Exercise 8.2

Consider a quantum measurement \(\{M_m\}\) with outcome labeled by \(m\), where \(M_m\) is not a unitary operator. For a pure state \(|\phi\rangle\), the probability of measuring outcome \(m\) is \(p(m) = \langle \phi|M^{\dagger}_mM_m|\phi\rangle\). Here, the expectation value of operator \(M^{\dagger}_m M_m\) can be written as

(1)\[ \langle \phi|M^{\dagger}_mM_m|\phi\rangle = {\rm tr}\left( M^{\dagger}_mM_m \rho_{\phi}\right) = p(m) \]

where \(\rho_{\phi} = |\phi\rangle\langle \phi|\). The post-measurement state \(|\phi'\rangle\) is given by

(2)\[ |\phi'\rangle = \frac{M_m|\phi\rangle}{\sqrt{p(m)}} = \frac{M_m|\phi\rangle}{\sqrt{\langle \phi|M^{\dagger}_mM_m|\phi\rangle}} \]

which can be also written in the form of density matrix,

(3)\[ \rho'_{\phi} = |\phi'\rangle \langle \phi'|= \frac{M_m|\phi\rangle \langle \phi| M^{\dagger}_m}{p(m)} = \frac{M_m\rho_{\phi} M^{\dagger}_m}{{\rm tr}\left( M^{\dagger}_mM_m \rho_{\phi}\right)} \]

Probability of measurement

Suppose density matrix \(\rho\) describes an ensemble of pure states \(\{q(j), |\psi_j\rangle\}\),

(4)\[ \rho = \sum_j q(j)|\psi_j\rangle\langle \psi_j | = \sum_j q(j)\rho_j \]

where \(\rho_j = |\psi_j\rangle\psi_j|\). The probability of outcome \(m\) after measuring the ensemble is given by

(5)\[ P(m) = \sum_j q(j) p(m|j) = \sum_j q(j) {\rm tr}\left( M^{\dagger}_mM_m \rho_{j}\right) \]

This is the product of probability of “getting pure state \(|\psi_j\rangle\)”, or \(q_j\), and the probability of “measuring \(m\) for \(|\psi_j\rangle\)”, or \(p(m|j)\). Here, \(p(m|j)\) is obtained from eq. (1). One can also rewrite eq. (5) as

(6)\[\begin{split} \begin{split} P(m) =& \sum_j q(j) {\rm tr}\left( M^{\dagger}_mM_m \rho_{j}\right) = \sum_j q(j) {\rm tr}\left(\rho_{j} M^{\dagger}_mM_m \right) \\ =& {\rm tr}\left[\sum_j q(j)\rho_{j} M^{\dagger}_mM_m \right] \\ =& {\rm tr}\left(\rho M^{\dagger}_mM_m \right) = {\rm tr}\left(M_m\rho M^{\dagger}_m \right) = {\rm tr}\left(\mathcal{E}(\rho) \right) \\ \end{split} \end{split}\]

where \(\mathcal{E}(\rho) = M_m\rho M^{\dagger}_m\).

Post-measurement state

After measurement, the new ensemble is described by a new probability distribution and post-measurement states. From eq. (3), the normalized post-measurement state \(\rho'_j\) for each \(|\psi_j\rangle\) is given by

(7)\[ \rho'_{j} = \frac{M_m \rho_j M^{\dagger}_m}{{\rm tr}\left( M^{\dagger}_mM_m \rho_{\phi}\right)} = \frac{M_m\rho_{j} M^{\dagger}_m}{p(m|j)} \]

The probability of getting \(\rho'_j\) is \(q(j)p(m|j)\). However, \(\{q(j)p(m|j)\}\) does not give a valid probability distribution since \(\sum_j q(j)p(m|j) = P(m) \) is NOT EQUAL TO ONE in general. As a result, we re-normalize and get a new probability distribution for post-measurement ensemble as \(\{q(j)p(m|j) / P(m)\}\). In this context, the density matrix that describes post-measurement ensemble is given by

(8)\[\begin{split} \begin{split} \rho' =& \sum_{j} \frac{q(j)p(m|j)}{P(m)} \rho'_{j} \\ =& \sum_{j} \frac{q(j)p(m|j)}{P(m)} \frac{M_m\rho_{j} M^{\dagger}_m}{p(m|j)} \\ =& \sum_{j} \frac{q(j)M_m\rho_{j} M^{\dagger}_m}{P(m)} = \frac{M_m \rho M^{\dagger}_m}{P(m)} \end{split} \end{split}\]

which is equivalent with \(\mathcal{E}_m (\rho) / {\rm tr}\left(\mathcal{E}(\rho) \right)\).