QFT for the Gifted Amateur Exercise 17.1

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In exercise 17.1 we are asked to show that the propagator:

$$G^+_o(p,t_x,q,t_y)=\theta(t_x-t_y)<0|\hat{a}_p(t_x)\hat{a}^\dagger_q(t_y)|0>$$ is the same as

$$\theta(t_x-t_y)e^{-i(E_pt_x-E_qt_y)}\delta^{(3)}(p-q)$$

so we can take the time dependence out of the creation and annihilation operators by using the time evolution operators giving us

$$\theta(t_x-t_y)<0|e^{iHt_x}\hat{a}_pe^{-iH(t_x-t_y)}\hat{a}^\dagger_q e^{-iHt_y}|0>$$

If I have this right, then the rightmost Hamiltonian acts on the ground state |0> to produce $$e^{-iE_gt_y}$$
The middle Hamiltonian acts on the ground state with a particle of momentum q added at times tx and ty to produce $$e^{-i(E_g+E_q)t_x-i(E_g+E_q)t_y}$$
The leftmost Hamiltonian acts on the ground state, the particle of momentum q and an annihilated particle p (which turns the energy negative?) at time tx producing $$e^{-i(E_g+E_q-E_q)t_x}$$
Putting this all together we arrive at the correct energy term $$e^{-i(E_pt_x-E_qt_y)}$$

Rewriting, $$G^+_o(p,t_x,q,t_y)=\theta(t_x-t_y)e^{-i(E_pt_x-E_qt_y)}<0|\hat{a}_p\hat{a}^\dagger_q|0>$$
So I am left with the questions:

1)what justifies the creation and annihilation operators combining to give the $$\delta^{(3)}(p-q)$$
2) where does the 3 come from on the $$\delta^{(3)}$$

 

[Anypodetos]

2) The (3) is just a shorthand for the 3-dimensional Dirac delta, $δ(p_x - q_x) δ(p_y - q_y) δ(p_z - q_z)$.

1) My knowledge here is rather shaky, but since no one has answered you, I'll give it a shot. Think of a superposition of 2 sine functions with frequencies p and q. For p≠q, they interfere constructively and destructively, so that the integral over all the reals vanishes. For p=q, they interfere constructively everywhere, giving a nonzero contribution. The time dependent ladder operators here do the same: their product vanishes for p≠q, leaving only the contribution for p=q, which is just what δ means.

So I suppose you'd have to replace the operators by the δ before you take out the time dependence, but at that point I'm lost as well.

Hope that helps.

 

[nrqed]

You still need to do the last step, right? You need to calculate \langle 0 | a_p a^\dagger_q |0 \rangle To calculate this, you need to express it in terms of a commutator and a product \langle 0 | a^\dagger_q a_p |0 \rangle which gives zero. The commutator piece will give you the three-dimensional delta function.