Why Equal-time commutation relation?

Question

Let $\phi(\mathbf{x},t)$ be a field, and $\pi(\mathbf{x},t)$ be the conjugate momentum field.

A standard practice is to apply the equal time commutation relation: $$[\phi(\mathbf{x_1},t), \pi(\mathbf{x_2},t)] = i\hbar\delta^3(\mathbf{x_1} - \mathbf{x_2}).$$

My question is, why have equal time at all? Why not say:

$$[\phi(\mathbf{x_1},t_1), \pi(\mathbf{x_2},t_2)] = i\hbar\delta^3(\mathbf{x_2} - \mathbf{x_1})\delta(t_2 - t_1)~?$$

What is the fundamental flaw with the above?

Answer 1

The problem is that once you define your operators at a fixed time $t_0$, your operators at all later times are immediately determined by the Heisenberg equations of motion. So you are not free to choose your commutation relations at all times at once.

It may be helpful to see this in practice, with the Klein-Gordon theory: $$ H = \frac{1}{2} \int d^d x \left[ \pi^2 + (\nabla \phi)^2 + m^2 \phi^2 \right] $$ Starting from the equal-time commutation relation, it is simple to derive the Heisenberg equations of motion for both $\phi$ and $\pi$. They are $$ \dot{\phi} = \pi, \quad \dot{\pi} = \nabla^2 \phi - m^2 \phi $$ In other words, $\phi$ satisfies the Klein-Gordon equation: $$ \left( \frac{\partial^2}{\partial t^2} - \nabla^2 + m^2 \right) \phi = 0 $$ Now, given $\phi(t,x)$ at $t=t_0$, you can solve the Klein-Gordon equation to calculate $\phi$ at some later time. And you can check explicitly that the result does not commute with $\pi$ at $t=t_0$. So, you are not free to demand your unequal-time commutation relation: it would contradict your Hamiltonian evolution.

Answer 2

1. OP's proposal $$\begin{align} [\hat{\phi}({\bf x}_1,t_1),\hat{\phi}({\bf x}_2,t_2)]~=~&0,\cr [\hat{\phi}({\bf x}_1,t_1),\hat{\pi}({\bf x}_2,t_2)]~=~&i\hbar\hat{\bf 1}~\delta^3({\bf x}_1-{\bf x}_2)\delta(t_1-t_2),\cr [\hat{\pi}({\bf x}_1,t_1),\hat{\pi}({\bf x}_2,t_2)]~=~&0, \end{align}\tag{Wrong}$$ would not be able to capture the usual causal interplay of operators in spacetime for non-zero but timelike separated events. E.g. the commutator $$[\hat{\phi}({\bf x}_1,t_1),\hat{\phi}({\bf x}_2,t_2)]$$ is known to have non-zero support for timelike separated events.

2. In ordinary QM, i.e. QFT in 0+1D, OP's proposal $$\begin{align} [\hat{q}(t_1),\hat{q}(t_2)]~=~&0,\cr [\hat{q}(t_1),\hat{p}(t_2)]~=~&i\hbar\hat{\bf 1}~\delta(t_1-t_2),\cr [\hat{p}(t_1),\hat{p}(t_2)]~=~&0, \end{align}\tag{Wrong}$$ would imply that operators at different times commute, which is patently false.

3. However, if we just consider the classical theory, then OP's proposal with a non-equal-time Poisson bracket $$\begin{align} \{\phi({\bf x}_1,t_1),\phi({\bf x}_2,t_2)\}~=~&0,\cr \{\phi({\bf x}_1,t_1),\pi({\bf x}_2,t_2)\}~=~&\delta^3({\bf x}_1-{\bf x}_2)\delta(t_1-t_2),\cr \{\pi({\bf x}_1,t_1),\pi({\bf x}_2,t_2)\}~=~&0, \end{align}\tag{1}$$ can work if we write Hamilton's equations as $$ \dot{\phi}({\bf x},t)~=~\{ \phi({\bf x},t), \mathbb{H}\}, \qquad \dot{\pi}({\bf x},t)~=~\{ \pi({\bf x},t), \mathbb{H}\}, \tag{2}$$ where $$ \mathbb{H}~:=~\int \! dt ~H(t) \tag{3}$$ is the time-integrated Hamiltonian $H(t)$. See also e.g. this related Phys.SE post. Hamilton's equations (2) implies the Klein-Gordon equation. So classically OP's proposal (1) is fine.

4. We observe that the non-equal-time Poisson bracket (1) does not fulfill Dirac's semiclassical correspondence principle $$ [\cdot,\cdot] \quad \longleftrightarrow\quad i\hbar\{\cdot,\cdot\} +{\cal O}(\hbar^2).\tag{4}$$

Answer 3

1. A factor $i$ is missing in your first equation on the RHS (a minor point).
2. The dimensions of the LHS and the RHS of your second equation differ, so this relation must be wrong already for this very trivial reason.
3. The fundamental flaw is, that the canonical commutation relations must be chosen in such a way that the correct equations of motion (in the case of a free scalar field $\phi(t, \vec{x})$ this would just be the Klein-Gordon equation) can be deduced from the Heisenberg equation of motion $\dot{A}(t) = i [H,A(t)]$, where $A(t)$ can be any operator in the Heisenberg picture.
4. This has nothing to do with (relativistic) quantum field theory. You have the same situation also in the simple case of a one-dimensional quantum system with canonical variables $Q(t)$ and $P(t)$: $[Q(t),P(t)]=i$ ($\hbar = 1$ as above), but $[Q(t_1),P(t_2)]\ne i$ in general (if $t_1\ne t_2$). Analogously, we obviously have $[Q(t), Q(t)]=0$, but $[Q(t_1),Q(t_2)] \ne 0$ in general.
5. It is a good (and simple) exercise to compute the above commutators for the case of a one-dimensional harmonic oscillator. It might even be helpful to discuss the situation in classical mechanics by computing the corresponding Poisson brackets!