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In Sakurai and Napolitano, chapter 2, there's a derivation of the QM Lorentz force.

Given $$H=\frac{1}{2m}\left(\mathbf{p}-\frac{e\mathbf{A}}{c}\right)^2+e\phi = \frac{\mathbf{\Pi}^2}{2m}+e\phi$$

the text claims

$$\frac{d\mathbf{\Pi}}{dt} = e\left[\mathbf{E}+\frac{1}{2c}\left(\frac{d\mathbf{x}}{dt}\times\mathbf{B}-\mathbf{B}\times\frac{d\mathbf{x}}{dt}\right)\right].$$

I'm having difficulty connecting these... I tried using the Heisenberg picture with

$$\begin{align} \frac{d\mathbf{\Pi}}{dt} &= \frac{1}{i\hbar}[\mathbf{\Pi},H] \\ &= \frac{1}{i\hbar}\left[\mathbf{\Pi},\frac{\mathbf{\Pi}^2}{2m}+e\phi\right] \\ &= \frac{e}{i\hbar}\left[\mathbf{\Pi},\phi\right] \\ &= \frac{e}{i\hbar}\left[\mathbf{p}-\frac{e\mathbf{A}}{c},\phi\right] \\ &= \frac{e}{i\hbar}\left[\mathbf{p},\phi\right]- \frac{e}{i\hbar}\left[\frac{e\mathbf{A}}{c},\phi\right] \\ &= -e\nabla\phi- \frac{e^2}{i\hbar c}\left[\mathbf{A},\phi\right] \\ &= e\mathbf{E} - \frac{e^2}{i\hbar c}\left[\mathbf{A},\phi\right] \end{align}$$

...and I'm stuck. It looks like this was the right direction, but I do not understand how to assess the remaining commutator. Aren't both $A$ and $\phi$ functions of $x$? I tried looking up this commutator, but my search skills did not procure it.

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    $\begingroup$ Note that $\mathbf{\Pi}$ does not commute with itself, cf. Eq. (2.7.25) in Sakurai. $\endgroup$
    – Ryan Unger
    Mar 11 '16 at 21:39
  • $\begingroup$ @ocelo7 Ahhh....wow. I missed that. Hopefully that clears things up. $\endgroup$
    – suneater
    Mar 11 '16 at 21:44
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    $\begingroup$ Why is this question off-topic? I had trouble with the computation when reading this same section of the book, and found this question and its answer useful (+1 to both, by the way). $\endgroup$
    – Tob Ernack
    Mar 4 '20 at 22:51
  • $\begingroup$ @TobErnack Physics SE is understandably rather strict on homework or study related questions. But I too disagree with the ruling on this question and similar ones discussing graduate level textbooks. $\endgroup$
    – suneater
    Mar 5 '20 at 23:16
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Following ocelo7's comment, the appropriate commutator relation is $[\Pi_i,\Pi_j]=i\hbar\frac ec \epsilon_{ijk}B_k$. Using index notation, I found the desired result:

$$\begin{align} \frac{d\Pi_i}{dt}&= \frac{1}{i\hbar}\left[\Pi_i,\frac{\Pi_j^2}{2m}-e\phi\right] \\ &= eE_i + \frac{1}{i\hbar2m}\left(\Pi_j[\Pi_i,\Pi_j]+[\Pi_i,\Pi_j]\Pi_j\right) \\ &= eE_i + \frac{e}{2mc}\left(\Pi_j\epsilon_{ijk}B_k+\epsilon_{ijk}B_k\Pi_j\right) \\ &= eE_i + \frac{e}{2c}\left(\left(\frac{d\mathbf{x}}{dt}\times \mathbf{B}\right)_i-\left(\mathbf{B}\times\frac{dx}{dt}\right)_i\right) \end{align}$$

where $\Pi_j = m\frac{dx_j}{dt}$.

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