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[/tex]1. Homework Statement [/b]

Given: [tex]\phi(x)=\phi^+(x)+\phi^-(x)[/tex]

Where:

[tex]\phi^+(x)=\sum_{\mathbf{k}} \sqrt{\frac{\hbar c^2}{2 V \omega_{\mathbf{k}}}} a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx}[/tex]

and

[tex]\phi^-(x)=\sum_{\mathbf{k}} \sqrt{\frac{\hbar c^2}{2 V \omega_{\mathbf{k}}}} a^\dagger (\mathbf{k}) \mathrm{e}^{\mathrm{i}kx}[/tex]

Show that:

[tex]a(\mathbf{k})=\frac{1}{\sqrt{2 \hbar c^2 V \omega_{\mathbf{k}}}} \int\mathrm{d}^3 \mathbf{x} \mathrm{e}^{\mathrm{i}kx} (\mathrm{i}\dot{\phi}(x)+\omega_{\mathbf{k}} \phi(x))[/tex]

See above

First differentiate [tex]\phi(x)[/tex] w.r.t. time:

[tex]\dot{\phi}(x) = -\mathrm{i} \omega_{\mathbf{k}} \phi^+(x) + \mathrm{i} \omega_{\mathbf{k}} \phi^-(x) = -\mathrm{i} \omega_{\mathbf{k}} (\phi^+(x)-\phi^-(x))[/tex]

Then add [tex]\mathrm{i}\dot{\phi}(x)[/tex] to [tex]\omega_\mathbf{k}\phi(x)[/tex] to give:

[tex]\omega_{\mathbf{k}} \phi(x)+\mathrm{i} \dot{\phi}(x) = 2 \omega_{\mathbf{k}} \sum_{\mathbf{k}} \sqrt{ \frac{\hbar c^2}{2 V \omega_{\mathbf{k}}}} a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx} = \sum_{\mathbf{k}} \sqrt{\frac{2 \omega_{\mathbf{k}} \hbar c^2}{V}} a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx}[/tex]

So:

[tex]\omega_{\mathbf{k}} \phi(x)+\mathrm{i} \dot{\phi}(x) = \sum_{\mathbf{k}} \sqrt{\frac{2 \omega_{\mathbf{k}} \hbar c^2}{V}} a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx}[/tex]

And then I get stuck, because I have no idea how to get rid of the summation sign. I'm not certain of what happens to the V either, though I suspect it has something to do with the integral in some way.

Does anyone have any hints?

Given: [tex]\phi(x)=\phi^+(x)+\phi^-(x)[/tex]

Where:

[tex]\phi^+(x)=\sum_{\mathbf{k}} \sqrt{\frac{\hbar c^2}{2 V \omega_{\mathbf{k}}}} a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx}[/tex]

and

[tex]\phi^-(x)=\sum_{\mathbf{k}} \sqrt{\frac{\hbar c^2}{2 V \omega_{\mathbf{k}}}} a^\dagger (\mathbf{k}) \mathrm{e}^{\mathrm{i}kx}[/tex]

Show that:

[tex]a(\mathbf{k})=\frac{1}{\sqrt{2 \hbar c^2 V \omega_{\mathbf{k}}}} \int\mathrm{d}^3 \mathbf{x} \mathrm{e}^{\mathrm{i}kx} (\mathrm{i}\dot{\phi}(x)+\omega_{\mathbf{k}} \phi(x))[/tex]

## Homework Equations

See above

## The Attempt at a Solution

First differentiate [tex]\phi(x)[/tex] w.r.t. time:

[tex]\dot{\phi}(x) = -\mathrm{i} \omega_{\mathbf{k}} \phi^+(x) + \mathrm{i} \omega_{\mathbf{k}} \phi^-(x) = -\mathrm{i} \omega_{\mathbf{k}} (\phi^+(x)-\phi^-(x))[/tex]

Then add [tex]\mathrm{i}\dot{\phi}(x)[/tex] to [tex]\omega_\mathbf{k}\phi(x)[/tex] to give:

[tex]\omega_{\mathbf{k}} \phi(x)+\mathrm{i} \dot{\phi}(x) = 2 \omega_{\mathbf{k}} \sum_{\mathbf{k}} \sqrt{ \frac{\hbar c^2}{2 V \omega_{\mathbf{k}}}} a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx} = \sum_{\mathbf{k}} \sqrt{\frac{2 \omega_{\mathbf{k}} \hbar c^2}{V}} a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx}[/tex]

So:

[tex]\omega_{\mathbf{k}} \phi(x)+\mathrm{i} \dot{\phi}(x) = \sum_{\mathbf{k}} \sqrt{\frac{2 \omega_{\mathbf{k}} \hbar c^2}{V}} a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx}[/tex]

And then I get stuck, because I have no idea how to get rid of the summation sign. I'm not certain of what happens to the V either, though I suspect it has something to do with the integral in some way.

Does anyone have any hints?

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