- #1
copernicus1
- 98
- 0
I'm trying to work out the total momentum operator on page 22 of Peskin/Schroeder for myself, and I'm a little confused about the last few steps.
Assuming I went through the first few steps correctly, I've arrive at this expression:
$${\bf P}=\frac12\int\frac{d^3p}{(2\pi)^3}{\bf p}\left(2a^\dagger_{\bf p}a_{\bf p}+\left[a_{\bf p},a_{\bf p}^\dagger\right]+a_{\bf p}a_{\bf -p}+a^\dagger_{\bf p}a^\dagger_{\bf -p}\right).$$
In the book they end up with
$${\bf P}=\int\frac{d^3p}{(2\pi)^3}{\bf p}\,a^\dagger_{\bf p}a_{\bf p},$$
so I'm obviously pretty close; I just have a couple extra terms.
I'm assuming the commutator just gets dropped, since it's the infinite term, but do the last two terms somehow cancel? I can see that they create and annihilate the same two states with opposite momenta, but I don't know if it's justifiable to just drop them...
Assuming I went through the first few steps correctly, I've arrive at this expression:
$${\bf P}=\frac12\int\frac{d^3p}{(2\pi)^3}{\bf p}\left(2a^\dagger_{\bf p}a_{\bf p}+\left[a_{\bf p},a_{\bf p}^\dagger\right]+a_{\bf p}a_{\bf -p}+a^\dagger_{\bf p}a^\dagger_{\bf -p}\right).$$
In the book they end up with
$${\bf P}=\int\frac{d^3p}{(2\pi)^3}{\bf p}\,a^\dagger_{\bf p}a_{\bf p},$$
so I'm obviously pretty close; I just have a couple extra terms.
I'm assuming the commutator just gets dropped, since it's the infinite term, but do the last two terms somehow cancel? I can see that they create and annihilate the same two states with opposite momenta, but I don't know if it's justifiable to just drop them...