Wavefunction of particle in a $2$D box with sides $L$ in the $x$ and $y$-axis: $$φ = \sqrt{\frac {4}{L^2}} \sin\frac{n_1πx}{L}\sin\frac{n_2πy}{L}$$

For the different states ($n_1=1$, $n_2=2$), ($n_1=2$, $n_2=2$) and ($n_1=1$, $n_2=3$), which has the highest probability density at the coordinate ($0.25L$, $0.25L$)

I know that the probability density is the integral of $φ^2$ but I don't know the interval for the integral. I would assume that it is $0<x<L$ but when I do this I get something completely different from the given answers. The given answers are:

For ($n_1=2$, $n_2=2$) $=>$ the value of the wavefunction is $\frac{4}{L^2}$

For ($n_1=1$, $n_2=2$) $=>$ the value of the wavefunction is $\frac{2}{L^2}$

For ($n_1=1$, $n_2=3$) $=>$ the value of the wavefunction is $\frac{1}{L^2}$

The dimension of the probability density must be $\frac{1}{length^2}$, so ($n_1=2$, $n_2=2$) is the correct answer.

I don't understand how they have gotten this and why ($n_1=2$, $n_2=2$) is the correct answer. Could someone please explain this to me. Thank you!

  • 1
    $\begingroup$ You are close. The probability of being in some area is the integral of $|\phi|^2$ over that area. The probability density at a point is just the value of $|\phi|^2$ at that point. In your case, you want the probability density at the point (0.25L,0.25L), so just plug this into $|\phi|^2$ with each choice of $n_1$ and $n_2$, no integration involved. $\endgroup$ – Tyberius Apr 20 '20 at 14:33
  • $\begingroup$ @Tyberius Oh, thank you very much! But why would (n$_1=2$, n$_2=2$) be the correct answer when $\frac{1}{length^2}$ corresponds to (n$_1=1$, n$_2=3$)? $\endgroup$ – katara Apr 20 '20 at 14:46
  • $\begingroup$ They phrase it confusingly here, but that is just the units of the probability density, which are the same for any choice of the quantum numbers. It does help to check that you got a reasonable answer, as it must have these units, but to choose between the options given you need to find the one with the largest value. $\endgroup$ – Tyberius Apr 20 '20 at 14:55
  • $\begingroup$ @Tyberius Okay, thank you very much for the help! :) $\endgroup$ – katara Apr 20 '20 at 14:58

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