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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.

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Um... Is this a homework problem set? Also, an arbitrary wavefunction won't repeat itself unless it is made up of a finite number of energy eigenstates. – Jerry Schirmer Dec 4 '11 at 23:36
@Jerry Schirmer : It is a mix of old examination problems. Are you saying that my attempt is wrong? – VVV Dec 4 '11 at 23:41
@Jerry: Why do you say this? It's not true--- all (normalizable) states repeat themselves after a period. – Ron Maimon Dec 5 '11 at 0:38
Maybe Jerry was talking about spatial periodicity? – David Z Dec 5 '11 at 22:27
No, see Jon's answer below. Exact recurrance doesn't occur unless the state is a finite sum of eigenstates. – Jerry Schirmer Dec 5 '11 at 22:41

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 $T$.

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+1: Ok, this is correct, although to be technically pedantic, the wavefunction goes to minus itself after period T, and then back to itself after 2T (because of the traditional zero point energy). – Ron Maimon Dec 6 '11 at 4:31
Just to be clear: $(n+1/2)\omega T$ is $2\pi$ times an integer plus a half-integer, i.e. its exponential is minus one. – Ron Maimon Dec 6 '11 at 4:48
Actually $(n+1/2)\omega T = 4\pi(n+1/2) = 2\pi(2n+1)$, so Steve was right about the repetition. – David Z Dec 6 '11 at 5:12
@David: No, no. Come on--- plug in n=0. $T={2\pi\over\omega}$. Is it April 1st? – Ron Maimon Dec 6 '11 at 5:38
$T = 4\pi/\omega$ as defined in the answer. – David Z Dec 6 '11 at 5:40

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.

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Fine. The minus sign, the phase, is mine $e^{-i\pi}$. – Jon Dec 6 '11 at 17:37
@Jon: It is now--- I removed the downvote after you fixed the answer. – Ron Maimon Dec 6 '11 at 18:03
Thank you a lot. Without your reprimenda I would not have seen my oversight. – Jon Dec 6 '11 at 18:10

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


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


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.

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Sorry for the nitpicking a fine answer(v2), but it should probably be stressed, that the theorem (which is referred to here) does not say that an exact repetition will appear after some time $T$ as OP originally wanted to show(v2), but only that an arbitrarily good approximation to it will occur some time in the future. – Qmechanic Dec 5 '11 at 15:19
@Qmechanic: You are right of course, but this is the best non-trivial answer to the original question that otherwise will boil down to a trivial but badly stated homework problem. – Jon Dec 5 '11 at 15:24
Jon, I edited parts 2 and 3 out of the original question because they were basically separate questions. I'm letting you know in case you would like to update your answer accordingly. – David Z Dec 5 '11 at 22:31
@Qmechanic: Can someone explain to me whats going on here? For a generic Hamiltonian with discrete spectrum one obtains almost periodicity since there is no rational relation between all the energies. But in the SHO all energies are multiples of the frequency $\omega$ and hence any normalizable solution should exactly recur with frequency $\omega$ (up to phase). How is this wrong? Can I see a counterexample? – BebopButUnsteady Dec 6 '11 at 3:50
-1: What is this answer? Any wavefunction repeats in one period, up to a global phase, which depends only on the zero point energy. – Ron Maimon Dec 6 '11 at 4:46

The point of your "Poincaré recurrence theorem analog in quantum mechanics" is that any sum of complex exponentials $$f(t) = \sum_{n=1}^N c_n e^{i \omega_n t}$$ with $\omega_n \in \mathbb{R}$ is almost periodic, i.e. $$\forall \epsilon > 0 ,\ \ \exists T \in \mathbb{R},\ \ \max_t |f(t) - f(t+T)| < \epsilon$$ The proof can be seen there : Almost Periodic Functions and Differential Equations, chapter I, property 6 which uses the Bochner's theorem.

Sketch of proof:

$f_T(t) = f(t+T)$, so $f_T$ is a notation for $f$ translated in time by a factor $T$

consider the $||.||_\infty$ norm : $||f-g|| = \max_t \ |f(t) - g(t)|$

If $f$ is almost periodic then for any sequence $(u_n)_{n\in \mathbb{N}}$ there exists a sub-sequence $(v_n)$ such that $$\lim_{n\to \infty} f_{v_n} = \hat{f}$$ converges. Reciprocally, if for any sequence there is a sub-sequence such that $\lim_{n\to \infty} g_{v_n}$ converges, then $g$ is almost periodic. Choose two almost periodic functions $f$ and $g$. Then, it follows that for any sequence $(u_n)$, $(f+g)_{u_n}$ has a convergent sub-sequence, so that $f+g$ is also almost periodic. Thus, $\sum_{n=1}^N c_n e^{i \omega_n t}$ is almost periodic. Finally, $$f(t) = \sum_{n=1}^\infty c_n e^{i \omega_n t}$$ is almost periodic if $\sum |c_n|$ exists.

The conclusion is that any wave-function solution of the Schrodinger equation with time independent potential energy is almost periodic : $$i \frac{\partial \varphi}{\partial t} = H \varphi$$ ${\Phi_n},E_n$ are the eigenvectors/eigenvalues of the time independent Hamiltonian $H$, $$\varphi(x,t) = \sum_n c_n e^{i E_n t} \Phi_n(x)$$

is almost periodic !


One consequence is that if the nucleus are fixed and there is no photon emission, then some atoms cannot react chemically. Chemical reaction involves nucleus moving and/or photon emission.

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