Channel-State Duality

Channel-state duality is the correspondence between linear maps on quantum states and bipartite operators. For completely positive maps, the corresponding bipartite operator is positive. For quantum channels, the normalized bipartite operator is a density matrix with a fixed maximally mixed marginal.

The compact version is:

$$\Phi \quad \longleftrightarrow \quad J_\Phi = (\Phi\otimes\mathrm{id}) \left( \lvert\Omega\rangle\langle\Omega\rvert \right),$$

where

$$\lvert\Omega\rangle = \sum_{i=1}^d \lvert i\rangle\otimes\lvert i\rangle$$

is the unnormalized maximally entangled vector on input and reference spaces. The first subsystem becomes the output after $\Phi$ acts; the second subsystem is the untouched input reference.

The Choi Matrix page owns the matrix representation and physicality tests. This page emphasizes the state interpretation:

$$\text{a channel} \quad \text{is a bipartite state with a constrained reference marginal}.$$

Choi State

If the input dimension is $d$, define the normalized maximally entangled state

$$\lvert\Omega_N\rangle = \frac{1}{\sqrt d} \sum_{i=1}^d \lvert i\rangle\otimes\lvert i\rangle.$$

Applying $\Phi$ to one half gives

$$\tau_\Phi = (\Phi\otimes\mathrm{id}) \left( \lvert\Omega_N\rangle\langle\Omega_N\rvert \right) = \frac{1}{d}J_\Phi.$$

When $\Phi$ is a channel, $\tau_\Phi$ is an ordinary density operator:

$$\tau_\Phi\ge0, \qquad \operatorname{Tr}\tau_\Phi=1.$$

It is called the Choi state or channel state of $\Phi$.

Marginal Constraint

Not every bipartite state is the Choi state of a channel from the chosen input system. Trace preservation imposes the marginal constraint

$$\operatorname{Tr}_{\mathrm{out}}J_\Phi = I_{\mathrm{in}}.$$

Equivalently, for the normalized Choi state,

$$\operatorname{Tr}_{\mathrm{out}}\tau_\Phi = \frac{I_{\mathrm{in}}}{d}.$$

Thus the reference/input marginal is maximally mixed. The output marginal is generally not maximally mixed:

$$\operatorname{Tr}_{\mathrm{in}}\tau_\Phi = \frac{\Phi(I_{\mathrm{in}})}{d}.$$

If the input and output dimensions are equal and $\Phi$ is unital, then the output marginal is also maximally mixed.

The map from channels to states is therefore:

$$\left\{ \text{channels } \Phi:\mathcal B(\mathcal H_{\mathrm{in}}) \to \mathcal B(\mathcal H_{\mathrm{out}}) \right\} \longleftrightarrow \left\{ \tau\ge0,\ \operatorname{Tr}_{\mathrm{out}}\tau=I_{\mathrm{in}}/d \right\}.$$

Reconstruction Formula

The correspondence is one-to-one once a basis is fixed. With the output-input convention used here,

$$\Phi(X) = \operatorname{Tr}_{\mathrm{in}} \!\left[ J_\Phi \left( I_{\mathrm{out}}\otimes X^T \right) \right].$$

In terms of the normalized Choi state,

$$\Phi(X) = d\, \operatorname{Tr}_{\mathrm{in}} \!\left[ \tau_\Phi \left( I_{\mathrm{out}}\otimes X^T \right) \right].$$

The transpose is basis dependent. This is not a physical transpose operation applied by the channel; it records the basis used to identify linear maps with bipartite operators.

Identity Channel

For the identity channel $\mathrm{id}$ on a $d$-dimensional system,

$$J_{\mathrm{id}} = \lvert\Omega\rangle\langle\Omega\rvert.$$

The normalized Choi state is

$$\tau_{\mathrm{id}} = \lvert\Omega_N\rangle\langle\Omega_N\rvert.$$

Thus the identity channel corresponds to a maximally entangled pure state. For qubits, this is the Bell state $\lvert\Phi^+\rangle$:

$$\lvert\Phi^+\rangle = \frac{ \lvert00\rangle+\lvert11\rangle }{\sqrt2}.$$

This example is the simplest way to remember the duality: perfect transmission is represented by perfect entanglement between output and reference.

Completely Depolarizing Channel

The completely depolarizing channel sends every input to the maximally mixed state:

$$\Delta(\rho) = \operatorname{Tr}(\rho)\frac{I}{d}.$$

Its Choi matrix is

$$J_\Delta = \frac{I_{\mathrm{out}}}{d} \otimes I_{\mathrm{in}}.$$

The normalized Choi state is therefore

$$\tau_\Delta = \frac{I_{\mathrm{out}}}{d} \otimes \frac{I_{\mathrm{in}}}{d}.$$

This is a product state. Complete depolarization destroys all correlation between output and reference.

Depolarizing Channel and Isotropic States

For the $d$-dimensional depolarizing channel

$$\mathcal D_\lambda(\rho) = \lambda\rho + (1-\lambda) \operatorname{Tr}(\rho)\frac{I}{d},$$

the normalized Choi state is

$$\tau_{\mathcal D_\lambda} = \lambda \lvert\Omega_N\rangle\langle\Omega_N\rvert + (1-\lambda) \frac{I_{\mathrm{out}}\otimes I_{\mathrm{in}}}{d^2}.$$

This is an isotropic state: a mixture of the maximally entangled state and the maximally mixed bipartite state. Positivity of this bipartite state is exactly the complete-positivity condition for the channel.

The depolarizing-channel page develops the allowed parameter range and channel action in detail: Depolarizing Channel.

Amplitude-Damping Channel

For the qubit amplitude-damping channel with decay probability $\gamma$, the Choi matrix is

$$J_{\mathcal A_\gamma} = \begin{pmatrix} 1&0&0&\sqrt{1-\gamma}\\ 0&\gamma&0&0\\ 0&0&0&0\\ \sqrt{1-\gamma}&0&0&1-\gamma \end{pmatrix}, \qquad 0\le\gamma\le1.$$

The normalized Choi state is

$$\tau_{\mathcal A_\gamma} = \frac{1}{2} J_{\mathcal A_\gamma}.$$

Its reference marginal is always maximally mixed:

$$\operatorname{Tr}_{\mathrm{out}}\tau_{\mathcal A_\gamma} = \frac{I_{\mathrm{in}}}{2}.$$

Its output marginal is

$$\operatorname{Tr}_{\mathrm{in}}\tau_{\mathcal A_\gamma} = \frac{1}{2} \left[ (1+\gamma) \lvert0\rangle\langle0\rvert + (1-\gamma) \lvert1\rangle\langle1\rvert \right].$$

The output marginal is not maximally mixed for $\gamma\ne0$, reflecting that amplitude damping is trace preserving but not unital.

Why the Duality Is Useful

Channel-state duality turns map questions into state questions:

1. complete positivity becomes positivity of $J_\Phi$;
2. trace preservation becomes a marginal constraint;
3. Kraus rank becomes rank of the Choi matrix;
4. process tomography becomes constrained state estimation of $J_\Phi$;
5. channel optimization often becomes semidefinite optimization over $J_\Phi$.

It also gives physical intuition. A channel preserves entanglement with a reference to the extent that its Choi state remains entangled. Noisy channels degrade that entanglement; sufficiently noisy channels can become entanglement breaking, corresponding to separable Choi states.

This last statement is a useful preview, but it belongs to quantum information and entanglement theory for full treatment.

Common Mistakes

Forgetting the marginal constraint

A positive bipartite state is not automatically a channel state. It must satisfy $\operatorname{Tr}_{\mathrm{out}}\tau=I_{\mathrm{in}}/d$ for a trace-preserving channel.

Mixing normalized and unnormalized conventions

Some formulas use $J_\Phi$; others use $\tau_\Phi=J_\Phi/d$. The former has trace $d$ for a channel, while the latter has trace $1$.

Ignoring tensor-factor order

This page uses output–input order. Some references use input–output order. Then reconstruction formulas and partial traces look transposed or swapped.

Treating the transpose as physical

The $X^T$ in the reconstruction formula is basis bookkeeping. It is not an additional physical operation applied by $\Phi$.

Assuming unitality from trace preservation

Trace preservation fixes the reference marginal of $\tau_\Phi$. Unitality fixes the output marginal when dimensions match. They are different conditions.

Exercises

Identity channel

Show that the Choi state of the identity channel is $\lvert\Omega_N\rangle\langle\Omega_N\rvert$.

Solution

For the identity channel,

$$(\mathrm{id}\otimes\mathrm{id}) \left( \lvert\Omega_N\rangle\langle\Omega_N\rvert \right) = \lvert\Omega_N\rangle\langle\Omega_N\rvert.$$

Therefore

$$\tau_{\mathrm{id}} = \lvert\Omega_N\rangle\langle\Omega_N\rvert.$$

Marginal condition

Starting from

$$J_\Phi = \sum_{i,j} \Phi(\lvert i\rangle\langle j\rvert) \otimes \lvert i\rangle\langle j\rvert,$$

show that trace preservation implies $\operatorname{Tr}_{\mathrm{out}}J_\Phi=I_{\mathrm{in}}$.

Solution

Taking the partial trace over the output factor gives

$$\operatorname{Tr}_{\mathrm{out}}J_\Phi = \sum_{i,j} \operatorname{Tr} \!\left[ \Phi(\lvert i\rangle\langle j\rvert) \right] \lvert i\rangle\langle j\rvert.$$

If $\Phi$ is trace preserving, then

$$\operatorname{Tr} \!\left[ \Phi(\lvert i\rangle\langle j\rvert) \right] = \operatorname{Tr} \lvert i\rangle\langle j\rvert = \delta_{ij}.$$

Hence

$$\operatorname{Tr}_{\mathrm{out}}J_\Phi = \sum_i \lvert i\rangle\langle i\rvert = I_{\mathrm{in}}.$$

Completely depolarizing channel

Compute the normalized Choi state of $\Delta(\rho)=\operatorname{Tr}(\rho)I/d$ and decide whether it is entangled.

Solution

For matrix units,

$$\Delta(\lvert i\rangle\langle j\rvert) = \delta_{ij}\frac{I}{d}.$$

Thus

$$J_\Delta = \sum_i \frac{I_{\mathrm{out}}}{d} \otimes \lvert i\rangle\langle i\rvert = \frac{I_{\mathrm{out}}}{d} \otimes I_{\mathrm{in}}.$$

Dividing by $d$ gives

$$\tau_\Delta = \frac{I_{\mathrm{out}}}{d} \otimes \frac{I_{\mathrm{in}}}{d}.$$

This is a product state, so it is not entangled.

Cross-Links

• Choi Matrix
• Process Tomography
• Operator Entanglement and Scrambling Preview
• Quantum Operations
• Trace-Preserving and Trace-Nonincreasing Maps
• Kraus Representation
• Depolarizing Channel
• Amplitude-Damping Channel
• Bell States
• Entangled States
• Entanglement Witnesses
• Glossary

References

• A. Jamiołkowski, “Linear transformations which preserve trace and positive semidefiniteness of operators,” Reports on Mathematical Physics 3, 275–278 (1972).
• M.-D. Choi, “Completely positive linear maps on complex matrices,” Linear Algebra and its Applications 10, 285–290 (1975).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 10th anniversary edition (2010).
• A. S. Holevo, Quantum Systems, Channels, Information: A Mathematical Introduction, De Gruyter (2012).
• M. M. Wilde, Quantum Information Theory, Cambridge University Press, 2nd edition (2017).
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018).