- #211
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I don't understand, where the problem might be. I was obviously wrong in trying to write down the operator for local thermal equilibrium. I'm even not sure anymore, whether it is possible to introduce such an idea, but first we should clarify, whether we agree on the general scheme. For simplicity let's work in the most simple example of a non-relativistic single particle in one dimension.
Let's write everything down in terms of a general picture. The time-dependence of the fundamental operators and are by some self-adjoint operator that of the states by some self-adjoint operator . Both can be also explictly time dependent, i.e., , and the Hamiltonian is given by .
The equations of motion for operators representing observables has a time dependence due to the time dependence of and and an explicit time dependence. and by definition have no explicit time-dependence. Thus we have
where refers only to the explicit time dependence. For the fundamental operators we have
The statistical operator, representing the state of the system is somewhat special. It's not special in fullfilling the equation of motion as any other operator,
On the other hand it must fulfill the von Neumann-Liouville equation,
and this makes it special in some sense.
Combining the last two equations by eliminating in (1), we find
This is the formalism in and arbitrary picture, usually called a Dirac picture. A often used special case is the interaction picture in QFT, where is the quadratic part of the hamilton functional of field operators and is the part with higher-order monomials in the hamilton density, the "interaction part".
The special cases we discussed so far are
The Heisenberg picture:
Thus the statistical operator is constant in time (and thus explicitly time dependent except for equilibrium states!):
The time dependence of observables' operators is given by
The Schrödinger picture:
The time dependence on the statistical operator (state) is given by
and that of observables' operators by
The equivalence of all pictures is proven by the fact that the transformation from one picture to another is given by a (time-dependent) unitary transformation.
So let's consider two pictures, labeled A and B. The time evolution for the operators of not explicitly time-dependent operators and that of the statistical operator are unitary transformations. We set withoug loss of generality the initial time (where the system is prepared) to and assume that both pictures coincide at this time.
Any not explicitly time-dependent observable operator fulfills the equation of motion as the fundamental operators. Let's take . For each picture we have
This can be (at least formally) solved by a unitary time-evolution operator , fulfilling the EoM
Then
The same holds true for the EoM of the statistical operator
The corresponding time-evolution operator is defined by the EoM
and
Now the transformation from one picture to the other is found by
Thus the unitary transformation from picture A to picture B is given by
In the same way one derives from the EoM of the observable operators that this transformation should be given by the operator
To have both pictures equivlent we should have . This can be proven as follows: Define
Then from the EoM of the time-evolution operators we get
Since also , fulfills the same EoM and initial condition for both and thus
This implies
and thus after some algebra
In addition we have shown that the physically, i.e., measurable quantities, of QT are picture independent, because the time evolution probability to find the value when measuring the observable is governed by the picture-independent operator . To see this let be a complete orthonormal set of eigenvectors of with eigenvalue . Then the said probability is
From this one reads off that the picture-independent unitary time-evolution operator can be interpreted as the time-evolution operator for the Schrödinger-picture or as the time-evolution operator for the eigenstates of observable operators (and thus the observable operators themselves) in the Heisenberg picture.
Let's write everything down in terms of a general picture. The time-dependence of the fundamental operators
The equations of motion for operators representing observables
where
The statistical operator, representing the state of the system is somewhat special. It's not special in fullfilling the equation of motion as any other operator,
On the other hand it must fulfill the von Neumann-Liouville equation,
and this makes it special in some sense.
Combining the last two equations by eliminating
This is the formalism in and arbitrary picture, usually called a Dirac picture. A often used special case is the interaction picture in QFT, where
The special cases we discussed so far are
The Heisenberg picture:
Thus the statistical operator is constant in time (and thus explicitly time dependent except for equilibrium states!):
The time dependence of observables' operators is given by
The Schrödinger picture:
The time dependence on the statistical operator (state) is given by
and that of observables' operators by
The equivalence of all pictures is proven by the fact that the transformation from one picture to another is given by a (time-dependent) unitary transformation.
So let's consider two pictures, labeled A and B. The time evolution for the operators of not explicitly time-dependent operators and that of the statistical operator are unitary transformations. We set withoug loss of generality the initial time (where the system is prepared) to
Any not explicitly time-dependent observable operator fulfills the equation of motion as the fundamental operators. Let's take
This can be (at least formally) solved by a unitary time-evolution operator
Then
The same holds true for the EoM of the statistical operator
The corresponding time-evolution operator is defined by the EoM
and
Now the transformation from one picture to the other is found by
Thus the unitary transformation from picture A to picture B is given by
In the same way one derives from the EoM of the observable operators that this transformation should be given by the operator
To have both pictures equivlent we should have
Then from the EoM of the time-evolution operators we get
Since also
This implies
and thus after some algebra
In addition we have shown that the physically, i.e., measurable quantities, of QT are picture independent, because the time evolution probability to find the value
From this one reads off that the picture-independent unitary time-evolution operator