# Variance in position for the infinite square potential well?

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This exercise pops up in the Cavendish Quantum Mechanics Primer (M. Warner and A. Cheung) but I can't seem to figure it out. So far, when looking at the infinite square well of width a (and at this point in the book complex wavefunctions are not yet considered), the general form of the wave function is $A_nsin(k_nx)$ where $k_n=\frac{\sqrt{2mE}}{\hbar}=\frac{n\pi}{a}=\frac{2\pi}{\lambda}$.
I am really at a loss here. I was thinking about $\langle x^2 \rangle - \langle x\rangle$ but I don't know how to apply it here.

The answer should be $a^2\left(\frac{1}{12}-\frac{1}{2\pi ^2n^2}\right)$...

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For a wavefunction ##\psi(x)##, the expectation value of an operator ##\hat{O}## is
##\langle \hat{O} \rangle \equiv \langle \psi | \hat{O} | \psi \rangle = \int_{-\infty}^{\infty} \psi^*(x) \hat{O} \psi(x) dx##

Hence, you need to compute this for ##x^2## and ##x##.

vanhees71
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This looks like a homework problem, and you should post it in the homework forum. You'll not get final answers, but we'll try to help you to find the answer yourself.

First of all, you should think about the question, how to evaluate expectation values from the given wave function, which is given up to the constant $A_n$. So you should check, how to determine it. Finally, the variance of the particle's position is
$$\Delta x^2=\langle x^2 \rangle - \langle x \rangle^2.$$

For a wavefunction ##\psi(x)##, the expectation value of an operator ##\hat{O}## is
##\langle \hat{O} \rangle \equiv \langle \psi | \hat{O} | \psi \rangle = \int_{-\infty}^{\infty} \psi^*(x) \hat{O} \psi(x) dx##

Hence, you need to compute this for ##x^2## and ##x##.
Thanks for your reply! Problem is, I am aware of these formulas but I don't know how to apply them here. For the non-complex case, as here $\psi=A_nsin(k_nx)$, how would you calculate $\int_{-\infty}^{\infty} \psi^*(x) \hat{O} \psi(x) dx$?

Thanks!

You do exactly the same thing you do with a complex wavefunction. ##\psi^*(x) = \psi(x)## if the wavefunction is real.

I have tried doing $\int_0^ax^2A_n^2sin^2(k_nx)dx-\left(\int_0^axA_n^2sin(k_nx)dx\right)^2$, but I got $\frac{1}{6}(a^2-9)$! Am I at least getting the method right?

If the missing square on the sine in the second integral is just a typo then, yes, that's the right method.

• 21joanna12
Apologies- it is a typo!