What is the explanation for the sequence in the Mandl QFT textbook (p.53)?

[intervoxel]

Could anyone please explain the sequence below taken from Mandl QFT textbook (p.53)?

1. i\hbar c\Delta^+(x-x')=[\phi^+(x),\phi^-(x')]

2. i\hbar c\Delta^+(x-x')=\langle 0|[\phi^+(x),\phi^-(x')]|0\rangle

3. i\hbar c\Delta^+(x-x')=\langle 0|\phi^+(x)\phi^-(x')|0\rangle

4. i\hbar c\Delta^+(x-x')=\langle 0|\phi(x)\phi(x')|0\rangle

From 1. to 2. does it mean that the vacuum expected value of the commutator is the commutator itself? How?

From 2. to 3. does it mean that the term \langle 0|\phi^-(x')\phi^+(x)|0\rangle is null? How?

From 3. to 4. does it mean that the terms

\langle 0|\phi^+(x)\phi^+(x')|0\rangle

\langle 0|\phi^-(x)\phi^+(x')|0\rangle

\langle 0|\phi^-(x)\phi^-(x')|0\rangle

are all null? How?

Thank you for any help.

 

[Bill_K]

Yes to 2, 3 and 4. φ⁺ is an absorption operator, so φ⁺|0> = 0 since there is nothing to absorb. Likewise <0|φ^- = 0.

 

[intervoxel]

Thank you. What about the first transformation?

 

[Bill_K]

I don't have a copy of Mandl to compare, but I'm looking at Weinberg section 6.1 where he says something similar. He's evaluating the S-matrix as a sum of terms, <0| ... |0> where ... is a string of creation and annihilation operators, and he's talking about rearranging the order of the operators. Every time you switch the order of two of them you get a numerical factor.

And on p262 he says: (f) Pairing of a field ψ with a field adjoint ψ^† in H(y) yields a factor -iΔ(x,y). (I'm leaving some subscripts out.) This is close to what you're saying. So I think the context is that Mandl's Eq (1) represents a subexpression that's eventually going to be placed between <0| |0>'s.

 

[Avodyne]

To go from eq.1 to eq.2, sandwich eq.1 between <0| and |0>. Use the fact that the left side of eq.1 is just a c-number (not an operator), and so on the left we just get that function times <0|0>=1.