Choi Matrix

The Choi matrix represents a linear map on operators as a bipartite operator. It is one of the most useful ways to test whether a proposed map is a physical quantum operation.

For finite-dimensional channels, the Choi matrix turns several structural questions into matrix questions:

$$\begin{array}{ccl} \text{complete positivity} &\Longleftrightarrow& J_\Phi\ge0,\\ \text{trace preservation} &\Longleftrightarrow& \operatorname{Tr}_{\mathrm{out}}J_\Phi=I_{\mathrm{in}},\\ \text{minimal Kraus number} &\Longleftrightarrow& \operatorname{rank}J_\Phi. \end{array}$$

This page uses the output-input convention:

$$J_\Phi\in \mathcal B(\mathcal H_{\mathrm{out}}\otimes\mathcal H_{\mathrm{in}}).$$

Other references sometimes reverse the two tensor factors. The formulas involving partial traces and transposes must then be adjusted.

Simulating quantum channels implements this convention numerically.

Helpful background. Kraus representation supplies an operator-sum form; the partial trace reduces a bipartite matrix. Kraus, Choi, and Stinespring views compares these descriptions.

Defining the output-input Choi matrix

Let

$$\Phi: \mathcal B(\mathcal H_{\mathrm{in}}) \to \mathcal B(\mathcal H_{\mathrm{out}})$$

be a linear map. Choose an orthonormal basis $\{\lvert i\rangle\}_{i=1}^{d_{\mathrm{in}}}$ of $\mathcal H_{\mathrm{in}}$ and define the unnormalized maximally entangled vector

$$\lvert\Omega\rangle = \sum_{i=1}^{d_{\mathrm{in}}} \lvert i\rangle_{\mathrm{in}} \otimes \lvert i\rangle_R,$$

where $R$ is a reference copy of the input space.

The Choi matrix is

$$J_\Phi = (\Phi\otimes\mathrm{id}_R) \left( \lvert\Omega\rangle\langle\Omega\rvert \right).$$

Equivalently,

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

The first tensor factor is the output system; the second is the input-reference system.

Recovering the Map

The Choi matrix determines the map. For any input operator $X$,

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

where the transpose is taken in the same input basis used to define $\lvert\Omega\rangle$.

This transpose is a common source of mistakes. It appears because the input matrix elements are read from the reference half of $\lvert\Omega\rangle$.

To verify the formula, write

$$X=\sum_{i,j}X_{ij}\lvert i\rangle\langle j\rvert.$$

Then

$$\operatorname{Tr}_{\mathrm{in}} \!\left[ \left( \Phi(\lvert i\rangle\langle j\rvert) \otimes \lvert i\rangle\langle j\rvert \right) \left( I_{\mathrm{out}}\otimes X^{\mathsf T} \right) \right] = X_{ij}\Phi(\lvert i\rangle\langle j\rvert),$$

and summing over $i,j$ gives $\Phi(X)$.

A sign-sensitive check is the identity channel acting on $|+y\rangle=(|0\rangle+i|1\rangle)/\sqrt2$:

$$X=\frac12\begin{pmatrix}1&-i\\i&1\end{pmatrix}.$$

The formula returns $X$. Replacing $X^{\mathsf T}$ with $X$ would return $X^{\mathsf T}=|-y\rangle\langle-y|$, an orthogonal state. Real diagonal test inputs would miss this error.

Complete Positivity Test

Choi’s theorem states that, in finite dimensions,

$$\Phi\ \text{is completely positive} \quad\Longleftrightarrow\quad J_\Phi\ge0.$$

One direction is immediate from the definition. If $\Phi$ is completely positive, then

$$(\Phi\otimes\mathrm{id}) \left( \lvert\Omega\rangle\langle\Omega\rvert \right) \ge0,$$

because $\lvert\Omega\rangle\langle\Omega\rvert$ is positive.

Conversely, if $J_\Phi\ge0$, diagonalize it as

$$J_\Phi = \sum_r \lambda_r \lvert v_r\rangle\langle v_r\rvert, \qquad \lambda_r>0.$$

Unvectorizing

$$\sqrt{\lambda_r}\,\lvert v_r\rangle$$

gives Kraus operators $K_r$ such that

$$\Phi(\rho)=\sum_rK_r\rho K_r^\dagger.$$

Thus $\Phi$ is completely positive.

Trace Preservation and Trace Nonincrease

With the output-input convention, trace preservation is

$$\Phi\ \text{trace preserving} \quad\Longleftrightarrow\quad \operatorname{Tr}_{\mathrm{out}}J_\Phi=I_{\mathrm{in}}.$$

Proof:

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

If $\Phi$ is trace preserving, then

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

so the partial trace is $I_{\mathrm{in}}$.

Trace nonincrease is the inequality

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

Together with $J_\Phi\ge0$, this describes a possible selected operation. More generally, $\operatorname{Tr}_{\mathrm{out}}J_\Phi=[\Phi^\dagger(I)]^{\mathsf T}$; transposition preserves the operator inequality against $I$.

Unital Maps

For a map from $\mathcal H_{\mathrm{in}}$ to the same output dimension, unitality means

$$\Phi(I_{\mathrm{in}})=I_{\mathrm{out}}.$$

In Choi form,

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

Therefore

$$\Phi\ \text{unital} \quad\Longleftrightarrow\quad \operatorname{Tr}_{\mathrm{in}}J_\Phi=I_{\mathrm{out}}.$$

Do not confuse this with trace preservation. Trace preservation traces out the output factor of $J_\Phi$; unitality traces out the input factor.

Kraus Rank

If

$$\Phi(\rho) = \sum_{\alpha=1}^r K_\alpha\rho K_\alpha^\dagger,$$

then

$$J_\Phi = \sum_{\alpha=1}^r \lvert K_\alpha\rangle\!\rangle \langle\!\langle K_\alpha\rvert,$$

where

$$\lvert K\rangle\!\rangle = \sum_{a,i} K_{ai} \lvert a\rangle_{\mathrm{out}} \otimes \lvert i\rangle_{\mathrm{in}}.$$

The minimal number of Kraus operators in any representation is

$$r_{\min} = \operatorname{rank}J_\Phi.$$

This is the same number as the minimal environment dimension in a finite-dimensional isometric, or pure-environment, Stinespring representation.

Identity Channel

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

$$\mathrm{id}(\lvert i\rangle\langle j\rvert) = \lvert i\rangle\langle j\rvert.$$

Thus

$$J_{\mathrm{id}} = \sum_{i,j} \lvert i\rangle\langle j\rvert \otimes \lvert i\rangle\langle j\rvert = \lvert\Omega\rangle\langle\Omega\rvert.$$

The identity channel has Choi rank $1$, matching its single Kraus operator $K=I$.

Since $\lvert\Omega\rangle$ is unnormalized,

$$\operatorname{Tr}J_{\mathrm{id}}=d.$$

The normalized state associated with the identity channel is $J_{\mathrm{id}}/d$.

Completely Depolarizing Channel

For the completely depolarizing channel on a $d$-dimensional system,

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

Since

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

the Choi matrix is

$$J_\Delta = \frac{I}{d}\otimes I.$$

It is positive, trace preserving, and full rank. Its minimal Kraus rank is $d^2$.

The interpolating Depolarizing Channel has Choi matrix

$$J_{\mathcal D_\lambda} = \lambda\lvert\Omega\rangle\langle\Omega\rvert + \frac{1-\lambda}{d}I\otimes I,$$

For $d\ge2$, the eigenvalue along $|\Omega\rangle$ is $[1+(d^2-1)\lambda]/d$, and each of the $d^2-1$ orthogonal eigenvalues is $(1-\lambda)/d$. Their nonnegativity gives $-1/(d^2-1)\le\lambda\le1$.

Transpose Map

The transpose map

$$T(X)=X^{\mathsf T}$$

has Choi matrix

$$J_T = \sum_{i,j} \lvert j\rangle\langle i\rvert \otimes \lvert i\rangle\langle j\rvert.$$

This is the swap operator:

$$J_T=S.$$

The swap operator has eigenvalue $+1$ on the symmetric subspace and eigenvalue $-1$ on the antisymmetric subspace. Therefore $J_T$ is not positive when $d\ge2$, and the transpose map is not completely positive.

This example is the Choi-matrix version of the standard partial-transpose test.

Qubit Dephasing Example

Let

$$\Phi_\eta \left( \begin{pmatrix} \rho_{00} & \rho_{01}\\ \rho_{10} & \rho_{11} \end{pmatrix} \right) = \begin{pmatrix} \rho_{00} & \eta\rho_{01}\\ \eta^*\rho_{10} & \rho_{11} \end{pmatrix}.$$

In the ordered output-input basis

$$\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle,$$

the Choi matrix is

$$J_{\Phi_\eta} = \begin{pmatrix} 1 & 0 & 0 & \eta\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \eta^* & 0 & 0 & 1 \end{pmatrix}.$$

Its nonzero block has eigenvalues

$$1+|\eta|, \qquad 1-|\eta|.$$

Therefore

$$J_{\Phi_\eta}\ge0 \quad\Longleftrightarrow\quad |\eta|\le1.$$

This is exactly the complete-positivity condition for the phase-damping channel.

Qubit Amplitude-Damping Example

For the Amplitude-Damping Channel with decay probability $\gamma$, the Choi matrix in the same output-input convention is

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

Its nonzero eigenvalues are $2-\gamma$ and $\gamma$, so the channel is completely positive for $0\le\gamma\le1$ and has Kraus rank $2$ when $\gamma>0$.

Channel-State Duality

If the normalized maximally entangled state is

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

then applying $\Phi$ to half of it gives

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

Every channel gives a density operator $J_\Phi/d$, but positivity and unit trace alone do not imply a channel. The necessary and sufficient conditions include the full marginal constraint:

$$J_\Phi\ge0, \qquad \operatorname{Tr}_{\mathrm{out}}J_\Phi=I_{\mathrm{in}}.$$

For example, the qubit map $\mathcal E(\rho)=K\rho K^\dagger$ with $K=\sqrt2|0\rangle\langle0|$ has $J_{\mathcal E}=2|00\rangle\langle00|$. Dividing by two gives a valid bipartite density operator, but its reference marginal is $|0\rangle\langle0|$, not $I/2$. The map would assign trace two to input $|0\rangle\langle0|$, so it is not a physical operation.

The constrained channel-state correspondence underlies process tomography and optimization over channels. Some channel simulations also use Choi states, with additional operational conditions.

For the normalized bipartite-state viewpoint, marginal constraints, and identity/depolarizing/amplitude-damping examples, see Channel-State Duality.

The correspondence is basis dependent in its formula because of the transpose in the reconstruction rule, but the underlying relation is invariant once a convention is fixed.

Process Tomography and Optimization

In process tomography, estimating a channel can be formulated as estimating $J_\Phi$ subject to physical constraints:

$$J_\Phi\ge0, \qquad \operatorname{Tr}_{\mathrm{out}}J_\Phi=I_{\mathrm{in}}.$$

These are semidefinite constraints. Many channel-optimization problems become semidefinite programs because probabilities and expectation values are linear functions of $J_\Phi$.

For example, if an experiment prepares input states $\rho_k$ and measures output effects $M_\ell$, the probability is

$$p(\ell|k) = \operatorname{Tr} \!\left[ M_\ell\,\Phi(\rho_k) \right].$$

The reconstruction formula makes the dependence explicit:

$$p(\ell|k)=\operatorname{Tr}[(M_\ell\otimes\rho_k^{\mathsf T})J_\Phi].$$

Thus the unknown Choi entries enter linearly. The input-first convention used in some measurement-tomography formulas swaps these tensor factors as well as those of the Choi matrix.

Process Tomography develops the prepare-and-measure and ancilla-assisted protocols, identifiability, physical estimators, SPAM limits, validation, and scaling built on this formula.

Exercises

Trace-preservation condition

Starting from

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

prove that $\operatorname{Tr}_{\mathrm{out}}J_\Phi=I$ if $\Phi$ is trace preserving.

Solution

Taking the partial trace over the output system gives

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

Trace preservation implies

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

Therefore

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

Choi matrix of the identity channel

Show that $J_{\mathrm{id}}=\lvert\Omega\rangle\langle\Omega\rvert$ and compute its rank.

Solution

For the identity channel,

$$\mathrm{id}(\lvert i\rangle\langle j\rvert) = \lvert i\rangle\langle j\rvert.$$

Thus

$$J_{\mathrm{id}} = \sum_{i,j} \lvert i\rangle\langle j\rvert \otimes \lvert i\rangle\langle j\rvert = \lvert\Omega\rangle\langle\Omega\rvert.$$

This is a rank-one positive operator, so the identity channel has Kraus rank $1$.

Complete positivity of dephasing

Use the dephasing Choi matrix to show that $\Phi_\eta$ is completely positive if and only if $|\eta|\le1$.

Solution

The only nonzero block of $J_{\Phi_\eta}$ is

$$\begin{pmatrix} 1 & \eta\\ \eta^* & 1 \end{pmatrix}.$$

Its eigenvalues are

$$1+|\eta|, \qquad 1-|\eta|.$$

The full Choi matrix is positive exactly when both are nonnegative, which is equivalent to

$$|\eta|\le1.$$

References

• M.-D. Choi, “Completely positive linear maps on complex matrices,” Linear Algebra and its Applications 10, 285–290 (1975). Article.
• 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), chapter 2. Author’s text.