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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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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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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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