Hamiltonian (electron in an electro-m field)

[AhmirMalik]

Homework Statement

Given H=\frac{1}{2m}\left[ \vec{P}-q\vec{A}\right] ^{2}+qU+\frac{q\hbar }{2m}\vec{\sigma}.\vec{B} ..(1)
show that it can be written in this form;
H=\frac{1}{2m}\left\{ \vec{\sigma}.\left[ \vec{P}-q\vec{A}\right] \right\}^{2}+qU ...(2)

Homework Equations

[/B]
In my case, I am going to use only;

\left( \vec{\sigma}.\vec{R}\right) ^{2}=\vec{R^2}+i\vec{\sigma}.\left( \vec{R}\times \vec{R}\right) ...(3)

The Attempt at a Solution

Writing equation (1) in this form;
H=\frac{1}{2m}\left[\left[ \vec{P}-q\vec{A}\right]^{2}-i\vec{\sigma}(iq\hbar\vec{B})\right]+qU

using the equation (3) and identifying terms;

\vec{R}=\vec{P}-q\vec{A}

then, to replace \left[\left[ \vec{P}-q\vec{A}\right]^{2}-i\vec{\sigma}(iq\hbar\vec{B})\right] by \left\{ \vec{\sigma}.\left[ \vec{P}-q\vec{A}\right] \right\}^{2}

the following must satisfy;

\left( \vec{R}\times\vec{R}\right)=iq\hbar\vec{B}

to do that, I write RxR like;
\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}+q^{2}\vec{A} \times \vec{A}-\vec{P}\times q\vec{A} -q\vec{A}\times \vec{P}

the first two term (PxP and AxA) on the right hand are equal to 0. But I don't know what to do with the rest;
\vec{P}\times q\vec{A} -q\vec{A}\times \vec{P}

 

[blue_leaf77]

the first two term (PxP and AxA) on the right hand are equal to 0. But I don't know what to do with the rest;
\vec{P}\times q\vec{A} -q\vec{A}\times \vec{P}

Remember that $\mathbf{p} = -i\hbar\nabla$ and that in the actual Schroedinger equation, the Hamiltonian will be multiplied by the wavefunction. Thus, instead of only \vec{P}\times q\vec{A} -q\vec{A}\times \vec{P}, you should consider (\vec{P}\times q\vec{A} -q\vec{A}\times \vec{P})\psi where $\psi$ is an arbitrary wavefunction.

 

[AhmirMalik]

Thank you for replying.

Actually, I knew that. And that's where I'm stucked. Could you please develop a bit more?

 

[blue_leaf77]

Consider the first term which is proportional to $\mathbf{p}\times (\mathbf{A}\psi) = i\hbar\nabla \times (\mathbf{A}\psi)$, At this point, you can use the identity involving the curl of a product between a scalar function and a vector like the one in https://en.wikipedia.org/wiki/Curl_(mathematics).

 

[AhmirMalik]

Ok, thank you !

 

[blue_leaf77]

Can I do the following? ;

-q\vec{A}\times\vec{P}\psi=i\hbar q\vec{A}\times\vec{\nabla}\psi=-i\hbar q \vec{\nabla}\psi\times\vec{A} and then the whole term;

-\vec{P}\times q\vec{A}\psi-q\vec{A}\times\vec{P}\psi=i\hbar q[\vec{\nabla}\psi\times\vec{A}+\psi\vec{\nabla}\times\vec{A}]-i\hbar q \vec{\nabla}\psi\times\vec{A}=i\hbar q\psi\vec{\nabla}\times\vec{A}=i\hbar q\psi\vec{B}

Yes, that's indeed how you should proceed.