# Boson in a box

• Hill

#### Hill

Gold Member
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?

Last edited:
vanhees71
Got it.

##\langle x|p \rangle= \frac 1 {\sqrt L} e^{ipx}##
and
##\langle p|y \rangle= \frac 1 {\sqrt L} e^{-ipy}##

OOH,
##\langle x|y \rangle = \delta(x-y)##

OTOH, inserting the resolution of identity,
##\langle x|y \rangle = \sum_p \langle x|p \rangle \langle p|y \rangle=\frac 1 L \sum_p e^{ip(x-y)}##

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

Last edited:
vanhees71