Physics Asked on April 28, 2021
Here’s what I understood. In quantum mechanics, there is two equivalent of treating time:
Scrödinger’s picture where operators are time-independent and states are time-dependent.
Heisenberg’s picture where operators are time-dependent and states are time-independent.
In both of these pictures:
time is seen as an external parameter on which depend observables or states (depending of the choice of formalism).
position is an observable, so there’s a position operator $hat{boldsymbol{X}}$.
As explained in Srednicki’s book, this difference in the treatment of time and position is why QM is hardly adaptable to special relativity, where time and space are treated on the same foot. One option is then to downgrade position to the status of "external variable". This gives raise to quantum fields wich are spacetime-dependent operators $hat{phi}(t,overrightarrow{x})$.
Now in QFT, we stil have Heisenberg’s and Scrödinger’s picture for the external variable $t$. But since position $overrightarrow{x}$ is now also an external parameter, could we also have an equivalent way of seeing fields as position independent and states as position-dependent?
If so, is this possible to work in this picture and in Schrödinger’s picture at same time, such that operators would be spacetime-independent?
No. This argument about down-grading the position operator to a variable is misleading about the structure of quantum mechanics and field theory, which are fundamentally just different realizations of the same generic structure.
In quantum mechanics (including field theory) Hilbert spaces are associated to constant-time slices of our spacetime. In a field theory, this means we have a single Hilbert space associated to all of space, the $boldsymbol{x}$'s. In QM, we are working in zero spatial dimensions, so we should really think of there being no position and instead the operator $boldsymbol X$ should be thought of as the analog of the field operators $phi$ in field theory.
So while the position dependence of fields $phi$ can be removed using the translation operator (write them all in terms of their value at the origin, say), it does not make sense to try and do the same for states in theory.
Correct answer by Richard Myers on April 28, 2021
Actually, there is currently a theory which treats position as the parameter and time as the operator. It is called "space-time-symmetric formalism." In it, the Heisenberg picture and Schrodinger picture basically take the position-dependence into account just as time-dependence is taken into account via those pictures in normal quantum mechanics. In other words, yes it is possible to work in the space-independent picture you describe.
However, in space-time-symmetric formalism you now have a time-dependence that won't be taken out of the operator via the Schrodinger or Heisenberg pictures...
Answered by PrawwarP on April 28, 2021
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