# Hermitian Operators in quantum mechanics

## Homework Statement

Within the framework of quantum mechanics, show that the following are Hermitian operators:

a) $$p=-i\hbar\bigtriangledown$$

b) $$L=-i\hbar r\times\bigtriangledown$$

Hint: In Cartesian form L is a linear combination of noncommuting Hermitian operators.

## Homework Equations

$$\int\psi_{1}^{*}\L\psi_{2}d\tau=\int(\L\psi_{1})^{*}\psi_{2}d\tau$$

## The Attempt at a Solution

I understand that a Hermitian operator is self-adjoint, and that it's eigenvalues are real, but as far as proving it, I'm not exactly sure how to use the formula above to do that.

Last edited:

Tom Mattson
Staff Emeritus
Gold Member
You need to insert the operator into each side of the identity separately and show that they both work out to the same quantity. That means you'll be manipulating one integrand until it looks like the other one.

Dick
Homework Helper
And consider integration by parts to move the derivative operators.

$$\int\psi^{*}(\frac{\delta\psi_{2}}{\delta x}+\frac{\delta\psi_{2}}{\delta y}+\frac{\delta\psi_{2}}{\delta z})d\tau$$

So do I start off like this? If $$\psi$$ is a function of x, y and z then how do I handle the integration by parts? (The textbook does it, but only dependent on one variable)

Dick
Homework Helper
You do it just the same way as with one variable. (?) I'm not sure what is confusing you. Split the integral into three separate parts if you need to.

Tom Mattson
Staff Emeritus
Gold Member
That expression in parentheses in the integrand doesn't look much like $\nabla(\psi_2)$ to me.

how do I handle the integration by parts? (The textbook does it, but only dependent on one variable)

You're working in Cartesian coordinates, which all appear symmetrically in your integrand. So you have the luxury of working out the 1D problem (say, in x) and then saying "Similarly, we have for y and z..."

I think I got it,

$$\int\psi_1^*(-i\hbar\bigtriangledown\psi_2)d\tau=-i\hbar\psi_1^*\psi_2+\int i\hbar\bigtriangledown\psi_1^*\psi_2 d\tau$$

Since $$\psi_1^*\psi_2=0$$ (eigenfunctions are orthogonal) then

$$\int\psi_1^* (-i\hbar\bigtriangledown\psi_2)d\tau=\int i\hbar\bigtriangledown\psi_1^*\psi_2 d\tau$$

Which is the same as above equation. Is this right?

Dick
Homework Helper
I'm afraid it's not orthogonality. The expression without an integral sign on it should be a difference betwen its values at + and - infinity. What is the the answer?

So it's zero because it's basically $$-\infty+\infty$$?

Dick
Homework Helper
Nooooo. You usually assume wave functions vanish at infinity or have some other similar boundary condition.

Tom Mattson
Staff Emeritus
Gold Member
Also, here's another nitpick.

This:

$$\int\psi_1^* (-i\hbar\bigtriangledown\psi_2)d\tau=\int i\hbar\bigtriangledown\psi_1^*\psi_2 d\tau$$

Should be carried one step further to this:

$$\int\psi_1^* (-i\hbar\bigtriangledown\psi_2)d\tau=\int (-i\hbar\bigtriangledown\psi_1)^*\psi_2 d\tau$$

I did part b) using the same method, but only the x component. (Since from cross product I get a vector.) Am I required to prove for each component, or is the linear combination hint enough to extend this proof to all components?

Tom Mattson
Staff Emeritus