# Field operator in Schrödinger picture

In Schrödinger picture operators do not depend on time explicitly. Consider a free scalar field with Lagrangian density

$${\cal L} ~=~ \frac{1}{2}\partial_{\mu} \phi\partial^{\mu}\phi-\frac{m^2}{2}\phi^2,$$

where $\phi=\phi(t,x,y,z)$. In quantum field theory we take it as an operator. Is this operator Schrödinger or Heisenberg one? How can we express free field Lagrangian in both pictures?

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You're asking if $L$ is an operator in Schrodinger or Heisenberg picture? Why would you care about this? – user1504 Mar 13 '13 at 12:33
Yes, if $L$ is in Heisenberg picture, how can we write it in Schrödinger picture and vice versa. It's important while discussing scattering theory. Thank you. – zoroastra Mar 13 '13 at 12:52
In the Schrodinger (functional) representation, aren't you more interested in the Hamiltonian, because that's what you'll use to evolve states? In the "coordinate" version of this, the canonical momenta go over into nice functional derivatives, explained in here just like p=-id/dx – twistor59 Mar 13 '13 at 13:04
Twistor59, thank you for your comment. You are right, but Lagrange's formalism is more fundamental than Hamilton's one. Without knowing Lagrangian we can not write down Hamiltonian (sometimes it even not exist). While jumping from quantum mechanics to quantum field theory we have direct correspondence, for ex. coordinate becomes field but(!) is this field in Schroedinger picture or in Heisenberg one. – zoroastra Mar 13 '13 at 13:20
@zoroastra it is not true to say that a Lagrangian operator for a given system is more fundimental than the Hamiltonian counterpart. In fact, I would say the opposit is true due to the phase space used in each representation. – Killercam Mar 13 '13 at 15:11

When the field $\phi$ is being written as a function of spacetime, with explicit dependence on time, then this field is said to be written in the Heisenberg picture. You can see how this is consistent with the terminology used in non-relativistic quantum mechanics by recalling that (See e.g. Peskin and Schroeder eq. 2.43) $$\phi(t,\mathbf x) = e^{i H(t-t_0)}\phi(\mathbf x, t_0)e^{-iH(t-t0)}.$$ This shows that we can choose some time slice, say at time $t_0$, in Minkowski space at which be define our usual Shrodinger picture operators (the canonical degrees of freedom of the field theory) and at later times, these operator degrees of freedom are related to those at time $t_0$ via the usual Hamiltonian, unitary time evolution above.

In order to use notation that is more in line with that used in non-relativistic quantum mechanics (e.g. see Sakurai p. 82 bottom), you might be inclined to define $$\phi^{(S)}(\mathbf x) = \phi(\mathbf x, 0), \qquad \phi^{(H)}(\mathbf x)(t)$$ in which case the above time evolution relation would take the familiar form $$\phi^{(H)}(\mathbf x)(t) = e^{iHt}\phi^{(S)}(\mathbf x)e^{-iHt}$$ but this would be somewhat non-standard notation in field theory as far as I am aware.

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Joshphysics, thank you very much for your answer. So, can we construct relativistic filed theory in Schroedinger picture? As we see there is an asymmetry between time and space in this picture. – zoroastra Mar 14 '13 at 19:31
@zoroastra Well I suppose it depends on what you mean by "we can construct...". In principle, I can't see a reason why one cannot write any "Heisenberg picture equation" in the Shrodinger picture using manipulations similar to those in my reply if that's what you mean? As for the "asymmetry" between space and time etc. I'd prefer to avoid commenting on qualitative statements like that in this case because I think it might just add to the confusion. – joshphysics Mar 14 '13 at 19:36
@zoroastra twistor59's response below addresses your last comment as well I'd say. – joshphysics Mar 14 '13 at 22:01

Like Josh said, if you want your operators $\hat{\phi}({\bf{x}},t)$to have a time dependence analogous to the time dependence of the classical fields $\phi({\bf{x}},t)$ which satisfy the equations of motion derived from the classical Lagrangian, then you're working in the Heisenberg picture, and the operator version of the classical equations of motion is just Heisenberg's equation $$\dot{\hat{\phi}}=-i[\hat{\phi},H]$$ If, however, you want to work in the Schroedinger picture, the operators will now be functions of position only, $\hat{\phi}({\bf{x}})$ and the time dependence is carried by the states. In the approach known as the functional Schroedinger picture, they pick a fixed timeslice and define the space of states to be space spanned by the eigenstates of the field operator $$\hat{\phi}({\bf{x}})|\phi({\bf{x}})\rangle= \phi({\bf{x}})|\phi({\bf{x}})\rangle$$ So much for the eigenstates. A generic state $|\Psi\rangle$ is a functional which maps a field $\phi({\bf{x}})$ to a number $$\Psi[\phi({\bf{x}})] = \langle \phi({\bf{x}})|\Psi \rangle$$ This is entirely analogous to the QM relation $$\Psi({\bf{x}}) = \langle {\bf{x}}|\Psi \rangle$$ The field $\phi({\bf{x}})$ plays the role of the coord $\bf{x}$ in QM. In this representation, just as we have the representation $$\hat{\phi} \leftrightarrow \phi({\bf{x}})$$ so we also represent the conjugate variable by a functional derivative $$\hat{\pi} \leftrightarrow -i\frac{\delta}{\delta\phi({\bf{x}})}$$ This ensures that the canonical commutation relations are respected. Moreover the time evolution of the states is now given by the functional Schroedinger equation

For standard perturbative calculations in scattering theory, however, it is much more convenient to use the slight modification of the Heisenberg picture known as the interaction picture.

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Right. This is probably a good occasion to emphasize that the Schrödinger picture should not be confused with the Schrödinger representation. The latter refers to the representation of the momentum operator $\hat{p}=\frac{\hbar}{i}\frac{\partial}{\partial x}$ as a position derivative (or its field theoretic generalization, cf. twistor59's answer). – Qmechanic Mar 13 '13 at 18:04
@twistor59: Dear twistor59, recently I am studying Path integrals for a scalar field theory. In some books the eigenstates of field operators are introduced. But I am facing difficulty understanding them. Can you give me a reference where this topic is clearly discussed? Secondly, You have written: $\hat{\Phi} (\textbf{x}) |\Phi(\textbf{x}) \rangle = \Phi(\textbf{x}) |\Phi(\textbf{x}) \rangle$ Does this mean that at a fixed time t we have different field operators for each position in the configuration space and for each field operator there is an eigenstate associated with an eigenvalue? – omephy Apr 21 '13 at 19:24
@Ome The eigenstates of the local field operators at a given spatial point i.e. $|\phi(\bf{x})\rangle$ aren't normalizable, but are still useful for formal manipulations. The rigorous way to interpret them is to consider them as operator valued distributions: true operators are recovered by smearing with test functions. This, together with the construction of the one particle state space and its extension to Fock space are discussed in Bob Geroch's notes – twistor59 Apr 21 '13 at 20:28