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.


1 Answer 1


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.


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