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 correspondence is between channels and a particular subset of bipartite states:

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

Helpful background. The Choi matrix gives the definition and reconstruction rule; Bell states illustrate entanglement with a reference. We use finite-dimensional input and output spaces and output-input tensor order.

The normalized 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, invariant under $U\otimes U^*$ for every unitary $U$. For $0\le\lambda\le1$ the formula is a convex mixture of two states; the physical interval also contains negative $\lambda$, where this expression is no longer a convex-mixture interpretation. For $d\ge2$, positivity gives $-1/(d^2-1)\le\lambda\le1$, exactly the channel’s complete-positivity range.

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

Unequal marginals under amplitude damping

The amplitude-damping Choi matrix gives a concrete nonunital example. For a qubit decay probability $0\le\gamma\le1$, its normalized marginals are

$$\operatorname{Tr}_{\mathrm{out}}\tau_{\mathcal A_\gamma}=\frac{I}{2}, \qquad \operatorname{Tr}_{\mathrm{in}}\tau_{\mathcal A_\gamma} =\frac12\begin{pmatrix}1+\gamma&0\\0&1-\gamma\end{pmatrix}.$$

At $\gamma=0.4$, the output marginal is $\operatorname{diag}(0.7,0.3)$, whereas the reference remains $I/2$. A local trace-preserving channel cannot change the untouched reference’s marginal, even though it changes the output and their correlations. This is also why trace preservation does not imply unitality.

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 $J_\Phi$.

The exact entanglement statement is that a channel is entanglement breaking if and only if its Choi state is separable (Horodecki, Shor, and Ruskai, 2003). An entangled Choi state establishes that at least one input’s entanglement survives, not that every entangled input is protected.

To see the separable direction constructively, suppose

$$\tau_\Phi=\sum_j p_j\sigma_j\otimes\omega_j, \qquad \sum_j p_j\omega_j=I/d.$$

The reconstruction formula gives

$$\Phi(\rho)=\sum_j\operatorname{Tr}(F_j\rho)\sigma_j, \qquad F_j=d p_j\omega_j^{\mathsf T}.$$

The $F_j$ are positive and sum to $I$. Thus this is a measure-and-prepare channel; applying it to one half of any state leaves a separable output-reference mixture.

Complete qubit dephasing illustrates why entanglement breaking does not mean erasing all correlations. Its Choi state is $\tfrac12|00\rangle\langle00|+\tfrac12|11\rangle\langle11|$, a separable state with perfectly correlated $z$ readings. Complete depolarization instead gives a product state.

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.

References

• M.-D. Choi, “Completely positive linear maps on complex matrices,” Linear Algebra and its Applications 10, 285–290 (1975). Article.
• M. Horodecki, P. W. Shor, and M. B. Ruskai, “Entanglement breaking channels,” Reviews in Mathematical Physics 15, 629–641 (2003). Author’s version.
• A. Jamiołkowski, “Linear transformations which preserve trace and positive semidefiniteness of operators,” Reports on Mathematical Physics 3, 275–278 (1972). Article.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), chapters 2 and 6. Author’s text.