# Equal time commutation relations in canonical quantization of relativistic free fields

Why is equal time commutation relation used in canonical quantization of relativistic free fields? In a relativistic theory, space and time are to be treated on equal footing.

• Maybe related to physics.stackexchange.com/q/90963 – Hunter Jan 29 '14 at 14:34
• I wouldn't say so, Hunter. – Danu Jan 29 '14 at 14:34
• When you say "why," are you asking how we can get away with equal time commutation relations? Why questions are tricky to interpret. One answer might be, for example, to note that we do the same thing in quantum mechanics with the operators representing the fundamental degrees of freedom on phase space, and we're just doing the same thing in QFT; it's. Would an answer of that type be satisfying? – joshphysics Jan 29 '14 at 17:18
• @RoopamSinha Poincare invariant field theories respect special relativity, not all field theories. In these theories, it is the invariance of the objects that define the theory under changes of frame that matters. The idea of treating space and time on the same footing is somewhat vague; would you say that the fact that the $tt$ component of the metric has a different sign than the spatial components is a violation of time and space being treated "on the same footing?" Time and space are not physically the same, they are simply being considered as two coordinates of a manifold. – joshphysics Jan 29 '14 at 18:12

As a matter of fact one could also discuss commutation relations at different time: $$[\phi(x),\phi(y)] = i E(x,y) I\quad (1)$$ for free fields $E$ is the so-called causal propagator or advanced minus retarded fundamental solution that depends on the free field equation satisfied by $\phi$.

The point is that, passing to considering interacting fields, at least formally, equal time commutation relations remain unchanged with respect to the free case, whereas the corresponding of (1) changes into an, in practice, unknown form as they include the full dynamics.

Actually even this idea does not work completely, as interacting fields $\Phi$ are affected by a renormalization constant $Z^{1/2}$: $$\Phi (t,\vec x) \to Z^{1/2} \phi(t, \vec x)\quad \mbox{in weak sense as } t \to \pm \infty$$ and, dealing with naively with renormalization procedure, it arises $Z=0$. So canonical commutation relations seem to be untenable for fields $\Phi(x), \partial_t \Phi(y) = \Pi(y)$. However all that sounds a bit academic as the renormalization procedure, in a sense, solves the problem.

I would like to stress that the fact that commutation relations are taken at equal time is not in contradiction with relativistic invariance: Covariance (i.e., the use of tensors and taking space and time on the same footing) is just one way to make explicit relativistic invariance, but it is by no means the unique one!

Hamiltonian formalism is not covariant, though it is relativistically invariant: all equations (including CCR) take the same form in every inertial reference frame.

Surely, the field and the momentum fail to commute when examined at the same instance, by extension of the logic of quantum mechanics. However, there is no reason why the field and momentum shouldn't commute when they are well-separated in time. Very loosely speaking, they just don't really have anything to do with each other. The logical conclusion is that one should use equal-time commutation relations that denote that the quantities do not commute at equal times but do commute when well-separated in time.

The field operators are defined in the Heisenberg picture, and the equal-time commutator in the Heisenberg picture is simply the usual commutator in the Schrodinger picture. Hence it is really the same as what we do in ordinary QM!

As you pointed out, the notion of "equal time" is not Lorentz invariant. Indeed, in the canonical quantization procedure, Lorentz invariance of the underlying field theory is obscured because we have to single out the time direction. Yet Lorentz invariance is not lost, although not manifest.

On the other hand, Lorentz invariance is manifest in path integral quantization, which is one of the advantages of this method.

Let me sketch an answer elaborated in the reference A First Book of Quantum Field Theory by Lahiri and Pal (Second edition, Page $$30$$).

According to this reference above, the commutator $$[\phi(t,\textbf{x}),\pi(t,\textbf{y})]=i\delta^{(3)}(\textbf{x}-\textbf{y})\tag{1}$$ is actually covariant though the covariance is not manifest!

If two spacetime points coincide in one inertial frame, they will also coincide in different inertial frame. Therefore, at least for the special case when $$x=y$$ (i.e., $$x^0=y^0=t$$ and $$\textbf{x}=\textbf{y}$$), the commutation relation $$(1)$$ is satisfied in the same form in a different inertial frame.

Now, let us ask whether the relation $$(1)$$, in general, transforms covariantly under Lorentz transformation. The first thing to note is that $$\phi$$ is a scalar, and $$\pi=\partial_0\phi$$. Therefore, the Lorentz transformation property of the LHS comes from $$\partial_0$$ which makes it clear that LHS must transform like the time component of a four-vector. Now, since $$\int dt\int d^3\textbf{y}\delta^{(3)}(\textbf{x}-\textbf{y})=\int dt\tag{2}$$ and $$d^4y=dtd^3\textbf{y}$$ is Lorentz invariant, $$\delta^{(3)}(\textbf{x}-\textbf{y})$$ must also transform like the time component of a four-vector. Therefore, $$(1)$$ is covariant.