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?


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 $t$.

  • $\begingroup$ So is the point that the canonical fields $\phi$ and $\pi_{\phi}$ along with time, $t$ are considered as the fundamental independent variables of the theory, and as such functions (or operators) are built out of these variables (hence these functions can be explicitly time dependent)? $\endgroup$ – user35305 Jun 10 '16 at 6:51
  • $\begingroup$ @user35305, well $t$ and $\phi$ are certainly not in the same class of objects. It is a formal distinction -- the time dependence can enter an observable through $\phi$ or $\pi$ or "explicitly". This distinction is made in order to write down an equation of motion valid for all observables. If you are concerned only with $\phi$ or $\pi$, you don't even have to think about it. $\endgroup$ – Peter Kravchuk Jun 10 '16 at 9:11
  • $\begingroup$ Is it analogous to Lagrangians in classical mechanics where the canonical variables are position and velocity, which aren't explicitly time dependent until one chooses a particular path through configuration space? So, it's simply by definition that $\phi$ and $\pi$ are not explicitly time dependent and there's no physical motivation? $\endgroup$ – user35305 Jun 10 '16 at 9:31
  • $\begingroup$ @user35305, yes, that is what I mean. $\endgroup$ – Peter Kravchuk Jun 10 '16 at 9:33

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.

  • $\begingroup$ But why is it often assumed that the canonical field operators do not explicitly depend on time? In many introductory texts that I've read the author states that the field operators in the Heisenberg picture evolve as $\frac{d}{dt}\hat{\mathcal{O}}(t)=i\left[\hat{H},\hat{\mathcal{O}}(t)\right]$, (i.e. $\partial_{t}\hat{\mathcal{O}}(t)=0$ and the operator doesn't explicitly depend on time). Why is this assumption made and what is its justification? $\endgroup$ – user35305 Jun 9 '16 at 14:19
  • $\begingroup$ first we have classical equation of motion.we have to quantize that motion.so we make field variables to operators.and from our study of QM(Schrodinger equation),we know that operators acting on states and give spectrum.and operators don't have explicit time dependence in QM(apart from time dependent perturbation theory) .first of all we have to impose canonical commutation relation at equal time because that only makes sense.but field variables is dynamical so to give time dependence we go to Heisenberg picture.you can go from Heisenberg picture to Schrodinger picture back and forth. $\endgroup$ – Hare Krishna Jun 9 '16 at 14:31

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.

  • $\begingroup$ There is no conflict between relativity and QM, a fact that is made manifest both by the Heisenberg picture and e.g. Feynman's path integral. What you want to make obvious is a fundamental falsehood. Moreover, your answer doesn't really answer the original question, anyway. $\endgroup$ – Luboš Motl Jun 25 '16 at 16:04
  • $\begingroup$ That you will not see the conflict is obvious. But this is not the place to discuss this, anyway you have not really made an argument. $\endgroup$ – Schmelzer Jun 27 '16 at 18:10
  • $\begingroup$ One can easily prove that there is no conflict. The Heisenberg equations of motion are demonstrably Lorentz-covariant and all probabilities may be calculated from the Heisenberg operators and the fixed density matrix, and nothing else than the outcomes whose probabilities are calculable may be measured. $\endgroup$ – Luboš Motl Jun 28 '16 at 5:11
  • $\begingroup$ This depends on your rejection of realism, and your positivist restriction of physics to what can be measured. $\endgroup$ – Schmelzer Jul 26 '16 at 20:10
  • $\begingroup$ It would be cool if the rejection etc. were "mine". In reality, it's a result of physics advances of the 1920s and the most important revolution in 20th century physics. $\endgroup$ – Luboš Motl Jul 27 '16 at 6:52

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