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

In summary, the Pauli equation contains its spin dependence in the term which reads \frac{1}{2m}\left[ \sigma\cdot\left(p-\frac{e}{c}A\right)\right]^2. This term can also be expressed as \left( \sigma\cdot B\right)^2, where B is any vector. However, when B is a vector operator, the term \left( \sigma\cdot B\right)^2 is not equal to (\sigma_1\sigma_2 + \sigma_2\sigma_1)B_1B_2, as the operators do not always commute. Additionally, there is an extra term in the final form of the equation that
  • #1
pellman
684
5
The Pauli equation (seen here) contains its spin dependence in the term which reads

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

So let B be any vector. Then

[tex]\left( \sigma\cdot B\right)^2[/tex]
[tex]=\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)[/tex]
[tex]=\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[/tex]
[tex]=1\cdot B_1^2 + 1\cdot B_2^2 + 1\cdot B_3^2 + 0 + 0 +0[/tex]
[tex]=B^2[/tex]

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

[tex]\frac{1}{2m}\left(p-\frac{e}{c}A\right)^2[/tex]

?

if yes, then in what sense is it spin-dependent? If no, then where did I go wrong?
 
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  • #2
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 [tex]\sigma_1\sigma_2B_1B_2 + \sigma_2\sigma_1B_2B_1 \neq (\sigma_1\sigma_2 + \sigma_2\sigma_1)B_1B_2[/tex], since you don't always have [tex][B_1,B_2] = 0[/tex].
 
  • #3
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.
 
  • #4
Ok. So I worked through it and have one more nagging question. The final form I get is

[tex]\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[/tex]

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?
 
  • #5
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.
 
  • #6
Thanks, Meir Achuz. But the term containing

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

when expanded results in both [tex]A\times\nabla[/tex] and [tex]\nabla\times A[/tex] terms (I think). I have the [tex]\nabla\times A[/tex] covered above in the term containing B. But I don't know what to do with the [tex]A\times\nabla[/tex] term
 
  • #7
Bump.

Just one little bump.
 
  • #8
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).
 
Last edited:
  • #9
yangjong said:
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).

thank you for your reply, yangjong. And welcome to the board!

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

That is, where I wrote

[tex]\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[/tex]

above, it should be instead

[tex]\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[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

That's it. Thank you!
 

1. Why do we have two different equations for quantum mechanics?

The Pauli equation and the Schrodinger equation are both formulations of quantum mechanics, but they describe different physical systems. The Pauli equation is used to describe the behavior of spin-1/2 particles, while the Schrodinger equation is used for particles with integer spin.

2. Is the Pauli equation more accurate than the Schrodinger equation?

No, both equations are equally accurate in their respective domains. The Pauli equation takes into account the spin of particles, while the Schrodinger equation does not. However, for particles with spin-1/2, the Pauli equation reduces to the Schrodinger equation.

3. Can we use the Pauli equation to describe particles with integer spin?

No, the Pauli equation is specifically designed for particles with spin-1/2 and cannot be applied to particles with integer spin. For these particles, the Schrodinger equation is the appropriate equation to use.

4. Why do we need the Pauli equation if the Schrodinger equation can also describe spin-1/2 particles?

The Pauli equation is necessary for describing systems with multiple spin-1/2 particles, such as atoms with multiple electrons. The Schrodinger equation cannot account for the interactions between these particles, but the Pauli equation can.

5. Are there any other differences between the Pauli equation and the Schrodinger equation?

Besides the inclusion of spin, the two equations also have different forms. The Schrodinger equation is a time-dependent equation, while the Pauli equation is a time-independent equation. Additionally, the Pauli equation has an additional term called the spin-orbit coupling term, which takes into account the interaction between a particle's spin and its orbital motion.

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