# Does Unitary operator take a pure state to a pure state or can it take a pure state to a mixed state? [closed]

Does Unitary operator take a pure state to a pure state or can it take a pure state to a mixed state? I think so but why? I assume the Unitary operator acts on a pure state only.

## closed as off-topic by Norbert Schuch, CuriousOne, user36790, Martin, David HammenJun 1 '16 at 18:40

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• It seems this is only possible with nonlinear QM. For example, in the D-CTC model, this is possible. – XXDD Jun 3 '16 at 2:54

A unitary operator $U$ can only take a pure state $|\psi \rangle$ to a pure state. Let $U | \psi \rangle = | \psi' \rangle$. Then acting the unitary $U$ on the pure state $| \psi \rangle$ yields

$$U \left( | \psi \rangle \langle \psi | \right) U^\dagger = | \psi' \rangle \langle \psi' |,$$

which is a pure state.

Another way to see this is that a (normalized) density matrix $\rho$ is pure iff its "purity" $\mathrm{Tr}\, \rho^2 = 1$. Acting with a unitary $U$ gives a new density matrix with

$$\mathrm{Tr}\, \left[ \left( U \rho U^\dagger \right)^2 \right] = \mathrm{Tr}\, \left[ U \rho U^\dagger U \rho U^\dagger \right] = \mathrm{Tr}\, \left[ U \rho^2 U^\dagger \right] = \mathrm{Tr}\, \left[ \rho^2 U^\dagger U \right] = \mathrm{Tr}\, \left[ \rho^2 \right],$$

so unitary operators preserve the purity of a state, and the new state is pure iff the original one was. (In fact, unitary operators preserve the entire eigenvalue spectrum of $\rho$, sometimes called the "Schmidt weights" or the "entanglement spectrum".)

(This simple fact lies at the heart of several open areas of physics research. For example, it is not a priori obvious how a system initially in a pure state could ever thermalize under (unitary) Schrodinger time evolution, since the thermal density matrix $e^{-\beta H}/Z$ is mixed at finite temperature. The "Eigenstate thermalization hypothesis" attempts to address this issue by postulating that a typical energy eigenstate of a non-integrable Hamiltonian "looks" like that of a thermal mixed state for local operators, even though globally it remains a pure state. Another apparent contradiction is the "black hole information paradox": the formation of a black hole in an initially pure state would naively seem to imply that it evolves into a mixed state as you lose track of the information that falls in - but this is impossible under unitary time evolution.)

It seems that by "operator" you mean a time evolution operator $\exp\left(\frac{i}{\hbar}\,H\,t\right)$ where $H$ is a quantum system's Hamiltonian, and such an operator by definition always maps (acts on) a pure quantum state to another pure quantum state. Unitary evolution is what happens whenever quantum measurement doesn't. So your statemtent "I assume the Unitary operator acts on a pure state only" is the correct one.

Having said this, the unitary time evolution operator can be thought of as implicitly at work in the calculation of the evolution of a mixed state. Conceptually (i.e. the practise is different), to calculate such a state's evolution, we calculate the evolution of all the pure states in the mixture separately. Suppose we have a system of mixed pure states $\psi_k$ with probability $p_k$ to be in each and we want to know the measurement statistics when we impart an observable $\hat{A}$ after time $t$. We calculate the evolution conditioned on the assumption that the system is in in state $k$ to be $\exp\left(\frac{i}{\hbar}\,H\,t\right)\,\psi_k$. Then we calculate the statistics: the conditional $n^{th}$ moment (conditional on the assumption of the $k^{th}$ pure state) of the measurement will be

$$M_{n\,k}=\left.\left<\psi_k\,\exp\left(\frac{i}{\hbar}\,H\,t\right)\dagger\right.\right|\hat{A}^n\left|\left.\,\exp\left(\frac{i}{\hbar}\,H\,t\right)\,\psi_k\right>\right.$$

and then we sum all these conditional moments up as we do in classical statistics to get the overall $n^{th}$ moment:

$$M_n=\sum\limits_k p_k\,M_{n\,k}$$

In practise, however, it is much easier to simply calculate the density matrix $\rho = \sum\limits_k p_k\,|\psi_k\rangle\langle\psi_k|$, calculate this object's evolution by the Liouville-von Neumann equation:

$$i \hbar \frac{\partial \rho}{\partial t} = [H,\rho]$$

and then use the trace formula to get the overall $n^{th}$ moment from the evolved density matrix:

$$M_n = \mathrm{tr}(\rho\,\hat{A}^n)$$

These two procedures are readily shown to be equivalent.

Remember: notwithstanding its slightly misleading name "matrix" (with the word's connotations of "mapping" and "operator"), the density matrix records a mixed quantum state.

An interesting aside is that every mixed state of a finite dimensional quantum system can also be thought of as part of a pure state (a "reduced pure state") of a larger finite dimensional quantum system; look up the notion of quantum purification for more information.