A particle of mass m in the infinite square well (of width a) started out in the left half of the well and is (at t=0) equally likely to be found at any point in that region.

What is its initial wave function $$\Psi\left ( x,0 \right )$$ (Assume the wave function is real.)

Getting started,

The solution to the time-independent Schrodinger's equation(or just the n stationary state) is $$\psi_{n}\left ( x,0 \right )=A_{n}Sin\left ( k_{n}x \right )$$ where $$k_{n}=\frac{n\pi}{a}$$

To obtain $$A_{n}$$, subject $$\psi\left ( x,0 \right ) $$to the normalisation condition.

Recall: $$\int_{0}^{a}\left | \Psi\left ( x,0 \right ) \right |^{2}dx=\int_{0}^{a}\Psi^{\dagger }\left ( x,0 \right )\Psi\left ( x,0 \right )dx=\int_{0}^{a}\psi^{\dagger}\left ( x,0 \right )\psi\left ( x,0 \right )dx=1$$

working out, the constant we get $$|A_{n}|=\sqrt{\frac{2}{a}}$$


$$\Psi_{n}\left(x,0\right)=\psi_{n}\left ( x,0 \right )=\sqrt{\frac{2}{a}}Sin\left ( k_{n}x \right )$$

I'd figured this was the solution on the domain of interest but looking at the solution sheet, the solution is given to be

$$\Psi\left( x,0 \right)=\left\{\begin{matrix} \sqrt{\frac{2}{a}} & x \in\left [ 0,\frac{a}{2} \right ]\\ 0 &everywhere \end{matrix}\right.$$

I think I might be close but missing a crucial piece of concept or an assumption.

Would appreciate a point in the right direction.

  • $\begingroup$ The solution you worked out does not have the property that $|\psi|^2=(\mathrm{constant})$ everywhere, so it doesn't meet the supposition of the problem. $\endgroup$
    – user12029
    Dec 19, 2015 at 2:50
  • $\begingroup$ Since the potential well exists for $x\in(0,\,a)$, the wave function should be 0 outside those bounds for sure. $\endgroup$
    – Kyle Kanos
    Dec 19, 2015 at 2:59
  • $\begingroup$ @KyleKanos To whom is this comment for? Indeed, the wavefunction is zero for anywhere else outside the potential well. I can see that the domain of my integration is correct. $\endgroup$
    – Physkid
    Dec 19, 2015 at 3:10
  • $\begingroup$ It was obviously addressed to you, since you neglect to include the 0 term in your solution. $\endgroup$
    – Kyle Kanos
    Dec 19, 2015 at 3:12
  • $\begingroup$ @KyleKanos I know to include to 0 term. Rather, I'm lost as to why the solution is $$\sqrt{\frac{2}{a}}$$ on the domain 0 to a/2 instead of $$\sqrt{\frac{2}{a}}Sin\left ( k_{n}x \right )$$ on the domain 0 to a. $\endgroup$
    – Physkid
    Dec 19, 2015 at 3:17

1 Answer 1


You're making it much harder than it is. It's actually an easy problem, and where you seem to be going wrong is assuming it's a hard problem and pulling out a bunch of extra tools that aren't needed.

Forget about the eigenvectors of the Hamiltonian (the $\psi_n$); they don't play a role in the thing you're trying to solve. Yes, the wave function can be expressed as a superposition of all of the $\psi_n$ functions (where the sum converges everywhere except where the wave function is discontinuous, as expected for a sine transform). But this is irrelevant to the problem you're trying to solve. Expressing the wave function as a superposition of the eigenvectors is a separate problem.

If we just rewrite the problem you're actually being asked to solve without the extraneous details, it's:

A particle is equally likely to be anywhere in the region $[0,a/2]$ and has zero probability of being found outside of that region. Its wave function is everywhere real and non-negative (yes, you do need this non-negativity assumption to answer this question, even though it's not in the problem statement). What's the wave function?

That's it. Forget about the Hamiltonian and its eigenvectors. Forget about the fact that it's in a square well at all. At this point you simply write down the answer, up to a normalization constant:

$\Psi\left( x,0 \right)=\left\{\begin{matrix} C & x \in\left [ 0,\frac{a}{2} \right ]\\ 0 &elsewhere \end{matrix}\right.$

for some constant $C$ (where I've corrected the typo "everywhere" to "elsewhere"). Then you just normalize it to find $C$ (which I see you already know how to do), taking $C$ to be real and positive.

There really isn't any more to it than that. That's the whole solution.

  • $\begingroup$ Thanks a lot! I think if the question had specified "up to a normalisation constant", it would have been clearer. Otherwise, the sine term would have to be included in the wave function. $\endgroup$
    – Physkid
    Dec 19, 2015 at 4:17
  • $\begingroup$ @Physkid, you always normalize wave functions, but as this answer says, the wave function isn't necessarily an eigenfunction of the system. $\endgroup$
    – Bill N
    Dec 19, 2015 at 4:22
  • $\begingroup$ That's just it, the sine functions play no role whatsoever in this particular problem. They're a red herring. Yes, you can express the answer as an infinite sum of sine functions. But it doesn't matter. You can express anything (within reason; there are some technical limitations, but don't worry about those) as an infinite sum of sine functions. $\endgroup$
    – elifino
    Dec 19, 2015 at 4:28
  • $\begingroup$ Said another way, you seem to be under the impression that the wave function is always one of the stationary states. It's not. It really can be anything at all. Any state can be expressed as a combination of the stationary states. $\endgroup$
    – elifino
    Dec 19, 2015 at 4:31
  • $\begingroup$ @elifino However, nothing is being specified about the boundary conditions in this question. So working from the time-independent SE and implementing the BC (with V(x)=0) $$\psi(0)=\psi(a)=0$$, I arrive at a solution with sine. $\endgroup$
    – Physkid
    Dec 19, 2015 at 4:48

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