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- Homework Statement
- Show for a boson particle in a box of volume ##V## that $$\frac 1 V \sum_{\mathbf{pq}} e^{i(\mathbf{px}-\mathbf{qy})}[\hat a_{\mathbf p},\hat a^\dagger_{\mathbf q}]=\delta^{(3)}(\mathbf x - \mathbf y)$$

- Relevant Equations
- ##[\hat a_{\mathbf p},\hat a^\dagger_{\mathbf q}]=\delta_{\mathbf{pq}}##

To simplify, I consider a one-dimensional box of the size L. I need to show in this case that

$$\frac 1 L \sum_{pq} e^{i(px-qy)}[\hat a_p,\hat a^\dagger_q]=\delta(x -y)$$

With the commutation relation above, it becomes

$$\frac 1 L \sum_p e^{ip(x-y)}=\delta(x -y)$$

p is quantized: ##p_m=\frac {2\pi m} L##

So I need to show that

$$\frac 1 L \sum_m e^{i \frac {2\pi m} L (x-y)}=\delta(x -y)$$

If ##x \neq y## the sum is ##0##, but I don't know how to proceed otherwise.

A hint?

$$\frac 1 L \sum_{pq} e^{i(px-qy)}[\hat a_p,\hat a^\dagger_q]=\delta(x -y)$$

With the commutation relation above, it becomes

$$\frac 1 L \sum_p e^{ip(x-y)}=\delta(x -y)$$

p is quantized: ##p_m=\frac {2\pi m} L##

So I need to show that

$$\frac 1 L \sum_m e^{i \frac {2\pi m} L (x-y)}=\delta(x -y)$$

If ##x \neq y## the sum is ##0##, but I don't know how to proceed otherwise.

A hint?

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