In areas of quantum mechanics related to information theory, open quantum systems, and foundations, I've seen the formalism of density matrices employed to model two distinct physical scenarios

  • Scenario 1: I have a quantum system in some pure state $|\psi\rangle\in\mathcal{H}_\text{system}$, corresponding to a density matrix $|\psi\rangle\langle\psi|$. This state can undergo decoherence by coupling with an environment, corresponding to a Hilbert space $\mathcal{H}_\text{env}$. Decoherence results in information loss, transforming the pure state into a mixed state $\rho$ by some CPTP map $\mathcal{E}$ that describes the decoherence $$ |\psi\rangle\langle\psi| \mapsto \rho := \mathcal{E}(|\psi\rangle\langle\psi|). $$ Here, a fundamental result is the dilation theorem, which implies the existence of a unitary $U$ acting on $\mathcal{H}_\text{system} \otimes \mathcal{H}_\text{env}$ whose restriction to $\mathcal{H}_\text{system}$ is the CPTP map $\mathcal{E}$.

  • Scenario 2: I have a quantum system in a state $|\psi\rangle \in \mathcal{H}$, and I apply a unitary $U$ from some set of unitaries $\{U_i\}_{i\in\mathcal{I}}$ (we'll assume $\mathcal{I}$ is a finite set for simplicity). However, for whatever reason, I don't know which unitary I applied, because maybe the unitary was chosen randomly or I simply forgot. Thus, the resulting state is now a mixed state $\rho$, which can be written as $$ \rho = \sum_{i\in\mathcal{I}}Pr(i) U_i |\psi\rangle\langle\psi|U_i^\dagger,$$ where $Pr(i)$ is the probability that $U_i$ was applied.

In other words, density matrices are used to model decoherence due to coupling with an environment, and simple epistemic certainty in an experiment.

My primary question is: is there any meaningful distinction in the application of the density matrix formalism to these two scenarios, either mathematically, physically, or otherwise? One of the reasons I am curious about this is, we can in principle apply the dilation theorem in scenario 2 to give us a unitary $U$ on some larger Hilbert space $\mathcal{H}\otimes\mathcal{H}_0$ whose restriction to $\mathcal{H}$ is the CPTP map that describes the application of a random unitary from $\{U_i\}_{i\in\mathcal{I}}$ to $|\psi\rangle$. This would seem to imply that both scenarios are physically the same, as the ''epistemic uncertainty'' map described in scenario 2 can be modelled as a decoherence process as described in scenario 1.


2 Answers 2


To elaborate on the other answer:

Both scenarios realize a quantum channel $$ \mathcal E:\rho\mapsto \sum M_k\rho M_k^\dagger\ . $$ In the first case, any trace-preserving completely positive map is allowed, the only condition is that $\sum M_k^\dagger M_k=I$.

The second case, however, is more special: There, the channel $\mathcal E$ can be written as a convex combinations of unitaries, $$ \mathcal E:\rho\mapsto p_k U_k\rho U_k^\dagger\ . $$ You can see that this is a special property already from the fact that $$ \mathcal E(I) = \sum p_k U_k U_k^\dagger = I\ , $$ that is, the identity matrix is a fixed point of $\mathcal E$: It is a unital channel. This is a special property (amounting to $\sum M_k M_k^\dagger=I$, which is not satisfied by a generic quantum channel).

You might wonder whether it the two cases are the same when you restrict case 1 to unital quantum channels, i.e. those with $\sum M_kM_k^\dagger=I$. This boils down to the question whether any unital channel is a convex combination of unitaries, termed the quantum Birkhoff conjecture (the quantum version of Birkhoff's theorem, which states that any doubly stochastic map is a convex combination of permutations). However, this conjecture turns out to be wrong except for qubits, see e.g. the introduction of https://arxiv.org/abs/1201.1172.

  • $\begingroup$ Just to confirm if my understanding of your response is correct, are you saying that any quantum channel satisfies $\sum_k E_k^\dagger E_k = I$, but only unital quantum channels satisfy both $\sum_k E_k^\dagger E_k = I$ and $\sum_k E_k E_k^\dagger = I$? $\endgroup$ May 16, 2021 at 17:53
  • 1
    $\begingroup$ Precisely._____ $\endgroup$ May 16, 2021 at 22:46

There is a small mathematical difference between the two situations.

In both situations, the evolution of the quantum system is given by a CPTP map. As you pointed out, the Stinespring dilation theorem ensures that both these evolutions can be represented as the partial trace of a unitary evolution in a larger Hilbert space. However, the Kraus decomposition reveals a difference. The CPTP map can be written :

$$\rho \mapsto \sum_k t_k \rho t_k^\dagger$$

where the $t_k$ satisfy the completeness relation $\sum_k t_k^\dagger t_k = 1$

In the second situation (epistemic uncertainty), $t_k = \sqrt{p_k} U_k$ are (proportional to) unitary operators, which need not be true in general. This is therefore a special case of the general CPTP evolution.


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