Dyson-Wick formalism in second-order QED - trouble with derivation

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I have in front of me Quantum Field Theory, Mandl & Shaw. Chapter 7 deals with the theoretical basis of Feynman Diagrams using the Dyson-Wick formalism.

The chapter begins with applying Wick's Theorem to produce six S-Matrix components with a variety of no-equal-time contractions. It then details the contribution of one:

[tex]S^{(2)}_{B}=-\frac{e^{2}}{2!}\int\frac{d^{4}}{dx_{1}}\frac{d^{4}}{dx_{2}}N(\overline{\psi}A\psi)_{x_{1}}(\overline{\psi}A\psi)_{x_{2}}+...][/tex]

LaTeX can't adequately represent contraction marks as far as I know, so imagine there's a cntraction between the first psi and the second psi-bar, and a repetition of same with a contraction between the first psi-bar and the second psi.

The derivation goes on to break this down for the case of Compton scattering by electrons; I am struggling to understand how the two complementary components for the two photon fields produce the final two equations; the contractions in each case are reduced to [tex]iS_{F\alpha\beta}(x_{1}-x_{2})[/tex] with, presumably, the A-slash 1 and 2 on each side. Somehow though the definitions of the slash operator migrate:

[tex]\gamma^{\alpha}iS_{F}(x_{1}-x_{2})\gamma^{\beta}A_{\alpha}^{-}(x_{1})A_{\beta}^{+}(x_{2})[/tex]

and

[tex]\gamma^{\alpha}iS_{F}(x_{1}-x_{2})\gamma^{\beta}A_{\beta}^{-}(x_{2})A_{\alpha}^{+}(x_{1})[/tex]

I am afraid I can't see how the normal ordering resolves to this sequence. Anyone help me revise my Einstein notation and normal-ordering?
 
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Nobody?

If it helps, the difficulty I have is that in the first resultant expression, the indices on the second gamma and the first A don't match. As far as I knew, this didn't mean anything as summation is over repeated indices.
 
Sojourner01 said:
Nobody?

If it helps, the difficulty I have is that in the first resultant expression, the indices on the second gamma and the first A don't match. As far as I knew, this didn't mean anything as summation is over repeated indices.

I don't have Mandl & Shaw, (though I have quite a few other QFT textbooks). If nobody
else gives you a useful answer within a reasonable time, you could try typing in more
of the context here and I'll try to say something helpful...
 
I'll see what I can do...

Given the definition of the S matrix:

[tex]S=\sum_{x=0}^{\infty}\frac{(-i)^{n}}{n!}\int...\int d^{4}x_{1}...d^{4}x_{n}T\{H_{I}(x_{1})...H_{I}(x_{n})\}[/tex]

and the interaction Hamiltonian being (for QED):

[tex]H_{I}(x)=-eN\{\overline{\psi}(x)\gamma_{i}A^{i}(x)\psi(x)\}[/tex]

The time-ordered product of two particles resolves, using Wick's Theorem, to six integrals of normal products, some of which aren't very interesting. The second one is used as an example:

[tex]S_{B}^{(2)}=-\frac{e^{2}}{2!}\int d^{4}x_{1}d^{4}x_{2}\left\{ N\left[(\overline{\psi}\gamma_{i}A^{i}\underbrace{\psi)_{x_{1}}(\overline{\psi}}\gamma_{i}A^{i}\psi)_{x_{2}}\right]+N\left[(\underbrace{\overline{\psi}\gamma_{i}A^{i}\psi)_{x_{1}}(\overline{\psi}\gamma_{i}A^{i}\psi})_{x_{2}}\right]\right\}[/tex]

where the underbraces represent contractions of [tex]\psi[/tex] and [tex]\overline{\psi}}[/tex].

The author states that the two normal products are equivalent and simply takes the first multiplied by two.

Using the identity:

[tex]\underbrace{\psi_{\alpha}(x_{1})\overline{\psi_{\beta}}(x_{2}})}=iS_{F\alpha \beta}(x_{1}-x_{2})[/tex]

where S is theFeynman propagator; this (apparently) resolves to the two counterpart expressions given in the original post. I don't see it, basically. There seems to have been a shuffling around of gamma matrices with no obvious justification - and you usually can't just change the order of matrix expressions.
 
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The two terms are the same, except for what is being contracted with what. In the first term, [tex]\psi(x_1)[/tex] is contracted with [tex]\overline\psi(x_2)[/tex]. In the second term, [tex]\psi(x_2)[/tex] is contracted with [tex]\overline\psi(x_1)[/tex]. But, [itex]x_1[/itex] and [itex]x_2[/itex] are both dummy integration variables, so swapping them doesn't change the result. Thus the two terms are equal after performing the integrations over [itex]x_1[/itex] and [itex]x_2[/itex].
 
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