In exercise 17.1 we are asked to show that the propagator:(adsbygoogle = window.adsbygoogle || []).push({});

$$G^+_o(p,t_x,q,t_y)=\theta(t_x-t_y)<0|\hat{a}_p(t_x)\hat{a}^\dagger_q(t_y)|0>$$ 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 $$\delta^{(3)}$$

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# I QFT for the Gifted Amateur Question (3)

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