Gradient of product of wave functions

[doktorglas]

Hi,

Short question: If you take the inner product of two arbitrary wave functions, and then the gradient of that, the result should be zero, right? (Since the product is just a complex number.) Am I missing something?

∇∫dΩψ_{1}*ψ_{2} = 0

 

[dextercioby]

Sure, you get the 0 vector in R^n, but what would you use that gradient for ?

 

[The_Duck]

Hi,

Short question: If you take the inner product of two arbitrary wave functions, and then the gradient of that, the result should be zero, right? (Since the product is just a complex number.) Am I missing something?

∇∫dΩψ_{1}*ψ_{2} = 0

I don't think this operation makes sense. The gradient is a derivative with respect to a spatial coordinate. What spatial coordinate are you differentiating with respect to? It must not be a coordinate you have already integrated over in the inner product.

 

[tom.stoer]

The_Duck is right. The expression is meaningless.

The gradient is related to the momentum operator. But the momentum operator acts on wave functions, not on scalar products i.e. not on integrals over wave functions.

 

[doktorglas]

Thanks for the answers, but I'm sorry, I made an error in the formulation of the question. What I really meant was with the gradient inside of the integral, that is <p psi1 | psi2> in Dirac's notation. What I need is the gradient of (psi1-conjugate times psi2) to be zero for a proof I was thinking about.

 

[The_Duck]

It's certainly not the case that $\langle \hat p \psi_1 | \psi_2 \rangle = 0$ in general. It's not hard to come up with counterexamples; try it out.

 

[doktorglas]

Well, it turns out that I partly misformulated myself again, and was somewhat right from the beginning. I will try to make sense now, and put it in a little more context. What I want to be zero is the following expression:

-i\hbar∫dΩ(\overline{ψ_{1}}∇ψ_{2}+(∇\overline{ψ_{1}})ψ_{2}) = -i\hbar∫dΩ∇(\overline{ψ_{1}}ψ_{2})

which means ∇(\overline{ψ_{1}}ψ_{2}) would have to be zero, which I figured it was since (\overline{ψ_{1}}ψ_{2}) is a constant - the product of the probability amplitudes.

 

[tom.stoer]

No.

$\nabla (\bar{\psi}_1 \psi_2) \neq 0$

in general (for arbitrary wave functions. The product of two wave functions is not a constant (only the integral over this product is a constant).

But

$\int dV \, \nabla (\bar{\psi}_1 \psi_2) = \oint dS \, (\bar{\psi}_1 \psi_2)$

and this vanishes for physically reasonable wave functions with

$\lim_{r\to\infty}\psi(r)= 0$