# Density Matrix

1. Jan 23, 2011

### Nusc

I have a question regarding the slide:

On page 18-21 it gives the proof of the theorem that $$| \psi_i^{~} \rangle$$ and $$|\phi_{i}^{~}\rangle$$ generate the same density matrix iff $$|\psi_{i}^{~}\rangle = \sum_{j} u_{ij} |\phi_{j}^{~}\rangle$$ assuming that $$| \psi_i^{~}\rangle$$ is not necessarily normalized.
What if $$| \psi_i^{~}\rangle$$ is normalized and $$| \phi_i^{~}\rangle$$ not independent?
Would the necessary condition for which $$p = | \psi_i \rangle \langle \psi_i |= q = | \phi_j \rangle \langle \phi_j |$$ require that you have $$|\psi_{i}^{~}\rangle = \sum_{j} u_{ij} |\phi_{j}^{~}\rangle$$ ?
We know for normalized states psi and phi that $$p = | \psi_i \rangle \langle \psi_i |= q = | \phi_j \rangle \langle \phi_j |$$ iff $$\sqrt{p_{i}} | \psi_i \rangle = \sum_j u_{ij} \sqrt{q_j} | \phi_j \rangle$$