Consider an interacting field theory with Hamiltonian


where $H_0$ is the Hamiltonian of the free theory and $V$ is the added interaction. Now, I know the full Hamiltonian $H$ should be time-independent. Indeed, from the Heisenberg equation of motion we have

$$i\frac{\partial}{\partial t}H=[H,H]=0$$

However, since $H$ and $H_0$ do not generally commute I must have some time dependence in $H_0$.

$$i\frac{\partial}{\partial t}H_0=[H_0,H]\neq 0.$$

In for example the books by Weinberg and Peskin & Schroeder, they do implicitly assume time-independence of $H_0$, i.e that

$$\frac{\partial}{\partial t}H_0=0$$

when showing that the operator


satisfies the Schrodinger Equation:

$$i\frac{\partial}{\partial t}U(t,t_0)=V_I(t)U(t,t_0), \quad V_I(t)=e^{iH_0(t-t_0)}Ve^{-iH_0(t-t_0)}$$

They write

$$\frac{\partial}{\partial t}U(t,t_0)=-ie^{iH_0(t-t_0)}(H-H_0)e^{-iH(t-t_0)}$$

which is what I would expect to get if I had $\frac{\partial}{\partial t}H_0=0$.

Can someone please tell me where I'm going wrong? I have a strong suspicion this boils down to me screwing up the differences between partial time derivatives and total time derivatives, so perhaps I should have

$$\frac{\partial}{\partial t}H_0=0, \quad \frac{d}{d t}H_0\neq 0$$

If this is the case I am then unsure whether the Heisenberg eom should involve a total or partial derivative (Peskin uses a partial).

  • $\begingroup$ Heisenberg's equations of motion is $i \frac{d}{dt} A = [ A , H ] + \frac{\partial}{\partial t} A$. P&S assume that there is no explicit time dependence in $H_0$, meaning $\frac{\partial}{\partial t} H_0 = 0$. This does not however mean that $frac{d}{dt} H_0 = 0$ which is true clearly due to Heisenberg's equations. $\endgroup$
    – Prahar
    Mar 9, 2016 at 19:23
  • $\begingroup$ @Prahar Directly related question - suppose we have some Noether current $j^\mu$ with $\partial_\mu j^\mu=0$. Then the conserved charge $Q$ given by the spatial integral of $j^0$ satisfies $\frac{dQ}{dt}=0$. Why does this have $\frac{d}{dt}$ when it was $\frac{\partial}{\partial t}$ in the continuity equation? $\endgroup$
    – Okazaki
    Mar 9, 2016 at 22:49
  • $\begingroup$ $Q(t)$ has no dependence on any other variable. Therefore, $\frac{dQ}{dt} = \frac{ \partial Q}{ \partial t}$. In the continuity equation, you are talking about charge density $\rho(t,\vec{x})$ which does have dependence on multiple variables so you want to use partial derivatives. $\endgroup$
    – Prahar
    Mar 10, 2016 at 0:29
  • $\begingroup$ @Okazaki My first hint (without seeing the specified lines in the book) would be that the calculations are performed in the Schroedinger Picture, and every Operator mentioned would be an Operator in the Schroedinger Picture. The Reason I think that this is the case is : a) It would make your calculation work (I don't spot any calculation mistakes on your side) b) It makes no sense in General to take the exponential of a time dependent operator, if it is not specified at what time this operator is evaluated at. $\endgroup$ Feb 21, 2019 at 14:02

1 Answer 1


An observable $A$ is explicitly time-dependent if it is time-dependent in the Schrödinger picture. This dependence is what we mean when we write $\frac{\partial A}{\partial t}$. It has nothing to do with the Heisenberg picture and everything to do with the way we defined the observable.

The Heisenberg equation of motion for explicitly time-dependent observables is $$ \left.\frac{\mathrm{d}}{\mathrm{d}t}A\right\vert_{t=t_0} = \mathrm{i}[A(t_0),H] + \left.\frac{\partial A}{\partial t}\right\vert_{t=t_0}$$ and so the free Hamiltonian $H_0$ has $\frac{\partial}{\partial t} H_0 = 0$ but $\frac{\mathrm{d}}{\mathrm{d}t} H_0 \neq 0$ since it does not, as you say, commute with the full Hamiltonian that appears in the Heisenberg equation of motion.

  • $\begingroup$ I think the sign of the commutator may be incorrect on this $\endgroup$ Jun 9, 2016 at 23:11
  • 1
    $\begingroup$ In field theory, and in P&S you tend to work in the Interaction Picture, which is a sort of mid-way between Heisenberg and Schrodinger pictures. $H=H_0+H_1$, $A_I=e^{iH_0 t}Ae^{-iH_0 t}$, and the equation for time dependence is a Heisenberg like equation. $\frac{dA_I}{dt}=i[H_0,A_I,] +\frac{\partial A}{\partial t}$. And in this case $\frac{d}{dt}H_0=0$, given that $H_0$ has no explicit time dependence. However $\frac{dH_1}{dt}\neq 0$ in general even if $H_1$ has no explicit time dependence. $\endgroup$
    – snulty
    Nov 29, 2016 at 8:38

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