Channel representations

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Next, we'll discuss mathematical representations of channels.

Linear mappings from vectors to vectors can be represented by matrices in a familiar way, where the action of the linear mapping is described by matrix-vector multiplication. But channels are linear mappings from matrices to matrices, not vectors to vectors. So, in general, how can we express channels in mathematical terms?

For some channels, we may have a simple formula that describes them, like for the three examples of non-unitary qubit channels described previously. But an arbitrary channel may not have such a nice formula, so it isn't practical in general to express a channel in this way.

As a point of comparison, in the simplified formulation of quantum information we use unitary matrices to represent operations on quantum state vectors: every unitary matrix represents a valid operation and every valid operation can be expressed as a unitary matrix. In essence, the question being asked is: How can we do something analogous for channels?

To answer this question, we'll require some additional mathematical machinery. We'll see that channels can, in fact, be described mathematically in a few different ways, including representations named in honor of three individuals who played key roles in their development: Stinespring, Kraus, and Choi. Together, these different ways of describing channels offer different angles from which they can be viewed and analyzed.

Stinespring representations

Stinespring representations are based on the idea that every channel can be implemented in a standard way, where an input system is first combined with an initialized workspace system, forming a compound system; then a unitary operation is performed on the compound system; and finally the workspace system is discarded (or traced out), leaving the output of the channel.

The following figure depicts such an implementation, in the form of a circuit diagram, for a channel whose input and output systems are the same system, $\mathsf{X}.$

In this diagram, the wires represent arbitrary systems, as indicated by the labels above the wires, and not necessarily single qubits. Also, the ground symbol commonly used in electrical engineering indicates explicitly that $\mathsf{W}$ is discarded.

In words, the way the implementation works is as follows. The input system $\mathsf{X}$ begins in some state $\rho,$ while a workspace system $\mathsf{W}$ is initialized to the standard basis state $\vert 0\rangle.$ A unitary operation $U$ is performed on the pair $(\mathsf{W},\mathsf{X}),$ and finally the workspace system $\mathsf{W}$ is traced out, leaving $\mathsf{X}$ as the output.

Note that we're presuming that $0$ is a classical state of $\mathsf{W},$ and we choose it to be the initialized state of this system, which will help to simplify the mathematics. One could, however, choose any fixed pure state to represent the initialized state of $\mathsf{W}$ without changing the basic properties of the representation.

A mathematical expression of the resulting channel, $\Phi,$ is as follows.

$$\Phi(\rho) = \operatorname{Tr}_{\mathsf{W}} \bigl( U (\vert 0\rangle \langle 0 \vert_{\mathsf{W}} \otimes \rho) U^{\dagger} \bigr)$$

As usual, we're using Qiskit's ordering convention: the system $\mathsf{X}$ is on top in the diagram, and therefore corresponds to the right-hand tensor factor in the formula.

In general, the input and output systems of a channel need not be the same. Here's a figure depicting an implementation of a channel $\Phi$ whose input system is $\mathsf{X}$ and whose output system is $\mathsf{Y}.$

This time the unitary operation transforms $(\mathsf{W},\mathsf{X})$ into a pair $(\mathsf{G},\mathsf{Y}),$ where $\mathsf{G}$ is a new "garbage" system that gets traced out, leaving $\mathsf{Y}$ as the output system. In order for $U$ to be unitary, it must be a square matrix. This requires that the pair $(\mathsf{G},\mathsf{Y})$ has the same number of classical states as the pair $(\mathsf{W},\mathsf{X}),$ and so the systems $\mathsf{W}$ and $\mathsf{G}$ must be chosen in a way that allows this.

We obtain a mathematical expression of the resulting channel, $\Phi,$ that is similar to what we had before.

$$\Phi(\rho) = \operatorname{Tr}_{\mathsf{G}} \bigl( U (\vert 0\rangle \langle 0 \vert_{\mathsf{W}} \otimes \rho) U^{\dagger} \bigr)$$

When a channel is described in this way, as a unitary operation along with a specification of how the workspace system is initialized and how the output system is selected, we say that it is expressed in Stinespring form or that it's a Stinespring representation of the channel.

It's not at all obvious, but every channel does in fact have a Stinespring representation, as we will see by the end of the lesson. We'll also see that Stinespring representations aren't unique; there will always be different ways to implement the same channel in the manner that's been described.

Remark

In the context of quantum information, the term Stinespring representation commonly refers to a slightly more general expression of a channel having the form

$$\Phi(\rho) = \operatorname{Tr}_{\mathsf{G}} \bigl( A \rho A^{\dagger} \bigr)$$

for an isometry $A,$ which is a matrix whose columns are orthonormal but that might not be a square matrix. For Stinespring representations having the form that we've adopted as a definition, we can obtain an expression of this other form by taking

$$A = U (\vert 0\rangle_{\mathsf{W}} \otimes \mathbb{I}_{\mathsf{X}}).$$

Completely dephasing channel

Here's a Stinespring representation of the qubit dephasing channel $\Delta.$ In this diagram, both wires represent single qubits — so this is an ordinary quantum circuit diagram.

To see that the effect that this circuit has on the input qubit is indeed described by the completely dephasing channel, we can go through the circuit one step at a time, using the explicit matrix representation of the partial trace discussed in the previous lesson. We'll refer to the top qubit as $\mathsf{X}$ — this is the input and output of the channel — and we'll assume that $\mathsf{X}$ starts in some arbitrary state $\rho.$

The first step is the introduction of a workspace qubit, $\mathsf{W}.$ Prior to the controlled-NOT gate being performed, the state of the pair $(\mathsf{W},\mathsf{X})$ is represented by the following density matrix.

$$\begin{aligned} \vert 0\rangle \langle 0 \vert_{\mathsf{W}} \otimes \rho & = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} \otimes \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle\\[1mm] \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix}\\[4mm] & = \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle & 0 & 0 \\[1mm] \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle & 0 & 0 \\[1mm] 0 & 0 & 0 & 0 \\[1mm] 0 & 0 & 0 & 0 \end{pmatrix} \end{aligned}$$

As per Qiskit's ordering convention, the top qubit $\mathsf{X}$ is on the right and the bottom qubit $\mathsf{W}$ is on the left. We're using density matrices rather than quantum state vectors, but they're tensored together in a similar way to what's done in the simplified formulation of quantum information.

The next step is to perform the controlled-NOT operation, where $\mathsf{X}$ is the control and $\mathsf{W}$ is the target. Still keeping in mind the Qiskit ordering convention, the matrix representation of this gate is as follows.

$$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0 \end{pmatrix}$$

This is a unitary operation, and to apply it to a density matrix we conjugate by the unitary matrix. The conjugate-transpose doesn't happen to change this particular matrix, so the result is as follows.

$$\begin{pmatrix} 1 & 0 & 0 & 0\\[1mm] 0 & 0 & 0 & 1\\[1mm] 0 & 0 & 1 & 0\\[1mm] 0 & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle & 0 & 0 \\[1mm] \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle & 0 & 0 \\[1mm] 0 & 0 & 0 & 0 \\[1mm] 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0\\[1mm] 0 & 0 & 0 & 1\\[1mm] 0 & 0 & 1 & 0\\[1mm] 0 & 1 & 0 & 0 \end{pmatrix}\\[3mm] = \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & 0 & 0 & \langle 0 \vert \rho \vert 1 \rangle\\[1mm] 0 & 0 & 0 & 0\\[1mm] 0 & 0 & 0 & 0\\[1mm] \langle 1 \vert \rho \vert 0 \rangle & 0 & 0 & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix}$$

Finally, the partial trace is performed on $\mathsf{W}.$ Recalling the action of this operation on $4\times 4$ matrices, which was described in the previous lesson, we obtain the following density matrix output.

$$\begin{aligned} \operatorname{Tr}_{\mathsf{W}} \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & 0 & 0 & \langle 0 \vert \rho \vert 1 \rangle\\[1mm] 0 & 0 & 0 & 0\\[1mm] 0 & 0 & 0 & 0\\[1mm] \langle 1 \vert \rho \vert 0 \rangle & 0 & 0 & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix} & = \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & 0 \\[1mm] 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\[1mm] 0 & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix}\\[3mm] & = \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & 0 \\[1mm] 0 & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix}\\[4mm] & = \Delta(\rho) \end{aligned}$$

We can alternatively compute the partial trace by first converting to Dirac notation.

$$\begin{pmatrix} \langle 0\vert \rho \vert 0\rangle & 0 & 0 & \langle 0\vert \rho \vert 1\rangle\\[1mm] 0 & 0 & 0 & 0\\[1mm] 0 & 0 & 0 & 0\\[1mm] \langle 1\vert \rho \vert 0\rangle & 0 & 0 & \langle 1\vert \rho \vert 1\rangle \end{pmatrix} = \begin{array}{r} \langle 0\vert \rho \vert 0\rangle \, \vert 0\rangle\langle 0\vert \otimes \vert 0\rangle\langle 0\vert \\[1mm] +\, \langle 0\vert \rho \vert 1\rangle \, \vert 0\rangle\langle 1\vert \otimes \vert 0\rangle\langle 1\vert \\[1mm] +\, \langle 1\vert \rho \vert 0\rangle \, \vert 1\rangle\langle 0\vert \otimes \vert 1\rangle\langle 0\vert \\[1mm] +\, \langle 1\vert \rho \vert 1\rangle \, \vert 1\rangle\langle 1\vert \otimes \vert 1\rangle\langle 1\vert \end{array}$$

Tracing out the qubit on the left-hand side yields the same answer as before.

$$\langle 0\vert \rho \vert 0\rangle \, \vert 0\rangle\langle 0\vert +\, \langle 1\vert \rho \vert 1\rangle \, \vert 1\rangle\langle 1\vert = \Delta(\rho)$$

An intuitive way to think about this circuit is that the controlled-NOT operation effectively copies the classical state of the input qubit, and when the copy is thrown in the trash the input qubit "collapses" probabilistically to one of the two possible classical states, which is equivalent to complete dephasing.

Completely dephasing channel (alternative)

The circuit described above is not the only way to implement the completely dephasing channel. Here's a different way to do it.

Here's a quick analysis showing that this implementation works. After the Hadamard gate is performed we have this two-qubit state as a density matrix:

$$\begin{aligned} \vert + \rangle\langle + \vert \otimes \rho & = \frac{1}{2}\begin{pmatrix} 1 & 1\\[1mm] 1 & 1 \end{pmatrix} \otimes \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle\\[1mm] \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix}\\[4mm] & = \frac{1}{2} \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle & \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle\\[1mm] \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle & \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle\\[1mm] \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle & \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle\\[1mm] \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle & \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix}. \end{aligned}$$

The controlled-$\sigma_z$ gate operates by conjugation as follows.

$$\frac{1}{2} \begin{pmatrix} 1 & 0 & 0 & 0\\[1mm] 0 & 1 & 0 & 0\\[1mm] 0 & 0 & 1 & 0\\[1mm] 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle & \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle\\[1mm] \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle & \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle\\[1mm] \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle & \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle\\[1mm] \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle & \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0\\[1mm] 0 & 1 & 0 & 0\\[1mm] 0 & 0 & 1 & 0\\[1mm] 0 & 0 & 0 & -1 \end{pmatrix}\\[3mm] = \frac{1}{2} \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle & \langle 0 \vert \rho \vert 0 \rangle & -\langle 0 \vert \rho \vert 1 \rangle\\[1mm] \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle & \langle 1 \vert \rho \vert 0 \rangle & -\langle 1 \vert \rho \vert 1 \rangle\\[1mm] \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle & \langle 0 \vert \rho \vert 0 \rangle & -\langle 0 \vert \rho \vert 1 \rangle\\[1mm] -\langle 1 \vert \rho \vert 0 \rangle & -\langle 1 \vert \rho \vert 1 \rangle & -\langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix}$$

Finally the workspace system $\mathsf{W}$ is traced out.

$$\frac{1}{2} \operatorname{Tr}_{\mathsf{W}} \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle & \langle 0 \vert \rho \vert 0 \rangle & -\langle 0 \vert \rho \vert 1 \rangle\\[1mm] \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle & \langle 1 \vert \rho \vert 0 \rangle & -\langle 1 \vert \rho \vert 1 \rangle\\[1mm] \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle & \langle 0 \vert \rho \vert 0 \rangle & -\langle 0 \vert \rho \vert 1 \rangle\\[1mm] -\langle 1 \vert \rho \vert 0 \rangle & -\langle 1 \vert \rho \vert 1 \rangle & -\langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix}\\[3mm] \begin{aligned} & = \frac{1}{2} \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle\\[1mm] \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix} + \frac{1}{2} \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & -\langle 0 \vert \rho \vert 1 \rangle\\[1mm] -\langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix}\\[4mm] & = \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & 0\\[2mm] 0 & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix} \end{aligned}$$

This implementation is based on a simple idea: dephasing is equivalent to either doing nothing (that is, applying an identity operation) or applying a $\sigma_z$ gate, each with probability $1/2.$

$$\begin{aligned} \frac{1}{2} \rho + \frac{1}{2} \sigma_z \rho \sigma_z & = \frac{1}{2} \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & \langle 0 \vert \rho \vert 1 \rangle\\[1mm] \langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix} + \frac{1}{2} \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & -\langle 0 \vert \rho \vert 1 \rangle\\[1mm] -\langle 1 \vert \rho \vert 0 \rangle & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix}\\[4mm] & = \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & 0\\[1mm] 0 & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix}\\[2mm] & = \Delta(\rho) \end{aligned}$$

That is, the completely dephasing channel is an example of a mixed-unitary channel, and more specifically, a Pauli channel.

Qubit reset channel

The qubit reset channel can be implemented as follows.

The swap gate simply shifts the $\vert 0\rangle$ initialized state of the workspace qubit so that it gets output, while the input state $\rho$ gets moved to the bottom qubit and then traced out.

Alternatively, if we don't demand that the output of the channel is left on top, we can take this very simple circuit as our representation.

In words, resetting a qubit to the $\vert 0\rangle$ state is equivalent to throwing the qubit in the trash and getting a new one.

Kraus representations

Now we'll discuss Kraus representations, which offer a convenient formulaic way to express the action of a channel through matrix multiplication and addition. In particular, a Kraus representation is a specification of a channel, $\Phi,$ in the following form.

$$\Phi(\rho) = \sum_{k = 0}^{N-1} A_k \rho A_k^{\dagger}$$

Here, $A_0,\ldots,A_{N-1}$ are matrices that all have the same dimensions: their columns correspond to the classical states of the input system, $\mathsf{X},$ and their rows correspond to the classical states of the output system, whether it's $\mathsf{X}$ or some other system $\mathsf{Y}.$ In order for $\Phi$ to be a valid channel these matrices must satisfy the following condition.

$$\sum_{k = 0}^{N-1} A_k^{\dagger} A_k = \mathbb{I}_{\mathsf{X}}$$

This condition is equivalent to the condition that $\Phi$ preserves trace. The other property required of a channel — which is complete positivity — follows from the general form of the equation for $\Phi,$ as a sum of conjugations.

Sometimes it's convenient to name the matrices $A_0,\ldots,A_{N-1}$ in a different way. For instance, we could number them starting from $1,$ or we could use states in some arbitrary classical state set $\Gamma$ instead of numbers as subscripts:

$$\Phi(\rho) = \sum_{a\in\Gamma} A_a \rho A_a^{\dagger} \quad \text{where} \quad \sum_{a\in\Gamma} A_a^{\dagger} A_a = \mathbb{I}.$$

These different ways of naming these matrices, which are called Kraus matrices, are all common and can be convenient in different situations — but we'll stick with the names $A_0,\ldots,A_{N-1}$ in this lesson for the sake of simplicity.

The number $N$ can be an arbitrary positive integer, but it never needs to be too large: if the input system $\mathsf{X}$ has $n$ classical states and the output system $\mathsf{Y}$ has $m$ classical states, then any given channel from $\mathsf{X}$ to $\mathsf{Y}$ will always have a Kraus representation for which $N$ is at most the product $nm.$

Completely dephasing channel

We obtain a Kraus representation of the completely dephasing channel by taking $A_0 = \vert 0\rangle\langle 0\vert$ and $A_1 = \vert 1\rangle\langle 1\vert.$

$$\begin{aligned} \sum_{k = 0}^1 A_k \rho A_k^{\dagger} & = \vert 0\rangle\langle 0 \vert \rho \vert 0\rangle\langle 0 \vert + \vert 1\rangle\langle 1 \vert \rho \vert 1\rangle\langle 1 \vert\\ & = \langle 0 \vert \rho \vert 0\rangle \, \vert 0\rangle\langle 0 \vert + \langle 1 \vert \rho \vert 1\rangle \, \vert 1\rangle\langle 1 \vert \\[2mm] & = \begin{pmatrix} \langle 0 \vert \rho \vert 0 \rangle & 0 \\[1mm] 0 & \langle 1 \vert \rho \vert 1 \rangle \end{pmatrix} \end{aligned}$$

These matrices satisfy the required condition.

$$\sum_{k = 0}^1 A_k^{\dagger} A_k = \vert 0\rangle\langle 0\vert 0\rangle\langle 0\vert + \vert 1\rangle\langle 1\vert 1\rangle\langle 1\vert = \vert 0\rangle\langle 0\vert + \vert 1\rangle\langle 1\vert = \mathbb{I}$$

Alternatively we can take $A_0 = \frac{1}{\sqrt{2}}\mathbb{I}$ and $A_1 = \frac{1}{\sqrt{2}}\sigma_z,$ so that

$$\sum_{k = 0}^1 A_k \rho A_k^{\dagger} = \frac{1}{2} \rho + \frac{1}{2} \sigma_z \rho \sigma_z = \Delta(\rho),$$

as was computed previously. This time the required condition can be verified as follows.

$$\sum_{k = 0}^1 A_k^{\dagger} A_k = \frac{1}{2} \mathbb{I} + \frac{1}{2} \sigma_z^2 = \frac{1}{2} \mathbb{I} + \frac{1}{2} \mathbb{I} = \mathbb{I}$$

Qubit reset channel

We obtain a Kraus representation of the qubit reset channel by taking $A_0 = \vert 0\rangle\langle 0\vert$ and $A_1 = \vert 0\rangle\langle 1\vert.$

$$\begin{aligned} \sum_{k = 0}^1 A_k \rho A_k^{\dagger} & = \vert 0\rangle\langle 0 \vert \rho \vert 0\rangle\langle 0 \vert + \vert 0\rangle\langle 1 \vert \rho \vert 1\rangle\langle 0 \vert\\ & = \langle 0 \vert \rho \vert 0\rangle \, \vert 0\rangle\langle 0 \vert + \langle 1 \vert \rho \vert 1\rangle \, \vert 0\rangle\langle 0 \vert\\[2mm] & = \operatorname{Tr}(\rho) \vert 0\rangle \langle 0 \vert \end{aligned}$$

These matrices satisfy the required condition.

$$\sum_{k = 0}^1 A_k^{\dagger} A_k = \vert 0\rangle\langle 0\vert 0\rangle\langle 0\vert + \vert 1\rangle\langle 0\vert 0\rangle\langle 1\vert = \vert 0\rangle\langle 0\vert + \vert 1\rangle\langle 1\vert = \mathbb{I}$$

Completely depolarizing channel

One way to obtain a Kraus representation for the completely depolarizing channel is to choose Kraus matrices $A_0,\ldots,A_3$ as follows.

$$A_0 = \frac{\vert 0\rangle\langle 0\vert}{\sqrt{2}} \quad A_1 = \frac{\vert 0\rangle\langle 1\vert}{\sqrt{2}} \quad A_2 = \frac{\vert 1\rangle\langle 0\vert}{\sqrt{2}} \quad A_3 = \frac{\vert 1\rangle\langle 1\vert}{\sqrt{2}}$$

For any qubit density matrix $\rho$ we then have

$$\begin{aligned} \sum_{k = 0}^3 A_k \rho A_k^{\dagger} & = \frac{1}{2} \bigl(\vert 0\rangle\langle 0\vert \rho \vert 0\rangle\langle 0\vert + \vert 0\rangle\langle 1\vert \rho \vert 1\rangle\langle 0\vert + \vert 1\rangle\langle 0\vert \rho \vert 0\rangle\langle 1\vert + \vert 1\rangle\langle 1\vert \rho \vert 1\rangle\langle 1\vert\bigr)\\ & = \operatorname{Tr}(\rho) \frac{\mathbb{I}}{2}\\[1mm] & = \Omega(\rho). \end{aligned}$$

An alternative Kraus representation is obtained by choosing Kraus matrices like so.

$$A_0 = \frac{\mathbb{I}}{2} \quad A_1 = \frac{\sigma_x}{2} \quad A_2 = \frac{\sigma_y}{2} \quad A_3 = \frac{\sigma_z}{2}$$

To verify that these Kraus matrices do in fact represent the completely depolarizing channel, let's first observe that conjugating an arbitrary $2\times 2$ matrix by a Pauli matrix works as follows.

$$\begin{aligned} \sigma_x \begin{pmatrix} \alpha_{0,0} & \alpha_{0,1}\\[1mm] \alpha_{1,0} & \alpha_{1,1} \end{pmatrix} \sigma_x & = \begin{pmatrix} \alpha_{1,1} & \alpha_{1,0}\\[1mm] \alpha_{0,1} & \alpha_{0,0} \end{pmatrix}\\[5mm] \sigma_y \begin{pmatrix} \alpha_{0,0} & \alpha_{0,1}\\[1mm] \alpha_{1,0} & \alpha_{1,1} \end{pmatrix} \sigma_y & = \begin{pmatrix} \alpha_{1,1} & -\alpha_{1,0}\\[1mm] -\alpha_{0,1} & \alpha_{0,0} \end{pmatrix}\\[5mm] \sigma_z \begin{pmatrix} \alpha_{0,0} & \alpha_{0,1}\\[1mm] \alpha_{1,0} & \alpha_{1,1} \end{pmatrix} \sigma_z & = \begin{pmatrix} \alpha_{0,0} & -\alpha_{0,1}\\[1mm] -\alpha_{1,0} & \alpha_{1,1} \end{pmatrix} \end{aligned}$$

This allows us to verify the correctness of our Kraus representation.

$$\begin{aligned} \sum_{k = 0}^3 A_k \rho A_k^{\dagger} & = \frac{\rho + \sigma_x \rho \sigma_x + \sigma_y \rho \sigma_y + \sigma_z \rho \sigma_z}{4} \\ & = \frac{1}{4} \begin{pmatrix} \langle 0\vert\rho\vert 0\rangle + \langle 1\vert\rho\vert 1\rangle + \langle 1\vert\rho\vert 1\rangle + \langle 0\vert\rho\vert 0\rangle & \langle 0\vert\rho\vert 1\rangle + \langle 1\vert\rho\vert 0\rangle - \langle 1\vert\rho\vert 0\rangle - \langle 0\vert\rho\vert 1\rangle \\[2mm] \langle 1\vert\rho\vert 0\rangle + \langle 0\vert\rho\vert 1\rangle - \langle 0\vert\rho\vert 1\rangle - \langle 1\vert\rho\vert 0\rangle & \langle 1\vert\rho\vert 1\rangle + \langle 0\vert\rho\vert 0\rangle + \langle 0\vert\rho\vert 0\rangle + \langle 1\vert\rho\vert 1\rangle \end{pmatrix} \\[4mm] & = \operatorname{Tr}(\rho) \frac{\mathbb{I}}{2} \end{aligned}$$

This Kraus representation expresses an important idea, which is that the state of a qubit can be completely randomized by applying to it one of the four Pauli matrices (including the identity matrix) chosen uniformly at random. Thus, the completely depolarizing channel is another example of a Pauli channel.

It is not possible to find a Kraus representation for the completely depolarizing channel $\Omega$ having three or fewer Kraus matrices; at least four are required for this channel.

Unitary channels

If we have a unitary matrix $U$ representing an operation on a system $\mathsf{X},$ we can express the action of this unitary operation as a channel:

$$\Phi(\rho) = U \rho U^{\dagger}.$$

This expression is already a valid Kraus representation of the channel $\Phi$ where we happen to have just one Kraus matrix $A_0 = U.$ In this case, the required condition

$$\sum_{k = 0}^{N-1} A_k^{\dagger} A_k = \mathbb{I}_{\mathsf{X}}$$

takes the much simpler form $U^{\dagger} U = \mathbb{I}_{\mathsf{X}},$ which we know is true because $U$ is unitary.

Choi representations

Now we'll discuss a third way that channels can be described, through the Choi representation. The way it works is that each channel is represented by a single matrix known as its Choi matrix. If the input system has $n$ classical states and the output system has $m$ classical states, then the Choi matrix of the channel will have $nm$ rows and $nm$ columns.

Choi matrices provide a faithful representation of channels, meaning that two channels are the same if and only if they have the same Choi matrix. One reason why this is important is that it provides us with a way of determining whether two different descriptions correspond to the same channel or to different channels: we simply compute the Choi matrices and compare them to see if they're equal. In contrast, Stinespring and Kraus representations are not unique in this way, as we have seen.

Choi matrices are also useful in other regards for uncovering various mathematical properties of channels.

Definition

Let $\Phi$ be a channel from a system $\mathsf{X}$ to a system $\mathsf{Y},$ and assume that the classical state set of the input system $\mathsf{X}$ is $\Sigma.$ The Choi representation of $\Phi,$ which is denoted $J(\Phi),$ is defined by the following equation.

$$J(\Phi) = \sum_{a,b\in\Sigma} \vert a\rangle\langle b \vert \otimes \Phi\bigl( \vert a\rangle\langle b \vert\bigr)$$

If we assume that $\Sigma = \{0,\ldots, n-1\}$ for some positive integer $n,$ then we can alternatively express $J(\Phi)$ as a block matrix:

$$J(\Phi) = \begin{pmatrix} \Phi\bigl(\vert 0\rangle\langle 0\vert\bigr) & \Phi\bigl(\vert 0\rangle\langle 1\vert\bigr) & \cdots & \Phi\bigl(\vert 0\rangle\langle n-1\vert\bigr) \\[1mm] \Phi\bigl(\vert 1\rangle\langle 0\vert\bigr) & \Phi\bigl(\vert 1\rangle\langle 1\vert\bigr) & \cdots & \Phi\bigl(\vert 1\rangle\langle n-1\vert\bigr) \\[1mm] \vdots & \vdots & \ddots & \vdots\\[1mm] \Phi\bigl(\vert n-1\rangle\langle 0\vert\bigr) & \Phi\bigl(\vert n-1\rangle\langle 1\vert\bigr) & \cdots & \Phi\bigl(\vert n-1\rangle\langle n-1\vert\bigr) \end{pmatrix}$$

That is, as a block matrix, the Choi matrix of a channel has one block $\Phi(\vert a\rangle\langle b\vert)$ for each pair $(a,b)$ of classical states of the input system, with the blocks arranged in a natural way.

Notice that the set $\{\vert a\rangle\langle b\vert \,:\, 0\leq a,b < n\}$ forms a basis for the space of all $n\times n$ matrices. Because $\Phi$ is linear, it follows that its action can be recovered from its Choi matrix by taking linear combinations of the blocks.

The Choi state of a channel

Another way to think about the Choi matrix of a channel is that it's a density matrix if we divide by $n = \vert\Sigma\vert.$ Let's focus on the situation that $\Sigma = \{0,\ldots,n-1\}$ for simplicity, and imagine that we have two identical copies of $\mathsf{X}$ that are together in the entangled state

$$\vert \psi \rangle = \frac{1}{\sqrt{n}} \sum_{a = 0}^{n-1} \vert a \rangle \otimes \vert a \rangle.$$

As a density matrix this state is as follows.

$$\vert \psi \rangle \langle \psi \vert = \frac{1}{n} \sum_{a,b = 0}^{n-1} \vert a\rangle\langle b \vert \otimes \vert a\rangle\langle b \vert$$

If we apply $\Phi$ to the copy of $\mathsf{X}$ on the right-hand side, we obtain the Choi matrix divided by $n.$

$$(\operatorname{Id}\otimes \,\Phi) \bigl(\vert \psi \rangle \langle \psi \vert\bigr) = \frac{1}{n} \sum_{a,b = 0}^{n-1} \vert a\rangle\langle b \vert \otimes \Phi\bigl(\vert a\rangle\langle b \vert\bigr) = \frac{J(\Phi)}{n}$$

In words, up to a normalization factor $1/n,$ the Choi matrix of $\Phi$ is the density matrix we obtain by evaluating $\Phi$ on one-half of a maximally entangled pair of input systems, as the following figure depicts.

Notice in particular that this implies that the Choi matrix of a channel must always be positive semidefinite.

We also see that, because the channel $\Phi$ is applied to the right/top system alone, it cannot affect the reduced state of the left/bottom system. In the case at hand, that state is the completely mixed state $\mathbb{I}_{\mathsf{X}}/n,$ and therefore

$$\operatorname{Tr}_{\mathsf{Y}} \biggl(\frac{J(\Phi)}{n}\biggr) = \frac{\mathbb{I}_{\mathsf{X}}}{n}.$$

Clearing the denominator $n$ from both sides yields $\operatorname{Tr}_{\mathsf{Y}} (J(\Phi)) = \mathbb{I}_{\mathsf{X}}.$

We can alternatively draw this same conclusion by using the fact that channels must always preserve trace, and therefore

$$\begin{aligned} \operatorname{Tr}_{\mathsf{Y}} (J(\Phi)) & = \sum_{a,b\in\Sigma} \operatorname{Tr}\bigl(\Phi( \vert a\rangle\langle b \vert)\bigr) \, \vert a\rangle\langle b \vert \\ & = \sum_{a,b\in\Sigma} \operatorname{Tr}\bigl(\vert a\rangle\langle b \vert\bigr) \, \vert a\rangle\langle b \vert \\ & = \sum_{a\in\Sigma} \vert a\rangle\langle a \vert \\ & = \mathbb{I}_{\mathsf{X}}. \end{aligned}$$

In summary, the Choi representation $J(\Phi)$ for any channel $\Phi$ must be positive semidefinite and must satisfy

$$\operatorname{Tr}_{\mathsf{Y}} (J(\Phi)) = \mathbb{I}_{\mathsf{X}}.$$

As we will see by the end of the lesson, these two conditions are not only necessary but also sufficient, meaning that any linear mapping $\Phi$ from matrices to matrices that satisfies these requirements must, in fact, be a channel.

Completely dephasing channel

The Choi representation of the completely dephasing channel $\Delta$ is

$$\begin{aligned} J(\Delta) & = \sum_{a,b = 0}^{1} \vert a\rangle\langle b \vert \otimes \Delta\bigl(\vert a\rangle\langle b \vert\bigr) \\ & = \sum_{a = 0}^{1} \vert a\rangle\langle a \vert \otimes \vert a\rangle\langle a \vert \\[4mm] & = \begin{pmatrix} 1 & 0 & 0 & 0\\[1mm] 0 & 0 & 0 & 0\\[1mm] 0 & 0 & 0 & 0\\[1mm] 0 & 0 & 0 & 1 \end{pmatrix}. \end{aligned}$$

Completely depolarizing channel

The Choi representation of the completely depolarizing channel is

$$\begin{aligned} J(\Omega) & = \sum_{a,b = 0}^{1} \vert a\rangle\langle b \vert \otimes \Omega\bigl(\vert a\rangle\langle b \vert\bigr)\\ & = \sum_{a = 0}^{1} \vert a\rangle\langle a \vert \otimes \frac{1}{2} \mathbb{I} \\[4mm] & = \frac{1}{2} \mathbb{I} \otimes \mathbb{I}\\[3mm] & = \begin{pmatrix} \frac{1}{2} & 0 & 0 & 0\\[1mm] 0 & \frac{1}{2} & 0 & 0\\[1mm] 0 & 0 & \frac{1}{2} & 0\\[1mm] 0 & 0 & 0 & \frac{1}{2} \end{pmatrix}. \end{aligned}$$

Qubit reset channel

The Choi representation of the qubit reset channel $\Phi$ is

$$\begin{aligned} J(\Lambda) & = \sum_{a,b = 0}^{1} \vert a\rangle\langle b \vert \otimes \Lambda\bigl(\vert a\rangle\langle b \vert\bigr)\\ & = \sum_{a = 0}^{1} \vert a\rangle\langle a \vert \otimes \vert 0\rangle \langle 0\vert\\[4mm] & = \mathbb{I} \otimes \vert 0\rangle \langle 0\vert\\[3mm] & = \begin{pmatrix} 1 & 0 & 0 & 0\\[1mm] 0 & 0 & 0 & 0\\[1mm] 0 & 0 & 1 & 0\\[1mm] 0 & 0 & 0 & 0 \end{pmatrix}. \end{aligned}$$

The identity channel

The Choi representation of the qubit identity channel $\operatorname{Id}$ is

$$\begin{aligned} J(\operatorname{Id}) & = \sum_{a,b = 0}^{1} \vert a\rangle\langle b \vert \otimes \operatorname{Id}\bigl(\vert a\rangle\langle b \vert\bigr) \\ & = \sum_{a,b = 0}^{1} \vert a \rangle\langle b \vert \otimes \vert a\rangle \langle b \vert \\ & = \begin{pmatrix} 1 & 0 & 0 & 1\\[1mm] 0 & 0 & 0 & 0\\[1mm] 0 & 0 & 0 & 0\\[1mm] 1 & 0 & 0 & 1 \end{pmatrix}. \end{aligned}$$

Notice in particular that $J(\operatorname{Id})$ is not the identity matrix. The Choi representation does not directly describe a channel's action in the usual way that a matrix represents a linear mapping.