The density operator is a representation of a state of a quantum system $\rho=|\psi\rangle\langle\psi|$, so it's just an alternative characterisation of a state (or more generally a statistical mixture of states). On the other hand, the density operator is a Hermitian operator on a Hilbert space and as such it can be thought of as representing some observable, i.e. some physical property. What is that physical property?


1 Answer 1


Here's a possible way of thinking of the measurement defined by a general density operator $\rho$, not necessarily one representing a pure state $\lvert\psi\rangle$. If $\mathcal H\simeq \mathbb C^d$ is the Hilbert space under consideration and $\rho$ is a density operator on $\mathcal H$, i.e. a $d\times d$ positive semidefinite Hermitian matrix with trace $1$, then it can be decomposed into rank-one matrices via its singular value decomposition like this: $$\rho = \sum_{i=1}^d \sigma_i \lvert u_i\rangle\langle u_i\rvert$$ where the $\sigma_i=\sqrt{\lambda_i}$ are its singular values and the $\lvert u_i\rangle$ are an orthonormal basis of singular vectors. Using the interpretation you mentioned of a density operator as a "statistical mixture" of states, the above expresses $\rho$ as the density matrix of a statistical mixture of the orthonormal states $\lvert u_i\rangle$ with respective statistical frequencies $\sigma_i^2 = \lambda_i$.

Now, performing a measurement with respect to the operator $\rho$ on a qubit in the state $\lvert\psi\rangle$ is almost like performing a full measurement with respect to the basis $\{\lvert u_i\rangle\}$, except that there's a possibility of some of the probabilities $\lambda_i$ being equal, meaning that performing a measurement with respect to $\rho$ gives us strictly less information than a full measurement with respect to this basis of singular vectors. Because this measurement "groups together" the singular vectors with the same statistical frequencies $\lambda_i$, making them indistinguishable given the result of the measurement, we might think of a measurement by $\rho$ as telling us how unlikely of a result we would get if we performed a full measurement, without actually performing one. To be more precise, the probability of observing $\lambda$ when measuring $\lvert\psi\rangle$ with $\rho$ is the same as the probability of finding $\lvert\psi\rangle$ to be in one of the states with statistical frequency $\lambda$ after doing a measurement in the full basis.

To give an example: consider $\mathcal H \simeq \mathbb C^3$ and the following density matrix: $$\rho = \tfrac{1}{100}\lvert 0\rangle\langle 0\rvert + \tfrac{1}{100}\lvert 1\rangle\langle 1\rvert + \tfrac{49}{50}\lvert 2\rangle\langle 2\rvert$$ which represents a statistical mixture consisting of $1\%$ of qutrits in state $\lvert 0\rangle$, $1\%$ of qutrits in state $\lvert 1\rangle$, and $98\%$ of qutrits in state $\lvert 2\rangle$. Performing a measurement by $\rho$ on a qutrit in state $\lvert\psi\rangle$ does not necessarily tell us what state it will end up in - in particular, if we obtain a result of $\tfrac{1}{100}$ from this measurement, we still do not know whether, say, it is in the state $\lvert 0\rangle$ or $\lvert 1\rangle$ (it could be any of infinitely many combinations of these two basis states, i.e. any unit eigenvector with eigenvalue $\tfrac{1}{100}$). However, in this case we would know that, should we proceed to perform a full measurement in the basis $\{\lvert u_i\rangle\}$ on this qutrit, we would find it to be one of the vectors with statistical frequency $\tfrac{1}{100}$ in the mixture represented by $\rho$, i.e. either $\lvert 0\rangle$ or $\lvert 1\rangle$.

To summarize, a measurement by $\rho$ on the state $\lvert\psi\rangle$ can tell us intermediate information about the common-ness (in the statistical mixture represented by $\rho$) of the state we will find it to occupy after measuring completely with respect to $\{\lvert u_i\rangle\}$.

NOTE: An analogous situation in classical probability would be to consider the how the distribution of a random variable $X$ (say, with a finite sample space) changes when you are given the probability of its result. The probabilities of each possible outcome without knowing this information are given by $p(x) = \mathbb P(X = x)$, but the probabilities of each possible outcome knowing the probability of the specific outcome are given by $\mathbb P(X = x ~ \lvert ~ p(X))$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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