Why is it not possible to describe a mixed quantum state by a Hilbert space vector?

Question

I read (for instance in Landau/Lifshitz III) that if I know the wave function of a quantum state, I have the maximal information of the state available, in different words, the description of the state is complete. I actually wondered what that means, but I guess, it means that if I apply a number of operators on the wave function of that state, like $\hat{H}$ or $\hat{{p}}$ or $\hat{L}_z$ or $\hat{s}_z$ I would get a (linear combination of) eigenvalue(s) of that operator ($\times$ the state vector), and indeed a lot of information of the state.

On the other hand, apparently I cannot do it with mixed state. Apparently I can't let act an operator on a mixed state. It already starts with the fact that a mixed state is mathematically described by a matrix, so it is no longer a (state) vector of a given Hilbert space. Why is it not possible to describe a mixed state by Hilbert state vector, why has it to be described by a matrix ? And what does it really mean to have a pure quantum state in difference to a mixed state ?

Answer 1

Imagine we have a system $X,$ which may be a subsystem of a larger composite system that is entangled, or simply a system about which we have insufficient information (or a noisy/lossy system), such that we don't really know in which state our system has been prepared in, we only know at most the set of states $|a\rangle, |b\rangle, \dots$ it could be in, each with some probability $p_{a}, p_{b}$ and so on, where $\sum_i p_i = 1.$ Now the question is, how do you characterize this system? Can you assign a definite state vector $|x\rangle$ to it? that is, describing the state in terms of a set of basis vectors of a Hilbert space. The short answer is, no: meaning that we cannot write down a ket/definite-state-vector $|x\rangle$ for $X$ in the form of $|x\rangle = \sum_n c_n |\phi_n\rangle$ where $\{|\phi_n\rangle\}$ is a basis and all $c_n$ coefficients are known and unique. Whenever we can write down such state, with all coefficients known, then we say our system is in a pure state, which means a definite ket in $\mathcal H,$ which means all the relative phases between the basis elements in the state are known, for instance it could be in a superposition of basis states with definite phase relations between the terms of the superposition (i.e., the $c_n$'s known).

The best one can do is describe $X$ as no more than what it is, namely, a mixture of states, which you can interpret as a statistical distribution, very much similar to the concept of ensembles in classical statistical mechanics, where adopting the same notation as before, $X$ is a mixed state which you can describe as an ensemble of various possible definite states $|a\rangle, |b\rangle, \dots,$ where then $p_a, p_b, \dots$ correspond to the fraction of the respective states in that ensemble. For mixed states, we have no choice but to describe the system in terms of density operators as opposed to kets in $\mathcal H,$ and to better understand density operators, let's assume we want to measure some observable $O$ of the system. We know the expectation value of $O$ for each state of the ensemble, e.g., for the state $|a\rangle,$ it is $\langle O\rangle_a = \langle a |O|a\rangle,$ extending this to the whole ensemble and knowing the fraction of times each state occurs in the ensemble, the average value of $O$ is the weighted average of all the expectation values: $$ \langle O\rangle = \sum_i p_i \langle O\rangle_i \tag{1} $$ which we can slightly rewrite using the trace (and using the linearity of the trace operation) \begin{align} \langle O\rangle &= \sum_i p_i \text{Tr }|i\rangle \langle i| O \tag{2} \\ &= \text{Tr } \sum_i p_i |i\rangle \langle i| O \\ &= \text{Tr } \rho O \tag{3} \end{align} $\rho$ is what we call the density operator, a description of $X$ that allows us to compute average values for any system observable. Note that the density operators are not limited to mixed states, we can also write down pure states in terms of them, e.g., if we know our system is in a definite state vector $|a\rangle,$ then the corresponding density operator is $\rho = |a\rangle \langle a|,$ which means the probability $p_a=1,$ in other words we know exactly in which state our system is, which you can further interpret as an ensemble where we only have copies of $|a\rangle.$ Inserting $\rho$ back into $(3),$ you can easily convince yourself that we retrieve the correct expectation value $\langle a |O|a\rangle$ for any observable $O.$

So you see with density operators, we have generalized our description of quantum systems to consistently include pure and mixed states. Now anything you know about statistical mixtures, e.g., in statistical mechanics, holds here as well. For instance, different mixtures of states (i.e., ensembles that involve different states) may lead to the same density operator, i.e., the different mixtures lead to the same expectation values for any observable, we say they are statistically indistinguishable. Take a simple qubit example, first an equiprobable mixture of states $|0\rangle, |1\rangle$ (with then $p_{0,1}=1/2$), and then an equiprobable mixture of superposition states $\frac{1}{\sqrt{2}}(|0\rangle \pm |1\rangle),$ and see what density operators you find for each mixture.

Answer 2

The state space of a mechanical system is a convex and compact set which is closed in the ultraweak topology. As such, the Krein-Milman theorem ensures that it is generated as the convex hull of the set of its extreme points. The extreme points correspond to pure states as they cannot be represented, by definition, as a non-trivial convex combination of two other points in the state space.

Now every state can be represented as a vector in a suitable representation (e.g. the GNS representation). However, pure states can be represented as vectors in irreducible representations. The Schroedinger representation, which is equivalent to the Heisenberg representation, is irreducible and therefore any mixed state, i.e. any non-trivial convex combination of pure states, cannot be represented by a single vector in the representation space. It is a general result that, for every state $\phi$ and an irreducible representation $\pi$ on a Hilbert space $H$, there is a trace-class operator with unital trace $\rho$ such that

$$\phi(O) = \operatorname{Tr}(\rho\pi(O))$$

for any observable $O$. It follows from the above that the rank of $\rho$ is 1 if and only if the state $\phi$ is pure.

Answer 3

You can think of a mixed state as a classical combination of quantum states. Let's say you have a system that flips a coin, if the coin is heads it produces an electron with $|z+ \rangle$ spin, if the coin is tails, it produces an electron with $|z-\rangle$ spin.

If you want to describe this system, you can not write it as a superposition of $|z+\rangle$ and $|z-\rangle$. You must write it as a density matrix $\frac{1}{2}\left(|z+ \rangle\langle z+| + |z- \rangle \langle z-|\right)$

For instance, you might want to try writing the following pure state to represent the electron:

$$|\psi\rangle = \frac{1}{\sqrt{2}} \left( |z+\rangle + |z-\rangle \right).$$

Measuring the $X$ operator, we'd find the expectation value is 1 because $|\psi\rangle$ is the $x+$ state.

Now consider the same measurement on the proper density matrix (i.e. 50% of the time you get $|z+\rangle$ and 50% of the time you get $|z-\rangle$), here the expectation value 0 because both $|z+\rangle$ and $|z-\rangle$ are equal superpositions of $|x+\rangle$ and $|x-\rangle$.

In fact, you can probably see with a bit of thought that the only way to get 50/50 measurements for the X component and 50/50 measurements for the Z component with a quantum superposition is to use the $|y+\rangle$ or $|y-\rangle$ states, but then we can extend this argument to include measurements of the Y component.

That is to say, there is no superposition state that gives 50/50 measurements for the X, Y, and Z spin operators, but the density matrix which encodes the equal classical mixture of spin z states can give such a distribution. If you'd like, you can take this as a proof by contradiction that you can represent any system as a pure quantum state.