] The state vector used to describe this system is m, m where mi can be equal to +/ or -/. Sˆ m, m = mħ m, m z Sˆ z m, m = mħ m, m C] Two different vectors can be associated with this physical system:, or, These two states span a two dimensional subspace whose normalized vectors are of the form: ψ = α, + β, with α + β = all of these vectors can be used to represent the same physical state (i.e. that of one particle with spin up and one with spin down). This is called the exchange degeneracy. D] The exchange degeneracy leads to fundamental problems. For instance, suppose we would like to predict the probability of obtaining a +/ measurement of the spin in the x direction. The spin operator associated with the spin measurement in an arbitrary direction is given by the Pauli matrices: Sˆ = ħ σˆ 5

6 0 0 i 0 σ =, σ, σ x 0 = y = i 0 z 0 0 The eigenvectors of S ˆz are S=+/ z = and S z=-/ =. The eigenvectors 0 of S ˆx are S=+/ x = = ( S=+/ z +S=-/ z ) - and S x=-/ = = ( S=+/ z +S=-/ z ) The state vector associated with the measurement of +/ in a spin measurement performed in the x direction for both particles is given by the tensor product of the two individual state vectors: S =+/ S =+/ x x = ( S z=+/ +S z=-/ ) ( S z=+/ +S z =-/ ) = ( +/,+/ + +/,-/ + -/,+/ + -/,-/ ) = S x =+/,S x =+/ Now considering that our state is: ψ = α, + β, The probability of obtaining the result of +/ in a spin measurement in the x direction is: P = ψ S x=+/,s x =+/ = α + β ψ = α /, / + β /,/ ( ) S x=+/,s x =+/ = ( +/,+/ + +/,-/ + -/,+/ + -/,-/ ) which depends on the choice of α and β! We must therefore specify unambiguously which state vector is to be chosen i.e. the exchange degeneracy has to be removed. Let us define a permutation operator: Pˆ u, u = u, u i j j i 6

7 Let us examine the spectrum of the permutation operator, ˆ P = Pˆ u =+ u S Pˆ ua = ua The e-functions of the permutation operator are functions that are symmetric or antisymmetric with respect to particle permutation. Operators of identical particles have to be symmetric with respect to particle number exchange and thus commute with the permutation operator. Therefore they should have common eigenfunctions. S The symmetrization Postulate (Postulate VIII): Only certain state vectors can be used to describe a physical system which consists of several identical particles. These vectors are, either completely symmetric or antisymmteric with respect to particle permutation. Particles for which the state vectors are symmetric are called Bosons and those for which they are antisymmetric are called Fermions. All currently known particles obey the following empirical rule: Particles of half-integral spin are all fermions while those with integral spin are bosons. There is a way to construct wavefunctions which satisfy the requirement: u u S A ( α β β α ) ( x, x ) = φ ( ) φ ( ) + φ ( ) φ ( ) ( α β β α ) ( x, x ) = φ ( ) φ ( ) φ ( ) φ ( ) Both of these solutions correspond to the same total energy eigenvalue and are thus degenerate this degeneracy is called exchange degeneracy since the difference between the eigenfunctions has to do with the exchange of the particle labels. Consequence: Pauli s Exclusion Principle Weaker form: In a multi electron atom there can never be more than one electron in the same quantum state. ua ( x, x) = ( φα ( ) φα ( ) φα ( ) φα ( ) ) = 0 Stronger from: A system describing several electrons must be described by an antisymmetric total eigenfunction. The eigenfunction must be antisymmetric with respect to simultaneous interchange of space and spin coordinates of electrons. 7

8 Finding the form of the normalized total electron wavefunction for a system of N non-interacting electrons The Slater determinant: It was pointed out by Slater that one can write write wavefunctions which are guaranteed to be antisymmetric for interchange of space and spin coordinates: Lithium in the ground state: Li : φ () φ () φ () 00, 00, 00, s s φα φα φα 3 s s s 3 ua = φβ φβ φβ 3 = s s s 3 3! 3! φ φ φ 3 s s s 3 α = 00, β = 00, γ = 00, γ γ γ Many Electron Atoms ˆ ħ H(,,3... N) = Ze + e m r r n n n n i i= i= i i= j=+ i ij Assuming fixed nucleus no kinetic energy term associated with the CM motion. 8

9 The Helium Atom The system: The Hamiltonian: ˆ ħ ħ e e e H = + + m m r r r which can be written as: ˆ ( ) ˆ ( ) ˆ e H, = H + H( ) + r There is no way to separate this Hamiltonian completely! independent electron approximation To simplify the analysis we neglect the e-e interaction define an approximate Hamiltonian: Hˆ, = Hˆ + Hˆ approx The spatial solutions of H() and H() will be just Hydrogen atomic orbitals scaled by a factor of Z ħ ao ao = µ e Z For example: E φ E I 4 µ e = ħ 3.6 = ev Z EI 3 r 00 = 3 8π a0 a0 e ( ) φ ( ) = Eφ ( ) Hˆ i i i r a0 The product of the atomic orbitals are eigenfunctions of the approximate Hamiltonian. ( ) φ ( ) φ ( ) = ( + ) φ ( ) φ ( ) Hˆ, E E approx i j i j i j 9

10 In the independent electron approximation stated above the ground state for the Helium atom is just: u, = s s ground state The electronic configuration for this state iss. What do we expect for the first excited state? u, = s s or (,) = ( ) ( ) u s s We have another case of exchange degeneracy! Obviously these functions are not eigenfunctions of the permutation operator and thus are not anti-symmetric with respect to exchange of particles violate the symmetrization postulate. The orbital symmetric and antisymmetric functions were described above: us ( x, x) = ( φα ( ) φβ ( ) + φβ ( ) φα ( ) ) ua ( x, x) = ( φα ( ) φβ ( ) φβ ( ) φα ( ) ) Let us now look at the spin eigenvectors, since the spins can take on discrete values there are four possibilities: Anti-symmetric (singlet) ( /, / /,/ ) ( ( ) ) a b b a Symmetric (triplet) ( ) a( ) a /,/ ( a( ) b( ) + b( ) a( ) ) ( /, / + /,/ ) b /, / ( ) b( ) Comment (addition of angular momentum see chapter X in CT) The first corresponds to a total spin s=0 obtained by adding two spins with antiparallel directions, while the second is a spin s= obtained by adding two spins that are parallel. 0

11 Exchange Forces Total eigenfunction=(space eigenfunction)x(spin eigenvector) How can we get an antisymmetric total eigenstate? Basically two solutions: (space antisymmetric) x (spin symmetric) (space symmetric) x (spin antisymmetric) Solution all possible antisymmetric eigenfunctions The general solution has the form: usinglet (,) = φα ( ) φβ ( ) + φβ ( ) φα ( ) i a b b a ( ) ( ( )) ( ) a( ) utriplet (,) = a ( φα ( ) φβ ( ) φβ ( ) φα ( ) ) i a b b( ) b( ) + b a The ground state: ( ( )) α 00 ( ) = 00 φ s usinglet (,) = s( ) s( ) + s( ) s( ) a b b a ( ) i ( ( )) is a triplet state possible for the ground state? ( ) a( ) a utriplet (,) = ( s( ) s( ) s( ) s( ) ) i ( a( ) b( ) + b( ) a( ) ) = 0 = 0! b( ) b( ) Conclusion: Ground state is a singlet! First excited state- ss: Both singlets and triplets are possible: usinglet (,) = s( ) s( ) + s( ) s( ) i a b b a ( ) ( ( )) ( ) a( ) a utriplet (,) = ( s( ) s( ) s( ) s( ) ) i ( a( ) b( ) + b( ) a( ) ) = 0 b( ) b( )

12 Which state has a lower energy? Remember that these are not exact eigenfunctions so we can only obtain an average energy: ħ ħ e e e usinglet + + usin glet dvdω triplet ˆ m m r r r triplet Esin glet = H = triplet u u dvdω singlet triplet sin glet triplet ħ e ħ e Esinglet = s s dv + s s dv triplet m r m r J Es Es e s( ) s( ) s( ) s( ) dv( ) dv( ) + r e ± s( ) s( ) s( ) s( ) dv( ) dv( ) = Es + Es + J ± K r K Physical interpretation of the energy terms. The first two terms are the average energy of He + atom in its s and s states.. The J integral can be viewed as a coulomb repulsion between two electrons in the s and s orbitals basically s*s interacts with s*s J is positive since this is a repulsive interaction. 3. K is called the exchange integral because the two product functions in the integrand differ by an exchange of the electrons.k is positive though smaller than J. Consequences Since K is positive we see that the triplet state is lower than the singlet state. Can we understand the lower energy of the triplet state? Lets suppose that both electrons are close together r r utriplet (,) = ( s( ) s( ) s( ) s( ) ) = 0 There is a zero probability of finding the electrons at the same location in the triplet state, this leads to minimization of the electrostatic repulsion term.

13 Conversely for two parallel spins we find that they tend to be close together i.e attract each other. singlet -60 triplet Energy (ev) ss singlet -09 ss The qualitative difference between the triplet and singlet levels can be observed when the He atom is put into a magnetic field which induces a splitting between the triplet states this is referred to as the fine structure. 3

14 Another example of the problems arising from the exchange degeneracy Let us consider a simple problem that of two non-interacting particles in an infinite potential well. ˆ ħ H( X, X, P, P) = + + V ( x) + V ( x) m x x x - are the coordinates of particle. x - are the coordinates of particle. We would like to look for eigenfunctions of the entire system. ˆ total ħ total total total Hu ( x, x) = + u ( x, x) + Vu ( x, x) = Eu ( x, x) m x x Since we are assuming that the particles do not interact they move independently. Through the method of separation of variables that the solutions are of the form: total u x, x = φ x φ x nn, n n The total eigenfunction is written as a product of two eigenfunctions describing the independent particles. In order to uniquely specify an eigenfunction we need to specify its quantum numbers for example in the Hydrogen atom α = ( n,l,m,m s ) such that: φ α ( x ) φ ( ) The eigenfunction uniquely defined by the set of quantum numbers - α evaluated at the coordinates of particle. The total eigenfunction can be written as: total u ( x, x) = φα ( ) φβ ( ) An eigenfunction indicating that particle is in state β and particle is in state α is written as: total u ( x, x) = φβ ( ) φα ( ) Since we assumed that the particles are indistinguishable we should be able to exchange their labels without changing any physically measurable quantity. In particular the probability density: total ( ) = φ ( ) φ ( ) φ ( ) φ ( ) total u u β α β α ( ) ( ) β α β α β α β α φ φ φ φ φ φ φ φ These two expressions are not identical. α 4