For a Klein-Gordon field, our QFT lecture notes say we use the following relationship to define the Heisenberg picture.

$$i \frac{dQ}{dt} = [Q,H]$$

which leads to

$$Q(t) = e^{iHT}Q(0)e^{-iHt}$$

However, for a Klein-Gordon field, shouldn't the Klein-Gordon equation replace the time-dependent Schrödinger equation, and therefore (since there is a double time derivative now), shouldn't the general solution not just be in the form $|\psi(t)\rangle = e^{-iHt} |\psi(0)\rangle$ (since $|\psi(t)\rangle = e^{+iHt} |\psi(0)\rangle$ would also be a valid solution)?

Since all the rest of QFT seems to use this Heisenberg picture for their operators, this seems like an important point for me to understand.


2 Answers 2


The Klein-Gordon equation does not replace the Schrödinger equation. The former is an equation for the field operators, the latter an equation governing the time evolution of the state.

See for example this, this or this.


In normal quantum mechanics the Heisenberg picture is defined as $${\displaystyle |\psi _{\rm {H}}\rangle=e^{iH_{\rm {S}}~t/\hbar }\displaystyle |\psi _{\rm {S}}(t)\rangle}$$ for states and $${\displaystyle A_{\rm {H}}(t)=e^{iH_{\rm {S}}~t/\hbar }A_{\rm {S}}e^{-iH_{\rm {S}}~t/\hbar }}$$ for operators

In QFT instead of the state, we will mainly deal with an operator field which in Heisenberg's picture is defined similarly to above $${\displaystyle \phi_{\rm {H}}(\vec{x},t)=e^{iH_{\rm {S}}~t/\hbar }\phi_{\rm {S}}e^{-iH_{\rm {S}}~t/\hbar }}$$ Here $\phi_{\rm {S}}$ is independent of time since in Schrödinger picture operators are constant.

The definition is not different. The difference is in QM we are not dealing with an operator but in QFT we deal with an operator.

Similarly in QFT the states in the Heisenberg picture are defined as $${\displaystyle |a_{\rm {H}}\rangle=e^{iH_{\rm {S}}~t/\hbar }\displaystyle |a_{\rm {S}}(t)\rangle}$$ here the state is constant. In QFT Heisenberg picture is more useful when there are no interactions. In QM Schrödinger picture is more useful when there are no interactions. When interactions are there Dirac picture is better in both cases.

  • $\begingroup$ Could you spell out exactly why there isn't an anaologous statement to $|/psi_H\rangle = e^{i H_s t/\hbar} |\psi_S(t)\rangle$ in QFT? Can't we have one for the field configuration states? $\endgroup$
    – Alex Gower
    Mar 25 at 17:15
  • $\begingroup$ @AlexGower I haven't written it earlier but there is a similar definition for Heisenberg picture states and I added it now. Read Cheng and Li Page number 6. $\endgroup$ Mar 25 at 17:22
  • $\begingroup$ Thanks, is there any way to solve the Klein Gordon equation in the Schrodinger picture? What would that look like? $\endgroup$
    – Alex Gower
    Mar 25 at 17:51
  • 2
    $\begingroup$ @AlexGower I guess this is a new question and should be treated separately, no? $\endgroup$ Mar 25 at 17:54
  • $\begingroup$ @AlexGower Read about Schrödinger functional they explain how the time-dependent state evolves in Schrödinger picture. I also didn't know about that, found it just now. $\endgroup$ Mar 25 at 17:57

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