The most generic quantum dynamics can be non-unitary. In fact, according to the standard interpretation of quantum mechanics, non-unitary dynamics plays an important role, since the process of "wave function collapse" is non-unitary.

My question is: how does the Heisenberg picture work if the dynamics is non-unitary?

For example, consider a two-state system with basis states $|0\rangle$ and $|1\rangle$. Suppose in the Schr$\ddot{\rm o}$dinger picture the initial state is $|0\rangle$, and the final state is expressed as a density operator as $\rho=\left(|0\rangle\langle 0|+|1\rangle\langle 1|\right)/2$. In the Heisenberg picture, how should the various operators change?

More generally, what is the rule to determine the operator evolution in the Heisenberg picture when the dynamics is non-unitary? Ideally we should be able to read off the evolved operator in the Heisenberg picture as long as the initial and final states in the Schr$\ddot{\rm o}$dinger picture are given (it is not necessary to know the entire time evolution, but only the initial and final states).

  • $\begingroup$ Collapse is not the only source of non-unitarity that I have in mind. If we only consider projective measurements, then the process of collapse can be modeled as $\rho\rightarrow\sum_iP_i\rho P_i$, where $\rho$ is the initial state and $P_i$ is the projector onto an eigensubspace of the operator to be measured. In this case, it makes sense to claim that the operator $O$ evolves as $O\rightarrow P_iOP_i$ in the Heisenberg picture, although I have not thought about how to fully justify it. However, I do not know how the Heisenberg picture works for a generic non-unitary process. $\endgroup$ Commented Apr 23, 2020 at 12:19
  • $\begingroup$ My rule of determining how the operators evolve is just that ${\rm Tr}(\rho O)$ can be viewed either in a picture where the state evolves while the operator is fixed, or in the picture where the operator evolves while the state is fixed. ${\rm Tr}(\rho O)$ can be viewed as an inner product in the operator space, and in Schrodinger's picture it becomes ${\rm Tr}(C(\rho)O)$, where $C$ is a quantum channel. So in Heisenberg picture, to meet my requirement, one can keep $\rho$ invariant and find $C'(O)$, where $C'$ is the adjoint of $C$. In you setting, I would say it is $\sum_iP_iABP_i$. $\endgroup$ Commented Apr 23, 2020 at 13:41
  • $\begingroup$ Now I see what you mean in your question. Thanks. $\endgroup$ Commented Apr 23, 2020 at 19:43

1 Answer 1


In the Heisenberg picture the states do not evolve, but the observables do. Any evolution of a quantum state can be described by a quantum channel

$$\rho\mapsto\mathcal E(\rho) $$

with $\mathcal E$ a completely positive and trace preserving map. But notice that the only thing we can observe about a quantum state are, well, observables, or more precisely statistics of observables, described by the Born rule

$$ \langle A\rangle_\rho=\mathrm{Tr}(A\rho)$$

In the Schroedinger picture the observables stay fixed while the states evolves, so you go from $\langle A\rangle_\rho$ to

$$\langle A\rangle_{\mathcal{E}(\rho)}=\mathrm{Tr}(A\mathcal{E}(\rho)) $$

under the evolution. But remember that $\langle S,T\rangle := \mathrm{Tr}(ST^\dagger)$ is an inner product (called Hilbert-Schmidt inner product) on the vector space of hermitian matrices (or on a Hilbert space of bounded operators, I will restrict the discussion to finite dimension to avoid mathematical details), and $\mathcal{E}$ is a linear operator on this space, hence $\mathcal E$ has an adjoint map $\mathcal E^\dagger$ such that $\langle S, \mathcal E(T)\rangle = \langle \mathcal E^\dagger (S),T \rangle$ and hence $$\langle A\rangle_{\mathcal{E}(\rho)}=\mathrm{Tr}(A\mathcal{E}(\rho))=\mathrm{Tr}(\mathcal E^\dagger(A)\rho) =\langle \mathcal{E}^\dagger(A)\rangle_\rho$$

so here you have your answer: in the Heisenberg picture we let $\rho$ stay fixed and evolve every observable with the adjoint channel $\mathcal{E}^\dagger$.

There are many ways to find the adjoint of a channel, it is particularly easy for example if you have a Kraus representation

$$\mathcal E(\rho)=\sum_j K_j\rho K^\dagger_j $$ in which case it is clear that $$\mathcal E^\dagger(A)=\sum_j K_j^\dagger A K_j $$

In your particular example we have

$$\mathcal{E}(|0\rangle\langle0|) =\frac12(|0\rangle\langle0|+|1\rangle\langle1|)$$

there isn't a unique channel with this effect, without knowing the effect on the other basis states. This might for example be simply be an "information eraser" channel that takes any state $\rho$ and returns the maximally mixed state

$$\mathcal{E}(\rho)=\mathrm{Tr}(\rho)\,\frac12 \mathbb 1 $$

The Kraus operators of this channel are simply $K_{ij}=\frac{1}{\sqrt{2}}|i\rangle\langle j|$ for $i,j\in\{0,1\}$, hence $K_{ij}^\dagger=K_{ji}$ and it is not difficult to see this channel is self adjoint, i.e. $\mathcal E^\dagger(A)=\mathcal E(A)=\mathrm{Tr}(A)\frac12\mathbb 1$ (this is not true in general!), i.e. this evolution corresponds to take all the observables to trivial observables the average of which is just the average of their eigenvalues, $\frac12 \mathrm{Tr}(A)$.


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