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There seem to be two contradictory values of this commutator. I will assume Minkowski spacetime here.

On one hand, since $v^{\mu}$ is space-like, there exists a frame $(x',t')$ where $v^{\mu} \partial _{\mu}=a \partial _{x'}, a\in R$. Then, using a Lorentz transform to $(x', t')$:

$$[\phi(x), v^{\mu} \partial _{\mu} \phi(x)]$$

$$= a [\phi(x'), \partial _{x'} \phi (x')]|_{x'=\Lambda x}$$

$$=0$$

using $[\phi(x'), \nabla '\phi(x')]=0$, which follows from the CCR in the $(x',t')$ frame

On the other hand, (taking spacetime as 1+1):

$$[\phi (x), v^{\mu} \partial _{\mu} \phi (x)] = [\phi (x) , v^0 \partial _t \phi (x) + v^1 \partial _x \phi (x)]= v^0[\phi (x) , \partial _t \phi(x)]+v^1[\phi(x), \partial _x \phi (x)]= v^0 i\hbar \delta (0)$$

These approaches give two different values of this commutator. How to resolve this contradiction?

Valter Moretti
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Ryder Rude
  • 6,915

2 Answers2

2

TL;DR: OP's paradox seems to be caused by ill-defined mathematical manipulations on a distributional object.

To give an idea of what may go wrong consider the following calculation for a scalar field $\phi$ with $|x|>c|t|$ spatially separated:

$$\begin{align}\overbrace{[\phi(x,t)\!-\!\phi(0,0),~\phi(0,0)]}^{=0\text{ spatially separated}} ~=~&\overbrace{[\underbrace{\phi(x,t)\!-\!\phi(x,0)}_{\approx t\partial_t\phi(x,0)+{\cal O}(t^2)},~\phi(0,0)]}^{=0\text{ spatially separated}}\cr\cr ~+~&\overbrace{[\phi(x,0)\!-\!\phi(0,0),~\phi(0,0)]}^{=0\text{ spatially separated}}. \end{align}$$

References:

  1. C. Itzykson & J.B. Zuber, QFT, 1985; eqs. (3-55) + (3-56) + (3-57).
Qmechanic
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2

There is no contraddiction.

First of all $[\phi(x), \partial_\mu \phi(x)]$ is not defined as the two arguments coincide. What we can do is to define the normal order of it $$:[\phi(x), \partial_\mu \phi(x)]: = :\phi(x)\partial_\mu \phi(x): - :(\partial_\mu\phi(x)) \phi(x):$$ A direct computation proves that $$:[\phi(x), \partial_\mu \phi(x)]:=0\:,$$ so no contraddiction arises through this way.

Instead CCR implies $$[\phi(t_x,\vec{x}), v^\mu\partial_\mu \phi(t_y,\vec{y})]|_{t_x=t_y}= i\hbar v^0I\delta(\vec{x}-\vec{y})\tag{1}$$ where both sides are intepreted as 3D distributions on test functions $f=f(\vec{x})$.

Assuming that $v$ is spacelike, when you change reference frame in order that the components of $v$ are spatial, you must drop the constraint $t_x=t_y$. So, again, no evident contraddiction takes place.

A deeper analysis should be performed referring to $4D$-distributions.

In that case (see this site) the general form of the commutator -- which is evidently $IO(1,3)_+$ invariant -- is $$[\phi(x),\phi(y)] = -\frac{1}{(2\pi)^3}\int_{\mathbb{R}^4}\delta(k_\mu k^\mu + m^2)\:\mbox{sign}(k_0) e^{i k_\mu(x^\mu-y^\mu)} d^4k\:.$$ It easily implies the standard commutation relations (also considering derivatives) when passing to the invariant measure on the mass shell and assuming $t_x=t_y$ in any given Minkowskian reference frame.

In summary, for a spacelike vector $v$ (which has only spatial components in the primed reference frame), both $$[\phi(t_x,\vec{x}), v^\mu \partial_\mu\phi(t_y,\vec{y})]|_{t_x=t_y} = v^0i\delta(\vec{x}-\vec{y})\quad \mbox{and} \quad [\phi(t'_{x'},\vec{x}'), \sum_{k=1}^3 v'^k\partial'_k\phi(t'_{y'},\vec{y})]|_{t'_{x'}=t'_{y'}}=0$$ are correct. There is no contraddiction: they refer to different Minkowskian frames.

Valter Moretti
  • 80,330