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The equation (2.34) of Negele and Orland has

$$\mathcal H_\text{A}(\hat{\mathbf p},\hat{\mathbf x}) = \frac{1}{2m}\left(\hat {\mathbf p} - \frac e c \mathbf A(\hat{\mathbf x})\right)^2.\tag{2.34a}$$

And it says in normal form,

$$\mathcal H_\text{A}(\hat{\mathbf p},\hat{\mathbf x}) = \frac{1}{2m}\left(\hat {\mathbf p}^2 - 2\frac e c \hat{\mathbf p}\mathbf A(\hat {\mathbf x}) - \frac e c i \nabla\cdot\mathbf{A}(\hat{\mathbf x})+(\frac e c \mathbf A(\hat{\mathbf x}))^2) \right).\tag{2.34b}$$

  1. The second term is particularly very odd for me. Since it is normal ordered, I understand I need to move $\mathbf p$ to left. But doesn't this act on $\mathbf A$ then? I also wonder why there is no dot product between $\mathbf p$ and $\mathbf A$ in the second term.

  2. Also the third term seems just $\frac e c \mathbf p \cdot \mathbf {A}(\hat{\mathbf x})$ and could be combined with the second term?

  3. Don't the sign for the third term need to be (+) given that $\hat{\mathbf p} = -i\nabla$?

    Because I think this should be (in a non-normal ordered form),

$$\mathcal H_\text{A}(\hat{\mathbf p},\hat{\mathbf x}) = \frac{1}{2m}\left(\hat {\mathbf p}^2 - 2\frac e c \mathbf A(\hat {\mathbf x}) \cdot \hat{\mathbf p} - \frac e c \mathbf p\cdot\mathbf{A}(\hat{\mathbf x})+(\frac e c \mathbf A(\hat{\mathbf x}))^2) \right).$$

Please let me know the answer. Any answer will be appreciated.

user3089810
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    It is a common shorthand to simply juxtapose vectors for representing the scalar product. That is, one writes $\vec a \vec b$ for $\vec a \cdot \vec b$ (just as for the product of numbers). Further I guess, they mean the $\hat{\vec p}$ not to act on the $\hat{\vec A}$ as they write this contribution explicitely ($\frac e c i \nabla \cdot \vec A(\hat{\vec x})$). I agree their notation seems somewhat sloppy. – Sebastian Riese May 10 '15 at 22:12

1 Answers1

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  1. Yes, the missing dot in the dot product of the second term $$\tag{2} -2\frac{e}{c}\hat{\mathbf{p}} \cdot \mathbf{A}(\hat{\mathbf{x}}) $$ of eq. (2.34b) is a typo. The operators $\hat{\mathbf{p}}$ and $\mathbf{A}(\hat{\mathbf{x}})$ do not commute, due to the CCRs $$\tag{CCR} [\hat{x}^i~,~ \hat{p}_j]_{-}~=~i\hbar\delta^i_j~{\bf 1}.$$

  2. The second and third term taken together are equal to the anticommutator $$\tag{2+3} -\frac{e}{c} \{ \hat{\mathbf{p}} ~;~\mathbf{A}(\hat{\mathbf{x}}) \}_{+}.$$ This becomes clear when one tries to derive (2.34b) from (2.34a). In particular, it is implicitly meant that the derivative $\nabla$ in the third term $$\tag{3} \frac{e}{c} \frac{\hbar}{i} \left(\nabla\cdot\mathbf{A}(\hat{\mathbf{x}})\right) ~=~\frac{e}{c}\frac{\hbar}{i} [\nabla~;~\mathbf{A}(\hat{\mathbf{x}})]_{-} ~=~\frac{e}{c} [\hat{\mathbf{p}} ~;~\mathbf{A}(\hat{\mathbf{x}})]_{-} $$ of eq. (2.34b) only acts on $\mathbf{A}(\hat{\mathbf{x}})$ and not beyond to the right. This notational issue is similar to my Phys.SE answer here.

  3. The sign in the third term of eq. (2.34b) is correct.

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