
#1
May2208, 04:00 AM

P: 932

I've been trying to work my way through some of my lecture notes, and have hit this snag. (n.b. I use [itex]k_0 \equiv +\sqrt{\vec{k}^2 + m^2}[/itex])
We have [tex]a(q) = \int d^3 x e^{iqx} \{ q_0 \phi(x) + i \pi(x) \} [/tex] [tex]a^{\dagger}(q) = \int d^3 x e^{iqx} \{ q_0 \phi(x)  i \pi(x) \}[/tex] To calculate the commutation relation between these operators, we simply multiply them out as required, and substitute the canonical commutation relation between fields and their conjugate momenta. I work through the relatively tedious steps and get [tex][a(q),a(p)] = \int d^3 x d^3 y e^{i(qxpy)} \delta^3(\vec{x}  \vec{y}) (q_0  p_0)[/tex] [tex]= \int d^3 x e^{i(qp)x} (q_0  p_0)[/tex] [tex]= \int d^3 x e^{i(q_0p_0)x^0} e^{i(\vec{q}\vec{p})\cdot\vec{x}} (q_0  p_0)[/tex] In my notes, the next step is to replace [itex]\vec{q}[/itex] with [itex]\vec{p}[/itex] and so get 0. However, if we integrate over x, surely we are left with a loose delta function outside an integral, which would mean that [itex][a(q),a(p)] = 0 \Leftarrow q=p[/itex] which I know is wrong. Can anyone explain that last step? Any textbooks I've seen assume this is trivial and just go on to state the commutation relation between the creation/annihilation operators rather than calculating it. 



#2
May2208, 01:57 PM

P: 932

Never mind. After doing the integral, I got
[tex] (q_0  p_0) \delta^3(\vec{q}\vec{p}) e^{i(q_0p_0)t} [/tex] which conspires to be zero when the delta function is nonzero because of the first term in brackets, and is zero everywhere else because of the delta function. 



#3
May2208, 03:27 PM

Sci Advisor
P: 1,185

Correct!



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