Harmonic Oscillator Physics

Physics 342 Lecture 9

Lecture 9

Friday, February 12th, 2010

For the harmonic oscillator potential in the time-independent Schr¨odinger equation: 2 1 2 d ψ(x) 2 2 2 −~ + m ω x ψ(x) = E ψ(x), (9.1) 2 m dx2 we found a ground state

2 − m ω x ψ0(x) = A e 2 ~ (9.2)

1 with energy E0 = 2 ~ ω. Using the raising and lowering operators 1 a+ = √ (−i p + m ω x) 2 m ω ~ (9.3) 1 a− = √ (i p + m ω x), 2 ~ m ω 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 ψ 1: 0 n 1 n a− ψ0 = 0 H (a+ψ0) = + n ~ ω a+ ψ0. (9.4) 2

The operators a+ and a− are Hermitian conjugates of one another – for any

1I 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 9.1. NORMALIZATION Lecture 9 f(x) and g(x) (vanishing at spatial inﬁnity), the inner product: Z ∞ Z ∞ ∗ ∗ ∂g(x) f(x) a± g(x) dx = f(x) ∓~ + m ω x g(x) dx −∞ −∞ ∂x Z ∞ ∗ ∂f(x) ∗ = ±~ g(x) + f(x) m ω x g(x) dx −∞ ∂x Z ∞ ∗ = (a∓ f(x)) g(x) dx, −∞ (9.5) (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.

9.1 Normalization

n The state ψn that comes from n applications of a+ is not normalized, nor does the eigenvalue form of the time-independent Schr¨odingerequation de- mand that it be. Normalization is the manifestation of our probabilistic interpretation of |Ψ(x, t)|2. Consider the ground state, that has an undeter- mined constant A. If we want |Ψ(x, t)|2 to represent a probability density, then Z ∞ 2 2 − m ω x A e ~ dx = 1, (9.6) −∞ and the left-hand side is a Gaussian integral:

Z ∞ 2 r 2 − m ω x 2 π ~ A e ~ dx = A , (9.7) −∞ m ω 1/4 so the normalization is A = m ω . Just because we have normalized the π ~ ground state does not mean that ψ1 ∼ a+ψ0(x) is normalized. Indeed, we have to normalize each of the ψn(x) separately. Fortunately, this can be done once (and for all). R ∞ 2 Suppose we have a normalized set of ψn, i.e. −∞ ψn dx = 1 for all n = 0 −→ ∞. We know that ψn±1 = α± a±ψn where the goal is to ﬁnd the constants α± associated with raising and lowering while keeping the wavefunctions normalized. Take the norm of the resulting raised or lowered state: Z ∞ Z ∞ 2 2 ∗ |ψn±1| dx = α± (a± ψn(x)) (a± ψn(x)) dx −∞ −∞ Z ∞ (9.8) 2 ∗ = α± (a∓ a± ψn(x)) ψn(x) dx, −∞

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and the operator a∓ a± is related to the Hamiltonian, as we saw last time: 1 H = ~ ω a± a∓ ± 2 . Then H 1 1 1 a∓ a± ψn = ± ψn = + n ± ψn, (9.9) ~ ω 2 2 2 so (9.8) becomes Z ∞ Z ∞ 2 ∗ 2 1 1 2 α± (a∓ a± ψn(x)) ψn(x) dx = α± + n ± |ψn(x)| dx. −∞ 2 2 −∞ (9.10) R ∞ 2 The integral −∞ |ψn(x)| dx = 1 by assumption, so we have 1 1 α+ = √ α− = √ , (9.11) n + 1 n and our ﬁnal relation is 1 1 ψn+1 = √ a+ ψn ψn−1 = √ a− ψn. (9.12) n + 1 n Starting from the ground state, for which we know the normalization,

ψ1 = a+ψ0

1 1 2 ψ2 = √ a+ψ1 = √ a+ ψ0 2 1 × 2 (9.13) 1 1 3 ψ3 = √ a+ψ2 = √ a+ ψ0. 3 1 × 2 × 3 The general case is 1 n ψn = √ a+ ψ0 (9.14) n! for the appropriately normalized ψ0,

1/4 2 m ω − m ω x ψ0(x) = e 2 ~ . (9.15) π ~

9.2 Orthonormality

The states described by ψn are complete (we assume) and orthonormal – take our usual inner product for ψm and ψn: Z ∞ Z ∗ 1 1 n ∗ m ψn · ψm = ψn(x) ψm(x) dx = √ √ (a+ψ0) (a+ ψ0) dx. −∞ m! n! −∞ (9.16)

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Now, suppose m > n, then using the fact that a+ and a− are Hermitian m conjugates, we can ﬂip the a+ onto the other term: Z Z 1 1 n ∗ m 1 1 m n ∗ √ √ (a+ψ0) (a+ ψ0) dx = √ √ (a− a+ψ0) (ψ0) dx, m! n! −∞ m! n! −∞ (9.17) n m and we know that a+ ψ0 ∼ ψn and a− ψn ∼ ψn−m – but for m > n, we m−n have a− ψ0 = 0, the deﬁning property of the ground state. In the case n > m, we just use the conjugate in the other direction, and make the same argument: Z Z 1 1 n ∗ m 1 1 ∗ n m √ √ (a+ψ0) (a+ ψ0) dx = √ √ (ψ0) (a−a+ ψ0) dx = 0. m! n! −∞ m! n! −∞ (9.18) R ∞ 2 Only when m = n will we get a non-zero result, and of course, −∞ ψm dx = 1 by construction. So

Z ∞ ∗ ψn(x) ψm(x) dx = δmn. (9.19) −∞

9.2.1 Hermite Polynomials

The prescription for generating ψn 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 Schr¨odinger’sequation 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):

2 1 ∂ − m ω x ψ1 = a+ ψ0 = √ −~ + m ω x A e 2 ~ 2 m ω ∂x ~ (9.20) r 2 A m ω − m ω x = √ 2 x e 2 ~ . 2 ~

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Consider the second state,

r 2 1 1 1 ∂ A m ω − m ω x ψ2 = √ a+ψ1 = √ √ −~ + m ω x √ 2 x e 2 ~ 2 2 2 m ω ~ ∂x 2 ~ 2 A − m ω x ∂ 2 x = √ e 2 ~ −~ 2 2 ∂x ~ r 2 1 m ω 1 ∂ A − m ω x + √ 2 x √ −~ + m ω x √ e 2 ~ 2 ~ 2 m ω ~ ∂x 2 2 2 r 2 A − m ω x A m ω − m ω x = √ e 2 ~ (−2) + √ 2 x e 2 ~ 2 2 2 2 ~ 2 A m ω 2 − m ω x = √ 4 x − 2 e 2 ~ . 2 2 ~ (9.21) The pattern continues – we always have some polynomial in x multiplying the exponential factor. That polynomial, for the nth wave function is called th p m ω Hn, the n Hermite polynomial. In the dimensionless variable ξ = x, 2 ~ we can read oﬀ the ﬁrst two – H1(ξ) = 2 ξ, and H2(ξ) = 4 ξ −2. Normalized, we have the expression

1/4 2 m ω 1 − ξ ψn(x) = √ Hn(ξ) e 2 (9.22) π ~ 2n n! with H0(ξ) = 1

H1(ξ) = 2 ξ 2 H2(ξ) = 4 ξ − 2 (9.23) 3 H3(ξ) = 8 ξ − 12 ξ . . 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 ψn, but that’s better than taking n successive derivatives of ψ0 – in essence, the Hermite polynomials have accomplished that procedure for you. Once again, we can plot the ﬁrst few wavefunctions (see Figure 9.1), 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).

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n =3

n =2 Energy

n =1

n =0

Figure 9.1: The ﬁrst four stationary states: ψn(x) of the harmonic oscillator.

9.3 Expectation Values

9.3.1 Classical Case

The classical motion for an oscillator that starts from rest at location x0 is

x(t) = x0 cos(ω t) . (9.24)

The probability that the particle is at a particular x at a particular time t is given by ρ(x, t) = δ(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 π cycle, take t = 0 → ω ,

π 1 Z ω ρ(x) = π δ(x − x0 cos(ω t)) dt. (9.25) ω 0

We perform the change of variables to allow access to the δ, let y = x0 cos(ω t) so that ω Z −x0 δ(x − y) ρ(x) = − dy π x0 x0 ω sin(ω t) 1 Z x0 δ(x − y) = dy p 2 π −x0 x0 1 − cos (ω t) (9.26) 1 Z x0 δ(x − y) = dy p 2 2 π −x0 x0 − y 1 = . p 2 2 π x0 − x

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This has R x0 ρ(x) dx = 1 as expected (note that classically, the particle re- −x0 mains between −x0 and x0). The expectation value for position is then zero, since ρ(x) is symmetric, x ρ(x) antisymmetric, and the limits of integration are symmetric. The variance is

Z x0 x2 1 σ2 = hx2i − hxi2 = dx = x2. (9.27) x p 2 2 0 −x0 π x0 − x 2

9.3.2 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 ω x = (a+ + a−) p = i (a+ − a−), (9.28) 2 m ω 2 and these facilitate the expectation value calculations. For example, we can find hxi for the nth stationary state:

r Z ∞ ~ ∗ hxi = ψn(x) (a+ + a−) ψn(x) dx = 0, (9.29) 2 m ω −∞ 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± Z ∞ 2 2 2 ~ ∗ σx = hx i − hxi = ψn(x) (a+ a+ + a+ a− + a− a+ + a− a−) ψn(x) dx 2 m ω −∞ Z ∞ ~ ∗ = ψn(x) (a+ a− + a− a+) ψn(x) dx 2 m ω −∞ Z ∞ Z ∞ ~ 2 2 = (n + 1) ψn+1(x) dx + n ψn−1(x) dx 2 m ω −∞ −∞ (2 n + 1) ~ = 2 m ω (9.30) using (9.12). It is interesting to compare the quantum variance with the classical one. 2 In the case of the above, we can write σx in terms of the energy En =

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1 ~ ω n + 2 , just En σ2 = . (9.31) x m ω2 2 1 2 For the classical variance, we had σx = 2 x0, but this is related to the classical energy. Remember we start from rest at x0, so the total energy 1 2 2 (which is conserved) is just E = 2 m ω x0, indicating that we can write the variance as E σ2 = . (9.32) x m ω2 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. Our classical temporal averaging is very diﬀerent in spirit than the statistical information carried in the quantum mechanical wavefunction – remember that “expectation values” 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 ω2 and 1 2 2 initial extension x0, then the total energy of the oscillator is E = 2 m ω x0. 1 For what n is this equal to En = n + 2 ~ ω? If we put an m = 1 kg mass 2 π on the end of a spring with frequency ω = 5 1/s and full extension at x0 = 1 m, then we can get a sense of the classical energy range (E ∼ .8 J) . At what value of n would a quantum mechanical system approach this −34 33 energy? Using ~ ∼ 1.05 × 10 J s, we have n ∼ 6 × 10 . We can compare the classical and quantum probability densities in the “large n” quantum limit, an example is shown in Figure 9.2.

9.4 Mixed States

The stationary states of the harmonic oscillator are useful in characterizing some of the peculiarities of the statistical interpretation, but as with the inﬁnite square well, the utility of these states is in their completeness, and our ability to construct arbitrary initial waveforms. Again, let’s look at

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ρ(x) 0.030

0.025 QM 0.020

0.015 CM

0.010

0.005

x !50 0 50

Figure 9.2: The quantum mechanical density (for n = 50) compared to the classical density (black). a two state admixture, just to get a feel for the time-dependence of the expectation values. Take

−i 3 ω t −i 5 ω t Ψ(x, t) = α e 2 ψ1(x) + β e 2 ψ2(x) (9.33) with α2 + β2 = 1. Then the expectation value for x is Z ∞ hxi = Ψ(x, t)∗ x Ψ(x, t) dx −∞ Z ∞ 2 −i ω t i ω t 2 = α ψ1 x ψ1 + α β e ψ1 x ψ2 + α β e ψ2 x ψ1 + β ψ2 x ψ2 dx. −∞ (9.34) Using our relation for x in terms of raising and lowering operators: x = q ~ 2 m ω (a+ + a−), we can see that the integral of ψ1 x ψ1 will be zero (we will end up with a term like ψ1 ψ0 which integrates to zero by orthogonality, and another term that goes like ψ1 ψ2, similarly zero). The same is true for ψ2 x ψ2, so we only need to evaluate the integral of ψ1 x ψ2

Z ∞ r Z ∞ ~ √ √ ψ1 x ψ2 dx = ψ1 2 + 1 ψ3 + 2 ψ1 dx −∞ 2 m ω −∞ r (9.35) ~ = m ω q R ∞ ~ using orthogonality. The integral −∞ ψ2 x ψ1 dx is also m ω , and we can put these together r ~ hxi = 2 α β cos(ω t). (9.36) m ω

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q ~ m ω A similar calculation for the operator p = i 2 (a+ − a−) gives √ hpi = −2 α β ~ m ω sin(ω t). (9.37)

Notice that this expectation values satisﬁes Newton’s second law:

dhpi dV = − (9.38) dt dx

1 2 2 for our potential V = 2 m ω x (so that the derivative is linear in x, and dV 2 h− dx i = −m ω hxi). This correspondence is an example of Ehrenfest’s theorem, relating expectation dynamics to classical dynamics.

Homework

Reading: Griﬃths, pp. 47–59 (we will return to the series method later on).

Problem 9.1

2 a. For the stationary states of the harmonic oscillator, we know σx from (9.31). Rewrite the momentum operator p in terms of the operators 2 a+ and a− and use that to calculate σp.

2 2 b. Evaluate σx σp, the product of the position and momentum variances – what is the minimum value this product can take?

Problem 9.2

Griﬃths 2.15. The probability of ﬁnding a quantum mechanically described particle in a harmonic well outside of the classically allowed region.

Problem 9.3

Griﬃths 2.13. Given the initial state, normalize the full wavefunction, and compute various expectation values for the harmonic oscillator.

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