Help with vector operator Del.

[cesaruelas]

Homework Statement

In the Pauli theory of the electron, one encounters the expresion:

(p - eA)X(p - eA)ψ

where ψ is a scalar function, and A is the magnetic vector potential related to the magnetic induction B by B = ∇XA. Given that p = -i∇, show that this expression reduces to ieBψ.

Homework Equations

pXp = 0 and AXA = 0

The Attempt at a Solution

I've come to this:

-e(pXA + AXp)ψ

but I don't even have a clue where to go next since, for all I know,

pXA + AXp = -(AXp) + AXp = 0

?

Someone's got a clue what I should do next? Am I missing something here?

 

[Dick]

Homework Statement

In the Pauli theory of the electron, one encounters the expresion:

(p - eA)X(p - eA)ψ

where ψ is a scalar function, and A is the magnetic vector potential related to the magnetic induction B by B = ∇XA. Given that p = -i∇, show that this expression reduces to ieBψ.

Homework Equations

pXp = 0 and AXA = 0

The Attempt at a Solution

I've come to this:

-e(pXA + AXp)ψ

but I don't even have a clue where to go next since, for all I know,

pXA + AXp = -(AXp) + AXp = 0

?

Someone's got a clue what I should do next? Am I missing something here?

xp-px isn't 0. That's not zero for the same reason. The vector potential is a function of x. x doesn't commute with the x component of the momentum operator.

 

[cesaruelas]

xp-px isn't 0. That's not zero for the same reason. The vector potential is a function of x. x doesn't commute with the x component of the momentum operator.

Thank you! I got it now. But now I have another question:

Is

pXA = (-AXp + pXA)/2 or -AXp + pXA

I need to use this result to end up the problem but it would only work if the sencod result is true. I used -AXp = pXA to develop it but the 1/2 is not letting me move on with the problem in question.

 

[Dick]

Thank you! I got it now. But now I have another question:

Is

pXA = (-AXp + pXA)/2 or -AXp + pXA

I need to use this result to end up the problem but it would only work if the sencod result is true. I used -AXp = pXA to develop it but the 1/2 is not letting me move on with the problem in question.

Don't know. I don't know why you think either one is true.

 

[cesaruelas]

Don't know. I don't know why you think either one is true.

pXA = -AXp → pXA = (pXA + pXA)/2 = (-AXp + pXA)/2

 

[Dick]

pXA = -AXp → pXA = (pXA + pXA)/2 = (-AXp + pXA)/2

But p x A isn't generally equal to -A x p. It's true for constant vectors because the components of constant vectors are constant and they commute. p is a vector of differential operators and A is a vector of functions of position. They don't commute.

 

[cesaruelas]

But p x A isn't generally equal to -A x p. It's true for constant vectors because the components of constant vectors are constant and they commute. p is a vector of differential operators and A is a vector of functions of position. They don't commute.

Maaaaaaan, you're right. I guess I'll have to find another way to solve this then. Thank you!