# 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 $(f(x)-f(a))\delta(x-a)$? Commented Oct 26, 2017 at 17:15
• It is actually a good question. Is it $0$? Commented Oct 26, 2017 at 17:40
• Obviously YES... Commented Oct 26, 2017 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. Commented Oct 26, 2017 at 17:41