Matrix Element of Position Operator

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The discussion focuses on calculating the matrix element of the position operator in the infinite square well's eigenstate basis. The integral for the matrix element involves substituting sine functions and applying trigonometric identities. A participant questions the validity of obtaining a negative result for off-diagonal elements, realizing it is not an expectation value issue. The conversation also touches on the momentum operator, clarifying that the resulting integral won't necessarily yield all zeros due to the dependence on the specific values of m and n. The key takeaway is that careful consideration of the integration limits and orthogonality is crucial in these calculations.
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Homework Statement



Calculate the general matrix element of the position operator in the basis of the eigenstates of the infinite square well.

Homework Equations



|\psi\rangle =\sqrt{\frac{2}{a}}\sin{\frac{n \pi x}{a}}

x_{n,m}=\langle\psi_{n}|\hat{x}|\psi_{m}\rangle=\int^a_0\psi^\star_{n} x \psi_mdx

The Attempt at a Solution



What I tried doing is substituting the sin eigenfections into the integral and then using the relation that sin(x)sin(y)=(cos(x-y)-cos(x+y))/2 and then integrating from 0 to a. I keep getting a negative answer, am I using the wrong approach? I know this doesn't work when m=n, but shouldn't it work when they aren't equal?

Thanks for the help
 
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Sounds like you're doing it right. Why do you think a negative answer for the off-diagonal elements is wrong?
 
I guess I was thinking of it as an expectation value which I now see is not correct.

When use this same method using the momentum operator, -i hbar d/dx, I get an integral with Sin(npix/a)Cos(npix/a), which must be zero when integrated from 0 to a by integral identities, correct? What does this mean then if the momentum operator is represented by a matrix with all zeros?
 
No, it won't be all zeros. Remember you're not always integrating over a full period (depending on what m and n equal), so you can't use orthogonality to say the integrals will all be zero.
 

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