# Generalized measurement interpreted as entanglement unitary + measurement. Why is it indeed a unitary?

A generalized measurement can be defined as follow:

It is a set of operators $$\{M_m \}$$ such that, given an initial quantum state $$|\psi\rangle$$, we have the state after measurement being:

$$|\psi\rangle \rightarrow \frac{M_m |\psi\rangle}{\langle \psi | M_m^{\dagger}M_m |\psi\rangle}$$

The probability to find the outcome $$m$$ is: $$p(m)=\langle \psi | M_m^{\dagger}M_m |\psi\rangle$$

We can interpret the generalized measurement as entanglement unitary between a system and an ancillary system, and then a measurement on this ancillary system via the following:

$$U |\psi \rangle |0 \rangle = \sum_m M_m |\psi \rangle |m \rangle$$

My question is: how to prove that this $$U$$ is indeed a unitary ? I guess we also have to define its action on other states than $$|0\rangle$$ in the ancillary system, but how to do it ?

It is not necessary to define the action of $$U$$ on other states. It is sufficient to know that the defined action is compatible with being a unitary (i.e. inner produces are preserved between all inputs) because there always exists an extension of the defined action such that it is unitary. So, all you really have to check is that $$\left(\langle\psi|\langle 0|\right)\left(|\phi\rangle|0\rangle\right)=\left(\langle\psi|\langle 0|U^\dagger\right)\left(U|\phi\rangle|0\rangle\right)=\sum_n\langle\psi|M_n^\dagger\langle n|\sum_mM_m|\phi\rangle|m\rangle.$$ The right-hand side simplifies to $$\sum_m\langle\psi|M_m^\dagger M_m|\phi\rangle=\langle\psi|\phi\rangle$$ thanks to the completeness relation. This is all you need. Of course, it can be a useful exercise, for a specific case, to find the extension.
• Hello. Thank you for your answer. I'm still a little confused. In your calculation you prove that the inner product is conserved on $\langle \psi, 0|\phi, 0 \rangle$. Said differently, on all $\langle i, 0 |, |j, 0 \rangle$ where $(i,j)$ indices represent orthonormal basis of the first system. Do you mean that then, we just have to make use of the fact it is always possible to "complete" the description of $U$ on other basis state such that inner product is conserved as well, thus ensuring unitary behavior on the full space ? Jan 6 '20 at 12:33
• I reformulate: Here you showed $U|i,0\rangle$ is an orthonormal basis. Then by taking enough vector $e_{k,l}$ such that associated to the $U|i,0\rangle$ we have an orthonormal basis, defining the column of $U$ as being: $(U|i,0\rangle,U|k,l\rangle=e_{k,l})$ we indeed have a unitary matrix. Jan 6 '20 at 12:40
Therefore, to check whether $$U$$ is unitary, it is enough to check that it sends a subset of a basis (i.e. a non-complete set of orthonormal vectors) into a set of orthonormal vectors. In the present case, this means to prove that orthonormal states in the first space are sent into orthonormal states into the larger space. Once this is proved, you can just extend the operation to one that acts unitarily on the rest of the space. You might notice that there are multiple (infinite) ways to do this.
For example, say $$U$$ acts on a two-qubit system as $$U|00\rangle=|00\rangle$$ and $$U|10\rangle=|11\rangle$$. This does not mean that $$U$$ is unitary, but it does mean that it can trivially be made into a unitary. You just need to define its action of the rest of the space (here $$|01\rangle$$ and $$|11\rangle$$) so that the overall action is unitary. A trivial way to do it here can be $$U|01\rangle=|01\rangle$$ and $$U|11\rangle=|10\rangle$$.