# What are pure states and density operators?

I am not sure that I precisely understand what pure states are. It is my understanding that:

The principle behind density operators is that the state of a quantum mechanical system is described by a probability distribution on some subset $\{\phi_i\}$ of its Hilbert space $\mathcal{H}$ called the pure states. The pure states are not necessarily an orthonormal basis for the space.

The density operator, $\rho$, is one where if $\psi\in \mathcal{H}$ is some state then $\langle \psi | \rho|\psi\rangle$ equals the probability of the system exhibiting the statistical behavior described by $\psi.$ $\rho$ is often represented in terms of the pure states: $$\rho = \sum_i p_i |\phi_i\rangle \langle \phi_i|$$ One special thing about this representation of $\rho$ that the $p_i$ are real, nonnegative probabilities. In particular, the above equation is not a complex linear combination. (In contrast to, say, its representation in terms of other bases for $\mathcal{H}$).

We say that a system is in a pure state if $\mathrm{Tr}\rho=1$ and it is in a mixed state if $\mathrm{Tr}\rho<1$

I think I might be missing the point of why some states are called 'pure' while others are identified as 'mixed.'

Is the choice of pure states for any given quantum mechanical system unique or at least canonical? If so then how?

## 1 Answer

Intuitively, a pure state is a state that "lies entirely inside" the Hilbert space, i.e. the system it describes is not entangled with anything outside of the Hilbert space. For example, for two spins entangled in the singlet configuration, the state of the two spins together is a pure state, because the two spins are not entangled with anything else. But if we consider the Hilbert space of just one of the spins, then we can't describe that one spin by a pure state in that reduced Hilbert space, because that spin is entangled with the other spin.

Mathematically, the pure states are just the individual vectors in the Hilbert space (or technically, the one-dimensional subspaces consisting of a vector and all its constant multiples, reflecting the freedom to choose a non-normalized wavefunction or to multiply by a constant phase). The mixed states correspond to positive-semidefinite Hermitian operators on the Hilbert space. Mixed states are sometimes described as "classical combinations" as opposed to "quantum superpositions" of pure states, but frankly that explanation never gave me any intuition until I understood the math behind it.

• Alright so if i understand properly, the 'state' of a system is actually a Hermitian operator on $\mathcal{H}$, not an element itself. A pure state is an operator of the form $|\psi \rangle \langle \psi |$ for some $\psi \in \mathcal{H}$. Is that correct? – enthdegree Nov 13 '16 at 4:27
• Yes, that's exactly right. In the special case of a pure state $| \psi \rangle$ (not entangled with anything external), the state $| \psi \rangle \langle \psi |$ is just the projection operator into the subspace of constant multiples of $| \psi \rangle$. But in the simple case of a pure state, you don't really need to use this projector - you can just get away with working with the ket $| \psi \rangle$ itself. – tparker Nov 13 '16 at 4:31
• @enthdegree If you found this answer satisfactory, you should accept it so that Stack Exchange changes recognizes it as "answered." – tparker Nov 22 '16 at 21:27