# Is $\rho | \psi \rangle$ invariant in the Wigners friend thought experiment?

## Background

Let's say I have a gas of $$N$$ particles where I cannot distinguish between the particles at a temperature $$T$$.

Its density matrix is given by $$\rho$$. Note, if my friend happens to measure the energy of all $$N$$ particles for large $$N$$ the system can still be described by the density matrix $$\rho$$.

To see this we can write the density matrix in the energy eigenbasis:

$$\rho =\sum_{i=0} p_i |E_i \rangle \langle E_i |$$

where $$p_i$$ is likelihood of eigenstate of the $$i$$'th energy eigenstate and $$|E_i \rangle$$ is the energy eigenket. The measurement of energy effectively takes $$|E_j \rangle \to |E_j \rangle$$ due to orthogonality. Think of $$p_i$$ as the number of particles with energy $$E_i$$ out of $$N$$ particles.

My friend however can use the wavefunction $$|\psi \rangle$$ to describe the system.

Claim: Both me and my friend will agree on the quantity $$\rho |\psi \rangle$$ and its time evolution:

$$i \hbar \frac{\partial \rho | \psi \rangle}{\partial t} = H \rho |\psi \rangle$$

(Use the product rule to prove the above).

## Question

In the thought experiment of Wigner's friend is $$\rho |\psi \rangle$$ an invariant where $$|\psi\rangle$$ is what Wigner's friend measures and $$\rho$$ is the pure state Wigner uses to describe the system?