# I QFT for the Gifted Amateur Question (3)

1. Oct 27, 2016

### 43arcsec

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 $$\delta^{(3)}$$

2. Oct 27, 2016

### vanhees71

I'd use the fact that $\hat{H}|0 \rangle=|0 \rangle$, then that $\hat{a}_q^{\dagger}|0 \rangle$ is a eigenstates of the Hamiltonian with eigenvalue $E_{\vec{q}}=\sqrt{m^2+\vec{q}^2}$ etc. etc. This should step by step lead to the equation you want to proof.

BTW: You should post that kind of questions to the homework forum! It doesn't help you, if we just present the solution. Hints for how to solve the problem youself is much better help (although less convenient for you ;-)).