# Determining the Wave Function From Initial Conditions

This is Problem 2.6 (b) in Griffiths, Intro to QM:

A particle in an infinite square well has its initial wave function an even mixture of the first two stationary states:

$$\Psi(x,0) = A[\psi_1(x) + \psi_2(x)]$$.

Here is the part of the problem that I am having a little trouble with:

(b) Find $$\Psi(x,t)$$ and $$|\Psi(x,t)|^2$$. Express the latter as a sinusoidal function of time, as in Example 2.1. To simplify the result, let $$\omega \equiv \frac{\pi^2 \hbar}{2ma^2}$$

According to the answer key, even after $$t=0$$, the wave function continues to be a mixture of the first two stationary states. Why is that? I am having a little difficulty understanding this. Why can't it be a new 'mixture?'

Are my questions sufficiently clear?

• Are you asking why there are no "new" stationary states in the mixture at later times? Just apply the time evolution operator to the initial state and see what happens. Aug 15 '14 at 19:50
• Yes, exactly. As far as I understand, the most general solution of the infinite well is $\Psi (x,t) = \sum_{n=1}^{\infty} c_n \psi_n(x) \phi_n(t)$, where the coefficients are $c_n = \sqrt{\frac{2}{a}} \int_{0}^{a} \sin(\frac{n \pi}{a} x) \Psi(x,t)$. Why wouldn't I use these two equations to calculate the wavefunction for all future times?
– Mack
Aug 15 '14 at 19:55
• @ACuriousMind I do not know of this time evolution operator of which you speak.
– Mack
Aug 15 '14 at 20:01
• Use equation (2.36) in the book. Aug 16 '14 at 14:16

The time evolution operator of a quantum system is (in units with $\hbar = 1$)

$$U(t_0,t) = \mathrm{e}^{\mathrm{i}H (t - t_0)}$$

and the "stationary states" are the eigenstates of this operator, i.e. eigenstates of the Hamiltonian. If you are given a collection of stationary states (not a basis of the space, mind you) $\{\lvert \psi_E \rangle\}$ with $H \lvert \psi_E \rangle = E \lvert \psi_E \rangle$, any sum of these will stay a sum of these under time evolution:

$$U(t_0,t)\sum_E \lvert \psi_E\rangle = \sum_E \mathrm{e}^{\mathrm{i}H (t - t_0)} \lvert \psi_E \rangle = \sum_E\mathrm{e}^{\mathrm{i}E (t - t_0)}\lvert \psi_E\rangle$$

There is simply no room for other stationary states to appear, since every single stationary state's time evolution is given by just a phase.

• So, what other justification could be used? The answer key does not use this operator, nor is it spoken of in the chapter from which this problem comes from.
– Mack
Aug 15 '14 at 20:10
• @Mack: How did you even define the stationary states if you do not know the time evolution operator? Aug 15 '14 at 20:12
• Well, I did not define them, but the textbook I am using does, which is Griffith's Introduction To Quantum Mechanics. Here is what he says regarding stationary states: "Although the wave function itself does (obviously) depend on $t$, the probability density, $|\Psi(x,t)|^2 = \Psi^* \Psi = \psi e^{i Et/\hbar} \psi e^{-i Et/\hbar} = |\psi(x)|^2$ does not.
– Mack
Aug 15 '14 at 20:19
• @Mack: But where did the factor $\mathrm{e}^{\mathrm{i}Et}$ come from? That is precisely time evolution applied to a stationary state! Aug 15 '14 at 20:26
• Oh, I see. You are using terminology that I am unfamiliar with. I still do not see how this relates to the wave function being only a mixture of the first two stationary states.
– Mack
Aug 15 '14 at 20:30

The superposition principle of quantum mechanics is not destroyed by quantum (hamiltonian) unitary evolution operator $U(t_0,t) = \mathrm{e}^{\mathrm{i}H (t - t_0)}$ as per @ACuriousMind's answer.

Event if you dont know of the evolution operator in terms of the hamiltonian (which can be derived easily from the Schrodiger equation), still the fact that the Schrodiger equation is linear (wrt to wavefunction $\psi$) (both time-independent and time-dependent equations), means the linearity of the wavefunction (or in QM terms the superposition) is not destroyed.

Superposition is destroyed only by a quatum measurement (as per the Copenhagen formalism)