If I have an ensemble distributed evenly among k states, I have k states {$|\psi_i \rangle$} where i$\in$ {1,...,k} with equal probabilities $$p_i = \frac{1}{k}$$ Defining the density matrix, $$\rho = \sum_{i=1}^k p_i|\psi_i\rangle \langle \psi_i| = \frac{1}{k}\sum_{i=1}^k |\psi_i\rangle \langle \psi_i|$$ My question is, can I say that the sum above is equal to the identity operator? Does the density operator only act on states within the ensemble or can it act on any state? Why do we even use such an operator (I.e. What is it's usefulness)?
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0 The density matrix is not really an observable and we rarely "act" with it on states . It is another representation of the state of the system which is also capable of handling statistical mixtures of states.
This is exactly the case here: Your state is randomly drawn from a number of states. You can't represent it by a vector in Hilbert space (those are only "pure states"). But you can do it using a density matrix - the one you wrote above.
Everything you usually do with a state vector, you can also do with the density matrix. For example, the expectation value of an observable $A$ given the state $\rho$ is given by $\mathrm{Tr}(\rho A)$. The Shrödinger equation is replaced by the von Neumann equation, etc.
Finally, $\rho$ is not neccessarily the identity. In fact, you can show that it's the identity if and only if the $\rho_i$ are an orthonormal basis.
