# Proving $[a_k^\dagger, a_q^\dagger]=0$

I am trying to prove the commutation relations between the creation and annihilation operators in field theory. I was already able to show that $[a_k, a_q^\dagger]=i\delta(k-q)$. I want to show that $[a_k^\dagger, a_q^\dagger]=0,$ but I am stuck on the last step. I think my confusion has to do with interpreting the delta function and I was hoping someone can help me out. So far I have shown that

$$a_k^\dagger = \int \! d^3\!x \: e^{-ikx}\left[ \sqrt{\frac{E_k}{2}} \phi(x) - i \sqrt{\frac{1}{2E_k}} \pi(x) \right],$$

where $\phi(x)$ and $\pi(x)$ are the field and conjugate momentum respectively. Based on this I was able to show that

$$[a_k^\dagger, a_q^\dagger]=\int \! d^3y \: d^3x \: e^{-i(kx+qy)} \left[ -i \sqrt{\frac{E_k}{E_q}} [\phi(x), \pi(y)] + i \sqrt{\frac{E_q}{E_k}} [\phi(y),\pi(x)] \right].$$

Using the commutation relation $[\phi(x), \pi(y)]=i \delta^3(x-y)$ this becomes

$$[a_k^\dagger, a_q^\dagger]=\int \! d^3y \: e^{-i (k+q) y} \left[ \sqrt{\frac{E_k}{E_q}} -\sqrt{\frac{E_q}{E_k}} \right].$$

And I believe the integral over y gives us a delta function. If I have done everything correctly then this should be

$$[a_k^\dagger, a_q^\dagger]=\delta^3(k+q) \left[ \sqrt{\frac{E_k}{E_q}} -\sqrt{\frac{E_q}{E_k}} \right].$$

Since I already know the commutation relations I know that the final expression should equal zero for any arbitrary value of $k$ and $q$. But my final expression doesn't really say that for $k=-q$. I don't really understand where this discrepancy is coming from, though. Is there some way to interpret this delta function (i.e. is $\infty \cdot 0=0$?) that I'm not understanding or have I made a computational error somewhere?

• What happens to the expression in the brackets when $k=-q$...? – Nephente Nov 25 '14 at 8:50
• I guess this is what I'm having a hard time interpreting. You get ∞⋅0, which I guess equals zero, but I'm not sure why. I edited the question to clarify what I'm asking. – Dargscisyhp Nov 26 '14 at 5:12
• Keep in mind $\int \delta(x-a) f(x) = f(a)$. To calculate anything physical you'll be integrating over this delta function, and the result will be zero. – adipy Nov 26 '14 at 6:41
• Another way to see it is to remember that $x\delta(x)=0$ (since when applied on any test-function, this will always give zero). – Adam Nov 26 '14 at 21:15

You can think of the continuos formalism as being the limiting case of the discrete-momentum one: if the momentum is taken as a discrete variable (which amounts to constraining the particles to be in some finite volume $V$) the fourier expansion of the (real, scalar) field is: $$\tag{1} \varphi(x) = \sum_{\textbf{k}} \frac{1}{\sqrt{2V \omega_{\textbf{k}}}} \left( a_\textbf{k} e^{-ikx} + a_\textbf{k}^\dagger e^{ikx} \right)$$ From this you can verify (similarly to what you did above) that $$\tag{2} [a_k,a_q^\dagger] = \delta_{k,q}, \qquad [a_k,a_q] = 0 = [a_k^\dagger,a_q^\dagger].$$ this times without the problems arising from handling the continuous Dirac's delta functions.