Density operator matrices of pure and completely mixed ensembles

Question

I want to confirm that the density matrix corresponding to a pure ensemble depends on the basis you choose in the following sense and then extend the idea to completely mixed states:

We have the property $\text{tr}(\rho) = 1$. We also have that $\rho$ is idempotent for a pure ensemble, hence we have that $\rho(\rho-1) = 0$. Since we know that $\rho$ is an Hermitian operator, we know that it's matrix is Hermitian and therefore diagonalizable with regard to some basis. Hence we consider this set of basis elements as $\{ |b_n \rangle \}_n$, we then get the following: $$\rho \bigg( \sum_{b'} |b' \rangle \langle b' | \bigg) (\rho - 1) = 0 \implies \sum_{b'}\bigg[ \rho |b' \rangle\langle b' | (\rho -1) \bigg] \implies \sum_{b'}\bigg[ b'\ |b' \rangle \langle b' | (b' - 1) \bigg] = 0 \implies b' = 0~~~\text{or}~~~b' = 1.$$

Since $tr(\rho) = 1$, it follows that the matrix density operator is given in the form where you have $1$ on the diagonal and $0$ everywhere else (imagine it being infinitly large) $$ \rho = \left[ {\begin{array}{cc} 0&&&&\\ &0&& \\ &&1&& \\ & & & 0& \\ &&&& 0\\ \end{array} } \right] $$ But if you consider a pure ensemble $|\psi \rangle = \sum_{j}c_j|j \rangle$ where your density operator would then be $\rho = | \psi \rangle \langle \psi|$, then the diagonal entries in the basis $\{ |j \rangle \}$ would be $\rho_{jj} = \langle j| \rho| j \rangle = |c_j|^2$, thus not necessarily zero or $1$ on the diagonal.

Hence we would have different density matrices. Is my reasoning correct or am I missing something? How would I generalize the first proof to account for the fact that for a completely mixed state we get a matrix of the form $\frac{I}{N}$ (where $I$ is the identity matrix $N$ is the dimension of the vector state)?

Answer 1

No. I think that part of the confusion is with your definition of "different density matrices".

Whether a state is pure or mixed does NOT depend on the basis. Since the density matrix is hermitian, you can always diagonalize it and all the information is then encoded in the eigenvalues of $\hat \rho$.

What depends on the basis is the form of the density matrix, i.e. whether or not it is diagonal.

The density matrix is the density matrix and it is the same irrespective of the basis you use. The expression (i.e. the matrix elements) of the density matrix will depend on the basis, just like the components of a vector depend on the basis, but the vector itself is independent of the basis.

For instance, consider the (pure) state $\vert +;z\rangle$, which is an eigenstate of $\hat \sigma_z$ with eigenvalue $+1$. In a basis spanned by $\{\vert \pm;z\rangle\}$, the density matrix for this pure state is $$ \hat \rho_{\vert +;z\rangle}=\left(\begin{array}{cc} 1 & 0 \\ 0 & 0\end{array}\right) $$ but if you use instead a basis spanned by eigenstates of $\sigma_x$ the density matrix will look like $$ \hat \rho_{\vert +;z\rangle} = \frac{1}{2}\left(\begin{array}{cc} 1 & 1 \\ 1 & 1\end{array}\right) $$ This is the same density matrix, just expressed in two bases.

In the case of a density matrix of the form you suggest, $\hat \rho=\hat 1/N$, the density matrix is proportional to the unit matrix so it will remain unaffected by a change of basis, i.e. if $U$ is a unitary matrix effecting this change of basis, then $U\cdot\rho\cdot U^{-1}=\rho$