For a particle in an infinite square-well potential in an energy eigenstate, the probability distribution relating to outcomes of position measurements vanishes outside the square well and takes a sinusoidal form inside the well, approaching zero at the edges of the well.

Suppose a particle trapped in an infinite square-well potential has an energy 10 eV in an energy eigenstate for which the position probability distribution has three maxima (or “humps”). If the particle is excited to an energy eigenstate in which the probability has five maxima, what is the energy of the particle in this new state?

I find the question to be rather vague. But here's my best attempt.

First begin by supposing the potential well is of length L.

Solving the Schrodinger equation with V=0,

we get

$$\frac{\mathrm{d}^{2}\psi }{\mathrm{d} x^{2}}=-\bar{k}^{2}$$ with


The general solution is $$\psi\left ( x \right )=Asin\left ( \bar{k}x \right )+B cos\left ( \bar{k}x \right ) for 0\leq x\leq L$$

On physical ground we require $$\psi\left ( 0 \right )=\psi\left ( L \right )=0$$

The physically acceptable solution is $$\psi\left ( x \right )=Asin\left ( \bar{k} x\right )$$ and$$ \bar{k}=\frac{n\pi}{L}$$

A three maxima implies

$$\left | \psi_{n=3}\left ( x \right ) \right |^{2}=\left | A_{n} \right |^{2}sin^{2}\left ( \frac{3\pi x}{L} \right )$$

The energy of the particle is quantised and is in general the formula $$E_{n}=\frac{n^{2}\pi^{2}h^{2}}{2mL^{2}}$$

For an n=3 state, $$E_{n=3}=10ev=\frac{9\pi^{2}h^{2}}{2mL^{2}}$$

This is as far as my attempt goes. For some reason, I do not understand the need for the information provided in the question as such the wave function having a sin and the physical condition-likely red herring? Obviously, if the mass m is known, I can solve for the energy level for which n=5.

  • $\begingroup$ The only unknown here is L, using the information given here you can solve for L. The fact that the distribution is sinusoidal and has 3 humps, must be used to determine that the you are in n=3 I think. Also, they are not saying that the wave function is a sin, just that it is sinusoidal, so it could have any superposition of sin/cos terms. $\endgroup$
    – Loonuh
    Commented Dec 30, 2015 at 3:56
  • $\begingroup$ @Loonuh m is unknown too $\endgroup$
    – Physkid
    Commented Dec 30, 2015 at 3:57
  • $\begingroup$ If $m$ is unknown that is still ok, because you can just solve for the constant $mL^2$ instead of just $L$. They are not asking you to solve for $m$ or $L$, so just solve as much as the problem asks; there very well might not be enough information to solve for $m$ or $L$ independently. $\endgroup$
    – Loonuh
    Commented Dec 30, 2015 at 3:58
  • 1
    $\begingroup$ I don't think that's necessary, just use the fact that $ E = \frac{n^2\pi^2\hbar^2}{2mL^2} $ . You already solved for the unknown $mL^2$, so just increase n to 5 and plug it in. :) $\endgroup$
    – Loonuh
    Commented Dec 30, 2015 at 4:09
  • 1
    $\begingroup$ No problem, the more problems you solve, the better you get at spotting these lazy tricks. xD $\endgroup$
    – Loonuh
    Commented Dec 30, 2015 at 4:12

1 Answer 1


So you know that $$10ev=\frac{9\pi^{2}h^{2}}{2mL^{2}}$$

You also know that $$E_{n=5}=\frac{25\pi^{2}h^{2}}{2mL^{2}}$$

Just divide the two equations, and you'll be able to get the energy $E_{n=5}$. If you feel that the question has given you more information than you need, that is not an issue.


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