Matrix Element of Position Operator

[phrygian]

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

 

[vela]

Sounds like you're doing it right. Why do you think a negative answer for the off-diagonal elements is wrong?

 

[phrygian]

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?

 

[vela]

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.

 

[paranormal]

!

I would like to clarify that the position operator is a fundamental concept in quantum mechanics and is represented by the operator ^x. It is defined as the observable that measures the position of a particle in space.

To calculate the matrix element of the position operator, we need to use the eigenstates of the infinite square well, which are given by |\psi_n\rangle = \sqrt{\frac{2}{a}}\sin{\frac{n \pi x}{a}}. These eigenstates form a complete orthonormal basis for the Hilbert space of the infinite square well.

Using the definition of the position operator, we can write the matrix element as x_{n,m} = \langle\psi_{n}|\hat{x}|\psi_{m}\rangle. Substituting the sin eigenfunctions into the integral and using the trigonometric identity mentioned in the problem, we get x_{n,m} = \frac{1}{2}\int^a_0 \psi_n^*(x)(\cos{\frac{(n-m)\pi x}{a}} - \cos{\frac{(n+m)\pi x}{a}})\psi_m(x)dx.

Since \psi_n(x) and \psi_m(x) are orthogonal for n \neq m, the second term in the integral vanishes and we are left with x_{n,m} = \frac{1}{2}\int^a_0 \psi_n^*(x)\cos{\frac{(n-m)\pi x}{a}}\psi_m(x)dx. This can be further simplified by using the trigonometric identity \cos{\alpha}\sin{\beta} = \frac{1}{2}(\sin{(\alpha+\beta)} + \sin{(\alpha-\beta)}), which gives us x_{n,m} = \frac{1}{4}\int^a_0 (\psi_n^*(x)\psi_{m+1}(x) + \psi_n^*(x)\psi_{m-1}(x))dx.

Finally, using the normalization condition of the eigenstates, we can simplify the integral to x_{n,m} = \frac{a}{4}\delta_{n,m\pm1}, where \delta_{n,m} is the Kronecker delta function.

Therefore, the general matrix element of the position operator in the basis of the eigenstates of the infinite square well is given