Harmonic Oscillator Physics - Reed College

1 Physics 342 Lecture 9. Harmonic Oscillator Physics Lecture 9. Physics 342. Quantum Mechanics I. Friday, February 12th, 2010. For the Harmonic Oscillator potential in the time-independent Schro dinger equation: 1 2 d (x). 2.. ~ + m x (x) = E (x), 2 2 2. ( ). 2m dx2. we found a ground state m x2. 0 (x) = A e 2~ ( ). with energy E0 = 1. 2 ~ . Using the raising and lowering operators 1. a+ = ( i p + m x). 2~m . ( ). 1. a = (i p + m x), 2~m . we found we could construct additional solutions with increasing energy using a+ , and we could take a state at a particular energy E and construct solutions with lower energy using a . The existence of a minimum energy state ensured that no solutions could have negative energy and was used to define 0 1 : 1.. a 0 = 0 H (a+ 0 ) =. n + n ~ an+ 0 . ( ). 2. The operators a+ and a are Hermitian conjugates of one another for any 1.

2 I am leaving the hats ` off, from here on we understand that H represents a differ- . ential operator given by H x, ~i x for a classical Hamiltonian. 1 of 10. NORMALIZATION Lecture 9. f (x) and g(x) (vanishing at spatial infinity), the inner product: g(x). Z Z . f (x) a g(x) dx = f (x) ~ + m x g(x) dx x f (x). Z .. = ~ g(x) + f (x) m x g(x) dx x Z . = (a f (x)) g(x) dx, . ( ). (integration by parts) or, in words, we can act on g(x) with a or act on f (x) with a in the inner product. Normalization The state n that comes from n applications of an+ is not normalized, nor does the eigenvalue form of the time-independent Schro dinger equation de- mand that it be. Normalization is the manifestation of our probabilistic interpretation of | (x, t)|2 . Consider the ground state, that has an undeter- mined constant A. If we want | (x, t)|2 to represent a probability density, then Z.

3 M x2. A2 e ~ dx = 1, ( ).. and the left-hand side is a Gaussian integral: r ~. Z . m x2. A2 e ~ dx = A2 , ( ). m . 1/4. so the normalization is A = m . Just because we have normalized the . ~. ground state does not mean that 1 a+ 0 (x) is normalized. Indeed, we have to normalize each of the n (x) separately. Fortunately, this can be done once (and for all). R . Suppose we have a normalized set of n , n2 dx = 1 for all n = 0 .. We know that n 1 = a n where the goal is to find the constants associated with raising and lowering while keeping the wavefunctions normalized. Take the norm of the resulting raised or lowered state: Z Z . | n 1 |2 dx = 2. (a n (x)) (a n (x)) dx . Z ( ).. = 2. (a a n (x)) n (x) dx, . 2 of 10. ORTHONORMALITY Lecture 9. and the operator a a is related to the Hamiltonian, as we saw last time: H = ~ a a 12 . Then H 1 1 1.

4 A a n = n = +n n , ( ). ~ 2 2 2. so ( ) becomes 1 1. Z Z .. 2. (a a n (x)) n (x) dx = . 2. +n | n (x)|2 dx. 2 2 . R ( ). The integral | n (x)|2 dx = 1 by assumption, so we have 1 1. + = = , ( ). n+1 n and our final relation is 1 1. n+1 = a+ n n 1 = a n . ( ). n+1 n Starting from the ground state, for which we know the normalization, 1 = a+ 0. 1 1. 2 = a+ 1 = a2+ 0. 2 1 2 ( ). 1 1. 3 = a+ 2 = a3+ 0 . 3 1 2 3. The general case is 1. n = an+ 0 ( ). n! for the appropriately normalized 0 , m 1/4 m x2. 0 (x) = e 2 ~ . ( ). ~. Orthonormality The states described by n are complete (we assume) and orthonormal . take our usual inner product for m and n : 1 1. Z Z. n m = n (x) m (x) dx = (an+ 0 ) (am + 0 ) dx. m! n! . ( ). 3 of 10. ORTHONORMALITY Lecture 9. Now, suppose m > n, then using the fact that a+ and a are Hermitian conjugates, we can flip the am + onto the other term: 1 1 1 1.

5 Z Z.. (a+ 0 ) (a+ 0 ) dx = . n m (am a+ 0 ) ( 0 ) dx, n m! n! m! n! . ( ). and we know that an+ 0 n and am n n m but for m > n, we have a 0 = 0, the defining property of the ground state. In the case m n n > m, we just use the conjugate in the other direction, and make the same argument: 1 1 1 1. Z Z.. (a+ 0 ) (a+ 0 ) dx = . n m ( 0 ) (an am + 0 ) dx = 0. m! n! m! n! . R 2 ( ). Only when m = n will we get a non-zero result, and of course, m dx =. 1 by construction. So Z . n (x) m (x) dx = mn . ( ).. Hermite Polynomials The prescription for generating n does not provide a particularly easy way to obtain the functional form for an arbitrary n we have to repeatedly apply the raising operator to the ground state. There is a connection between the Hermite polynomials and our procedure of lifting up the ground state. Using the Frobenius method, it is possible to solve Schro dinger's equation as a power series expansion (described in Griffiths), and we won't re-live that argument (yet).

6 But it is important to understand the connection between the algebraic, operational approach and the brute force series expansion. Starting from the ground state, let's act with a+ (call the normalization constant A again, just to make the expressions more compact): 1 . 2.. m2 ~x 1 = a+ 0 = ~ + m x Ae 2m ~ x ( ). A m . r 2. m2 ~x = 2 x e . 2 ~. 4 of 10. ORTHONORMALITY Lecture 9. Consider the second state, 1 1 1 A m . r . m x2. 2 = a+ 1 = ~ + m x 2 x e 2 ~. 2 2 2m ~ x 2 ~. A 2x 2.. m x = e 2 ~ ~. 2 2 x ~. 1 m 1 A m x2. r . + 2 x ~ + m x e 2 ~. 2 ~ 2m ~ x 2. 2. A A m . 2. r 2. m x m x = e 2 ~ ( 2) + 2 x e 2 ~. 2 2 2 2 ~. A m 2 m x2. = 4 x 2 e 2 ~ . 2 2 ~. ( ). The pattern continues we always have some polynomial in x multiplying the exponential factor. That polynomial, for the nth wave function p is called Hn , the nth Hermite polynomial.

7 In the dimensionless variable = m~ x, we can read off the first two H1 ( ) = 2 , and H2 ( ) = 4 2 2. Normalized, we have the expression m 1/4 1 2. n (x) = Hn ( ) e 2 ( ). ~ 2n n! with H0 ( ) = 1. H1 ( ) = 2 . H2 ( ) = 4 2 2 ( ). H3 ( ) = 8 12 . 3.. This set of polynomials is well-known, and they have a number of interesting recursion and orthogonality properties (many of which can be developed from the a+ and a operators). One still needs a table of these in order to write down a particular n , but that's better than taking n successive derivatives of 0 in essence, the Hermite polynomials have accomplished that procedure for you. Once again, we can plot the first few wavefunctions (see Figure ), and as we increase in energy, we see a pattern similar to the infinite square well case (note that for the Harmonic Oscillator , we start with n = 0 as the ground state rather than 1).

8 5 of 10. EXPECTATION VALUES Lecture 9. n=3. n=2. Energy n=1. n=0. Figure : The first four stationary states: n (x) of the Harmonic Oscillator . Expectation Values Classical Case The classical motion for an Oscillator that starts from rest at location x0 is x(t) = x0 cos( t) . ( ). The probability that the particle is at a particular x at a particular time t is given by (x, t) = (x x(t)), and we can perform the temporal average to get the spatial density. Our natural time scale for the averaging is a half cycle, take t = 0 , . 1. Z.. (x) = (x x0 cos( t)) dt. ( ). 0. We perform the change of variables to allow access to the , let y = x0 cos( t). so that x0 (x y). Z. (x) = dy x0 x0 sin( t). 1 x0 (x y). Z. = dy x0 x0 1 cos2 ( t). p ( ). 1 x0 (x y). Z. = dy x0 x20 y 2. p 1. = p 2 . x0 x2. 6 of 10. EXPECTATION VALUES Lecture 9. Rx This has x0 0 (x) dx = 1 as expected (note that classically, the particle re- mains between x0 and x0 ).

9 The expectation value for position is then zero, since (x) is symmetric, x (x) antisymmetric, and the limits of integration are symmetric. The variance is x2 1. Z x0. x = hx i hxi =. 2 2 2. dx = x20 . ( ). x0 x0 x 2. p 2 2. Quantum Case Referring to the definition of the a+ and a operators in terms of x and p, we can invert and find x and p in terms of a+ and a these are all still operators, but we are treating them algebraically. The inversion is simple r r ~ ~m ( ). x= (a+ + a ) p = i (a+ a ), 2m 2. and these facilitate the expectation value calculations. For example, we can find hxi for the nth stationary state: r Z . ~. hxi = n (x) (a+ + a ) n (x) dx = 0, ( ). 2 m . by orthogonality. Similarly, hpi = 0. Those are not particularly surprising. The variance for position can be calculated by squaring the position operator expressed in terms of a.

10 Z . ~. x = hx i hxi =. 2 2 2. n (x) (a+ a+ + a+ a + a a+ + a a ) n (x) dx 2 m . Z . ~. = n (x) (a+ a + a a+ ) n (x) dx 2 m . Z Z . ~. = (n + 1) n+1 (x) dx +. 2. n n 1 (x) dx 2. 2m . (2 n + 1) ~. =. 2m . ( ). using ( ). It is interesting to compare the quantum variance with the classical one. In the case of the above, we can write x2 in terms of the energy En =. 7 of 10. MIXED STATES Lecture 9. ~ n + 1. , just . 2. En x2 = . ( ). m 2. For the classical variance, we had x2 = 12 x20 , but this is related to the classical energy. Remember we start from rest at x0 , so the total energy (which is conserved) is just E = 21 m 2 x20 , indicating that we can write the variance as E. x2 = . ( ). m 2. This is interesting, but we must keep in mind a number of caveats: 1. the classical density is time-dependent, and we have chosen to average over the natural timescale in the system, if no such scale presented itself, we would be out of luck making these comparisons, 2.