# What are constraints on a “purity” operator in quantum mechanics?

Consider the normalized state, written in some orthonormal basis as: $$\psi = A |0\rangle + B |1\rangle$$

Let's define a "purity operator" for a basis as any operator whose expectation value gives 1 for a pure state in this basis, and 0 for the most mixed state in this basis. Inbetween states should give between 0 and 1, although the specific value doesn't matter.

One possible example (please note my question is on the general case though, this specific example is just to aid discussion), is $$\langle \mathcal{O} \rangle = 1 - 4 \frac{|A||B|}{|A|+|B|}$$

What mathematically prevents such measurements in quantum mechanics?

• There is obviously no nontrivial "purity operator" on the Hilbert space because all states $|\psi\rangle$ in the Hilbert space are pure. You could define a "purity operator" on the space of density matrices but it wouldn't be a linear operator because one may get (and usually gets) mixed states by making superpositions of pure states. These non-existences shouldn't be surprising because "purity" is not an observable. An observable is something that you may measure in 1 repetition of an experiment, to get 1 number. Purity can't be measured in this sense. – Luboš Motl Apr 27 '11 at 5:07
• First of all, as others have mentioned below purity and mixedness are properties of density matrices, not pure states. I would think that the von Neumann entropy $S = - Tr \rho \ln (\rho)$ and related measures such as the mutual information $I(\rho_A,\rho_B)$ (a measure of the information shared by two systems $A$ and $B$) would be the starting point for finding a measure of the purity of such systems. – user346 Apr 27 '11 at 8:32

## 2 Answers

From the example you've given, it's clear that you're using the wrong terms to describe what you want. Purity and mixedness apply to density operators and not to state vectors -- if your system is described by a state vector $|\psi\rangle$, it is already pure.

What you seem to want to know is whether there is an operator that, in a particular basis, has an expectation value that lets you know to what degree the state is in a superposition of basis states. I think the uncertainty operator might work for you here. If $S$ is the operator whose eigenstates are $|0\rangle$ and $|1\rangle$, then the operator $(S-\langle S\rangle)^2$ has zero expectation value for the basis states and a non-zero value for all superpositions. You could scale this in some way to give you a value between 0 and 1.

The question (v1) abuses language slightly no matter how one understands it. Here I sketch three interpretations.

1) Interpretation in terms of a density matrix $\rho$:

Assume the operator ${\cal O}$ is diagonalizable. Since the eigenstates are pure states, the eigenvalues must be $1$. In other words, ${\cal O}$ is the identity operator. Then $\langle {\cal O} \rangle = \mathrm{tr}{\cal O}\rho=\mathrm{tr}\rho=1$ for any mixed or pure density matrix $\rho$.

2) Interpretation in terms of an operator ${\cal O}$ that satisfies

$$\langle 0 |{\cal O}| 0\rangle =1,$$ $$\langle 1 |{\cal O}| 1\rangle =1,$$ $$\langle \psi |{\cal O}| \psi\rangle =0 \qquad \mathrm{for~~all} \qquad |A|=|B|=\frac{1}{\sqrt{2}}.$$

Here

$$|\psi\rangle = A| 0\rangle+B| 1\rangle, \qquad 1=||\psi||^2 = \langle \psi|\psi\rangle =|A|^2+|B|^2 .$$

$|A|=|B|=\frac{1}{\sqrt{2}}$ is not what is traditionally meant by being "the most mixed state". Note that there are not just one of these states, but infinitely many pairs of complex numbers $(A,B)$ that satisfy $|A|=|B|=\frac{1}{\sqrt{2}}$. Even after removing an overall phase, there is a relative phase left.

Assume ${\cal O}$ is a Hermitian operator. Then ${\cal O}$ has a matrix of the form

$${\cal O} = \left[\begin{array}{cc} 1 & c \\ c^* & 1 \end{array}\right].$$

So

$$0=\langle \psi |{\cal O}| \psi\rangle = |A|^2+|B|^2+A^*cB+B^*c^*A = 1+2\mathrm{Re}(A^*cB).$$

But this is impossible for general complex phases of $A$ and $B$ with $|A|=|B|=\frac{1}{\sqrt{2}}$.

3) Interpretation in terms of an operator ${\cal O}$ that satisfies

$$\langle 0 |{\cal O}| 0\rangle =1,$$ $$\langle 1 |{\cal O}| 1\rangle =1,$$ $$\langle \psi |{\cal O}| \psi\rangle =0 \qquad \mathrm{for~at~least~one} \qquad |A|=|B|=\frac{1}{\sqrt{2}}.$$

The characteristic polynomial $p_{\cal O}(\lambda)$ for the matrix ${\cal O}$ reads

$$p_{\cal O}(\lambda)=\det({\cal O}-\lambda {\bf 1})= (\lambda-1)^2-|c|^2= (\lambda-1-|c|) (\lambda-1+|c|).$$

So the two eigenvalues of ${\cal O}$ are $\lambda=1\pm|c|$. Since the expectation values for ${\cal O}$ should be between $0$ and $1$, we must demand that $c=0$. In other words, ${\cal O}$ is the identity operator, which leads to the following inconsistency

$$0=\langle \psi |{\cal O}| \psi\rangle =\langle \psi |\psi\rangle =1.$$

• For (1), better yet, the expectation value of $\rho$ can be used as the desired measure since $\mathrm{tr} \rho\rho = 1$ iff $\rho$ is pure. And you get values between 0 and 1 for mixed states. But this is probably not what the OP is asking for. – dbrane Apr 26 '11 at 23:21
• The question is not ill-posed; instead it abuses the language slightly. Why don't you edit it so that it's correct? By the way, +1 for the correct solution. – Carl Brannen Apr 27 '11 at 3:00
• Dear @dbrane. Neat observation, although in interpretation (1), we first construct an operator ${\cal O}$ (which could be a density matrix itself), and we then probe with arbitrary $\rho$'s. Therefore the operator ${\cal O}$ is not allowed to depend on $\rho$. – Qmechanic Apr 28 '11 at 10:33