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.

For a small numerical notebook contract that tests this convention, see Simulating Quantum Channels.

Kraus, Choi, and Stinespring Views uses Choi factorization as one side of the focused unread-map and minimal-realization crosswalk; this page retains the Choi definition, convention-sensitive reconstruction, CP, TP, and TNI criteria, channel-state duality, and worked examples.

Definition

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 $\Phi(X)$.

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

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

This is the Choi form of a selected measurement branch or probabilistic operation.

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 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

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

which gives the complete-positivity interval $-1/(d^2-1)\le\lambda\le1$.

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

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

Thus $J_\Phi/d$ is a bipartite density operator exactly when $\Phi$ is a channel from a $d$-dimensional input:

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

This is the channel-state correspondence. It underlies process tomography, teleportation-based channel simulation, and semidefinite optimization over channels.

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

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

Using the reconstruction formula, this is a linear function of $J_\Phi$.

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

Common Mistakes

Mixing normalized and unnormalized conventions

This page uses $\lvert\Omega\rangle=\sum_i\lvert i\rangle\lvert i\rangle$. If a source uses $\lvert\Omega_N\rangle=d^{-1/2}\sum_i\lvert i\rangle\lvert i\rangle$, its Choi matrix differs by a factor of $d$.

Reversing the tensor factors

Some references define $J_\Phi$ on input–output rather than output–input. Then the trace-preservation and reconstruction formulas look different.

Forgetting the transpose

The recovery formula uses $X^{\mathsf T}$ in the chosen input basis:

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

Dropping the transpose silently changes the convention.

Checking only trace preservation

Trace preservation is not enough. A trace-preserving linear map is a channel only if its Choi matrix is also positive.

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).
• A. Jamiołkowski, “Linear transformations which preserve trace and positive semidefiniteness of operators,” Reports on Mathematical Physics 3, 275–278 (1972).
• E. C. G. Sudarshan, P. M. Mathews, and J. Rau, “Stochastic dynamics of quantum-mechanical systems,” Physical Review 121, 920–924 (1961).
• K. Kraus, States, Effects, and Operations: Fundamental Notions of Quantum Theory, Springer (1983).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press (2010).
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018).