df(x) = h(x) dx Chemistry 4531 Mathematical Preliminaries Spring 2009 I. A Primer on Differential Equations Order of differential equation

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1 Chemistry 4531 Mathematical Preliminaries Spring 009 I. A Primer on Differential Equations Order of differential equation Linearity of differential equation Partial vs. Ordinary Differential Equations Quantum Mechanics involves primarily nd order linear partial differential equations First order linear ordinary df(x) = h(x) dx one arbitrary constant

2 d f(x) + a f(x) = 0 dx Second order linear ordinary differential equation There must be two arbitrary constants in the general solution. The solution is fx ( ) = Asin ax+ Bcos ax or ( ) ( ) ( ) fx ( ) = Csin ax+ D or ( ) f ( x ) = E cos ax + F or even f(x) = Ge i ( ax ) + He ( ax ) The values of the constants are determined by boundary conditions on the differential equations. How do we know which form to use?? It cannot matter, so we use the one that is most convenient.

3 f(x) f(x ) + 0 Taylor Series expansion of a function f(x) about x = x 0 df(x) (x x )+ d f(x) (x x 0 ) dx dx! x=x Convergence Radius of Convergence Absolute convergence expand e ix about x =0 x=x 0 0 ix ix i0 e = e + de dx x=0 ix x + de dx x=0 3 ix x 3! + de dx x=0 3 x 3! +... sin(x) = cos(x) = ix SO e = cos(x) + i sin (x) The EULER FORMULA What is the magnitude of e ix?

4 Odd and Even Functions Odd: F(x) = -F(-x) Even: F(x) = F(-x) This definition refers to the symmetry behavior (parity) about the origin, x=0. The definition can be readily generalized to describe symmetry properties about any point, say x = a. e.g. Odd: F(x-a) = -F(-[x-a]) Even: F(x-a) = F(-[x-a]) Using Mathcad, let us look at examples and also some integrals. x 10, F( x ) ( x ) The function is even about x=. It is neither odd nor even about any other point.

5 Let's look at an odd function about x=: Gx ( ) = 64 sin( x ) 3 64 [ ] What can we say about integrals centered at x=? Let us evaluate some integrals involving F and G. We do so by inspection!! 8 8 F( x) dx 4 ( ) 8 8 F( x) dx Gxdx ( ) Gxdx ( ) F( xgxdx ) ( ) F( xgxdx ) ( )

6 OPERATORS An operator is an instruction telling you what to do with a function, much like a function is an instruction telling you what to do with a number. Examples of operators include d/dx, *,,, etc. We will denote a general symbolic operator with a "hat" over the symbol, e.g.  So if  =, then  f(x) = f( x) An entire operator algebra can be derived. d Notation: ÂĈf(x) Â(Ĉf(x)) Ĉ f(x) Ĉ [Ĉf(x)] ( + Ĉ) f(x)  f(x) + Ĉ f(x) The ORDER of operators can make a difference! A very important concept! ÂĈ f(x) =? Ĉ  f(x) DEFINITION:  Ĉ - Ĉ  [ Ĉ] the COMMUTATOR of  and Ĉ DEFINITION:  Ĉ Ĉ  [Â, Ĉ], the COMMUTATOR of  and Ĉ. If the commutator is zero for any function f, then the operators are said to commute. Namely, the order of carrying out the operations  and Ĉ does not matter. Operators and commutation are critical concepts in quantum mechanics and are directly related to measurement. Example: let  = x and Ĉ = d/dx. All Quantum Mechanical Operators are linear.

7 Definition:  is linear if and only if  (c 1 f + c g) = c 1  f + c  g where c 1 and c are constants and f and g are any two functions. Which of these operators is linear?? R = ( ) ( ) d dx P = x d( ) 3 dx d( ) ˆM = x 3 dx

8 Eigenvalue equations and eigenvalues The operator equation Â(f(x)) = c f(x), where c is a constant, is called an eigenvalue equation. The function f(x) is called the eigenfunction of the operator Â, with eigenvalue c. It turns out that t in quantum mechanics all physical properties are represented by operators and that the possible values of that property are related to the eigenvalues of that operator. Thus much of our work will be related to finding eigenvalues and eigenfunctions of operators. Examples: I. Let Â= d/dx. Then we must find f(x) such that df dx = c f(x) The solution in this case is f(x) = e cx, and the eigenvalue is c, where c is any number. II. Find the eigenvalues and eigenfunctions of  = -d /dx. That is, find all a i and f i (x) such that dfi( x) = af i i( x) dx. df i (x) Rewriting, dx equation with solution: ( ) ( ) i i i f (x) = Asin a x + Bcos a x, where a i is real, and A and B are any numbers. af i i(x) = 0, a nd order homogeneous differential

9 More about eigenvalue equations and eigenvalues Linear momentum, p Kinetic energy, T Total energy, H = T + V

10 POTENTIAL ENERGY We will consider the case that the force F on a particle depends only on its position (coordinate). As an aside, when might the force depend on velocity? Conservative system?? If this is the case then one can prove the following: Let the forces acting on a particle be dv given by a vector F. If there exists a function V(x) such that F = dx, then the force field is said to be describable by the potential energy function, V(x). More generally, if there is a function V(x, y, z) such that V Fx,y,z ( ) = F i+ F j+ F k = x i V y j V x y z k = V(x, y, z) z then the force field F is said to be conservative. Here is the gradient operator and i, j and k are unit vectors in the x, y and z directions.

11 Let's look at several examples to clarify this concept. Gravitational potential for a particle of mass m: z m V(z) z Since F = mgk, then V(z) = mg z

12 An ideal (Hookes Law) Spring Since F= kxi, we have V(x) = k x +C The quadratic potential energy curve also reminds us how we generally can visualize the forces acting on a particle. How would we modify the potential energy for a diatomic molecule??

13 -e r m +e For the hydrogen atom, e m p The force is = qq 1 e F = attractive r r and the potential energy is V(r) = e, the Coulomb attractive Potential r

14 Schrödinger equation for a particle in one dimension Consider a particle of mass m confined to move in one dimension but with NO forces acting on the particle. Schrödinger postulated that the amplitude of the matter waves, ψ(x), must satisfy the differential equation, d ψ = Eψ m dx where E is the total energy and = h/ π More generally, if the particle is in a potential energy field V(x), then d ψ(x) + V(x) ψ (x) = E ψ (x) m dx Before developing formal postulates, let s see what this can mean. A solution to the first equation can be written in terms of complex exponentials or sines and cosines: ikx ψ(x) = e = cos kx + isin kx, where k = me This is a sine wave of wavelength λ = π/k. Since there are no forces (no potential energy), the total energy E is also the kinetic energy. Thus E = p /m and k= p/. The wavelength λ associated with the solution is thus h/p, the debroglie wavelength!!

15 In the more general second case, the only difference is that E in the solution is m(e V) ikx ψ(x) = e = cos kx + i sin kx, where k = replaced by E-V: ( ) ( ) The kinetic energy of the particle is (E-V), and the wavelength associated with the matter waves is now given by the relation λ = h m(e V) Thus, the slower the particle, the longer (and less wiggly) is the matter wave wavelength. Notice that when the particle is at rest, the wavelength is infinite, and we cannot tell where the particle is. This is the beginnings of the Uncertainty Principle at work!.

16 In the more general second case, the only difference is that E in the solution is m(e V) ikx ψ(x) = e = cos kx + i sin kx, where k = replaced by E-V: ( ) ( ) The kinetic energy of the particle is (E-V), and the wavelength associated with the matter waves is now given by the relation λ = h m(e V) What happens if the potential is infinitely high, V =?. What happens if the potential is just high, V > E?

17 Example: A Potential Step E > V V E 1 < V V=0 We will come back to this free particle shortly, but we first will look at the postulates of quantum mechanics and then solve the particle in a box problem with renewed understanding

18 Chemistry 4531 Postulates of Quantum Mechanics Spring 009 Consider a system with N degrees of freedom. 1. There exists a function Ψ(q 1... q N,t) that.contains all information available for this system. It is called the state function or wave function. Here q represents coordinate(s). Properties of Ψ: a. Ψ * Ψ is a probability density Born hypothesis z b. Ψ * Ψ dq Ψ Ψ 1 dq N = 1 (normalized) Ψ Ψ c. Ψ(q), and must be continuous. q q d. Ψ(q 1... q N,t) must be a single valued function of its arguments

19 Chemistry 4531 Postulates of Quantum Mechanics Spring 009 Consider a system with N degrees of freedom. 1. There exists a function Ψ(q 1... q N,t) that.contains all information available for this system. It is called the state function or wave function. Here q represents coordinate(s). Properties of Ψ: a. Ψ * Ψ is a probability density Born hypothesis z b. Ψ* Ψdq 1 dq N Ψ Ψ = 1 (normalized) Ψ Ψ c. Ψ(q), and must be continuous. q q d. Ψ(q( 1... q N,t) must be a single valued function of its arguments. For every physical observable A, there corresponds a linear Hermitian operator Â. More later on Hermitian operators. The operator  is obtained by writing A classically ll in terms of cartesian coordinates and momenta, and then replacing x x, y y, z z, p x i, p y i, pz i x y z 3. If  corresponds to physical observable A, then for a system with state function Ψ, every measurement of the observable A must yyield one of the eigenvalues g of Â. ˆ Ψ(q, t) 4. The time evolution of Ψ(q, t) is given by HΨ(q, t) = i where Ĥ is the Hamiltonian t operator. The classical Hamiltonian Ĥ corresponds to the total energy (kinetic and potential) of an isolated system. This equation is the time-dependent Schrödinger equation.

20 (All symbols and operations should become clear in future lectures) We will see shortly that when we wish to calculate the stationary states of a system, the time-dependent Schrödinger equation for a single particle of mass m moving in one dimension reduces to the simpler time-independent d Schrödinger equation, d (x) m ψ dx + V(x) ψ(x) = E ψ(x) Here, V(x) is the potential energy of the particle. This expression is identical to the eigenvalue equation for the total energy (postulate 3), ˆΨ = E Ψ H.

21 PARTICLE IN A BOX a x) =, x < 0 or x > a (Regions I,III) x) = 0, 0 < x < a (Region II) m d ψ (x) +V(x) ψ(x) = E ψ (x) dx he formal QM postulates recognize that the solution to the Schrödinger equation, x), has a connection with physical observables. In order to represent a physical ality the postulates insist that ψ(x) be a) continuous, and b) single valued. e will solve the equation in regions I, II and III and then put the pieces together to ey the above constraints.

22 REGIONS I and III In these regions, the Schrödinger equation is given by m d ψ (x) dx + ψ(x) = E ψ(x) which implies ψ(x) = 0 for x<0 or x>a Now consider Region II, where V(x) = 0 d m ψ(x) dx = E ψ (x) so a solution in this region is me me ψ(x) ) = Asin x + B cos x for 0 x a. We now have the total solution, and must apply the boundary conditions. The state function ψ(x) is certainly well behaved except possibly at x = 0 or x = a. To be continuous, we must require that the Region II solution vanish at these points THUS at x = 0, me me 0= Asin 0 + B cos 0 (1)

23 THUS at x = 0, me me 0= Asin 0 + B cos 0 (1) and at x = a, me me 0= Asin a + Bcos a () From the first equation, we have or 0 = A (0) + B(1) so B = 0 and from the second, me me me 0 = Asin a + Bcos a = Asin a There can only be TWO possibilities: 1) A = 0, which says ψ(x)= 0 for all x (no good) Why?? or ) me sin 0 a =

24 Can we make this happen?? What parameters are available to us? ONLY the Energy, E!!!! HOW? Where are the zeros of the sin function? We know that sin z = 0 for z = n π, where n = 1,, 3,.. So we must have me a Solving for E, = nπ where n = 1,,3 and the state function is ψ E n n n π n h = = where n = 1,,3, ma 8ma ( x) = A sin for 0 x a and 0 otherwise n nπ x a The imposition of boundary conditions and periodic motion has forced quantized energy states upon us!!

25 The Born Postulate says that ψ * ψ is a probability density and thus its integral over all space must equal 1. A so A n = a a * nπ x ( x) ψ( x) dx= 1= An sin dx= An a 0 ψ a What do these functions look like??? Particle in a Box ψ n and ψ n We plot ψ n and ψ n for the first few states in the following picture.

26 ψ n (x) ψ n *(x) ψ n (x) 16 4 ts of π / m a Energy, units of / m a Energy, unit n Classical vs. Quantum Behavior 1. Nodes 4. Localization in part of box?. Non-uniform ψ n 5. "getting across the zeroes 3. Zero point energy 6. Classical Limit - Large n What is very misleading about this picture?? Placing the state functions on a potential energy plot! We need to think and talk about this!

27 More on Postulates Bracket Notation Definitions: Let φ and ψ be arbitrary complex-valued functions of coordinate(s) q. z z φ A ψ φ* A ψ dq and φ ψ φ* ψ dq The * implies taking the complex conjugate of the function. The integration dq is over all values of all coordinates. φ φ * is often called a bra and ψ ψ is similarly called a ket.. For every physical observable A, there corresponds a linear Hermitian operator Â, whose mathematical properties es tell us about the physical quantity ty A. a.  is Hermitian if and only if φa ψ = A φ ψ We will see that the Hermitian property of an operator guarantees that its eigenvalues are real. This Hermitian property also guarantees that eigenfunctions corresponding This Hermitian property also guarantees that eigenfunctions corresponding to different eigenvalues are orthogonal, ie, φ i φ j = 0, if a i a j (if  is Hermitian). We will prove this statement shortly.

28 More on Postulates 3. If  corresponds to physical observable A, then for a system with state function Ψ, every measurement of the observable A must yield one of the eigenvalues of Â. 3a) If a system is characterized by state function ψ, and ψ is an eigenfunction of the operator  with eigenvalue a, then every measurement of the property corresponding to  will give exactly a for the result. 3b) If a system is characterized by state function ψ, and ψ is an not an eigenfunction of the operator Â, then repeated measurements of the property corresponding to  will result in a distribution of results. The average value of the measurements, a, is given by a = Ψ*AˆΨ dq= Ψ Aˆ Ψ q Actually, 3b) encompasses all of the other versions

29 Examples of Significance of Postulates Let us look at some properties of the particle in a box, using the postulates to interpret the results. Let ψ n (x) be the normalized eigenfunction for the particle in a box extending from 0 to a. 1) Suppose the state function is ψ s = C(ψ 1 +5ψ ). Find C such that ψ s is normalized. N.B., we will soon see that this orthogonality of ψ 1 and ψ extends to any eigenfunctions with different eigenvalues. See Atkins,.1, c & d for more.

30 ) Let the state function be ψ Is it normalized?? Calculate E, p and p. Interpret the results. 3) Given ψ s = C(1ψ 1 +3ψ ), find E. Interpret the result.

31 Consider a function W x = 5 Another Example 30 ψ b g x(a x) for 0<x<a, and 0 otherwise. Is ψ W an a acceptable state function? a) continuous and continuous derivatives? b) single valued? c) normalized? a 30 a 5 0 a x (a x dx = 1 ) Thus ψ W fulfills the requirements for a state function, but it is NOT an eigenfunction. Why?? Let's plot ψ W. x x(a-x) 0 0 a 0 a/ a /4 a/4, 3a/4 3 a /16 a/10, 9a/10 9 a /100 So ψ W is symmetric about a/ and has no nodes. It is an acceptable state function. So ψ W is symmetric about a/ and has no nodes. It is an acceptable state function. It looks a lot like ψ 1. Is its energy close to that of ψ 1?

32 a a ψ * ˆ ψ ψ * ψ 0 0 m x E = H dx = dx = 30 5 a L NM m O a x(a x) d QP z x(a x)dx 0 dx Finally, E = 5 h = m a 7.9m a h 1 = Recall for ψ 1, E 8ma, very close to that above!! That makes sense!! What will we observe if we make repeated measurements of E W? A distribution of energies, since ψ W is not an eigenfunction of E Each member of the distribution will be one of the energy eigenvalues E Finally E h for ψ W, expected, as ψ W looks a lot like ψ 1 = 79. ma n = nh 8ma

33 QUANTUM MECHANICAL THEOREMS Remember postulate For every physical observable A, there corresponds a linear Hermitian operator Â, whose mathematical properties tell us about the physical quantity A. Remember:  is Hermitian if and only if Bracket Notation Definition: Many properties of Hermitian operators follow from a single derivation. Thus, let ψ n and ψ m be eigenfunctions of an arbitrary operator  with eigenvalues a n and a m. Then  ψ m = a m ψ m (1) And Âψ n = a n ψ n () PROOF: Multiply (1) by ψ n* and integrate ψ n*  ψ m = ψ n* a m ψ m = a m ψ n* ψ m (3) Take * of (), multiply by ψ m and integrate  * ψ n* = a n* ψ n * ψ m  * ψ n * = ψ m a n* ψ n* = a n* ψ m ψ n * (4)

34 Now subtract (4) from (3) ψ n*  ψ m - ψ m  * ψ n * = a m ψ m ψ n* -a n* ψ m ψ n * (5) = ( a m -a n* ) ψ m ψ n * This result is valid for ANY operator Â. Suppose that  is Hermitian. Then the left side of (5) vanishes and we have that 0 = ( a m -a n* ) ψ m ψ n * There are two possibilities involving the indices n and m: Case I. n = m. Since ψ * * n ψ n = 1, then ( a n - a n ) = 0. Thus : "The eigenvalues of a Hermitian operator are real." Case II. n m 0 = ( a m -a n ) ψ m ψ n* = ( a m -a n ) ψ n ψ m The a n* became a n because it is real. If ( a m -a n ) 0, then I can divide by this term to obtain the following: ψ m ψ n* = 0 if a n a m. Thus: "Eigenfunctions corresponding to different eigenvalues of a Hermitian p operator are orthogonal". This vanishing of the integral of two functions is the definition of orthogonality, as we shall soon see.