# Pauli Hamiltonian

1. Dec 11, 2008

### diegzumillo

1. The problem statement, all variables and given/known data
The Hamiltonian of an electron with mass m, electric charge q and spin
of $$\frac{\hbar }{2}\vec{\sigma}$$ in a magnetic field described by the
potential vector $$\vec{A}\left( \vec{r},t\right)$$ and a scalar potential $$U\left( \vec{r},t\right)$$ is given by

$$$H=\frac{1}{2m}\left[ \vec{P}-q\vec{A}\right] ^{2}+qU-\frac{q\hbar }{2m}\vec{ \sigma}.\vec{B}$$$

where $$\vec{B}=\vec{\nabla}\times \vec{A}$$. Show that this Hamiltonian can
also be obtained from Pauli Hamiltonian:

$$$H=\frac{1}{2m}\left\{ \vec{\sigma}.\left[ \vec{P}-q\vec{A}\right] \right\}^{2}+qU$$$

2. Relevant equations

I belive this one is useful here:

$$$\left( \vec{\sigma}.\vec{A}\right) \left( \vec{\sigma}.\vec{B}\right) =\vec{A}.\vec{B}I+i\vec{\sigma}.\left( \vec{A}\times \vec{B}\right)$$$

Wich in our case, we can rewrite it as

$$$\left( \vec{\sigma}.\vec{A}\right) ^{2}=A^{2}I+i\vec{\sigma}.\left( \vec{A}\times \vec{A}\right)$$$

(it's not the same vector A of the problem statement, of course)

3. The attempt at a solution

Using the above identity, we end up with a term like this:

$$$\left[ \vec{P}-q\vec{A}\left( \vec{R},t\right) \right] \times \left[ \vec{P}-q\vec{A}\left( \vec{R},t\right) \right] =\vec{P}\times \vec{P}-\vec{P}\times q\vec{A}\left( \vec{R},t\right) -q\vec{A}\left( \vec{R},t\right)\times \vec{P}+q^{2}\vec{A}\left( \vec{R},t\right) \times \vec{A}\left( \vec{R},t\right)$$$

Wich is... almost nice. If I knew what to do with all of these guys! I can see that if we consider only the second term we can solve the problem. What does this mean?..

Using
$$$\vec{P}\rightarrow i\hbar \vec{\nabla}$$-\vec{P}\times q\vec{A}\left( \vec{R},t\right) =-i\hbar q\vec{\nabla}\times \vec{A}\left( \vec{R},t\right) =-i\hbar q\vec{B}$$$

And using this result in the Hamiltonian..

$$$H=\frac{1}{2m}\left\{ \left[ \vec{P}-q\vec{A}\left( \vec{R},t\right) \right] ^{2}+i\vec{\sigma}.\left[ -i\hbar q\vec{B}\right] \right\} +qU\left( \vec{R},t\right)$$$$ $$H=\frac{1}{2m}\left\{ \left[ \vec{P}-q\vec{A}\left( \vec{R},t\right) \right]^{2}+\hbar q\vec{\sigma}.\vec{B}\right\} +qU\left( \vec{R},t\right)$$$
$$$H=\frac{1}{2m}\left[ \vec{P}-q\vec{A}\left( \vec{R},t\right) \right] ^{2}+ \frac{\hbar q}{2m}\vec{\sigma}.\vec{B}+qU\left( \vec{R},t\right)$$$

2. Dec 11, 2008

### dextercioby

Actually

$$\hat{\vec{P}} = - i\hbar \vec{\nabla} \hat{1}$$ in ondulatory mechanics.

3. Dec 11, 2008

### diegzumillo

Oops!!
Thanks :] that lead to the correct answer, but still no reason why to cancel all those other terms!

I mean, the last term of the first equation in the 'attempt at a solution' is easy.. any regular vector (not the case with the derivative operator) the cross product with itself is 0.

I'm not sure this is the right path.. but I think there must be an argumento to justify this equation:

$$\vec{P}\times \vec{P}-q\vec{A}\times \vec{P}=0$$

4. Dec 12, 2008

### turin

Yeah, it's what you said. You're making the vanishing of the first and last terms too complicated. Of course, you have probably been viciously trained by now to hesitate when applying an old math rule to a new situation, and, for instance, you are probably afraid of things like PxP because you know that P is an operator, and you also remember that JxJ is NOT zero. OK, so good for you; this should be your initial attitude. In fact, calling A a "regular vector" (i.e. c-number vector) is now a big no-no. However, now let's decide what these cross products are.

Hints:
What are the commutators of X, Y, and Z among themselves?
What are the commutators of Px, Py, and Pz among themselves?
Does A depend on any operators other than X, Y, and Z?
Does P depend on any operators other than Px, Py, and Pz?
Write the cross products in terms of components and relate this expression to commutators.
Is it possible for any of these commutators to be nonzero?

OK, that takes care of first and last term. However, there are TWO terms left over, not just the one. You seem to have no problem with this. Can you explain?

Last edited: Dec 12, 2008
5. Dec 12, 2008

### diegzumillo

Hi Turin!
Thanks for taking the time :]

Well, I was indeed a little afraid of P! Because there was a little contradiction in my head... While I considered P as an general operator that its components commute with each other, the cross product with itself should vanish. But When I considered it to be the gradient operator the commutation didnt seem obvious anymore.

But now that you rubbed this in my nose :D I can think clearer. Partial derivatives on different variables will commute if the function has continuous second partial derivatives. (there's probably a theorem for that) So, PXP will vanish in every way I approach this.

Now, for the term qAXP, I don't really know. Wich lead me to your third question: "Does A depend on any operators other than X, Y, and Z?". er... I dont know! The only thing I know about it is that it's curl is the magnetic field... Sometimes I feel the answer is right below my nose.

6. Dec 15, 2008

### turin

Hint 1: You are definitely on the right track when you consider that P is actually a derivative operator, and how it should operate on functions. This gets at a very important point in QM: the operators are to some extent arbitrary, but their matrix elements had better behave. In particular, try to consider not only the right-action, but also the left-action of the operators.

Hint 2: Is it really true that PxA is proportional to B? (Use hint 1 to approach this question.)

BTW, I feel your pain. This is on my top ten list for most difficult issues in QM. Fortunately, this one has an explanation that I (think I) understand.

7. Feb 12, 2009

### Jezuz

You must be very careful when working with operators. Remember that you should always assume that it is multiplied with a function. This means that

$$\mathbf p \times \mathbf A \neq -i\hbar (\nabla \times \mathbf A).$$
Rather we have
$$\mathbf p \times \mathbf A \psi = - i\hbar \nabla \times (\mathbf A \psi) = - i\hbar \mathb B \psi - i\hbar \mathbf A \times \nabla \psi .$$