Quantized Dirac Field Interacting with a Classical Potential

  • #1
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Main Question or Discussion Point

Hi,

I'm working through Section 4-3 of Itzykzon and Zuber's QFT textbook, but I am a bit stuck while trying to understand some of the quantities and equations.

First of all, what is this "one-body scattering operator [itex]\mathcal{F}(A)[/itex]"? It is defined (eqn 4-89, page 188) as

[tex]\mathcal{F}(A) = e\gamma^{\mu}A_{\mu}(x) + e\gamma^{\mu}A_{\mu}(x)\frac{1}{\gamma^{\mu}P_{\mu}-m+i\epsilon}\mathcal{F}(A)[/tex]

What does this operator physically represent? For most of this section, it seems to be a trick to go through with the calculation. But the pair production probability in eqn 4-99 (page 191) is expressed in terms of an operator [itex]t[/itex] which is in turn defined in terms of the Hermitian adjoint of [itex]\mathcal{F}[/itex], in equation 4-98 (page 191).

Secondly, given a particular [itex]A^{\mu}(x)[/itex], I suppose I have to find [itex]\mathcal{F}[/itex] somehow. How do I do this? The defining expression above is recursive, and the only two situations for which the authors have actually gone through with the calculation are the lowest order emission rate and the constant field emission rate.

Thirdly, the 'lowest order' emission probability is linear in [itex]\alpha[/itex] (the coupling constant) [equation 4-105, page 192]. But so is the 'exact' emission probability for a constant electric field (equation 4-118, page 195). So, if I understand correctly, the higher order contributions vanish for a constant electric field. Without actually computing the rate, can this be justified physically?

I want to compute higher order corrections to 4-105. So, I need to figure out what [itex]\mathcal{F}(A)[/itex] really is, and how I can obtain it for an arbitrary [itex]A^{\mu}(x)[/itex].

Would appreciate your thoughts...

Thanks in advance.
 

Answers and Replies

  • #2
Avodyne
Science Advisor
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What does this operator physically represent?
It's a one-particle-in one-particle-out scattering amplitude, with incoming and outgoing spinor factors removed, and then Fourier transformed to position space. If you generate an infinite series by recursion, the [itex]n[/itex]th term corresponds to a Feynman diagram with [itex]n[/itex] interactions with the external field.

Secondly, given a particular [itex]A^{\mu}(x)[/itex], I suppose I have to find [itex]\mathcal{F}[/itex] somehow. How do I do this?
No one knows. Only special cases have been solved.

Thirdly, the 'lowest order' emission probability is linear in [itex]\alpha[/itex] (the coupling constant) [equation 4-105, page 192]. But so is the 'exact' emission probability for a constant electric field (equation 4-118, page 195).
I don't have I&Z in front of me, but there is certainly a [itex]1/\alpha[/itex] (or [itex]1/e^2[/itex]) in the exponent of the exponential factor of the exact solution.
 
  • #3
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Thanks for your reply Avodyne.

No one knows. Only special cases have been solved.
Okay, so given a general time dependent (and space independent) field [itex]A^{\mu}(x)[/itex] which corresponds to a time-dependent electric field, how does one go about finding the production rate from this method, if there is no known way to write down the scattering operator?

The total pair production probability is given in I&Z by

[tex]W_{tot} = \int d^{4}x\,w(x) = \int d^{3}p\ln(1+tt^{\dagger})[/tex]

where [itex]t[/itex] is an operator defined by

[tex]\langle\boldsymbol{p},a|t|\boldsymbol{p}',b\rangle = 2\pi\sum_{\alpha,\beta}\frac{m}{\sqrt{\omega_P}\sqrt{\omega_P'}}\bar{u}_{\alpha}^{(a)}(p)\langle p,\alpha|\bar{\mathcal{F}}|p',\beta\rangle v_{\beta}^{(b)}(-p')[/tex]

They write down the emission rate to lowest order apparently by evaluating the trace in the definition for [itex]W_{tot}[/itex], to get

[tex]W^{(1)} = -Tr(e\gamma^\mu A_\mu \rho^{(+)}e\gamma^\nu A_\nu \rho^{(-)})[/tex]

where

[tex]\rho^{(\pm)}(P) \equiv 2\pi(\gamma^\mu P_\mu + m)\theta(\pm P^0)\delta(P^2-m^2)[/tex]

I don't understand how the lowest order contribution is obtained. Can you explain this? I want to try and compute higher order corrections. By compute, of course I mean that I want to be able to write them down as integrals involving |E|^2-|B|^2 as done for the lowest order case.

I don't have I&Z in front of me, but there is certainly a [itex]1/\alpha[/itex] (or [itex]1/e^2[/itex]) in the exponent of the exponential factor of the exact solution.
You're right of course...the production probability p.u. time p.u. volume is

[tex]w = \frac{\alpha E^2}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}\exp\left(-\frac{n\pi m^2}{|eE|}\right)[/tex]

for a constant electric field. So, the exponential does involve [itex]1/e[/itex] or equivalently, [itex]1/\sqrt{\alpha}[/itex].
 
  • #4
Avodyne
Science Advisor
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how does one go about finding the production rate from this method, if there is no known way to write down the scattering operator?
Um, you can't, unless you come up with some new method.
I don't understand how the lowest order contribution is obtained. Can you explain this? I want to try and compute higher order corrections.
They expanded the log, [itex]\ln(1+tt^\dagger)=tt^\dagger-{1\over2}tt^\dagger tt^\dagger + \ldots[/itex], and kept only the first term. I don't know how feasible it would be to compute the next term in the general case.
 
  • #5
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They expanded the log, [itex]\ln(1+tt^\dagger)=tt^\dagger-{1\over2}tt^\dagger tt^\dagger + \ldots[/itex], and kept only the first term. I don't know how feasible it would be to compute the next term in the general case.
How does [itex]ln(1+tt^{\dagger})[/itex] give

[tex]-Tr(e\gamma^\mu A_\mu \rho^{(+)}e\gamma^\nu A_\nu \rho^{(-)})[/tex]

? This part isn't obvious to me..
 
Last edited:
  • #6
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How does [itex]ln(1+tt^{\dagger})[/itex] give

[tex]-Tr(e\gamma^\mu A_\mu \rho^{(+)}e\gamma^\nu A_\nu \rho^{(-)})[/tex]

? This part isn't obvious to me..
Okay, I think I got it.
 

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