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Notes and problems for Desai Chapter VI.

Posted by peeterjoot on November 29, 2010

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Motivation.

Chapter VI notes for [1].

Notes

section 6.5, interaction with orbital angular momentum

He states that we take

\begin{aligned}\mathbf{A} = \frac{1}{{2}} (\mathbf{B} \times \mathbf{r})\end{aligned} \hspace{\stretch{1}}(2.1)

and that this reproduces the gauge condition \boldsymbol{\nabla} \cdot \mathbf{A} = 0, and the requirement \boldsymbol{\nabla} \times \mathbf{A} = \mathbf{B}.

These seem to imply that \mathbf{B} is constant, which also accounts for the fact that he writes \boldsymbol{\mu} \cdot \mathbf{L} = \mathbf{L} \cdot \boldsymbol{\mu}.

Consider the gauge condition first, by expanding the divergence of a cross product

\begin{aligned}\boldsymbol{\nabla} \cdot (\mathbf{F} \times \mathbf{G})&=\left\langle{{ \boldsymbol{\nabla} -I \frac{ \mathbf{F} \mathbf{G} - \mathbf{G} \mathbf{F} }{2} }}\right\rangle \\ &=-\frac{1}{{2}} \left\langle{{ I \boldsymbol{\nabla} \mathbf{F} \mathbf{G} - I \boldsymbol{\nabla} \mathbf{G} \mathbf{F} }}\right\rangle \\ &=-\frac{1}{{2}} \left\langle{{I \mathbf{G}(\stackrel{ \rightarrow }{\boldsymbol{\nabla}} \mathbf{F}) - I \mathbf{F} (\stackrel{ \rightarrow }{\boldsymbol{\nabla}} \mathbf{G})+I (\mathbf{G} \stackrel{ \leftarrow }{\boldsymbol{\nabla}}) \mathbf{F} - I (\mathbf{F} \stackrel{ \leftarrow }{\boldsymbol{\nabla}}) \mathbf{G}}}\right\rangle \\ &=-\frac{1}{{2}} \left\langle{{I \mathbf{G}(\stackrel{ \rightarrow }{\boldsymbol{\nabla}} \wedge \mathbf{F}) - I \mathbf{F} (\stackrel{ \rightarrow }{\boldsymbol{\nabla}} \wedge \mathbf{G})+I (\mathbf{G} \wedge \stackrel{ \leftarrow }{\boldsymbol{\nabla}}) \mathbf{F} - I (\mathbf{F} \wedge \stackrel{ \leftarrow }{\boldsymbol{\nabla}}) \mathbf{G}}}\right\rangle \\ &=\frac{1}{{2}} \left\langle{{\mathbf{G} (\stackrel{ \rightarrow }{\boldsymbol{\nabla}} \times \mathbf{F}) - \mathbf{F} (\stackrel{ \rightarrow }{\boldsymbol{\nabla}} \times \mathbf{G})+(\mathbf{G} \times \stackrel{ \leftarrow }{\boldsymbol{\nabla}}) \mathbf{F} - (\mathbf{F} \times \stackrel{ \leftarrow }{\boldsymbol{\nabla}}) \mathbf{G}}}\right\rangle \\ &=\frac{1}{{2}} \left(\mathbf{G} \cdot (\boldsymbol{\nabla} \times \mathbf{F}) - \mathbf{F} \cdot (\boldsymbol{\nabla} \times \mathbf{G})-\mathbf{F} \cdot (\boldsymbol{\nabla} \times \mathbf{G}) + \mathbf{G} \cdot (\boldsymbol{\nabla} \times \mathbf{F} )\right) \\ \end{aligned}

This gives us

\begin{aligned}\boldsymbol{\nabla} \cdot (\mathbf{F} \times \mathbf{G})&=\mathbf{G} \cdot (\boldsymbol{\nabla} \times \mathbf{F}) - \mathbf{F} \cdot (\boldsymbol{\nabla} \times \mathbf{G})\end{aligned} \hspace{\stretch{1}}(2.2)

With \mathbf{A} = (\mathbf{B} \times \mathbf{r})/2 we then have

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{A} =\frac{1}{{2}} \mathbf{r} \cdot (\boldsymbol{\nabla} \times \mathbf{B}) - \frac{1}{{2}} \mathbf{B} \cdot (\boldsymbol{\nabla} \times \mathbf{r})=\frac{1}{{2}} \mathbf{r} \cdot (\boldsymbol{\nabla} \times \mathbf{B})\end{aligned} \hspace{\stretch{1}}(2.3)

Unless \boldsymbol{\nabla} \times \mathbf{B} is always perpendicular to \mathbf{r} we can only have a zero divergence when \mathbf{B} is constant.

Now, let’s look at \boldsymbol{\nabla} \times \mathbf{A}. We need another auxillary identity

\begin{aligned}\boldsymbol{\nabla} \times (\mathbf{F} \times \mathbf{G})&=-I \boldsymbol{\nabla} \wedge (\mathbf{F} \times \mathbf{G}) \\ &=-\frac{1}{{2}} {\left\langle{{I \stackrel{ \rightarrow }{\boldsymbol{\nabla}} (\mathbf{F} \times \mathbf{G})- I (\mathbf{F} \times \mathbf{G}) \stackrel{ \leftarrow }{\boldsymbol{\nabla}}}}\right\rangle}_{1} \\ &=\frac{1}{{2}} \left(-\stackrel{ \rightarrow }{\boldsymbol{\nabla}} \cdot (\mathbf{F} \wedge \mathbf{G})+ (\mathbf{F} \wedge \mathbf{G}) \cdot \stackrel{ \leftarrow }{\boldsymbol{\nabla}}\right) \\ &=\frac{1}{{2}} \left(-(\stackrel{ \rightarrow }{\boldsymbol{\nabla}} \cdot \mathbf{F}) \mathbf{G}+(\stackrel{ \rightarrow }{\boldsymbol{\nabla}} \cdot \mathbf{G}) \mathbf{F}+ \mathbf{F} (\mathbf{G} \cdot \stackrel{ \leftarrow }{\boldsymbol{\nabla}} )- \mathbf{G} (\mathbf{F} \cdot \stackrel{ \leftarrow }{\boldsymbol{\nabla}} )\right)\\ &=\frac{1}{{2}} \left(-(\boldsymbol{\nabla} \cdot \mathbf{F}) \mathbf{G}+(\boldsymbol{\nabla} \cdot \mathbf{G}) \mathbf{F}+ (\boldsymbol{\nabla} \cdot \mathbf{G} ) \mathbf{F}- (\boldsymbol{\nabla} \cdot \mathbf{F} ) \mathbf{G}\right)\end{aligned}

Here the gradients are all still acting on both \mathbf{F} and \mathbf{G}. Expanding this out by chain rule we have

\begin{aligned}2 \boldsymbol{\nabla} \times (\mathbf{F} \times \mathbf{G})=&-(\mathbf{F} \cdot \boldsymbol{\nabla}) \mathbf{G}-\mathbf{G} (\boldsymbol{\nabla} \cdot \mathbf{F}) +\mathbf{F} (\boldsymbol{\nabla} \cdot \mathbf{G})+(\mathbf{G} \cdot \boldsymbol{\nabla} ) \mathbf{F} \\ \quad&+\mathbf{F} (\boldsymbol{\nabla} \cdot \mathbf{G} )+ (\mathbf{G} \cdot \boldsymbol{\nabla} ) \mathbf{F} - (\mathbf{F} \cdot \boldsymbol{\nabla} ) \mathbf{G}- \mathbf{G} (\boldsymbol{\nabla} \cdot \mathbf{F} )\end{aligned}

or

\begin{aligned}\boldsymbol{\nabla} \times (\mathbf{F} \times \mathbf{G})&=\mathbf{F} (\boldsymbol{\nabla} \cdot \mathbf{G}) -(\mathbf{F} \cdot \boldsymbol{\nabla}) \mathbf{G}+(\mathbf{G} \cdot \boldsymbol{\nabla} ) \mathbf{F} -\mathbf{G} (\boldsymbol{\nabla} \cdot \mathbf{F})\end{aligned} \hspace{\stretch{1}}(2.4)

With \mathbf{F} = \mathbf{B}/2, and \mathbf{G} = \mathbf{r}, we have

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{A}&=\frac{1}{{2}}\mathbf{B} (\boldsymbol{\nabla} \cdot \mathbf{r}) -\frac{1}{{2}}(\mathbf{B} \cdot \boldsymbol{\nabla}) \mathbf{r}+\frac{1}{{2}}(\mathbf{r} \cdot \boldsymbol{\nabla} ) \mathbf{B} -\frac{1}{{2}}\mathbf{r} (\boldsymbol{\nabla} \cdot \mathbf{B})\end{aligned}

We note that \boldsymbol{\nabla} \cdot \mathbf{r} = 3, and

\begin{aligned}(\mathbf{B} \cdot \boldsymbol{\nabla} ) \mathbf{r} &=B_k \partial_k x_m \mathbf{e}_m \\ &=B_k \delta_{km} \mathbf{e}_m \\ &=\mathbf{B}\end{aligned}

If \mathbf{B} is constant, we have

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{A} = \frac{3\mathbf{B}}{2} - \frac{\mathbf{B}}{2} = \mathbf{B},\end{aligned} \hspace{\stretch{1}}(2.5)

as desired. Now this would all likely be a lot more intuitive if one started with constant \mathbf{B} and derived from that what the vector potential was. That’s probably worth also thinking about.

Problems

Problem 1.

Statement.

Solution.

TODO.

Problem 3.

Statement.

Solution.

TODO.

Problem 3.

Statement.

Solution.

TODO.
…

References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

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