# Stuck, learning QFT

1. Jan 13, 2012

### spacelike

solving for commutator in QFT

I am just starting to learn quantum field theory. I'm still an undergraduate but I love particle physics and I want to get ahead, so I got the book "introduction to quantum field theory by peskin and schroeder"
I got to page 21 and I'm already stuck at an integral that I've been staring at for a few days now. The book says
"If computations such as this one are unfamiliar to you, please work them out carefully; they are quite easy after a little practice, and are fundamental to the formalism of the next two chapters."

So I don't want to to just move on, I'd like to understand how they do it. It is quite a big integral, but I'm really just hoping that if it really is fundamental to the theory then someone may recognize it and just be able to give me a push in the right direction.

So without further ado, here is the integral:
$$\int\frac{d^{3}pd^{3}p^{\prime}}{\left( 2\pi\right)^{6}}\frac{-i}{2}\sqrt{\frac{\omega_{p^{\prime}}}{\omega_{p}}}\left( \left[ a_{-\textbf{p}}^{\dagger},a_{\textbf{p}^{'}}\right] -\left[ a_{\textbf{p}},a_{-\textbf{p}^{'}}^{\dagger}\right]\right) e^{i\left( \textbf{p}\cdot\textbf{x}+\textbf{p}^{\prime}\cdot\textbf{x}^{\prime}\right)}=i\delta^{\left( 3\right)}\left(\textbf{x}-\textbf{x}^{\prime}\right)$$
where:
$$\left[ a_{\textbf{p}},a_{\textbf{p}^{'}}^{\dagger}\right]=\left( 2\pi\right)^{3}\delta^{\left( 3\right)}\left(\textbf{p}-\textbf{p}^{\prime}\right)$$
and
$$\omega_{\textbf{p}}=\sqrt{\left|\textbf{p}\right|^{2}+m^{2}}$$

Basically all the integral is doing is to find the commutator between two fields (sorry if my wording is off, I'm still trying to understand). So in normal quantum mechanics we have the commutator between 'x' and 'p', $[x_{i},p_{j}]=i\delta_{ij}$ (with h-bar set to 1). But in relativistic quantum mechanics I guess we need to use fields so they "quantize" x and p, into $\phi(\textbf{x})$ and $\pi(\textbf{y})$, and they express them in terms of integrals. So then to verify that we get a similar commutator as the one for non-relativistic quantum mechanics they multiply those integrals accordingly to find $[\phi(\textbf{x}),\pi(\textbf{y})]$, and that is exactly what the integral above is, and as shown above they get the answer $i\delta^{(3)}(\textbf{x}-\textbf{x}^{\prime})$.

It makes sense because the answer is very similar to the non-relativistic commutator.. But, I just don't see how the above integral works out to this.

Last edited: Jan 13, 2012
2. Jan 13, 2012

### petergreat

$$\int\frac{d^{3}pd^{3}p^{\prime}}{\left( 2\pi\right)^{6}}\frac{-i}{2}\sqrt{\frac{\omega_{p^{\prime}}}{\omega_{p}}}\left( \left[ a_{-\textbf{p}}^{\dagger},a_{\textbf{p}^{'}}\right] -\left[ a_{\textbf{p}},a_{-\textbf{p}^{'}}^{\dagger}\right]\right) e^{i\left( \textbf{p}\cdot\textbf{x}+\textbf{p}^{\prime}\cdot\textbf{x}^{\prime}\right)}$$
$$=\int\frac{d^{3}pd^{3}p^{\prime}}{\left( 2\pi\right)^{6}}\frac{-i}{2}\sqrt{\frac{\omega_{p^{\prime}}}{\omega_{p}}}\left[-(2\pi)^3\delta(-p-p^\prime)-(2\pi)^3\delta(p+p^\prime)\right] e^{i\left( \textbf{p}\cdot\textbf{x}+\textbf{p}^{\prime}\cdot\textbf{x}^{\prime}\right)}$$
$$=\int\frac{d^{3}p}{\left( 2\pi\right)^{3}}i\sqrt{\frac{\omega_{p^{\prime}}}{\omega_{p}}}\left[\delta(p+p^\prime)\right] e^{i\left( \textbf{p}\cdot\textbf{x}+\textbf{p}^{\prime}\cdot\textbf{x}^{\prime}\right)}$$
$$=\int\frac{d^{3}p}{\left( 2\pi\right)^{3}}i\sqrt{\frac{\omega_{p}}{\omega_{-p}}}e^{i\left( \textbf{p}\cdot\textbf{x}-\textbf{p}\cdot\textbf{x}^{\prime}\right)}$$
$$=\int\frac{d^{3}p}{\left( 2\pi\right)^{3}}i\cdot 1\cdot e^{i \textbf{p}\cdot\left(\textbf{x}-\textbf{x}^{\prime}\right)}$$
$$=i\delta(x-x^\prime)$$

Last edited: Jan 13, 2012
3. Jan 13, 2012

### petergreat

By $\delta$ I mean $\delta^{(3)}$.

4. Jan 14, 2012

### spacelike

Ahh, now I'm going to kick myself! I actually got exactly that:
$$=\int\frac{d^{3}p}{\left( 2\pi\right)^{3}}i\cdot 1\cdot e^{i \textbf{p}\cdot\left(\textbf{x}-\textbf{x}^{\prime}\right)}$$

But that's where I stopped because I didn't see how to get the delta function from that, and I still don't.. Now I feel bad because it's probably just a definition or something.. I was trying to mess around with nascent delta functions and seeing if taking a limit of some sort would make any sense.. But no luck

Could you just elaborate a little bit on that last step please?
and thank you very much for your time, I hope it wasn't too much trouble to type all that out.

5. Jan 14, 2012

### petergreat

You're welcome! The last step will become trivial after you've learned Fourier transform. Basically, it says that the Fourier transform of the constant function is the delta function. You need to know Fourier transform and complex analysis before tackling QFT.

6. Jan 14, 2012

### spacelike

Great, that makes sense now thank you.

I have taken some complex analysis, I've just never had a really formal introduction to Fourier analysis, but it's good to know that it will be important for this subject so I can get caught up on that first.