Time dependence of canonical variables

Question

As far as I understand it, at least in scalar QFT, the canonical variables are the field operator $\hat{\phi}(x)$ and its conjugate momentum $\hat{\pi}_{\phi}(x)=\frac{\partial\mathcal{L}}{\partial\dot{\hat{\phi}}}$ (where $x=(t,\mathbf{x})$ and $c=\hbar =1$).

Someone has recently told me that it is usually assumed that these have no explicit time dependence, i.e. $\partial_{t}\hat{\phi}(x)=0$ and $\partial_{t}\hat{\pi}_{\phi}(x)=0$ (where $\partial_{t}:=\frac{\partial}{\partial t}$). As such, in the Heisenberg picture, their time evolution is governed by $$\frac{d}{dt}\hat{\phi}(x)=i\left[\hat{H},\hat{\phi}(x)\right]\, ,\quad \frac{d}{dt}\hat{\pi}_{\phi}(x)=i\left[\hat{H},\hat{\pi}_{\phi}(x)\right]$$ where $\hat{H}$ is the Hamiltonian of the theory.

If this is indeed the case, what is the argument (rationale) for why this is a valid assumption?

If I've understood things correctly, in the "standard" case of canonical quantisation the fields are quantised in the Schrödinger picture, where they have no time-dependence, and then through mapping to the Heisenberg picture, they pick up time-dependence through the unitary transformation $\hat{U}(t)=e^{-i\hat{H}t}$, hence I can see why, in this case, they have no explicit time dependence (since $\partial_{t}\hat{\phi}_{H}(t,\mathbf{x})=e^{i\hat{H}t}\partial_{t}\hat{\phi}_{S}(\mathbf{x})e^{-i\hat{H}t}$ and $\partial_{t}\hat{\phi}_{S}(\mathbf{x})=0$).

But what about the case where the Hamiltonian (or Lagrangian) has explicit time dependence? Won't operators have explicit time dependence even in the Schrödinger picture in this case?

Answer 1

The fields $\pi$ and $\phi$ are quantum fields which satisfy equations of motion, given classically by $$ d_t \pi=\{\pi,H\},\\ d_t q=\{q,H\}, $$ and these classical equations are established in any text on Hamiltonian mechanics. Heisenberg equations are just the quantization of those (replacing the Poisson bracket by the commutator). It is therefore not any special quantum magic.

The "explicit time dependence" is an artificial construct, which means basically that $\phi$ and $\pi$ by definition do not have explicit time dependence, while $A=\phi+\alpha t\pi$ does, because $t$ enters explicitly in the definition of this quantity. Therefore the time evolution of this quantity is given not only by evolution of $\pi$ and $\phi$, but also be the explicit dependence on time. Namely, you can write $$ \frac{dA}{dt}=\dot\phi\frac{\partial A}{\partial \phi}+\dot\pi\frac{\partial A}{\partial \pi}+\frac{\partial A}{\partial t}=\{A,H\}+\frac{\partial A}{\partial t}, $$ where the last derivative is equal to $\alpha \pi$, i.e. differentiates only the explicit $t$ in $A$.

In other words, if you continue to call the dependence $\phi(t)$ explicit, then the "explicit time dependence" you are worried about you should call "super-explicit time dependence". The fact that $\phi$ has no explicit time dependence is not something to be motivated, it is the definition -- an observable is said to have no explicit time dependence if it can be written as a function of the canonical variables only, not using $t$.

Answer 2

in the Schrodinger picture operator are time independent and states evolve with time.but in the Heisenberg picture operators are time dependent that are given by unitary transformation which you mentioned. when H is time dependent then integral in the exponent will come and time ordering also matters.Dyson formula work there. In the interaction picture, the evolution operator $U$ defined by the equation

$${\displaystyle \Psi (t)=U(t,t_{0})\Psi (t_{0})}$$ is called the Dyson operator. This leads to the following Neumann series:

$${\displaystyle {\begin{array}{lcl}U(t,t_{0})&=&1-i\int _{t_{0}}^{t}{dt_{1}V(t_{1})}+(-i)^{2}\int _{t_{0}}^{t}{dt_{1}\int _{t_{0}}^{t_{1}}{dt_{2}V(t_{1})V(t_{2})}}+\cdots \\&&{}+(-i)^{n}\int _{t_{0}}^{t}{dt_{1}\int _{t_{0}}^{t_{1}}{dt_{2}\cdots \int _{t_{0}}^{t_{n-1}}{dt_{n}V(t_{1})V(t_{2})\cdots V(t_{n})}}}+\cdots .\end{array}}}$$ Summing up all the terms, we obtain the Dyson series:

$${\displaystyle U(t,t_{0})=\sum _{n=0}^{\infty }U_{n}(t,t_{0})={\mathcal {T}}e^{-i\int _{t_{0}}^{t}{d\tau V(\tau )}}.}$$see interacting fields(phi 4,phi 3)theory form more example . $${\displaystyle {\frac {d}{dt}}A(t)={\frac {i}{\hbar }}[H,A(t)]+\left({\frac {\partial A}{\partial t}}\right)_{H},} $$this is the exact Heisenberg equation of motion and explicit time dependence of operator is also there.

Answer 3

I think this can be explained if one considers the Schrödinger picture to be more fundamental. There one has to define some configuration space, and this configuration space is something which does not change in time at all - only trajectories in the configuration space change in time. And the configuration of a field is described by all the field values in fixed points, $\varphi(x)$.

The reason why the Heisenberg picture is preferred in RQFT is because the conflict between relativity and quantum theory becomes much less visible in the Heisenberg picture, once one has, in this picture, operators depending on x and t, and in this operator algebra the non-relativistic parts related with the wave functional - a functional defined on the configuration space Q which changes in some absolute time t - plays no role.

But this hides it only, it does not make it disappear. So, if you define the equation, it appears that it is not about operators $\varphi(x^\mu)$ defined on some spacetime, but that there is a well-defined operator $\frac{d}{dt}$ which connects different events, and this operator plays a role in the equation. And all you can hope for is to show that in another frame you can define other equations which give results which are indistinguishable by observation.