How to show a density matrix is in a pure/mixed state?

Say we have a single qubit with some density matrix, for example lets say we have the density matrix $$\rho=\begin{pmatrix}3/4&1/2\\1/2&1/2\end{pmatrix}$$. I would like to know what is the procedure for checking whether this state is pure or mixed.

I know that a pure state is one which can be written $$\rho=|\psi\rangle\langle\psi|$$, and that a mixed state is one which can be written $$\rho=\sum_{k=1}^Np_k|\psi_k\rangle\langle\psi_k|$$. But I have two problems. Firstly I'm not sure what the $$p_k$$ is, I know it's called the weight function but I don't understand it significance, how to find it, or its role mathematically. Secondly I don't see how it is possible to determine from these definition whether the sate I mentioned above is pure or mixed. How can we? Or perhaps there's some method that doesn't use the definitions directly?

Could anyone please clear up this problem for me ?

• Is it possible that you instead mean the density matrix $\rho=\begin{pmatrix}3/4&1/2\\1/2&1/4\end{pmatrix}$? – Mithrandir24601 Apr 22 at 9:35
• @Mithrandir24601 That $\rho$ isn't positive semi-definite. – DaftWullie Apr 23 at 7:31

However, if all you want to do is determine if the state is mixed, there’s a simpler way: calculate the trace of the square of the density matrix, $$\text{Tr}(\rho^2)$$. That's called the purity. If it’s 1, the state is pure. If it’s less than 1, the state is mixed. If it’s more than 1, you’ve messed up.
By spectral theorem density matrices are diagonizable, since they are hermitian (also they are positive semi-definite and have trace 1). That means that there is a set of $$n$$ non-negative eigenvalues $$\lambda_i$$ with $$n$$ corresponding mutually orthogonal eigenvectors $$|v_i\rangle$$ such that $$\rho = \sum_{i=1}^n{\lambda_i |v_i\rangle \langle v_i|}$$ This matrix represents pure state only if it has exactly one non-zero eigenvalue (it must be equal to 1 since $$\mathrm{tr}(\rho)=\sum_{i=1}^n{\lambda_i}=1$$).
Though, to check if $$\rho$$ is pure it is enough to verify the equality $$\mathrm{tr}(\rho^2)=1$$