# Retrieving the non-relativistic Hamiltonian from Relativistic QM

I'm trying to follow section 15.5 here, which derives the low-energy limit of the Dirac equation for an electron in a EM-field.

After some manipulations (which I think I follow alright) the author arrives at this result for the non-relativistic Hamiltonian:

$$H_\text{non-rel} = \frac{1}{2m}\left[ \vec\sigma\cdot \left(\vec p - \frac{q}{c}\vec A\right) \right]^2 + q\phi$$

The author then says that by the help of the relation for Pauli matrices, $\sigma_i\sigma_j=\delta_{ij} + i\epsilon_{ijk}\sigma_k$ we get the result:

$$H_\text{non-rel} = \frac{1}{2m}\left(\vec p - \frac{q}{c}\vec A\right)^2 - \frac{q\hbar}{2mc}\vec\sigma\cdot(\nabla\times\vec A) +q\phi$$

I was trying to follow this derivation, and thought it would be smart to use another identity for Pauli matrices, namely:

$$(\vec \sigma\cdot \vec a)(\vec\sigma\cdot\vec b) = (\vec a\cdot\vec b)\mathbf{I} + i\vec\sigma\cdot(\vec a\times\vec b)$$

My thinking: If I take $\vec a \equiv \vec p - \frac{q}{c}\vec A$, then the squared factor in the first equation is just $(\vec\sigma\cdot\vec a)^2$. But since $\vec a\times\vec a=0$, the above identity should imply that the whole cross term disappears. Since the authors result obviously does not agree we with me, where is the error in my approach?

Update: Making sure to respect what does and doesn't commute as suggested in comments and answer: $$(\vec p -\frac{q}{c}\vec A)\times(\vec p -\frac{q}{c}\vec A) = -\frac{q}{c}(\vec p\times\vec A + \vec A\times\vec p)$$ Looking at the first term, letting it act on a test function $f$: $$(\vec p\times\vec A)f=(-i\hbar\nabla\times\vec A)f = -i\hbar( (\nabla\times \vec A)f +(\nabla f)\times\vec A) = -i\hbar(\nabla\times\vec A - \vec A\times\nabla )f$$ Putting this back in cancels the second term and yields nicely the expected result.

• akhmeteli's answer is I think correct but I just wanted to observe that both identities that you're citing are absolutely identical. The author's relation is just your relation but leaving out the details of $\vec a$ and $\vec b$ by using a tensor-component notation. – CR Drost Sep 16 '17 at 4:39
• The claim is that $(p - q/c A)\times(p-q/c A)$ should be broken down into its parts, $p\times p$ being identically zero in the momentum basis and therefore zero in the position basis, and $A \times A$ being identically zero in the position basis and therefore zero in the momentum basis -- but since they are noncommuting operators one has $p\times A - A \times p \ne 0$ in general. – CR Drost Sep 16 '17 at 4:42
• Sorry, that should read $p\times A + A \times p \ne 0$ in general. Basically there is no basis which simultaneously diagonalizes both $p$ and $A$ to make the argument go through, because they don't commute; $A$ is a function of $x$ and $p$ is $-i\hbar\partial_x.$ – CR Drost Sep 16 '17 at 4:54

I am not sure $\overrightarrow{a}\times\overrightarrow{a}=0$ in your case, you should take into account commutators of $p_i$ and $A_j$.
• I thought this was a general result, holding for any arbitrary vector? I guess it may not be when operators are involved... I guess the same might be true for things like $\vec a\times\vec b = -\vec b\times\vec a$ - I can see how things won't hold anymore then. I'll try and work it out explicitly (again) and be extra mindful of possible non-zero commutators. – Bendik Sep 16 '17 at 4:27