0

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 !

StarBucK
  • 1,688

1 Answers1

3

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.

L. Werneck
  • 248
  • 1
  • 10