# What physical quantity does a density operator represent as an observable?

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?

• Mar 11, 2023 at 13:23

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 $$\mathbb P(X = x ~ \lvert ~ p(X))$$.