# Measurement and Evolution

## Generalised Measurement Operators

The extension of measurement beyond orthogonal projectors alone arises from the extension of quantum mechanics to open systems, where we are performing the measurement on a subsystem S alone. We can understand this as allowing our system to interact with a 'pointer' or measurement device; consider a particle of mass $m$, helping us to measure an observable $\mathbf{M}$. The interaction is turned on during a window defined by the function $\lambda(t): \lambda(t) \in \{0,1\}.$ Thus, the system and pointer have a Hamiltonian

We assume that $\commutator{\mathbf{M}}{\mathbf{H_{0}}}=0$ for simplicity. We have evolution during the measurement $\mathbf{U}(T)\approx \exp(-i\lambda T\mathbf{M}\otimes\mathbf{P})$, which we can rewrite as using the spectral decomposition of $\mathbf{M}.$

The operator $\mathbf{P}$ translates the position of the particle, and the 'amount' of evolution depends on the eigenvalues $M_{a}$. More abstractly, given a set of orthogonal projectors $\{\mathbf{E}_{a}, a=0,1,2\cdots N-1\}$ which satisfy $\mathbf{E}_{a}\mathbf{E}_{b}=\delta_{a,b}\mathbf{E}_{a}, \sum_{a}\mathbf{E}_{a}=\mathbf{I}.$

We can measure these projectors by coupling to a system with n basis states $\{\ket{a}\}$, and then doing a transformation $\mathbf{U}=\sum_{a,b}\mathbf{E}_{a}\otimes\ketbra{b+a}{b}$.

This is a unitary $\mathbf{U}:\ket{\Psi}=\ket{psi}\otimes\ket{0}\rightarrow\ket{\Psi'}=\sum_{a}\mathbf{E}_{a}\ket{psi}\otimes\ket{a}.$

Thus, the state of the system before the measurement outcome is known is given by and thus we can rewrite this in the density matrix formalism as $\rho\rightarrow\sum_{a}\mathbf{E}_{A}\rho\mathbf{E}_{a}.$

### Beyond Orthogonality

We can extend this description of measurement beyond the case where the measurements we are doing are all orthogonal. This is called a 'generalised measurement' made up of a set of operators $\{\mathbf{M}_{a}\}.$ For normalisation, these operators need to satisfy a completeness relation $\sum_{a}\mathbf{M}_{a}^{\dagger}\mathbf{M}_{a}=1$, such that the evolution is given by

We have that $Prob(a)=\|\mathbf{M}_{a}\ket{\psi}\|^{2}$, and $Prob(b|a)=\frac{\|\mathbf{M}_{b}\mathbf{M}_{a}\ket{\psi}\|^{2}}{\|\mathbf{M}_{a}\ket{\psi}\|^{2}}.$

For each $\mathbf{M}_{a}$ there is an associated Hermitian, positive operator $\mathbf{E}_{a}\equiv\mathbf{M}_{a}^{\dagger}\mathbf{M}_{a}$ such that $Prob(a)=tr\left(\rho\mathbf{E}_{a}\right).$ These are also called Positive Operator Valued Measurements (POVMS).

## Quantum Channels

Quantum channels are a method of modelling the evolution of one subsystem, based on unitary evolution of the composite system. In fact, the mathematics of quantum channels is identical to the evolution of the state under a POVM before collapse given above:

$\mathcal{E}:\rho\rightarrow\rho'$ is a linear map acting on the density operator of the subsystem. These channels are also called superoperators, or Complete Positive Trace-Preserving (CPTP) Maps. They obey 4 simple properties

1. Linearity: $\mathcal{E}(\alpha\rho_{1}+\beta\rho_{2})=\alpha\mathcal{E}(\rho_{1})+\beta\mathcal{E}(\rho_{2})$
2. Hermiticity Preserving: $\rho=\rho^{\dagger}\implies\mathcal{E}(\rho)=\mathcal{E}(\rho^{\dagger})=\mathcal{E}(\rho)^{\dagger}$
3. Positivity Preserving: $\rho\geq 0 \implies \mathcal{E}(\rho)\geq 0$
4. Trace Preserving: $tr\left(\mathcal{E}(\rho)\right)=tr(\rho)$

We can prove that any quantum channel admits a decomposition in to a set of operators $\{\mathbf{M}_{a}\}$ called 'Kraus' operators, introudcing a unitary acting on the system $\ket{\psi}_{A}\otimes\ket{0}_{B}$ and tracing out B. These operator sum rotations are not unique, and we can see that the action of a channel is invariant under a chnge of basis $\ket{a}=\sum_{u}\ket{\mu}V_{\mu a}$, giving new operators $\mathbf{N}_{\mu}=\sum_{a}V_{\mu a}\mathbf{M}_{a}.$

Channels can be compose $\mathcal{E}' = \mathcal{E}_{2}\circ\mathcal{E}_{1}: \mathcal{E}'(\rho) = \sum_{\mu a}\mathbf{N}_{\mu}\mathbf{M}_{a}\rho\left(\mathbf{N}_{\mu}\mathbf{M}_{a}\right)^{\dagger}$ where $\mathbf{M}_{a},\mathbf{N}_{\mu}$ are the Kraus operators of channel 1 and 2 respectively.

### Reversibility

Under what conditions is a Quantum channel reversible? This means the channel has an inverse $\mathcal{E}_{2}$ such that

This imposes the condition that $\mathbf{N}_{\mu}\mathbf{M}_{a}=\lambda_{\mu a}\mathbf{I}\;\forall\mu, a.$ Using the completeness relation for Kraus operators, we can show

Using the polar decomposition of $\mathbf{M}_{a}$, we can thus see that the Kraus operators are proprotional to unitary operators, and thus $\mathcal{E}_{1}$ is a simple unitary map.

### The Dual of a Channel

The Heisenberg representation of quantum mechanics is one in which the time dependence of the evolution is carried by the opertors, instead of the states as in the 'Schrödinger' picture. In this case, we have the Heisenberg equaiton of evolution

The two pictures are defined such that they agree on the expectation values at any time, and that the two picutres are indistinguishable for $t=0.$

We can define an alternative depiction of a channel where general observables evolve under a channel $\mathcal{E}^{*}:\mathbf{A}'=\sum_{a}\mathbf{M}^{\dagger}_{a}\mathbf{A}\mathbf{M}_{a}.$ Thus, we have

In this case, instead of requiring the dual channel be trace-preserving, ew require it is 'unital' Not all channels are unital, but all dual channels are. A channel is unital if the completeness relation is satisfied for both 'permutations' of $\mathbf{M}_{a}, \mathbf{M}_{a}^{\dagger}.$ These channels are analagous to doubly stochastic maps.

### Quantum Operations

Channels sharing a mathematical form with maps allow us to consider the more general class of 'quantum operations'. For example, consider applying a POVM that can potentially give information about two observables $a,\mu$, but discarding the value of $\mu.$ This gives us an effective channel which has outcome $a$ with probability $tr\mathcal{E}_{a}(\rho).$ Just summing over $\mu$, the Kraus operators of $\mathcal{E}_{a}$ will not satisfy the completeness relation, and so we need to renormalise the state by the probability of the outcome. We can use this channel form to concatenate successive measurements too.

### Linearity

Why are quantum channels generlly linear maps? One note is that the description of a density operator as an ensemble of states is incompatible with a nonlinaer-evolution; indeed, given the description of the density matrix as a convex sum of points, a quantum channel must be linear.

Consider an initial state $\rho_{i}$ prepared with probability $p_{i}$, and then subsequently measured to obtain outcome $a$ ith probability $p(a|i).$ We have the post measurement state $\mathcal{E}_{a}(\rho_{i})/ p(a|i)$, and thus the ensemble post measurement is given by: where $p(i|a)$ is the probability that $i$ was prepared given outcome a. We can also write this as using a convex combination of initial states $\{\rho_{i}\}.$

Using Bayes' rule, $p_{i}p(a|i)=p_{a}p(i|a)$, and thus $\mathcal{E}_{a}$ is a linear map:

### Complete Positivity

Complete positivity is the stronger requirement that a quantum channel is positive, even if acting on only part of a composite system.

For example, consider a channel $\mathcal{E}: \rho\in\mathcal{H}_{A}\rightarrow\rho'\in\mathcal{H}_{A'}.$ What happens if we append the input space to $\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ and apply the operation $\mathcal{E}\otimes\mathbf{I}$ i.e. do nothing to this additional space? As long as this extended operation is positive, then the channel is called completely positive.

This is easy to show for a channel in the Kraus representation: the extended map $\mathcal{E}\otimes\mathbf{I}$ simply has Kraus operators $\mathbf{M}_{a}\otimes\mathbf{I}.$ The complete positivity requirement is a fairly loose constraint; physically, it can be understood as saying it is possible to manipulate systems in isolation.

Not all positive maps are complete positive. An interesting example is the Transpose map, which is defined as a map $T:\ketbra{i}{j}\rightarrow\ketbra{j}{i}.$ This map is positive, as

But, if we perform the 'partial transpose' $T\otimes\mathbf{I}$, this map is not complete positive. In fact, it is negative for entangled states! This is called the Peres-Horodecki criterion for entanglement.

## Alternative Representations of Channels

A quantum channel acting on a subsystem $A$ derives from a unitary map acting on an extension of A, and is itself a CPTP map in the space of density operators. We can thus interpret any evolution of A as a unitary evolution of its purification AR.

This can be understood using the Choi-Jamiolkowski isomporhism. Consider the action of an extended channel $\mathbf{I}\otimes\mathcal{E}$ acting on a maximally entangled state $\ket{\Phi}_{RA}=\sum_{i=0}^{d-1}\ket{i}_{R}\ket{i}_{A}\rightarrow \sum_{a}\left(\ketbra{\Psi_{a}}{\Psi_{a}}\right)_{RA'}.$

We note that $\ket{\phi}_{A}=\sum_{i}\phi_{i}\ket{i}_A= \sum_{i}\phi_{i} \left( _{R}\braket{i}{\Phi}_{RA}\right) = _{R}\braket{\phi^{*}}{\Phi}_{RA}.$ Using the linearity of $\mathcal{E}$, we can thus write

What this tells is that the mapping of a state $\ket{\phi}\in\mathcal{H}_{A}\rightarrow\ket{\phi'}\in\mathcal{H}_{A'}$ can be obtained using the state $\ket{\Psi}_{RA'}$ as $\mathbf{M}_{A}= _{R}\braket{\phi^{*}}{\Psi_{a}}_{RA'}.$ These operators $\mathbf{M}_{a}$ form the Kraus operators of the channel, and thus there is a direct connection between the state $\ket{\Psi}_{RA'}$ and $\mathcal{E}.$

Thus, we can draw an equivalence between the freedom to choose a set of Kraus oeprators, and the freedom to choose a decomposition of $\ket{\Psi}_{RA'}$ into an ensemble of pure states. Recall that two ensembles are related by a unitary change of basis $\ket{\gamma_{\mu}}=\sum_{a}V_{\mu a}\ket{\Psi}_{a}$, and so are two sets of Kraus operators $\mathbf{N}_{\mu}=\sum_{a}V_{\mu a}\mathbf{M}_{a}.$

This also makes it clear that, for a mapping from $A:|A|=d \rightarrow A':|A'|=d'$, we need $d^{2}(d'^{2}-1)$ parameters.

### Stinespring dilation

Consider a channel $\mathcal{E}:A\rightarrow A' : |A|>|A'|.$ In this case we can define an 'environment' E, such that the actionof the channel can be described by an inner-product preserving map or isometry V

We can see that thus is a 'unitary' isometry $: V^{\dagger}V=\mathbf{I}_{A}.$ This isometry is called the 'Stinespring Dilation' of a channel $\mathcal{E}=\{\mathbf{M}_{a}\}.$ Another way of understanding this is to conceptualise the Stinespring dilation as the purification of the Choi state.

It can be shown that