# Regarding the commutator of ladder operator in QFT

I am trying to verify the computation of the commutator of the ladder operator for Klein-Gordon solutions, but it seems like I am unable to do it properly. Here is what I do:

For, $$\varphi(x^\mu)=\int\frac{\mathrm{d}^3p}{(2\pi)^{3/2}\sqrt{2p_0}}\left(a(\vec p)e^{-ip_\mu x^\mu}+a^\dagger(\vec p)e^{ip_\mu x^\mu}\right)\nonumber\\ \Pi(x^\mu)=\int\frac{\mathrm{d}^3p}{(2\pi)^{3/2}}(-i)\sqrt{\frac{p_0}{2}}\left(a(\vec p)e^{-ip_\mu x^\mu}-a^\dagger(\vec p)e^{ip_\mu x^\mu}\right),\nonumber$$ write: $$a(\vec p)=\int \frac{\mathrm{d}^3\vec x}{(2\pi)^{3/2}\sqrt{2p_0}}(p_0\varphi(\vec x,t)+i\Pi(\vec x,t))e^{ip_\mu x^\mu}\nonumber\\ a^\dagger(\vec p)=\int \frac{\mathrm{d}^3\vec x}{(2\pi)^{3/2}\sqrt{2p_0}}(p_0\varphi(\vec x,t)-i\Pi(\vec x,t))e^{-ip_\mu x^\mu}\nonumber.$$ From this compute the following: $$\left[a(\vec p),a^\dagger(\vec q)\right]{=\int\frac{\mathrm{d}^3\vec x\>\mathrm{d}^3\vec y}{(2\pi)^32\sqrt{p_0q_0}}\left[p_0\varphi(\vec x,t)+i\Pi(\vec x,t), q_0\varphi(\vec y,t)-i\Pi(\vec y,t)\right]e^{i(p_\mu x^\mu-q_\mu y^\mu)}\nonumber\\ =\int\frac{\mathrm{d}^3\vec x\>\mathrm{d}^3\vec y}{(2\pi)^32\sqrt{p_0q_0}}e^{i(p_\mu x^\mu-q_\mu y^\mu)}(-ip_0\left[\varphi(\vec x,t),\Pi(\vec y,t)\right]+iq_0\left[\Pi(\vec x,t),\varphi(\vec y,t)\right])\nonumber\\ =\int\frac{\mathrm{d}^3\vec x\>\mathrm{d}^3\vec y}{(2\pi)^32\sqrt{p_0q_0}}e^{i(p_\mu x^\mu-q_\mu y^\mu)}i(-ip_0\delta^3(\vec x-\vec y)-iq_0\delta^3(\vec y-\vec x))\nonumber\\ =\int\frac{\mathrm{d}^3\vec x\>\mathrm{d}^3\vec y}{(2\pi)^32\sqrt{p_0q_0}}e^{i(p_\mu x^\mu-q_\mu y^\mu)}\delta^3(\vec x-\vec y)(p_0+q_0)\nonumber\\ =\int\frac{\mathrm{d}^3\vec x}{(2\pi)^3}e^{i(p-q)\cdot x}\frac{p_0+q_0}{2\sqrt{p_0q_0}}\nonumber\\ =\delta^3(\vec p-\vec q)\nonumber\frac{p_0+q_0}{2\sqrt{p_0q_0}}e^{i(p_0-q_0)t}.}$$ And I don't understand why I have this strange factor which should not be there.

Remark: The factor is one if $p_0=q_0$ which is of course what we want. Also, one can see that the last line can only be true if the preceding condition holds, because of the argument of the exponential.

• What is the value of $(f(x)-f(a))\delta(x-a)$? – Valter Moretti Oct 26 '17 at 17:15
• It is actually a good question. Is it $0$? – Soap312 Oct 26 '17 at 17:40
• Obviously YES... – Valter Moretti Oct 26 '17 at 18:28

You need to think about what $p_0$ actually is here. When going from the Schrödinger to the Heisenberg picture, one simplifies the expressions for the mode expansions by introducing* $$p_0 = p_0(\vec p) = E_{\vec p} = \sqrt{\vec p^2 + m^2} \;.$$ Hence, your delta function does in fact set $p_0 = q_0$.
By the way, in your second-to-last line (v2) you have $$\mathrm e^{\mathrm i(p_\mu x^\mu - q_\mu y^\mu)} = \mathrm e^{\mathrm i (p_0 - q_0) t}\, \mathrm e^{\mathrm i(\vec x \cdot \vec p - \vec y \cdot \vec q)} \;.$$ You left out the first factor -- I assume, accidentally -- but it vanishes for the same reason ($p_0 = q_0$).
Edit as response to comment: In general, $$f(\vec p, \vec q) \delta(\vec p - \vec q) = f(\vec p, \vec p) \delta(\vec p - \vec q)$$ because $f(\vec p, \vec q) \delta(\vec p - \vec q) = 0$ whenever $\vec q \neq \vec p$. That means you can replace all $q_0 = \sqrt{m^2 + \vec q^2}$ with $p_0 = \sqrt{m^2 + \vec p^2}$.
• Yes but I'm not sure however why this should imply $p_0=q_0$. And yes, I corrected it. – Soap312 Oct 26 '17 at 17:41