# Why isn't the Pauli equation equivalent to the Schrodinger equation?

The Pauli equation (seen here) contains its spin dependence in the term which reads

$$\frac{1}{2m}\left[ \sigma\cdot\left(p-\frac{e}{c}A\right)\right]^2$$

So let B be any vector. Then

$$\left( \sigma\cdot B\right)^2$$
$$=\left(\sigma_1 B_1 +\sigma_2 B_2 + \sigma_3 B_3\right)\left(\sigma_1 B_1 +\sigma_2 B_2 + \sigma_3 B_3\right)$$
$$=\sigma_1^2 B_1^2 +\sigma_2^2 B_2^2+\sigma_3^2 B_3^2 + (\sigma_1\sigma_2+\sigma_2\sigma_1)B_1 B_2 + (\sigma_1\sigma_3+\sigma_3\sigma_1)B_1 B_3 + (\sigma_3\sigma_2+\sigma_2\sigma_3)B_3 B_2$$
$$=1\cdot B_1^2 + 1\cdot B_2^2 + 1\cdot B_3^2 + 0 + 0 +0$$
$$=B^2$$

So isn't the sigma-dependent term in the Pauli equation identically equal to

$$\frac{1}{2m}\left(p-\frac{e}{c}A\right)^2$$

?

if yes, then in what sense is it spin-dependent? If no, then where did I go wrong?

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I have only briefly looked at your calculation and it looks correct, but I don't think you are calculating what you really want. (By the way, I assume that the identity matrix is implied in your notation.)

Try to do the same, but now assume that B is a vector operator. So in general $$\sigma_1\sigma_2B_1B_2 + \sigma_2\sigma_1B_2B_1 \neq (\sigma_1\sigma_2 + \sigma_2\sigma_1)B_1B_2$$, since you don't always have $$[B_1,B_2] = 0$$.

Thanks, element4. I get it now. The momentum operator and the EM potential do not commute since the latter is a function of position, so the B^2 terms in my derivation are not so simple. Haven't worked it out yet, but I'm pretty sure that's it.

Ok. So I worked through it and have one more nagging question. The final form I get is

$$\left\{\frac{1}{2m}\left(\vec{p}-\frac{e}{c}\vec{A}\right)^2-\frac{e\hbar}{2mc}\vec{\sigma}\cdot\vec{B}+e\Phi-i\hbar\frac{\partial}{\partial t}\right\}\Psi-\frac{e\hbar}{mc}\vec{\sigma}\cdot\left(\vec{A}\times\nabla\right)\Psi=0$$

The part in the braces is familiar and can also be seen on the wikipedia page: http://en.wikipedia.org/wiki/Pauli_equation#Special_Cases What about the last term? Is it something typically neglected? Why? Or did I screw up the derivation?

Meir Achuz
Homework Helper
Gold Member
The correct result is not AXdel. It should be curl A, which equals B because the curl does not act on psi. This just gives -mu.B, the interaction of the electron magnetic moment with B. This result also derives the g=2 for electron magnetic moment. Be more careful with the combination -[del.A + A.del]psi.

Thanks, Meir Achuz. But the term containing

$$\frac{1}{2m}\left[ \sigma\cdot\left(p-\frac{e}{c}A\right)\right]^2$$

when expanded results in both $$A\times\nabla$$ and $$\nabla\times A$$ terms (I think). I have the $$\nabla\times A$$ covered above in the term containing B. But I don't know what to do with the $$A\times\nabla$$ term

Bump.

Just one little bump.

I did not work this through, but I think the del X A term does not equal B. It's the same way with [P, X] = (d/dx)*x - x*(d/dx) != 1 - x*(d/dx).

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I did not work this through, but I think the del X A term does not equal B. It's the same way with [P, X] = (d/dx)*x - x*(d/dx) != 1 - x*(d/dx).

I think what you are suggesting is that when I see $$\nabla\times A$$ I should not read this as "multiply by B" but rather consider it an operator: when acting on a function $$\Psi$$ it yields $$\nabla\times (A\Psi)$$.

That is, where I wrote

$$\left\{\frac{1}{2m}\left(\vec{p}-\frac{e}{c}\vec{A}\right)^2-\frac{e\hbar}{2mc}\vec{\sigma}\cdot\vec{B}+e\Phi \Psi -i\hbar\frac{\partial}{\partial t}\right\}\Psi-\frac{e\hbar}{mc}\vec{\sigma}\cdot\left(\vec{A}\times\nabla\right)\Psi=0$$

$$\left\{\frac{1}{2m}\left(\vec{p}-\frac{e}{c}\vec{A}\right)^2\Psi-\frac{e\hbar}{2mc}\vec{\sigma}\cdot\left(\nabla\times [\vec{A}\Psi]\right)+e\Phi\Psi -i\hbar\frac{\partial\Psi}{\partial t}\right\}-\frac{e\hbar}{mc}\vec{\sigma}\cdot\left(\vec{A}\times\nabla\right)\Psi=0$$

$$\left\{\frac{1}{2m}\left(\vec{p}-\frac{e}{c}\vec{A}\right)^2\Psi-\frac{e\hbar}{2mc}\vec{\sigma}\cdot\left(\nabla\Psi\times \vec{A} +\Psi\nabla\times \vec{A}\right)+e\Phi\Psi -i\hbar\frac{\partial\Psi}{\partial t}\right\}-\frac{e\hbar}{mc}\vec{\sigma}\cdot\left(\vec{A}\times\nabla\right)\Psi=0$$

$$\left\{\frac{1}{2m}\left(\vec{p}-\frac{e}{c}\vec{A}\right)^2\Psi-\frac{e\hbar}{2mc}\vec{\sigma}\cdot\left(\Psi\nabla\times \vec{A}\right)+e\Phi\Psi -i\hbar\frac{\partial\Psi}{\partial t}\right\}=0$$

$$\left\{\frac{1}{2m}\left(\vec{p}-\frac{e}{c}\vec{A}\right)^2-\frac{e\hbar}{2mc}\vec{\sigma}\cdot\vec{B}+e\Phi-i\hbar\frac{\partial}{\partial t}\right\}\Psi =0$$

That's it. Thank you!