# Comparing the purity of the unconditional state of a system (post projective measurement) with that of the a-priori state

The context of the question:

This is exercise 1.9 in the book Quantum Measurement and Control (Wiseman, H. and Milburn, G., 2010).

We have some operator acting on our system, $$\hat{\Lambda}$$ corresponding to a physical observable. We also have our projection operators $$\hat{\Pi}_\lambda$$ which project onto the subspace of eigenstates with eigenvalue $$\lambda$$. The state of the system before measurement is given by the density matrix $$\rho(t)$$ and the unconditional state post measurement (i.e. the result of the measurement isn't recorded) is given by: $$\rho(t+T) = \sum_{\lambda}\hat{\Pi}_{\lambda}\rho(t)\hat{\Pi}_{\lambda}$$.

We have to show that the projective measurement of $$\hat{\Lambda}$$ decreases the purity $$Tr[\rho^2]$$ of the unconditional state, unless if the a-priori state $$\rho(t)$$ can be diagonalized using the same basis $$\hat{\Lambda}$$ can.

My attempt at the solution:

So my understanding is that we have to show that $$Tr[\rho(t)^2] \geq Tr[\rho(t+T)^2]$$, where equality only holds if $$\rho(t)$$ can be diagonalized using the same basis $$\hat{\Lambda}$$ can be diagonalized with. The hint we are given is to first of all define $$p_{\lambda, \lambda^{'}} = Tr[\hat{\Pi}_{\lambda}\rho(t)\hat{\Pi}_{\lambda^{'}}\rho(t)]$$, show that this is greater than or equal to zero across all values of $$\lambda, \lambda^{'}$$ and then write the respective traces in terms of these $$p_{\lambda, \lambda^{'}}$$.

I am able to derive the following:

$$Tr[\rho(t+T)^2] = Tr[\sum_{\lambda,\lambda^{'}}\hat{\Pi}_{\lambda}\rho(t)\hat{\Pi}_{\lambda}\hat{\Pi}_{\lambda^{'}}\rho(t)\hat{\Pi}_{\lambda^{'}}] = Tr[\sum_{\lambda,\lambda^{'}}\hat{\Pi}_{\lambda}\rho(t)\hat{\Pi}_{\lambda}\delta_{\lambda,\lambda^{'}}\rho(t)\hat{\Pi}_{\lambda^{'}}] = Tr[\sum_{\lambda}\hat{\Pi}_{\lambda}\rho(t)\hat{\Pi}_{\lambda}\rho(t)\hat{\Pi}_{\lambda}]$$

The trace would be over the orthonormal eigenvectors of the physical observable hence I arrive at:

$$Tr[\rho(t+T)^2]= \sum_{\lambda}^{}{ p_{\lambda,\lambda}}$$

I struggle however to write $$Tr[\rho(t)^2]$$ in terms of these $$p_{\lambda, \lambda^{'}}$$ and am wondering if I'm missing a really obvious trick? Or whether the result I have obtained above is incorrect and I'm approaching this in a wrong manor?

Any help and guidance would be very appreciated, if more detail is needed please don't hesitate to ask.

Hint: Write the projectors in terms of vectors, like this: $$\hat\Pi_\lambda=\sum_n|\lambda_n\rangle\langle\lambda_n|$$, where the $$|\lambda_n\rangle$$ are orthonormal basis vectors for the subspace onto which $$\hat\Pi_\lambda$$ projects.
The identities $$\sum_\lambda\hat\Pi_\lambda=1$$ and $$\text{Tr}(\hat\Pi_\lambda X)=\sum_n\langle\lambda_n|X|\lambda_n\rangle$$ are also useful.