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. I am leaving the hats ` off, from here on we understand that H represents a differ.

2 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 . 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.

3 ~. 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.. 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.

4 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. 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 =.

5 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). 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.

6 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. 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.

7 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). 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).

8 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 ). 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.

9 The variance for position can be calculated by squaring the position operator expressed in terms of a . 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.

10 Our classical temporal aver- aging is very different in spirit than the statistical information carried in the quantum mechanical wavefunction remember that expectation val- ues and variances refer to observations made multiple times on identically prepared systems, and most definitely not observations made over time for a single system. We will return to this point later on, for now, the com- parison between quantum and classical probabilities is mainly a vehicle for motivating the notion of density as a good descriptor of Physics . Finally, we should ask where n needs to be to achieve a particular classical energy. For example, if we have a mass m with spring constant k = m 2 and initial extension x0 , then the total energy of the Oscillator is E = 12 m 2 x20 . For what n is this equal to En = n + 12 ~ ? If we put an m = 1 kg mass . on the end of a spring with frequency = 25 1/s and full extension at x0 = 1 m, then we can get a sense of the classical energy range (E.)