Probability that the outcome of a measurement coincides with a simple eigenvalue

Question

I have been working on the following problem during a Quantum Mechanics course I am taking as part of my postgraduate in Mathematics, and I feel I am missing something very simple.

Let $\xi\in\mathbb{C}^n$, $||\xi||=1$ and $P_\xi:\mathbb{C}^n\to\mathbb{C}^n$ be the projector on $\xi$. That is, $P_\xi\psi=(\psi,\xi)\xi$ for any $\psi\in\mathbb{C}^n$.

1. Show that $P_\xi$ is a density matrix.

2. Let $\omega$ be the state described by $P_\xi$. Show that the expectation $\langle A\rangle_\omega=(A\xi,\xi)$ for any $A\in\mathcal{A}$, where $\mathcal{A}$ is the space of linear Hermitian operators on $\mathbb{C}^n$.

3. Let $\lambda_j$ be a simple eigenvalue of $A$ and $\psi_j$, $||\psi_j||=1$ be an eigenvector corresponding to $\lambda_j$. Suppose that the observable $A$ is measured when the quantum system is in the state $\omega$ described in part $2$. Show that the probability that the outcome of the measurement coincides with $\lambda_j$ is equal to $|(\psi_j,\xi)|^2$.

I have managed to work my way through parts $1$ and $2$ but am unsure on how to proceed with part 3. Any hints would be greatly appreciated.

Answer 1

The system is in a pure state with density matrix $P_\xi$ (It's a pure state because $P_\xi=P^2_\xi$)

Working with density matrices is very useful to express in a compact form the result from a measurement for a wide class of initial states. Before the measurement it's said that the system is in the state $\omega$ described by $P_\xi$. In general, after a measurement of the observable represented by the hermitian operator $A$, the system is in a mixed state described by the new density matrix.

$\displaystyle P_{A}=\sum_i P_iP_\xi P_i$

Where $P_i$ are the projectors on $\psi_i$, the eigenstates of $A$, so is $P_i\psi=(\psi,\psi_i)\psi_i$

We now evaluate $P_{A}$ on some state $\psi$ to find some equivalence,

$\displaystyle P_{A}\psi=\left(\sum_i P_iP_\xi P_i\right)\psi=\sum_i P_iP_\xi P_i\psi=$

$=\displaystyle\sum_i P_iP_\xi(\psi,\psi_i)\psi_i=\sum_i(\psi,\psi_i)P_iP_\xi\psi_i=$

$=\displaystyle\sum_i(\psi,\psi_i)P_i(\psi_i,\xi)\xi=\sum_i(\psi,\psi_i)(\psi_i,\xi)P_i\xi=$

$\displaystyle=\sum_i(\psi,\psi_i)(\psi_i,\xi)(\xi,\psi_i)\psi_i=\sum_i(\psi,\psi_i)(\psi_i,\xi)(\psi_i,\xi)^*\psi_i$

$\displaystyle=\sum_i\vert(\psi_i,\xi)\vert^2(\psi,\psi_i)\psi_i=\sum_i\vert(\psi_i,\xi)\vert^2P_i\psi=\left(\sum_i\vert(\psi_i,\xi)\vert^2P_i\right)\psi$

So is $\displaystyle P_{A}=\sum_i\vert(\psi_i,\xi)\vert^2P_i$

We have as result of this measurement a mixed state of eigenstates of $A$ with weights $\vert(\psi_i,\xi)\vert^2$

From the definition of density operator, $\displaystyle P_{A}=\sum_ip_iP_i$, understood the weight $p_i$ as the probability for the system of being in the state $\psi_i$, we conclude that $p_i=\vert(\psi_i,\xi)\vert^2$