What does it mean for a projection operator to represent a state? I can understand that an idempotent operator can be represented as a projection operator, such as $|x\rangle\langle x|$. But some authors seem to use projection operators, instead of vectors, to represent states (presumably only mixed states?).
I would like to have an intuitive grip on this because it just looks to me like a category mistake. How can an operator represent a state? Is there any easier way to make a good sense of this?
 A: You are correct that the density operator is most useful for mixed states, but it can also be used for pure states and it then contains the same information as a state vector. Below, I will show how to build the density operator for pure states, and give a few examples of what quantum mechanics looks like in terms of this operator compared to using state vectors.
Consider a state $|\psi\rangle$ written in the $\{u_i\}$ basis:
$$
|\psi\rangle=\sum_ic_i|u_i\rangle,
$$
where $c_i=\langle u_i|\psi\rangle$. I assume the basis is discrete for simplicity, but the argument can be extended to continuous bases. The corresponding representation of an operator is given by:
$$
\hat{A}=\sum_i\sum_jA_{ij}|u_i\rangle\langle u_j|,
$$
where $A_{ij}=\langle u_i|\hat{A}|u_j\rangle$ are the matrix elements of the operator in the $\{u_i\}$ basis. Therefore, the "projection operator" or "density operator" in the case of pure states, given by $\hat{\rho}=|\psi\rangle\langle\psi|$, has matrix elements:
$$
\langle u_i|\psi\rangle\langle\psi|u_j\rangle=c_i^{\ast}c_j=\rho_{ij}
$$
This density operator contains the same information as the state vector, and it is sufficient to fully characterize the state of the system. To see this, consider the following examples:

*

*Normalization. For the state vector normalization reads $\sum_i|c_i|^2=1$. This becomes:

$$
\sum_i|c_i|^2=\sum_i\rho_{ii}=\mathrm{Tr}(\hat{\rho})=1.
$$


*Expectation value of an operator. For the state vector, this reads $\langle\psi|\hat{A}|\psi\rangle=\sum_{ij}c_i^{\ast}c_jA_{ij}$. This becomes:

$$
\langle\psi|\hat{A}|\psi\rangle=\sum_{ij}c_i^{\ast}c_jA_{ij}=\sum_{ij}\langle u_j|\hat{\rho}|u_i\rangle\langle u_i|\hat{A}|u_j\rangle=\sum_j\langle u_j|\hat{\rho}\hat{A}|u_j\rangle=\mathrm{Tr}(\hat{\rho}\hat{A}),
$$
where I used the fact that $\sum_i|u_i\rangle\langle u_i|$ is the identity operator to remove the summation over $i$.


*As a final example, you can also show that the time evolution of the system becomes:

$$
i\frac{d\hat{\rho}}{dt}=[\hat{H},\hat{\rho}].
$$
