# In terms of covariance matrices, are partial measurement and partial trace equivalent?

Partial measurement and partial trace

There is a connection between a measurement of a part of a system and tracing this subsystem out. Say, we have a system composed of subsystems $A$ and $B$ in a pure state $|\psi\rangle = \sum_{i,j}a_{ij}|i\rangle_A|j\rangle_B$ and do a projective measurement on part $B$, projecting it on a state $|k\rangle_B$. The resulting state of subsystem $A$, conditioned on the measurement result, will be (up to normalization) $$|\psi^{(k)}\rangle_A = \sum_i a_{ik}|i\rangle_A.$$ Taking partial trace over the subsystem $B$ then corresponds to taking a statistical mixture of all possible measurement outcomes (this can be shown by a simple explicit calculation), $$\mathrm{Tr}_B(|\psi\rangle\langle\psi|) = \sum_{k}|\psi^{(k)}\rangle_A\langle\psi^{(k)}|.$$

Gaussian states

When dealing with Gaussian states in quantum optics [i.e., states with Gaussian phase space representation, where the phase space is spanned by quadrature operators $x = a+a^\dagger$, $p = i(a^\dagger-a)$], one often uses the description by first and second statistical moments. The first moment is, as far as entanglement is concerned, irrelevant and one then works just with the covariance matrix, i.e., the second moment. It can then be shown that a partial measurement affects the covariance matrix of the remaining subsystem in such a way that the resulting covariance matrix does not depend on the particular measurement result.

The question

Does this mean that, when dealing with covariance matrices of Gaussian states, partial measurement and partial trace are effectively the same? This is something one could intuitively (and perhaps naively) expect from the observations above. However, there is a difference in the purity of the state after partial measurement (pure for pure initial state) and partial trace (generally mixed). For Gaussian states, state purity is directly connected to determinant of the covariance matrix (the purity scales as $P\propto 1/\sqrt{\det\gamma}$, where $\gamma$ is the covariance matrix). Therefore, there must be a difference between covariance matrix after a partial measurement and partial trace. What am I missing?

-

## 1 Answer

The problem with the reasoning above is that it considers only the effect the measurement has on the covariance matrix and does not consider the displacement vector. While the covariance matrix of the resulting state does not depend on the particular measurement result, this is not true for the displacement vector. As a result, when doing the partial trace, one has to average over all possible measurement results with their respective weights. If we start with a Gaussian state, the measurement outcomes will have a Gaussian probability distribution, resulting in a Gaussian envelope or "blob" of identical Gaussian states. The properties of this Gaussian envelope (particularly its variance) will then affect the covariance matrix of the resulting state after partial trace, leading to a decrease in state purity.

In fact, the covariance matrix after partial trace is given by the respective block in the original covariance matrix. If we have a covariance matrix of the composite system in the block form given by $$\gamma_\mathrm{AB} = \left(\begin{array}{cc} A&C\\C^T&B\end{array}\right),$$ the covariance matrix of subsystem A after tracing out subsystem B is given by the block $A$.

-
Is there a place where this calculation is done explicitly? In particular, I'd like a rigid framework for translating back and forth between Gaussian POVMs and operations on the $(d,\gamma)$ representation, where $d$ is the first moment and $\gamma$ is the second. – Jess Riedel May 27 '14 at 20:52
@JessRiedel There are some excellent papers on the topic of general completely positive transformations of Gaussian states, see, e.g., this one or this one. Hope that helps! – Ondřej Černotík May 28 '14 at 7:18