# Two distinct uses for the density matrix formalism

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.

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.

• 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 $\sum_k E_k E_k^\dagger = I$? – Solarflare0 May 16 at 17:53
• Precisely._____ – Norbert Schuch May 16 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.