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

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masudr
#1
May22-08, 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(qx-py)} \delta^3(\vec{x} - \vec{y}) (q_0 - p_0)[/tex]
[tex]= \int d^3 x e^{i(q-p)x} (q_0 - p_0)[/tex]
[tex]= \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)[/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.
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masudr
#2
May22-08, 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_0-p_0)t}
[/tex]

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.
Avodyne
#3
May22-08, 03:27 PM
Sci Advisor
P: 1,199
Correct!


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