How can I show that an arbitrary wavefunction in a 1D SHO is periodic in time?

Question

I want to show that an arbitrary wavefunction $f$ in a one dimensional harmonic potential reproduces itself after a period T up to a phase factor: $f(x,t+T)=Af(x,t)$, $|A|=1$

I am not sure if this is the correct way to begin, i googled the following, I do not know how to obtain it algebraically:

$$f(x,t)= B\exp\biggl(\frac{-x^{2}}{2a^{2}}-iw_{0}\frac{t}{2}\biggr)$$

Then $t\rightarrow t+T$ gives:

$$\begin{align}f(x,t+T) &= B\exp\biggl(\frac{-x^{2}}{2a^{2}}-iw_{0}\frac{t+T}{2}\biggr)\\ &= B\exp\biggl(\frac{-x^{2}}{2}-iw_{0}\frac{t}{2}\biggr)\exp\biggl(-iw_{0}\frac{T}{2}\biggr)\\ &= \exp\biggl(-iw_{0}\frac{T}{2}\biggr)f(x,t)\\ &= Af(x,t) \Rightarrow A=e^{-iw_{0}\frac{T}{2}}, |A|=1\end{align}$$

Does anybody see if this is correct so far? Thank you.

Answer 1

As in Jon's answer

$$\psi(x,t)=\sum_{n=0}^\infty c_n e^{-i(n+1/2)\omega t}\phi_n(x).$$

where $\omega$ is the classical oscillator angular frequency. Let $T=4\pi/\omega$.

$$e^{-i(n+1/2)\omega (t+T)} = e^{-i(n+1/2)\omega t}e^{-i(n+1/2)\omega T} = e^{-i(n+1/2)\omega t}$$

because $(n+1/2)\omega T$ is $2\pi$ times an integer. Therefore $$\psi(x,t)=\psi(x,t+T)$$ So the whole wavefunction repeats itself exactly after period $T$.

Answer 2

I am mystified by the confusion regarding this, so I will give a sketch of an answer to the problem, and the proof of the related theorem that has been quoted in another answer.

The Problem

Let the coefficients of the expansion of the initial wavefunction in energy eigenstates be $c_n$, so that at time 0:

$$\psi(x) = \sum_n c_n \phi_n(x) $$

Where $\phi_n(x)$ are the energy eigenstates. Then at time t,

$$c_n(t) = e^{-i(n+{1\over 2})\omega t} c_n(0)$$

by the time dependence of the eigenstates--- eigenstates just get a constant phase per unit time, proportional to the energy. So at $t=T={2\pi\over\omega}$, all the phases are $\pi$ plus an integer multiple of $2\pi$, and

$$ c_n(T) = - c_n(0)$$

So that any (normalized) sum, finite or infinite, of eigenfunctions comes back to minus itself, so that up to a global phase, it is exactly the same, while at time $t=2T$, the wavefunction comes back to itself including phases. In terms of the time-dependent wavefunction $\psi(t,x)$,

$$ \psi(T,x) = \sum c_n(T) \phi_n(x) = - \psi(0,x)$$ $$ \psi(2T,x) = \psi(0,x)$$

The reasons for the silly phase reversal is the zero point energy. If you subtract the zero point energy from the Hamiltonian, you make the energy of the n-th level an integer multiple of $\omega$, and then the wavefunction precisely repeats every T.

When the energy levels are equally spaced, the wavefunction always repeats up to a global phase, after a time equal to h over the energy spacing (or $2\pi$/spacing in units where hbar is 1, as above).

The General Recurrence Theorem

Jon mentioned the quantum analog of the Poincare recurrence theorem. I will give the proof of this theorem below, because the proof is a consequence of the argument above. It is circular to argue using the theorem, because the proof of this theorem is based on the special case of equally spaced levels, on the harmonic oscillator.

The general analog of the Poincare recurrence theorem in quantum mechanical systems with discrete energy levels $E_n$ is proved a little more simply than the classical Poincare recurrence (although both are simple).

First set the ground state energy to zero by adding a constant to H (so that you throw away any overall phase in the wavefunction), set the first excited state energy to 1 by redefining the unit of energy/time, and then approximate the i-th energy level by a rational number $N_i/D$ with error $\epsilon_i$ so that $\epsilon_i\le {1\over 2D}$ (this is just the stupidest approximation--- pick any denominator, and choose the closest numerator. You do this with the same denominator D for all the energies).

Now the first N energy levels are, by necessity, close to integer multiple of $1/D$, so that any superposition of these states will recur (up to a small controlled error which shrinks with D) after a time which is the reciprocal of this common denominator, $T={2\pi D}$, by the harmonic oscillator repeating shown above.

Make N big enough so that superposing the first N levels with the best coefficients gives you a state which is $\epsilon/3$ close to the initial state (this is possible because the energy eigenstates are complete). and then make D big enough so that the period T makes the state recur with error $\epsilon/3$ (which is possible, because the recurrence of any superposition of the first n energy states gets more and more perfect as D gets large). Then the initial state is a distance $\epsilon/3$ from a state which repeats itself within an $\epsilon/3$ to something which is again $\epsilon/3$ from the original state, so that the state has come back within $\epsilon$ of itself.

Answer 3

About question 1) one can state the following. If you have a discrete set of eigenstates, as happens for the harmonic oscillator, if the system is in one of these eigenstates, then the evolution in time gives

$$\psi(x,t)=e^{-i(n+1/2)\omega t}\phi_n(x)$$

and so the conclusion follows straightforwardly. But, for a generic initial wavefunction, you can express it through the eigrnstates of the harmonic oscillator as follows

$$\psi(x,0)=\sum_{n=0}^\infty c_n\phi_n(x)$$

provided you know how to compute $c_n=\int_{-\infty}^\infty dx\phi_n(x)\psi(x,0)$. Then, time evolution will give you

$$\psi(x,t)=\sum_{n=0}^\infty c_n e^{-i(n+1/2)\omega t}\phi_n(x).$$

For a general quantum system, a recurrence theorem exists in quantum mechanics (see here), similar to the Poincaré recurrence theorem in classical mechanics, that states that

Let us consider a system with discrete energy eigenvalues $E_n$, if $\psi(t_0)$ is its state vector in the Schrodinger picture at the time to and $\epsilon$ is any positive number, at least one time $T$ will exist such that the norm $||\psi(t)-\psi(t_0)||$ of the vector $\psi(t)-\psi(t_0)$ is smaller than $\epsilon$.

and one can answer yes to this question but in a very good approximation. In the case of the harmonic oscillator, one is in the lucky situation such that

$$\psi(x,t+2\pi/\omega)=e^{-i\pi}\psi(x,t).$$

This complete the answer.

For the other two questions, you are able to prove that $\Delta x\Delta p=\hbar/2$ only if if your initial state is a coherent state, that is

$$\psi(x,0)=\frac{1}{(2\pi\sigma^2)^\frac{1}{4}}e^{-\frac{1}{4}\frac{(x-x_0)^2}{\sigma^2}}$$

but this is nothing else than the ground state of your harmonic oscillator, with the rest position shifted from zero, and so, all your conclusions follow immediately, being a simple homework problem.