# Computation of $[P_{\mu},\phi(x)]$ with $\phi$ : real scalar field

In my course, we said that : $[P_{\mu},\phi(x)] = -i \partial_{x_{\mu}} \phi(x)$

Where : $\phi(x)$ is a real scalar field that we can write as :

$$\phi(x) = \int d\widetilde{k} ~ a(k,t) e^{-ikx}+a^{\dagger}(k,t)e^{+ikx}$$

(It is an interacting field that is why I have a time dependance on my fock operators).

How can we prove the commutation relation ?

For me the only thing we know is that :

$$\langle x | P_{\mu} |\psi \rangle = -i \partial_{x_{\mu}} \psi(x)$$ by definition of the momentum operator acting on a wavefunction.

Extra info about what confuses me a lot (but what I say here is not needed to answer the question):

I tried to do things but I'm a lot confused with the two significations $x$ can have. It can refer to the argument of the field (because at each point of the space we have an observable $\phi(x)$) AND it refers to the amplitude of the wavefunction in $x$.

More precisely, what I mean is that in the following quantity :

$$\phi(x) | \psi \rangle$$

I have the $x$ variable in the field and a "hidden" x variable in the ket.

Thus, if someone could help me to prove the commutation relation by taking care of according to which x we do the derivative would be nice !

• Is your confusion regarding the domain of $P_{\mu}$ (single particle hilbert space) and $\phi(x)$ (fock space)? – Sunyam Oct 26 '17 at 13:04
• Well there is probably some of it also. However, I think the most part of it lies in the fact as I have written above, $x$ has two significations : the position representation and the point of $\mathbb{R}^3$ where we evaluate the operator. But I think that a clear proof of the commutation relation would solve my problems. – StarBucK Oct 26 '17 at 13:29
• Isn't $P_{\mu}$ by definition the generator for (infinitesimal) translations in the Poincare group...? – Alex Nelson Oct 26 '17 at 13:44
• @AlexNelson well for me the only thing we have by using this definition is $\langle x | P_{\mu} |\psi \rangle = -i \partial_{x_{\mu}} \psi(x)$, but not the commutation relation (at least as far from how I understand things which may be wrong...) – StarBucK Oct 26 '17 at 13:46
• Write down the operator P as defined in your course, and perform the commutation. Do not use any bras or kets, or connection to nonrelativistic QM. This is a strict operator identity. – Cosmas Zachos Oct 26 '17 at 13:50

## 1 Answer

Define the translation operator

$$T(a)\equiv\exp\left(-iP^{\mu}a_{\mu}\right)\ ,$$

where $a_{\mu}$ are transformation parameters and $P^\mu$ are the generators of translations (which turn out to be the linear momentum operator). Then, a scalar field $\phi(x)$ transforms under translations as (remember quantum mechanics)

$$T(a)\phi(x)T^{-1}(a)=\phi(x-a)\ . \quad(\star)$$

Now, consider an infinitesimal translation by a small parameter $\epsilon_\mu$. Then the LHS of $(\star)$ becomes

\begin{align*} T(\epsilon)\phi(x)T^{-1}(\epsilon)&=\exp\left(-iP^{\mu}\epsilon_{\mu}\right)\phi(x)\exp\left(iP^{\nu}\epsilon_{\nu}\right)\\ &=\left(1-iP^{\mu}\epsilon_{\mu}\right)\phi(x)\left(1+iP^{\nu}\epsilon_{\nu}\right)\\ &=\phi(x)-i\epsilon^{\mu}\left[P_{\mu},\phi(x)\right]+\mathcal{O}\left(\epsilon^2\right)\ ,\quad (1) \end{align*}

where from the first to the second line we have used the fact that $\epsilon$ is infinitesimal and from the second to the third line we have done a few index relabels. The RHS of $(\star)$ becomes

$$\phi(x-\epsilon)=\phi(x)-\epsilon^{\mu}\partial_{\mu}\phi(x)+\mathcal{O}\left(\epsilon^2\right)\ .\quad (2)$$

Then, setting (1) = (2) and keeping only terms linear in $\epsilon$ we get

\begin{gather} \phi(x)-i\epsilon^{\mu}\left[P_{\mu},\phi(x)\right]=\phi(x)-\epsilon^{\mu}\partial_{\mu}\phi(x)\ , \end{gather}

which after a trivial manipulation yields the desired result

$$\boxed{\left[P_{\mu},\phi(x)\right]=-i\partial_{\mu}\phi(x)}\ .$$

Please let me know if you still have doubts! Cheers.

EDIT: A nice reference to understand how fields transform under Lorentz transformations is Tong's lecture notes on QFT. On this website you can also watch a few of his lectures where he talks about these transformations (I think somewhere in the first three videos).

Extra

In quantum mechanics, consider operators $\hat{\mathcal{O}}(\hat{x})$ and states $|x\rangle$, such that

$$\hat{\mathcal{O}}(\hat{x})|x\rangle=\mathcal{O}(x)|x\rangle\ .$$

A trivial example is $\hat{\mathcal{O}}(\hat{x})=\hat{x}$ such that

$$\hat{\mathcal{O}}(\hat{x})|x\rangle=\hat{x}|x\rangle=x|x\rangle\ .$$

Now consider the following procedure

\begin{align} \hat{\mathcal{O}}(\hat{x})|x\rangle&=\mathcal{O}(x)|x\rangle\ ,\\ \hat{\mathcal{O}}(\hat{x})T(a)|x\rangle&=\mathcal{O}(x+a)|x+a\rangle\ ,\\ T^{-1}(a)\hat{\mathcal{O}}(\hat{x})T(a)|x\rangle&=\mathcal{O}(x+a)|x\rangle\ ,\\ \Rightarrow T^{-1}(a)\hat{\mathcal{O}}(\hat{x})T(a) &= \hat{\mathcal{O}}(\hat{x}+a)\ ,\\ \Rightarrow \hat{\mathcal{O}}(\hat{x})=T(a)&\hat{\mathcal{O}}(\hat{x}+a)T^{-1}(a)\ ,\\ \Rightarrow \hat{\mathcal{O}}(\hat{x}-a)=T(a)&\hat{\mathcal{O}}(\hat{x})T^{-1}(a)\ , \end{align}

although one must keep in mind that the last three lines only make sense if they are acting on states.

Now, in the context of quantum mechanics this should be enough to convince you that operators should transform in this way (this is essentially how they transform in the Heisenberg picture). In QFT everything is always a bit more subtle/tricky, but the argument still holds and you can conclude that $(\star)$ is valid.

• Thank you for your very detailed answer. Just a little question about your equation ( * ). For a ket in quantum mechanics, we just multiply by one matrix the ket under a group transformation like translations. But for operator when we have representations of groups we always have to multiply by $U$ and $U^{-1}$ to understand how the group act on the Hilbert space right ? It is a general property that we have to multiply on both side by $U$ and $U^{-1}$ ? – StarBucK Oct 27 '17 at 11:20
• @StarBucK I've added an "extra" section to the answer trying to make this last point clear. Hope it helps! Also, on your question regarding $\phi(x)|\psi\rangle$, you need to keep in mind that in QFT $x$ is merely a label, though a continuous one. So states $|x\rangle$ are not well defined anymore, but $|\psi(x)\rangle$ still are, much like $|x_{\mu}\rangle$ are well defined but $|\mu\rangle$ don't make any sense (this is a sloppy argument, but I hope you can understand it). – L. Werneck Oct 27 '17 at 16:58
• Really thank you a lot for your answer. It was really good and accurate, well done. – StarBucK Oct 28 '17 at 13:15