Question about some math during wavefunction renormalization

[Chopin]

I've been going through Sidney Coleman's QFT video lectures (http://www.physics.harvard.edu/about/Phys253.html, with notes at http://arxiv.org/PS_cache/arxiv/pdf/1110/1110.5013v1.pdf). I'm up to the part on fixing counterterms for wavefunction renormalization (page 179 in the notes), and have what's hopefully just a stupid little math question.

We have a field [itex]\phi(x)[/itex], which obeys the CCR, and a renormalized field [itex]\phi'(x)=Z^{1/2}_3\phi(x)[/itex], which obeys the equation [itex]\langle k|\phi'(0)|0\rangle=1[/itex]. We now use the Lehmann-Kallen decomposition to determine that:

[tex]\langle 0|\phi'(x)\phi'(y)|0\rangle = \Delta_+(x-y,\mu^2) + \int^\infty_0{da^2\sigma(a^2)\Delta_+(x-y,a^2)}[/tex]

The text then goes on to use this to make a statement about [itex]Z_3[/itex] using the equal-time commutators, by saying that:

(1) [itex]\langle 0|[\phi'(x),\dot{\phi'}(y)]|0\rangle = iZ^{-1}_3\delta^{(3)}(x-y)[/itex]
(2) [itex]\langle 0|[\phi'(x),\dot{\phi'}(y)]|0\rangle = \delta^{(3)}(x-y) + \int^\infty_0{da^2\sigma(a^2)i\delta^{(3)}(x-y)}[/itex]

where (1) is by the definition of [itex]\phi'(x)[/itex], and (2) is by manipulation of the decomposition derived above. My question is about this second statement--the text essentially handwaves its derivation, but I'm not quite following it. It sounds like they're just differentiating the RHS of it, but I don't understand how that gives you [itex]\langle 0|[\phi'(x),\dot{\phi'}(y)]|0\rangle[/itex] on the left.

I'm sure it's just some kind of silly property of commutators or propagators or something that I'm not getting, but I'd appreciate if somebody could help me connect the dots. Can anybody show how to go through this derivation in slow motion?

 

[eljose79]

Thank you for sharing your progress and question regarding Sidney Coleman's QFT video lectures. I am always excited to see others engaging with and learning about quantum field theory.

To answer your question, let's first define the equal-time commutator of two fields \phi(x) and \phi(y) as:

[\phi(x),\phi(y)] = \phi(x)\phi(y) - \phi(y)\phi(x)

Now, using the definition of \phi'(x) and \dot{\phi'}(y), we can rewrite the first statement as:

\langle 0|[\phi'(x),\dot{\phi'}(y)]|0\rangle = \langle 0|[\phi(x),\phi(y)]|0\rangle

This is because \phi'(x) and \dot{\phi'}(y) only differ from \phi(x) and \phi(y) by a constant factor Z_3^{1/2}, and the vacuum expectation value of a constant is just 1.

Now, let's focus on the second statement. We can rewrite it as:

\langle 0|[\phi(x),\phi(y)]|0\rangle = \langle 0|[\phi'(x),\dot{\phi'}(y)]|0\rangle - \int^\infty_0{da^2\sigma(a^2)i\langle 0|[\phi'(x),\dot{\phi'}(y)]|0\rangle}

In other words, we have used the definition of \phi'(x) and \dot{\phi'}(y) to replace the first term in the integral with the first statement, and the second term represents the contribution from the integral in the Lehmann-Kallen decomposition.

Now, if we differentiate both sides of this equation with respect to x, we get:

\langle 0|[\phi(x),\dot{\phi'}(y)]|0\rangle = \langle 0|[\phi'(x),\dot{\phi'}(y)]|0\rangle - \frac{d}{dx}\left(\int^\infty_0{da^2\sigma(a^2)i\langle 0|[\phi'(x),\dot{\phi'}(y)]|0\rangle}\right)

But since the integrand is independent of x, the integral on the right-hand side