- #211

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Let's write everything down in terms of a general picture. The time-dependence of the fundamental operators ##\hat{x}## and ##\hat{p}## are by some self-adjoint operator ##\hat{H}_0## that of the states by some self-adjoint operator ##\hat{H}_1##. Both can be also explictly time dependent, i.e., ##\hat{H}_j=\hat{H}_j(\hat{x},\hat{p},t)##, ##j \in\{0,1 \}## and the Hamiltonian is given by ##\hat{H}=\hat{H}_0+\hat{H}_1##.

The equations of motion for operators representing observables ##\hat{O}(\hat{x},\hat{p},t)## has a time dependence due to the time dependence of ##\hat{x}## and ##\hat{p}## and an explicit time dependence. ##\hat{x}## and ##\hat{p}## by definition have no explicit time-dependence. Thus we have

$$\mathrm{d}_t \hat{O}=\frac{1}{\mathrm{i} \hbar} [\hat{O},\hat{H}_0] + \partial_t \hat{O},$$

where ##\partial_t## refers only to the explicit time dependence. For the fundamental operators we have

$$\mathrm{d}_t \hat{x}=\frac{1}{\mathrm{i} \hbar} [\hat{x},\hat{H}_0], \\

\mathrm{d}_t \hat{p}=\frac{1}{\mathrm{i} \hbar} [\hat{p},\hat{H}_0].$$

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,

$$\mathrm{d}_t \hat{\rho}=\frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}_0]+\partial_t \hat{\rho}. \qquad (1)$$

On the other hand it must fulfill the von Neumann-Liouville equation,

$$\mathring{\hat{\rho}}=\frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}] +\partial_t \hat{\rho}=0,$$

and this makes it special in some sense.

Combining the last two equations by eliminating ##\partial_t \hat{\rho}## in (1), we find

$$\mathrm{d}_t \hat{\rho} = \frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}_0-\hat{H}] = -\frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}_1].$$

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 ##\hat{H}_0## is the quadratic part of the hamilton functional of field operators and ##\hat{H}_1=\hat{H}_I## 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**:

$$\hat{H}_0=\hat{H}_{\text{H}} \; \Rightarrow \; \hat{H}_1=0.$$

Thus the statistical operator is constant in time (and thus explicitly time dependent except for equilibrium states!):

$$\mathrm{d}_t \hat{\rho}_{\text{H}}=0.$$

The time dependence of observables' operators is given by

$$\mathrm{d}_t \hat{O}_{\text{H}}=\frac{1}{\mathrm{i} \hbar} [\hat{O}_{\text{H}},\hat{H}_{\text{H}}]+\partial_t \hat{O}_{\text{H}}.$$

**The Schrödinger picture**:

$$\hat{H}_0=0 \; \Rightarrow \; \hat{H}_1=\hat{H}_{\text{S}}.$$

The time dependence on the statistical operator (state) is given by

$$\mathrm{d}_t \hat{\rho}_{\text{S}}=-\frac{1}{\mathrm{i} \hbar} [\hat{\rho}_{\text{S}},\hat{H}_{\text{S}}],$$

and that of observables' operators by

$$\mathrm{d}_t \hat{O}_{\text{S}}=\partial_t \hat{O}_{\text{S}}.$$

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 ##t_0=0## 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 ##x##. For each picture ##j \in \{A,B \}## we have

$$\mathrm{d}_t \hat{x}^{(j)}=\frac{1}{\mathrm{i} \hbar} [\hat{x}^{(j)},\hat{H}_0^{(j)}].$$

This can be (at least formally) solved by a unitary time-evolution operator ##\hat{C}^{(j)}##, fulfilling the EoM

$$\mathrm{d}_t \hat{A}^{(j)}=\frac{\mathrm{i}}{\hbar} \hat{A}^{(j)}, \quad \hat{A}^{(j)}(0)=\hat{1}.$$

Then

$$\hat{x}^{(j)}(t)=\hat{A}^{(j)\dagger}(t) \hat{x}^{(j)}(0) \hat{A}^{(j)}(t).$$

The same holds true for the EoM of the statistical operator

$$\mathrm{d}_t \hat{\rho}^{(j)}=-\frac{1}{\mathrm{i} \hbar} [\hat{\rho}^{(j)},\hat{H}_1^{(j)}].$$

The corresponding time-evolution operator is defined by the EoM

$$\mathrm{d}_t \hat{C}^{(j)}=-\frac{\mathrm{i}}{\hbar} \hat{H}_1^{(j)} \hat{C}^{(j)}, \quad \hat{C}^{(j)}(0)=\hat{1}$$

and

$$\hat{\rho}^{(j)}(t)=\hat{C}^{(j)\dagger}(t) \hat{\rho}^{(j)}(0) \hat{C}^{(j)}(t).$$

Now the transformation from one picture to the other is found by

$$\hat{\rho}^{(B)}(t)=\hat{C}^{(B)\dagger}(t) \hat{\rho}^{(B)}(0) \hat{C}^{(B)}(t) = \hat{C}^{(B)\dagger}(t) \hat{\rho}^{(A)}(0) \hat{C}^{(B)}(t) = \hat{C}^{(B)\dagger}(t) \hat{C}^{(A) \dagger}(t) \hat{\rho}^{(A)}(t) \hat{C}^{(A) \dagger}(t) \hat{C}^{(B)}(t).$$

Thus the unitary transformation from picture A to picture B is given by

$$\hat{B}^{(AB)}(t)=\hat{C}^{(A) \dagger}(t) \hat{C}^{(B)}(t).$$

In the same way one derives from the EoM of the observable operators that this transformation should be given by the operator

$$\hat{D}^{(AB)}(t)=\hat{A}^{(A) \dagger}(t) \hat{A}^{(B)}(t).$$

To have both pictures equivlent we should have ##\hat{B}^{(AB)}=\hat{D}^{(AB)}##. This can be proven as follows: Define

$$\hat{G}^{(j)}=\hat{A}^{(j) \dagger} \hat{C}^{(j)}, \quad j \in \{A,B \}.$$

Then from the EoM of the time-evolution operators we get

$$\mathrm{d}_t \hat{G}^{(j)}=-\frac{\mathrm{i}}{\hbar} \hat{A}^{(j) \dagger} [\hat{H}_0^{(j)}(t)+\hat{H}_1^{(j)}(t)] \hat{C}^{(j)}(t) \\

=-\frac{\mathrm{i}}{\hbar} \hat{A}^{(j) \dagger} \hat{H}^{(j)}(t) \hat{C}^{(j)}(t) \\

=-\frac{\mathrm{i}}{\hbar} \hat{A}^{(j) \dagger} \hat{H}^{(j)}(t) \hat{A}^{(j)}(t) \hat{G}^{(j)}(t) \\

=-\hat{H}(0) \hat{G}^{(j)}(t).$$

Since also ##\hat{G}^{(j)}(0)=\hat{1}##, ##\hat{G}^{(j)}## fulfills the same EoM and initial condition for both ##j \in \{A,B\}## and thus

$$\hat{G}^{(A)}=\hat{G}^{(B)}$$

This implies

$$\hat{A}^{(A)\dagger} \hat{C}^{(A)} = \hat{A}^{(B)\dagger} \hat{C}^{(B)},$$

and thus after some algebra

$$\hat{B}^{(AB)}=\hat{C}^{(A) \dagger} \hat{C}^{(B)} = \hat{A}^{(A)} \hat{A}^{(B) \dagger} = \hat{D}.$$

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 ##o## when measuring the observable ##O## is governed by the picture-independent operator ##\hat{G}=\hat{G}^{(A)}=\hat{G}^{(B)}##. To see this let ##|o,\alpha \rangle## be a complete orthonormal set of eigenvectors of ##\hat{O}## with eigenvalue ##o##. Then the said probability is

$$P(o,t)={\sum_{\alpha}} {_{(j)}\langle o,\alpha,t|}\hat{\rho}^{(j)}(t)|\langle o,\alpha,t \rangle_{(j)} \\

=\sum_{\alpha} \langle o,\alpha,0|\hat{A}^{(j) \dagger}(t) \hat{C}^{(j)}(t) \hat{\rho}(0) \hat{C}^{(j) \dagger}(t) \hat{A}^{(j)}(t)|o,\alpha,0 \rangle \\

=\sum_{\alpha} \langle o,\alpha,0|\hat{G}(t) \hat{\rho}(0) \hat{G}^{\dagger}(t)|o,\alpha,0 \rangle.$$

From this one reads off that the picture-independent unitary time-evolution operator ##\hat{G}## can be interpreted as the time-evolution operator for the Schrödinger-picture ##\hat{\rho}^{(S)}## or ##\hat{G}^{\dagger}## as the time-evolution operator for the eigenstates of observable operators (and thus the observable operators themselves) in the Heisenberg picture.