Pauli matrix with del operator

[KleZMeR]

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

The Attempt at a Solution

I'm first working this matrix determinant out without substitution, although I know the answer should include the B vector, and look something like this: ieBψ

I split (p - eA) into its vector components and used the determinant, but it's just getting messy and I don't know what to do.

My first question is, do I need to work this matrix out or is there an identity I should use to avoid all this? I've done this problem almost 10 times by brute force and can't seem to get it.

This problem has been asked about previously but it wouldn't let me re-open the thread.

 

[KleZMeR]

Ok, I got it into this form -e(pXA + AXp)ψ, trying to work it out from here.

 

[KleZMeR]

Ok, having some issues. I used the identity on the first cross product PxA and got

$\frac{1}{2}B ( \nabla \bullet r) + \frac{1}{2} ( r \bullet \nabla )B$

But I don't know what to do with the second cross product, it's not simplifying well.

I am wondering, can I treat these three as scalars because the del operator is not operating on anything? Can I use the scalar triple product identity?

 

[vela]

How is $\vec{r}$ appearing?

You start with
$$(-i\nabla - e\vec{A})\times (-i\nabla - e\vec{A})\psi = (i\nabla + e\vec{A})\times (i\nabla \psi + e\vec{A}\psi).$$ What terms do you get when you multiply that out?

 

[KleZMeR]

Ohh, I wrote something down wrong in the question,

I was given $A = \frac{1}{2}(B \times r)$

I managed to get the whole thing down to $ei(3/2)B\psi$ but that was using the scalar triple product identity for

$Axp = (\frac{1}{2}B \times r) \times p$

 

[KleZMeR]

How is $\vec{r}$ appearing?

You start with
$$(-i\nabla - e\vec{A})\times (-i\nabla - e\vec{A})\psi = (i\nabla + e\vec{A})\times (i\nabla \psi + e\vec{A}\psi).$$ What terms do you get when you multiply that out?

Multiply it out? You mean matrix cross-products component-wise?

 

[vela]

The simplification holds in general. You don't need to know $\vec{A}$ to show it, so I'd ignore that info for now.

 

[vela]

Multiply it out? You mean matrix cross-products component-wise?

No, nothing that painful. Just distribute the cross product across the sums.

 

[KleZMeR]

I have gone through painful cross product component wise stuff for a whole day and still can't get it,

If I just foil I get

$-\nabla^2 \psi + i\nabla eA\psi + eAi\nabla\psi + e^2 AA\psi$ ?

 

[KleZMeR]

This is the last problem in some excruciating HW that's due today and I feel like I'm losing some marbles...

 

[vela]

If I just foil I get

$-\nabla^2 \psi + i\nabla eA\psi + eAi\nabla\psi + e^2 AA\psi$ ?

You need to be more careful. The first term isn't $\nabla^2\psi = \nabla \cdot \nabla\psi$. Likewise, the other terms should still be cross products.

 

[KleZMeR]

So I see that $\nabla \times \nabla = 0 , and, A \times A = 0$ ?

 

[vela]

Yup. For the second term, remember that $\nabla$ acts on the product $\vec{A}\psi$.

 

[KleZMeR]

I am getting $ie((\nabla \times A) + (A \times \nabla))\psi$ and it seems I should find an identity now but I can't seem to find it.

 

[KleZMeR]

Ohhh, I see, product rule via the determinant. I am hoping some of this cancels out.

 

[KleZMeR]

And my $i \nabla \times eA \psi$ should relinquish two terms, one of them being a $-(eA \times i \nabla \psi)$?

 

[vela]

Ohhh, I see, product rule via the determinant. I am hoping some of this cancels out.

You need to get over your obsession with determinants. ;)

http://en.wikipedia.org/wiki/Vector_calculus_identities

And my $i \nabla \times eA \psi$ should relinquish two terms, one of them being a $-(eA \times i \nabla \psi)$?

Right.

 

[KleZMeR]

lol, Ok, I'm getting a result as $i \nabla \times eA \psi$

I think my extra linear algebra classes made me WANT to do all this algebra :/

Now I will plug in for A and use an identity again!?

Which gives me $ie \frac{1}{2}(B( \nabla r)+(r \nabla)B) \psi$

 

[vela]

Do you mean $i(\nabla\times (e\vec{A}))\psi$ as opposed to $i(\nabla\times (e\vec{A}\psi))$? Recall that $\vec{B} = \nabla \times \vec{A}$.