QFT for the Gifted Amateur Exercise 17.1

AI Thread Summary
The discussion centers on demonstrating the equivalence of two expressions for the propagator in exercise 17.1, specifically showing that the propagator can be expressed in terms of time evolution operators. It highlights how the Hamiltonians act on the ground state to yield the correct energy terms, leading to the final form of the propagator. Questions arise regarding the justification for the creation and annihilation operators combining to yield the three-dimensional Dirac delta function, with explanations involving interference of sine functions. The discussion also notes that the three-dimensional delta function arises from the properties of the operators when evaluated at equal momenta. Ultimately, the calculation of the ground state expectation value is crucial for confirming the results.
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43arcsec said:
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) 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.
 
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.
 
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