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GCUEasilyDistracted
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- Homework Statement
- Quantum Field Theory for the Gifted Amateur Exercise 3.1 (reworded a bit): Suppose we are working with a system defined in a volume of space ##V'##. For boson operators satisfying ##[\hat{a}_p, \hat{a}^{\dagger}_q] = \delta_{pq}## show that $$\frac{1}{V}\sum\limits_{pq}e^{i(p \cdot x - q \cdot y)}[\hat{a}_p, \hat{a}^{\dagger}_q] = \delta^{(3)}(x - y)$$, where ##V## is the magnitude of ##V'##. Show the same with fermion operators.
- Relevant Equations
- Seen above.
I'm studying Quantum Field Theory for the Gifted Amateur and feel like my math background for it is a bit shaky. This was my attempt at a derivation of the above. I know it's not rigorous, but is it at least conceptually right? I'll only show it for bosons since it's pretty much the same for fermions except the commutator is replaced with the anticommutator.
First note that
$$\frac{1}{V}\sum\limits_{pq}e^{i(p \cdot x - q \cdot y)}[\hat{a}_p, \hat{a}^{\dagger}_q]$$
$$= \frac{1}{V}\sum\limits_{pq}e^{i(p \cdot x - q \cdot y)}\delta_{pq}$$
$$= \frac{1}{V}\sum\limits_p e^{ip \cdot (x - y)}$$
To prove that the sum above is equivalent to ##\delta^{(3)}(x - y)##, we must show 2 things:
1. ##\frac{1}{V}\sum\limits_p e^{ip \cdot (x - y)} = 0## whenever ##x \neq y##
2. $$\int\limits_{V'} d^3x \frac{1}{V}\sum\limits_p e^{ip \cdot (x - y)} = 1$$
The first follows from the periodicity of ##e^{ip \cdot (x - y)}## [EDIT: the periodicity and the fact that for each value, the negative of that value will also appear within one period]. When ##x - y## is nonzero, each ##e^{ip \cdot (x - y)}## term will be cancelled out by some other term ##e^{ip' \cdot (x - y)}##.
For the second,
$$\int\limits_{V'} d^3x \frac{1}{V}\sum\limits_p e^{ip \cdot (x - y)}$$
$$= \frac{1}{V}\sum\limits_p \int\limits_{V'} d^3x e^{ip \cdot (x - y)}$$
Note that due to the same periodicity mentioned above, ##\int\limits_{V'} d^3x e^{ip \cdot (x - y)} = 0## when ##p \neq 0##. So
$$\frac{1}{V}\sum\limits_p \int\limits_{V'} d^3x e^{ip \cdot (x - y)}$$
$$= \frac{1}{V} \int\limits_{V'} d^3x e^{i(0) \cdot (x - y)}$$
$$= \frac{1}{V} \int\limits_{V'} d^3x$$
$$= \frac{1}{V}(V)$$
$$= 1$$
Therefore we have shown that
$$\frac{1}{V}\sum\limits_{pq}e^{i(p \cdot x - q \cdot y)}[\hat{a}_p, \hat{a}^{\dagger}_q] = \delta^{(3)}(x - y)$$
Does this make sense? Is there anything that should be corrected or made more rigorous? Thanks for your help!
First note that
$$\frac{1}{V}\sum\limits_{pq}e^{i(p \cdot x - q \cdot y)}[\hat{a}_p, \hat{a}^{\dagger}_q]$$
$$= \frac{1}{V}\sum\limits_{pq}e^{i(p \cdot x - q \cdot y)}\delta_{pq}$$
$$= \frac{1}{V}\sum\limits_p e^{ip \cdot (x - y)}$$
To prove that the sum above is equivalent to ##\delta^{(3)}(x - y)##, we must show 2 things:
1. ##\frac{1}{V}\sum\limits_p e^{ip \cdot (x - y)} = 0## whenever ##x \neq y##
2. $$\int\limits_{V'} d^3x \frac{1}{V}\sum\limits_p e^{ip \cdot (x - y)} = 1$$
The first follows from the periodicity of ##e^{ip \cdot (x - y)}## [EDIT: the periodicity and the fact that for each value, the negative of that value will also appear within one period]. When ##x - y## is nonzero, each ##e^{ip \cdot (x - y)}## term will be cancelled out by some other term ##e^{ip' \cdot (x - y)}##.
For the second,
$$\int\limits_{V'} d^3x \frac{1}{V}\sum\limits_p e^{ip \cdot (x - y)}$$
$$= \frac{1}{V}\sum\limits_p \int\limits_{V'} d^3x e^{ip \cdot (x - y)}$$
Note that due to the same periodicity mentioned above, ##\int\limits_{V'} d^3x e^{ip \cdot (x - y)} = 0## when ##p \neq 0##. So
$$\frac{1}{V}\sum\limits_p \int\limits_{V'} d^3x e^{ip \cdot (x - y)}$$
$$= \frac{1}{V} \int\limits_{V'} d^3x e^{i(0) \cdot (x - y)}$$
$$= \frac{1}{V} \int\limits_{V'} d^3x$$
$$= \frac{1}{V}(V)$$
$$= 1$$
Therefore we have shown that
$$\frac{1}{V}\sum\limits_{pq}e^{i(p \cdot x - q \cdot y)}[\hat{a}_p, \hat{a}^{\dagger}_q] = \delta^{(3)}(x - y)$$
Does this make sense? Is there anything that should be corrected or made more rigorous? Thanks for your help!
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