Quantum purity $\gamma$ of a general state $\rho$, where $\rho$ is a density matrix, is defined as $\gamma = \text{tr} (\rho^2)$.

I'd like to understand how $\gamma$ can be measured experimentally. Is there a simple example of a system in which we can understand how $\gamma$ is measured by performing a set of more standard measurements (like measuring position, or momentum, or energy for instance)?

An equivalent way of asking this question is: what set of experiments would allow me to measure the eigenvalues of $\rho$?

  • $\begingroup$ Look up quantum state tomography. The details depend on what kind of system you have in consideration. For a two-level system it's easy. For a harmonic oscillator it's considerably more complicated. $\endgroup$ – DanielSank Oct 14 '15 at 7:40

It holds that $$ \mathrm{tr}(\rho^2)=\mathrm{tr}((\rho\otimes\rho)\mathbb F)\ , $$ where $\mathbb F$ ("flip" or "swap") is the operator which swaps two systems. Note that in order to measure the purity with a single measurement setting, it is necessary to measure two copies of $\rho$ simultaneously, as $\mathrm{tr}(\rho^2)$ is quadratic in $\rho$. (Alternatively, one can use tomography, which hoever requires many different measurement settings.)

A measurement of $\mathbb F$ can e.g. be realized by preparing a control qubit in a $\vert+\rangle =(\vert0\rangle + \vert1\rangle)/\sqrt{2}$ state, performing a controlled swap conditional on the control qubit being $1$, and subsequently measuring the control qubit. It is straightforward to see that the measurement outcome is directly related to the purity.

EDIT: A similar scheme is e.g. described in http://arxiv.org/abs/1201.2736, where the authors also quote some other papers describing ways to measure the purity in the introduction.

  • 2
    $\begingroup$ I wish I could give this more than one upvote. The only thing I don't like is "with a single measurement"; while certainly an average of these measurements is the purity, $\hat F$ is its own adjoint (hence unitary+Hermitian) with $\hat F^2=\hat1:$ its eigenvalues are $\pm1,$ and those are what you'll measure in one measurement. To measure that something's purity is in $[1-\epsilon, 1]$ you will therefore need something like $1/\epsilon$ measurements, but you can get away with less if it's considerably mixed. $\endgroup$ – CR Drost Oct 14 '15 at 16:05
  • $\begingroup$ @ChrisDrost I fully agree with you. What I meant to say was that a single measurement setting is sufficient, but of course, one has to measure repeatedly. This is in contrast with a tomography scheme, where one needs to measure in many different settings (and in each repeatedly). $\endgroup$ – Norbert Schuch Oct 14 '15 at 16:08

There is no way to measure $\rho$ or any function of its matrix elements such as $\gamma$ by a single measurement (in one repetition of the situation) because $\rho$ is a quantum counterpart of the probability distribution. It describes the observer's knowledge of the properties of the physical system. And the knowledge is intrinsically subjective, not objective, so it can't be measured.

The same comments apply to the wave function $\psi$. The wave function – and similarly the density matrix (which is the tensor square of the wave function or a linear combination of such squares) – isn't an observable so it isn't observable (without "a"), either. At most, one can make measurements and see that they statistically agree or statistically disagree with a given particular form of the density matrix. But without knowing a good candidate form of the density matrix, one can't measure its properties in isolation.

According to basic postulates of QM, everything that can be measured may be interpreted as an observable and each observable has to be represented by a well-defined particular linear Hermitian operator acting on the Hilbert space. Possible results of measurements are the eigenvalues and the probabilities of different outcomes are calculable by the usual formulae. The density matrix isn't a fixed linear operator, however, so it can't be measured. This rule, that only linear operators are physically meaningful and "measurable", is a totally important and universal axiom of all of quantum physics and is often ignored, like most other facts about quantum mechanics, but it's still totally right and crucial.

One may measure all of $\rho$ by doing infinitely many measurements on the repetitions of the same situation, however. This general strategy is referred to as the quantum state tomography. Just like all features of a probabilistic distribution, $\rho$ may be fully (increasingly accurately) reconstructed from the frequency counts resulting from such infinitely many measurements. $\gamma$ may be calculated from such a $\rho$ at the end.

There exist situations in which "well-behaved" states of some kind are mixed and measurements of $\rho$ or some features of $\rho$ similar to $\gamma$ may be "effectively" approximated by the measurements of some observables, and therefore become doable.

For example, one may assume that $\rho$ for a given system is thermal. And one may measure some $\gamma$-like function of $\rho$, the temperature, by the thermometer because this temperature may be transformed to the height of the level of mercury, or something like that. Quantum gravity is full of important examples like that because by the Hawking-Bekenstein insights, the entropy and energy are "equivalent" (for black holes) to geometric quantities such as the surface area of the event horizon and the gravitational acceleration at the event horizon.

But whenever some features of $\rho$, especially the $U(\infty)$-invariant features such as $\gamma$ (and the set of eigenvalues of $\rho$), may be measured without repeating the same situation, one needs to be sure that we're only working with some very specific forms of $\rho$, e.g. the thermal ones, described by a small number of parameters. If $\rho$ is a priori completely general, nothing about it may be measured by a single measurement.

  • $\begingroup$ $\rho$ is a linear, self-adjoint operator. Therefore it is an observable; there is no extra conditions in the definition of observable. When you say "The density matrix isn't a fixed linear operator, however, so it can't be measured.", can you make precise what you mean by "fixed"? Expectation value of an observable $A$ is $\text{tr} \rho A$, so taking $A=\rho$ looks fine to me. I feel like there must be an experiment in which the outcome is one of the eigenvalue of $\rho$. $\endgroup$ – Bru Oct 14 '15 at 8:22
  • $\begingroup$ No, Bru, $\rho$ isn't an observable or an operator on the space of states because an operator has to be well-defined, a particular one, that has the same matrix entries regardless of the states on which it may act. But by its essence, $\rho = |\psi \rangle \langle \psi |$ or a combination is state-dependent, it is a description of a (mixed) state, so even though it's mathematically a matrix, it can't be interpreted as a map from the Hilbert space to the Hilbert space. By "fixed", I mean independent of $\psi$ or $\rho$. $\endgroup$ – Luboš Motl Oct 14 '15 at 8:24
  • $\begingroup$ Measuring $\rho$ or its functions like $\gamma$ is exactly like "measuring" probabilities. Roll a dice, the six probabilities are 1/6 each, right? You can't measure those numbers 1/6 by rolling the dice once, can you? You can't determine whether all results are equally likely by one rolling, either. The case of $\rho$ and $\gamma$ is the same. $\endgroup$ – Luboš Motl Oct 14 '15 at 8:27
  • $\begingroup$ I agree with @Bru on the fact that, indeed, any density matrix $\rho$ is a linear self-adjoint operator. And therefore, at least theoretically, it could perfectly qualify as an observable. And, again theoretically, there exists a measuring process for it. It is actually a rigorous result: for any observable there exists at least one measuring process (à la von Neumann). This has been proved by Ozawa in a 1984 paper. Therefore, you can always make a thought experiment that measure density matrices. Now how this can be done in practice is all another, more complicated, problem. $\endgroup$ – yuggib Oct 14 '15 at 8:54
  • 3
    $\begingroup$ @Bru You can only measure an observable $A=\rho$ if you know what $\rho$ is. Your goal is however to determine the purity of an unknown $\rho$. As I have described above, this requires measuring an observable on $\rho\otimes\rho$, which is quadratic in $\rho$ just as $\mathrm{tr}(\rho^2)$. $\endgroup$ – Norbert Schuch Oct 14 '15 at 11:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.