Proving [itex][a_k^\dagger, a_q^\dagger]=0[/itex]

[mjordan2nd]

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 operators 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 that I'm not understanding or have I made a computational error somewhere? Thanks

 

[Demystifier]

But my final expression doesn't really say that for k=−q.

Actually it does, because the square bracket vanishes in that case and $\delta(0)\cdot 0 =0$.

 

[mjordan2nd]

I don't really understand why. I guess the $\infty \cdot 0 = 0$ is what's throwing me off. Is there a way to put this on firmer mathematical footing?

 

[Matterwave]

Just take it a step back to the integral. The integral of 0 is 0, even if the integral is over all of space. :)

 

[strangerep]

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

Hint: what's the formula for $E_k$ ... ? ;)

 

[bhobba]

I don't really understand why. I guess the $\infty \cdot 0 = 0$ is what's throwing me off. Is there a way to put this on firmer mathematical footing?

You need to study distribution theory - get the following book - you won't regret it:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

The Dirac Delta function only has a meaning in integrals. When integrated it gives the value of the function before it at zero. In your case that's zero, so, when integrated gives zero. Hence, on a distribution basis is the same as zero.

Its a bit strange on first acquaintance, but once you get used to it its easy - which is why I really do suggest getting the book I mentioned - it will make everything clear - and much much more besides eg Fourier transforms become a doodle.

Thanks
Bill

 

[Demystifier]

I don't really understand why. I guess the $\infty \cdot 0 = 0$ is what's throwing me off. Is there a way to put this on firmer mathematical footing?

The $\delta$-function has a mathematical meaning only inside an integral:
$$\int dx\, \delta(x) f(x) =f(0)$$
which is $0$ if $f(0)=0$.

Another, more heuristic way to think of a product $\delta(k-q) f(k,q)$ is to replace the Dirac $\delta(k-q)$ with the Kronecker $\delta_{kq}$. In this way you can put
$$\delta_{kq}f(k,q)=f(k,k)=0$$
without doing the integral/sum.

 

[mjordan2nd]

Thank you for the replies. I will check out that book, Bill!