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## Notes and problems for Desai Chapter VI.

Posted by peeterjoot on November 29, 2010

# 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.

TODO.

TODO.

TODO.

# References

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