# Homework Help: Pauli matrix with del operator

1. Oct 14, 2014

### KleZMeR

1. The problem statement, all variables and given/known data
In the Pauli theory of the electron, one encounters the expresion:

(p - eA)X(p - eA

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.

2. Relevant equations

3. 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.

2. Oct 14, 2014

### KleZMeR

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

3. Oct 14, 2014

### 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?

4. Oct 14, 2014

### vela

Staff Emeritus
How is $\vec{r}$ appearing?

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

5. Oct 14, 2014

### 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

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

6. Oct 14, 2014

### KleZMeR

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

7. Oct 14, 2014

### vela

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

8. Oct 14, 2014

### vela

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

9. Oct 14, 2014

### 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

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

10. Oct 14, 2014

### KleZMeR

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

11. Oct 14, 2014

### vela

Staff Emeritus
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.

12. Oct 14, 2014

### KleZMeR

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

13. Oct 14, 2014

### vela

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

14. Oct 14, 2014

### 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.

15. Oct 14, 2014

### KleZMeR

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

16. Oct 14, 2014

### KleZMeR

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

17. Oct 14, 2014

### vela

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

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

Right.

18. Oct 14, 2014

### 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 $ie \frac{1}{2}(B( \nabla r)+(r \nabla)B) \psi$

Last edited: Oct 14, 2014
19. Oct 14, 2014

### vela

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