# Explicit expressions for creation/annihilation operator of the free scalar field

by masudr
Tags: explicit, expressions, field, free, operator, scalar
 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 $k_0 \equiv +\sqrt{\vec{k}^2 + m^2}$) We have $$a(q) = \int d^3 x e^{iqx} \{ q_0 \phi(x) + i \pi(x) \}$$ $$a^{\dagger}(q) = \int d^3 x e^{-iqx} \{ q_0 \phi(x) - i \pi(x) \}$$ 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 $$[a(q),a(p)] = \int d^3 x d^3 y e^{i(qx-py)} \delta^3(\vec{x} - \vec{y}) (q_0 - p_0)$$ $$= \int d^3 x e^{i(q-p)x} (q_0 - p_0)$$ $$= \int d^3 x e^{i(q_0-p_0)x^0} e^{i(\vec{q}-\vec{p})\cdot\vec{x}} (q_0 - p_0)$$ In my notes, the next step is to replace $\vec{q}$ with $\vec{p}$ 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 $[a(q),a(p)] = 0 \Leftarrow q=p$ 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.
 P: 932 Never mind. After doing the integral, I got $$(q_0 - p_0) \delta^3(\vec{q}-\vec{p}) e^{i(q_0-p_0)t}$$ which conspires to be zero when the delta function is non-zero because of the first term in brackets, and is zero everywhere else because of the delta function.