# Density operators in 2 dimension

Consider any density operator in two dimension. Call it $A$. Let $I$ be the identity matrix, and $\sigma_i, i=x,y,z$ be the Pauli Matrices. Then we have to show that $$A=\frac{1}{2}(I+n\cdot \sigma)$$ where $n=(n_1,n_2,n_3)$ satisfies $|n|\le 1$. Also, show that $A$ is idempotent iff $|n|=1$.

My attempt: The fact that $A$ is self adjoint and has trace $1$ is trivial. To show that $A$ is positive definite, I multiplied $(x,y)A(x,y)^T$ and this comes out to be $x^2(\frac{1+n_3}{2})+y^2(\frac{1-n_3}{2})+n_1xy$ which has to be positive for all $x,y$. Then we must have $n_1^2-4\cdot \frac{1-n_3}{2}\cdot \frac{1+n_3}{2}\le 0\implies n_1^2+n_3^2\le 1$

I don't understand why don't I have any factor of $n_2$ here, I believe I am almost there, but I don't understand my mistake.

It's a much simpler issue actually:

The hermitian conjugate of $(x\; y)^T$ is $(x^*\; y^*)$ not $(x\; y)$. Substitute and your result should read $$(x^*\; y^*) A (x \; y)^T = \frac{1+n_3}{2}|x|^2 + \frac{1-n_3}{2}|y|^2 + \frac{n_1 - i n_2}{2}x^*y + \frac{n_1 + i n_2}{2}xy^*\; \ge 0$$ etc.

You're missing the additional requirement that

$$\mathrm{Tr}\left(\rho^2\right) \leq 1$$

This becomes an equality for pure states, since for mixed states $\rho_{ii} < 1$ but for pure states $\rho_{ii} = 1$ or $0$.

Squaring your density matrix adds $n_2$ terms to the diagonals, so the trace has an $n_2$ term, from which your result follows.

The requirement of idempotence then follows easily by requiring that all elements in the squared matrix equal those of the original matrix.