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Given a neutron (mass$\approx$939 MeV/c$^2$) in an infinite square well of size $a$, the value of the expectation value for position should be in the range $[0-a]$. I know that the general form of the expectation value for position is $$\langle X\rangle=\int_{-\infty}^{\infty}\psi^*x\psi dx=\int^a_0\psi^*x\psi dx \, ,$$ My wave function is given by: $$\psi[x]=\sqrt[]{\frac{2}{7a}}\sin{\frac{x\pi}{a}}+\sqrt[]{\frac{4}{7a}}\sin{\frac{2x\pi}{a}}+\sqrt[]{\frac{8}{7a}}\sin{\frac{3x\pi}{a}}\, ,$$ which is a superposition of wavefunctions of the form $\sin(n\pi x / a)$. Because all of these $\sin$ functions are orthogonal, the expectation value can be written: $$\sum_n[p_n\int^a_0\phi_n^*x\phi_ndx]$$ For the probabilities of each $n$ state given by $p_n$. However, for all $n$, the above integrals all evaluate to $\frac{a^2}{4}$, which not only has the wrong units, but for $a>4$ gives a magnitude larger than the size of the well. How can this be?

EDIT: this was poorly worded, and under-explained due to a mixture of pressure and lack of sleep (not that anyone on stack exchange cares). I think I've fixed it.

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    $\begingroup$ There is an error in your result as you haven't normalized the wavefunction before doing the integral. But even then, the expectation value can be more than one. It dosen't give the probability but the average value of position. The average value can take any value from $0$ to $a$ $\endgroup$ – biryani Feb 10 '16 at 7:33
  • $\begingroup$ Apart from the issues already mentioned, your integral should be only from $-a$ to $a$. $\endgroup$ – Prahar Feb 10 '16 at 14:21
  • $\begingroup$ So, the problem I was actually having I'll cover in an answer that I'll post (it boils down to biryani being right, I didn't normalize), but with regard to the points you brought up: The expectation value for the position of a particle in an infinite square well should be within that well. Apart from the magnitude of $\frac{a^2}{2}$ being larger than $a$ for $a>4$, the units are of length$^2$, which is incorrect. Furthermore, I've defined the well to have the origin at one side, which is why the integral goes from $0$ to $a$. $\endgroup$ – ocket8888 Feb 11 '16 at 2:12
  • $\begingroup$ Wow, seriously downvoted me again. This is why I hate this place. $\endgroup$ – ocket8888 Feb 11 '16 at 2:40
  • $\begingroup$ @ocket8888 Nothing forces you to post on this site. But if you do decide to, please make sure you write clear, well worded questions that amount to more than just "here is what I did, and I get the wrong answer: What did I do wrong?" $\endgroup$ – Danu Feb 11 '16 at 22:54
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The problem I was having is that the integral must be taken of the normalized wavefunction, not the orthonormal wavefunction. It's incorrect to multiply by $p_n$, that's not what the expectation value entails. Specifically, for this case: $$\Psi[x]=\sqrt[]{\frac{4}{7a}}\sin[\frac{\pi x}{a}]+\sqrt[]{\frac{2}{7a}}\sin[\frac{2\pi x}{a}]+\sqrt[]{\frac{8}{7a}}\sin[\frac{3\pi x}{a}]\, .$$ Note that the coefficients are neither the probabilities of each state, nor their square, they are simply the normalization coefficients. Because $$\Psi\in\mathbb{R}\, ,$$ The expectation value evaluates to: $$\langle X\rangle=\int_0^ax\Psi^2dx\\ ~\\ =\frac{4}{7}a\left(\frac{7}{8}-\frac{8(54+25\sqrt[]{2})}{255\pi^2}\right)\\ ~\\ \approx0.316054\space a\, ,$$ within the box, more or less what you'd expect.

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    $\begingroup$ You've forgotten the cross product terms like $\int \phi_1 x \phi_2 dx$, etc. those aren't all zero. $\endgroup$ – Bill N Feb 11 '16 at 3:51
  • $\begingroup$ Actually, you'll find they are for the specified interval of integration. I made a typo in equation 2 of my answer, but you can see now the correct form and if you don't believe me I'm sure you can look it up or at the very least put it into wolframalpha/mathematica. $\endgroup$ – ocket8888 Feb 11 '16 at 5:18
  • $\begingroup$ Actually, Mathematica disagrees with you. $\int_0^b x \sin[x\pi/b]\sin[2x\pi/b]\mathrm{d}x = \frac{-8b^2}{9\pi^2}.$ $\endgroup$ – Bill N Feb 11 '16 at 18:26
  • $\begingroup$ $\int_0^b x \sin [2x\pi/b]\sin [3x\pi/b]\,\mathrm{d}x = \frac{-24b^2}{25\pi^2}.$ $\endgroup$ – Bill N Feb 11 '16 at 18:30
  • $\begingroup$ You were correct, I was confused because it seemed to me like I was multiplying by zero, and not getting zero. I took a step back and figured out why that assumption was stupid. $\endgroup$ – ocket8888 Feb 19 '16 at 19:35

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