Prove that the following integral vanishes

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  • #1
diracsgrandgrandson
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Homework Statement
Prove that the integral $$\int d^3p(\xi^r)^T\mathbf{p}\mathbf{\sigma}\xi^s(a^{r\dagger}_pa^s_p+a^{s\dagger}_pa^r_p)$$ vanishes.
Relevant Equations
Here, ##\xi^{1,2} = (1,0)^T,(0,1)^T## are spinors. We are working with Dirac particles, so the anticommutation relations are given by ##\{a^{r\dagger}_p,a^s_q\} = \delta^{rs}\delta(p-q)##. Note that all in my notation here ##\mathbf{p} = p## (I am lazy but there are no four-momenta in this question, so no risk of confusion).
We use the invariance of the measure under ##p\rightarrow -p## to get $$-\int d^3p\xi^{rT}\mathbf{p}\mathbf{\sigma}\xi^s(a^{r\dagger}_{-p}a^s_{-p}+a^{s\dagger}_{-p}a^r_{-p}) = -\int d^3p\xi^{rT}\mathbf{p}\mathbf{\sigma}\xi^sA(-p).$$ If this pesky ##A(-p)## can be shown to be equal to ##A(p)## or somehow ##p##-independent, then it all works. But I don't know how. I know the anticommutation relations are somehow involved but the complication is that ##p=q## which gives a delta function evaluated at zero.
 
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  • #2
I'd say you're missreading your anticommutation relation in a sense. If you want to do this the physicist way(which means by being less formal about math) you would say:
$$\{a^{r\dagger}_p, a^s_p\} = \delta^{rs}\delta(0)$$
This, as it stands clearly doesn't depend on ##p##, and so you have your answer that ##A(p)## shouldn't depend on ##p##, so from there it holds that this integral is zero by symmetry. If you wanted to define these operators at a fixed momentum, you'd have this same relation without ##\delta(0)##. But these are field operators and so you have to define it as a distribution, where Dirac delta generalizes Kronecker delta.

But this delta function bothers you because, I assume you find it as infinite, or more precisely, it could be said that it's undefined in this point. If you wanted to do it the formal way, without this simple deduction that is not exactly clean mathematically, you would define the delta function as a limit of an array of functions, for example normalized square impulses whose width tends to zero with normalization held constant. Or similarly using an array of gaussians with the same property. If you do it that way, you'd work out through this limit to see that the array of functions you're looking at in the integral is even around zero, so the whole integral is odd, and it would go to zero before evaluating the limit(which would be ##\delta(0)##). More precisely, you'd find a divergent array which doesn't even depend on ##p##.

In short, ##\delta(0)## is a particular limiting value, it does not equal infinity, but your symmetry property with respect to ##p## makes the integral exactly zero regardless of this limit. Hope that makes a bit of sense, I don't know how far you worked into the theory of distributions, since in QFT you usually don't go that far in a standard course, so I'm providing you a way to think about it this way.

It is a similar situation to a situation where you for example have a limit ##\lim_{n \rightarrow \infty} a \cdot b_n## where ##\lim_{n \rightarrow \infty} b_n = \infty## but ##a=0##. Exact zero times a divergent array is zero.
 
  • #3
Antarres said:
I'd say you're missreading your anticommutation relation in a sense. If you want to do this the physicist way(which means by being less formal about math) you would say:
$$\{a^{r\dagger}_p, a^s_p\} = \delta^{rs}\delta(0)$$
This, as it stands clearly doesn't depend on ##p##, and so you have your answer that ##A(p)## shouldn't depend on ##p##, so from there it holds that this integral is zero by symmetry. If you wanted to define these operators at a fixed momentum, you'd have this same relation without ##\delta(0)##. But these are field operators and so you have to define it as a distribution, where Dirac delta generalizes Kronecker delta.

But this delta function bothers you because you, I assume you find it as infinite, or more precisely, it could be said that it's undefined in this point. If you wanted to do it the formal way, without this simple deduction that is not exactly clean mathematically, you would define the delta function as a limit of an array of functions, for example normalized square impulses whose width tends to zero with normalization held constant. Or similarly using an array of gaussians with the same property. If you do it that way, you'd work out through this limit to see that the array of functions you're looking at in the integral is even around zero, so the whole integral is odd, and it would go to zero before evaluating the limit(which would be ##\delta(0)##).

In short, ##\delta(0)## is a particular limiting value, it does not equal infinity, but your symmetry property with respect to ##p## makes the integral exactly zero regardless of this limit. Hope that makes a bit of sense, I don't know how far you worked into the theory of distributions, since in QFT you usually don't go that far in a standard course, so I'm providing you a way to think about it this way.

It is a similar situation to a situation where you for example have a limit ##\lim_{n \rightarrow \infty} a \cdot b_n## where ##\lim_{n \rightarrow \infty} b_n = \infty## but ##a=0##. Exact zero times a divergent array is zero.
But the anticommutator that is independent of ##p## is not the same expression as ##A(p)##, and therefore it being independent does not say anything about ##A(p)##. Am I missing something?
 
  • #4
All I'm saying is, ##\delta(0)## is independent of ##p## obviously, and your ##A(p)## is proportional to this, so it is also independent of ##p##. The question of being mathematically rigorous is basically about rewriting this in such a way so as to evade writing ##\delta(0)##, an undefined value. But it is a situation where you'd have a divergent array multiplied by a zero integral, which is zero.

If you're not asked to rigorously deal with a Dirac delta here, even writing ##\delta(0)## and knowing that this is not an exact value(because Dirac delta is not a function but a distribution), you'd get zero next to it.

To add, in case you didn't notice ##A(p)## is precisely the anticommutator I wrote in the first equation of my previous post. Right hand side of this equation isn't dependent on ##p##.

Edit: nevermind, you're correct, it's not exactly that expression. I'll add the derivation in the next post.
 
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  • #5
diracsgrandgrandson said:
Homework Statement: Prove that the integral $$\int d^3p(\xi^r)^T\mathbf{p}\mathbf{\sigma}\xi^s(a^{r\dagger}_pa^s_p+a^{s\dagger}_pa^r_p)$$ vanishes.
It might be helpful to know the context of this exercise. Is it part of a larger calculation, or is it a stand-alone problem?

Also, what is the meaning of the symbol ##\sigma## in the integrand?
 
  • #6
Yes, it is part of a larger exercise. It's Peskin&Schroeder 3.4 (e). One has to construct the diagonalized Hamiltonian. I have arrived at the following expression: $$H = \frac{1}{2}\int d^3p\xi^{rT}p\sigma\xi^{s}(a^{r\dagger}_pa^s_p+a^{s\dagger}_pa^r_p),$$ where ##p\sigma = p^{\mu}\sigma_{\mu}##, ##\sigma_{\mu} = (1,\sigma_i)## with ##\sigma_i## being the Pauli matrices.

The correct Hamiltonian is $$H = \int d^3pa^{s\dagger}_pa^{s}_p$$ as one would expect. While the exercise does not guarantee anywhere that the spatial part of the expression above must vanish, if it does, it works out perfectly and I get the correct Hamiltonian. It is entirely possible that the integral does not vanish but it did make sense to me that an integrand that's odd in ##\mathbf{p}## should vanish.
 

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