Quantum Channels for QI

A quantum channel describes one input–output transformation, but there are several useful ways to calculate with it. Kraus operators propagate states; a Choi matrix tests complete positivity; a transfer matrix makes circuit composition ordinary matrix multiplication. Choosing coordinates should not change the predicted probabilities. This page explains the conversions in finite dimensions, including rectangular maps and selected measurement outcomes, using small examples that can be checked by hand.

Helpful background. Density operators describe mixed and subnormalized states; quantum channels introduces complete positivity and instruments. Familiarity with tensor-product order and Pauli matrices helps with the matrix examples.

Channels act on operators

Let the input and output dimensions be $d_{\rm in}$ and $d_{\rm out}$. A channel is a linear, completely positive, trace-preserving map

$$\Phi:\mathcal B(\mathcal H_{\rm in})\longrightarrow \mathcal B(\mathcal H_{\rm out}).$$

The dimensions need not match: an encoding can map a qubit into a qutrit. Although density matrices are the physical inputs, linearity defines the map on all operators. In particular, its action on every matrix unit $E_{ij}=|i\rangle\langle j|$ determines the whole map, including coherence. Checking only computational-basis states tests $E_{ii}$ and misses the off-diagonal $E_{ij}$.

A selected outcome instead gives a completely positive, trace-nonincreasing operation $\mathcal E_y$. Its output and probability are

$$\begin{aligned} \widetilde\rho_y&=\mathcal E_y(\rho),\\ p_y&=\operatorname{Tr}\widetilde\rho_y,\\ \rho_y&=\widetilde\rho_y/p_y\quad(p_y>0). \end{aligned}$$

An instrument is a collection of such operations whose sum is a channel. The conditional state alone loses $p_y$ and generally depends nonlinearly on $\rho$. Keep the subnormalized branches when composing measurements and feedforward. The same coordinate methods below apply to each linear branch, with its trace condition changed appropriately.

Kraus operators and the output–input Choi matrix

Kraus operators have shape $d_{\rm out}\times d_{\rm in}$ and give

$$\Phi(A)=\sum_\alpha K_\alpha A K_\alpha^\dagger.$$

Complete positivity is automatic in this form. Trace preservation requires $\sum_\alpha K_\alpha^\dagger K_\alpha=I_{\rm in}$; trace nonincrease requires the corresponding operator inequality $\preceq I_{\rm in}$. Kraus families are not unique: isometric mixing of their labels leaves the unread map unchanged. A label is therefore not automatically an experimentally observed environmental event.

We use the unnormalized, output–input Choi convention

$$J_\Phi=\sum_{i,j}\Phi(E_{ij})\otimes E_{ij}.$$

Rows are ordered as $(a,i)$ and columns as $(b,j)$, with output indices $a,b$ and input indices $i,j$. Thus its flattened row is $a d_{\rm in}+i$. For a trace-preserving channel, $\operatorname{Tr}J=d_{\rm in}$. Using a normalized maximally entangled input instead gives $J/d_{\rm in}$; all partial-trace and reconstruction formulas must carry that factor.

The action can be recovered without Kraus operators:

$$\begin{aligned} \Phi(A)&=\operatorname{Tr}_{\rm in} \left[J_\Phi(I_{\rm out}\otimes A^{\mathsf T})\right],\\ [\Phi(A)]_{ab}&=\sum_{i,j}(J_\Phi)_{ai,bj}A_{ij}. \end{aligned}$$

The transpose in the first expression produces the ordinary $A_{ij}$ contraction in the second. Both refer to the same chosen input basis.

For finite-dimensional maps, the Choi criterion gives complete positivity exactly when $J\succeq0$ (Choi; Jamiołkowski). Among CP maps,

$$\begin{aligned} \text{trace preservation: }&\operatorname{Tr}_{\rm out}J=I_{\rm in},\\ \text{trace nonincrease: }&\operatorname{Tr}_{\rm out}J\preceq I_{\rm in}. \end{aligned}$$

The partial trace equals the transpose of the input effect:

$$\begin{aligned} F&=\sum_\alpha K_\alpha^\dagger K_\alpha\\ &=(\operatorname{Tr}_{\rm out}J)^{\mathsf T},\\ \operatorname{Tr}\Phi(\rho)&=\operatorname{Tr}(F\rho). \end{aligned}$$

Transposing the identity leaves the TP condition unchanged, which can hide this distinction in deterministic examples. For the filter $K=|0\rangle\langle +y|$, where $|+y\rangle=(|0\rangle+i|1\rangle)/\sqrt2$, one has $F=|+y\rangle\langle+y|$ but $\operatorname{Tr}_{\rm out}J=|-y\rangle\langle-y|$. The filter accepts $|+y\rangle$ with probability one; using the untransposed partial trace as its effect would incorrectly predict zero.

To obtain Kraus operators from $J$, diagonalize $J=\sum_\alpha\lambda_\alpha|v_\alpha\rangle\langle v_\alpha|$ and set

$$(K_\alpha)_{ai}=\sqrt{\lambda_\alpha}\,(v_\alpha)_{ai}.$$

Here reshaping uses the output–input order just defined. The smallest number of Kraus operators is $\operatorname{rank}J$. Their individual entries remain nonunique, especially in degenerate eigenspaces. See Kraus, Choi and Stinespring views for the corresponding environmental realization.

Column stacking and channel composition

Column stacking means

$$\operatorname{vec}_{\rm col} \begin{pmatrix}a&b\\c&d\end{pmatrix} =\begin{pmatrix}a\\c\\b\\d\end{pmatrix}.$$

The Liouville superoperator satisfies

$$\begin{aligned} \operatorname{vec}_{\rm col}\Phi(A)&=S_\Phi\operatorname{vec}_{\rm col}A,\\ S_\Phi&=\sum_\alpha K_\alpha^*\otimes K_\alpha. \end{aligned}$$

It has shape $d_{\rm out}^2\times d_{\rm in}^2$. Notice the complex conjugate, not adjoint, in the first Kronecker factor. Also notice that column stacking a Kraus matrix orders its entries differently from the output–input Choi ket above. A real symmetric example may conceal either mistake; complex rectangular examples are better checks.

Choi and Liouville matrices contain the same four-index coefficients, but group them differently. Using zero-based array indices,

$$J[a d_{\rm in}+i,\,b d_{\rm in}+j] =S[a+d_{\rm out}b,\,i+d_{\rm in}j].$$

This reshuffling is not a matrix similarity transformation and does not preserve eigenvalues. Testing $S\succeq0$ is not a complete-positivity test. For example, unitary $Z$ conjugation has Liouville eigenvalues $1,-1,-1,1$ while its Choi matrix is positive with rank one. Conversion procedures are discussed by Havel and by Wood, Biamonte and Cory.

If $\Phi$ acts first and $\Psi$ second, then

$$S_{\Psi\circ\Phi}=S_\Psi S_\Phi.$$

Kraus operators compose as $\{L_\beta K_\alpha\}_{\alpha,\beta}$. For independent local channels, the state map is $\Phi_A\otimes\Psi_B$, but the numerical Liouville matrix is generally a permuted version of $S_A\otimes S_B$ under global column stacking. The subsystem and operator indices must be regrouped. Directly applying local Kraus operators to a matrix unit is a simple way to determine the correct placement.

Noise placement matters too: $\mathcal N\circ\mathcal U$ and $\mathcal U\circ\mathcal N$ are unequal unless they commute. A fitted channel must be tied to its circuit location, frame and duration.

Process matrices, Pauli coordinates and qubit geometry

Choose a complete, linearly independent operator basis $\{A_m\}$ of $d_{\rm out}\times d_{\rm in}$ matrices. A process matrix writes

$$\Phi(\rho)=\sum_{m,n}\chi_{mn}A_m\rho A_n^\dagger.$$

Let $B$ have columns equal to the output–input flattened $A_m$. Then

$$J=B\chi B^\dagger,\qquad \chi=B^{-1}J(B^{-1})^\dagger.$$

This formula works for a nonorthogonal basis as well; replacing $B^{-1}$ by $B^\dagger$ is justified only for an orthonormal basis. In this independent basis, $\chi\succeq0$ is equivalent to complete positivity. An overcomplete operator list makes coefficients nonunique and needs separate care.

For a qubit, fix $P=(I,X,Y,Z)$ and $Y=\begin{pmatrix}0&-i\\i&0\end{pmatrix}$. Define the Pauli transfer matrix

$$R_{\mu\nu}=\tfrac12\operatorname{Tr}[P_\mu\Phi(P_\nu)].$$

If $q_\nu=\operatorname{Tr}(P_\nu A)$, then $q'=Rq$. For a density operator, $q=(1,r_x,r_y,r_z)^{\mathsf T}$. Trace preservation fixes the first row, so a qubit channel has the block form

$$R=\begin{pmatrix}1&\boldsymbol0^{\mathsf T}\\ \boldsymbol t&T\end{pmatrix},\qquad \boldsymbol r'=T\boldsymbol r+\boldsymbol t.$$

Unitality, $\Phi(I)=I$, means $\boldsymbol t=0$. Amplitude damping translates the Bloch ball; Pauli noise is unital. The detailed geometry is developed by Ruskai, Szarek and Werner.

Mapping the Bloch ball into itself establishes positivity, not complete positivity. Transposition sends $(r_x,r_y,r_z)$ to $(r_x,-r_y,r_z)$ and preserves the whole ball, yet its Choi matrix is the swap operator, with an eigenvalue $-1$ on the antisymmetric two-qubit state. It cannot act on one part of every entangled state. Correct TP and unital PTM rows and columns do not remove this obstruction.

Representation	Particularly useful for
Kraus operators	direct propagation, rectangular maps and selected branches
Choi matrix	complete positivity, trace constraints and minimum Kraus rank
Process matrix $\chi$	coefficients in a specified operator basis
Pauli transfer matrix $R$	qubit coordinates and circuit composition
Liouville matrix $S$	linear action and composition in a vectorization convention

Worked example: one Pauli channel in several coordinates

Take probabilities $(p_I,p_X,p_Y,p_Z)=(0.55,0.15,0.10,0.20)$ and $K_\alpha=\sqrt{p_\alpha}P_\alpha$. Each Kraus operator contributes $p_\alpha I$ to the completeness sum, so this is a channel. In the unnormalized Pauli basis,

$$\chi=\operatorname{diag}(0.55,0.15,0.10,0.20).$$

Conjugating $X$ by $I,X$ preserves it, whereas conjugating by $Y,Z$ reverses it. Thus its contraction is $p_I+p_X-p_Y-p_Z=0.4$. The same calculation for $Y,Z$ gives

$$R=\operatorname{diag}(1,0.4,0.3,0.5).$$

The other two arrays are

$$S=\begin{pmatrix} 0.75&0&0&0.25\\0&0.35&0.05&0\\ 0&0.05&0.35&0\\0.25&0&0&0.75 \end{pmatrix},$$ $$J=\begin{pmatrix} 0.75&0&0&0.35\\0&0.25&0.05&0\\ 0&0.05&0.25&0\\0.35&0&0&0.75 \end{pmatrix}.$$

The vectors obtained by flattening $I,X,Y,Z$ and dividing by $\sqrt2$ are orthonormal Choi eigenvectors, with eigenvalues $2p_\alpha$. They are nonnegative and sum to two; tracing out the output gives $I$.

For a genuinely complex input,

$$\rho=\begin{pmatrix}0.6&0.3+0.2i\\0.3-0.2i&0.4\end{pmatrix},$$

all representations give

$$\Phi(\rho)=\begin{pmatrix} 0.55&0.12+0.06i\\0.12-0.06i&0.45 \end{pmatrix}.$$

Its Bloch vector changes from $(0.6,-0.4,0.2)$ to $(0.24,-0.12,0.10)$. The minus sign on $r_y$ follows from the stated $Y$ convention.

A small calculation to check the conversions

This NumPy example constructs the Choi matrix by applying the map to matrix units, independently of its Kraus-vector factorization. The final loop checks action on every matrix unit, not only on one density matrix.

import numpy as np

I = np.eye(2, dtype=complex)
X = np.array([[0, 1], [1, 0]], dtype=complex)
Y = np.array([[0, -1j], [1j, 0]])
Z = np.diag([1, -1])
K = [np.sqrt(p)*P for p, P in zip([.55, .15, .10, .20], [I, X, Y, Z])]

def apply(A):
    return sum(k @ A @ k.conj().T for k in K)

units = [np.outer(I[:, i], I[:, j]) for i in range(2) for j in range(2)]
J = sum(np.kron(apply(E), E) for E in units)
S = sum(np.kron(k.conj(), k) for k in K)
np.testing.assert_allclose(J, J.conj().T, atol=1e-12)
assert np.linalg.eigvalsh(J).min() >= -1e-12
np.testing.assert_allclose(np.einsum('aiaj->ij', J.reshape(2,2,2,2)), I)
for E in units:
    via_S = (S @ E.flatten(order='F')).reshape(2,2,order='F')
    via_J = np.einsum('aibj,ij->ab', J.reshape(2,2,2,2), E)
    np.testing.assert_allclose(via_S, apply(E), atol=1e-12)
    np.testing.assert_allclose(via_J, apply(E), atol=1e-12)
print('All four matrix units agree.')

For a fitted matrix, first check Hermiticity, then use a Hermitian eigensolver and tolerances reflecting numerical and experimental uncertainty. Clipping substantial negative eigenvalues changes an estimate; it does not demonstrate that the original map was CP. Constraining a fit to CPTP maps likewise does not establish its predictive accuracy.

Dense superoperators require $d_{\rm out}^2d_{\rm in}^2$ complex entries. For $n$ qubits that is $2^{4n}$: complex128 storage is 16 MiB at five qubits and 16 TiB at ten, before temporary arrays. Direct Kraus action can avoid this storage. For pure-state trajectories, a Kraus branch has probability $\langle\psi|K_\alpha^\dagger K_\alpha|\psi\rangle$, generally dependent on the current state; only special random-unitary decompositions have fixed branch probabilities. Noise simulation develops scalable choices.

What a channel representation does not determine

A matrix describes a particular input–output map, not its microscopic cause. The same channel can arise from different environments or controls. Common noise models connects standard parameterizations to assumptions, and noise mechanisms explains context and temporal correlations.

Composing fixed powers $S^L$ assumes that the same map remains valid across uses, including after intermediate operations. Initial correlations, environmental memory, drift or neighboring drives may invalidate that assumption. A process matrix $\chi$ describes one map; a process tensor accepts a sequence of interventions and contains additional temporal information (Pollock et al.). A coordinate change cannot supply it. See Markovian and non-Markovian noise.

Finally, process tomography estimates channel coordinates from preparations and measurements. Their calibration, gauge freedom and uncertainty remain part of the inference problem, whichever coordinates are reported (Nielsen et al.). Compare predictions with circuits excluded from fitting; a positive Choi matrix alone cannot show that the experimental model is correct.

Exercises

1. Test a dephasing Choi family

For real $t$, consider

$$J_t= \begin{pmatrix} 1&0&0&t\\ 0&0&0&0\\ 0&0&0&0\\ t&0&0&1 \end{pmatrix}.$$

Determine when it defines a CPTP map under this page’s Choi convention. Test unitality, find its spectrum and a Kraus family, and give its PTM.

Solution

Tracing over the output gives $I$ for every real $t$, so the represented map is TP whenever it is CP. Tracing over the input also gives $I$, so it is unital. The only nonzero block is

$$\begin{pmatrix}1&t\\t&1\end{pmatrix},$$

with eigenvalues $1+t$ and $1-t$ and normalized eigenvectors $(\lvert00\rangle\pm\lvert11\rangle)/\sqrt2$. The other two eigenvalues are zero. Hence $J_t\succeq0$ exactly when $|t|\le1$.

Reshaping the two weighted eigenvectors gives the Kraus family

$$K_0=\sqrt{\frac{1+t}{2}}I, \qquad K_1=\sqrt{\frac{1-t}{2}}Z.$$

The weights are nonnegative precisely in the CP interval and sum to one. Conjugation by $I$ leaves every Pauli component fixed, while conjugation by $Z$ reverses $X$ and $Y$. Therefore

$$R=\operatorname{diag}(1,t,t,1).$$

At $t=1$ the map is identity; at $t=-1$ it is $Z$ conjugation; and at $t=0$ it completely dephases in the computational basis.

2. A rectangular encoding

Let

$$K=\begin{pmatrix}1&0\\0&1/\sqrt2\\0&i/\sqrt2\end{pmatrix}.$$

Show that $\Phi(A)=KAK^\dagger$ is a qubit-to-qutrit channel. Find the Choi rank, Choi trace and Liouville dimensions. What does it do to $|+y\rangle$? Is it unital onto the qutrit space?

Solution

The columns are orthonormal, so $K^\dagger K=I_2$. The output–input Choi ket is $(1,0,0,1/\sqrt2,0,i/\sqrt2)^{\mathsf T}$; its outer product is a $6\times6$ positive matrix of rank one and trace two. The Liouville matrix has $9$ rows and $4$ columns. The input becomes

$$K|+y\rangle=\frac{|0\rangle}{\sqrt2} +\frac{i|1\rangle}{2}-\frac{|2\rangle}{2}.$$

The map is not unital onto the qutrit space: $KK^\dagger$ is a rank-two projector, not $I_3$. Preserving input trace does not require filling every output level.

3. Reshuffle rather than relabel

Starting from the worked Pauli example superoperator, reconstruct the output–input Choi matrix using $(J_\Phi)_{ai,bj}=(S_\Phi)_{ab,ij}$. Explain why this is not direct equality and verify the entries $J_{03}$ and $S_{03}$.

Solution

Interpret each Liouville row as a pair $(a,b)$ and each column as $(i,j)$. Move the middle indices so that the Choi row is $(a,i)$ and its column is $(b,j)$. Index explicitly as J[a*dIn+i][b*dIn+j] = S[a+dOut*b][i+dIn*j]. Equivalently, a row-major reshape views $S$ with axes $(b,a,j,i)$, which are transposed to $(a,i,b,j)$; a column-major reshape uses $(a,b,i,j)\to(a,i,b,j)$.

For Choi entry $(0,3)$, the composite indices are $(a,i)=(0,0)$ and $(b,j)=(1,1)$. The corresponding Liouville entry is row $(a,b)=(0,1)$ and column $(i,j)=(0,1)$, which carries $0.35$. Hence $J_{03}=0.35$. By contrast, $S_{03}$ directly means row $(a,b)=(0,0)$ and column $(i,j)=(1,1)$, and equals $0.25$.

The mismatch is expected: reshuffling changes which elementary indices are paired into a row and column. Applying the inverse reshuffle recovers $S$, so no information was lost. Testing $J=S$ would confuse coordinate layout with channel equality.

4. Compose dephasing and a Hadamard

For $\eta=0.6$ and input $\lvert0\rangle\!\langle0\rvert$, compute dephasing after a Hadamard and a Hadamard after dephasing. Find both purities, their trace distance, and the corresponding superoperator products.

Solution

Hadamard first produces $\lvert+\rangle\!\langle+\rvert$. Dephasing preserves its populations and multiplies its off-diagonal entries by $0.6$, so

$$\mathcal D\circ\mathcal H(\rho_0) = \begin{pmatrix}0.5&0.3\\0.3&0.5\end{pmatrix}.$$

Dephasing first leaves the computational-basis state $\rho_0$ unchanged; Hadamard then produces the pure plus state,

$$\mathcal H\circ\mathcal D(\rho_0) = \begin{pmatrix}0.5&0.5\\0.5&0.5\end{pmatrix}.$$

For the first matrix, $\operatorname{Tr}(\rho^2)=2(0.5^2)+2(0.3^2)=0.68$. The second is a rank-one projector and has purity one. Their difference has eigenvalues $\pm0.2$, so half its trace norm is $0.2$.

Because the rightmost operation acts first, the first result is generated by $S_{\mathcal D}S_{\mathcal H}$ and the second by $S_{\mathcal H}S_{\mathcal D}$. The products are unequal. This output difference reveals a reversed composition order immediately.

5. Retain instrument probabilities

Apply the operators $M_0=\operatorname{diag}(\sqrt{0.9},\sqrt{0.2})$ and $M_1=\operatorname{diag}(\sqrt{0.1},\sqrt{0.8})$ to $\rho_i=\begin{pmatrix}0.3&0.1\\0.1&0.7\end{pmatrix}$. Compute both subnormalized branches, probabilities, conditional states, and the unread channel. State exactly what premature normalization erases.

Solution

Direct multiplication gives

$$\begin{aligned} \widetilde\rho_0&= \begin{pmatrix} 0.27&3/(50\sqrt2)\\ 3/(50\sqrt2)&0.14 \end{pmatrix},\\ p_0&=0.41, \end{aligned}$$

and

$$\begin{aligned} \widetilde\rho_1&= \begin{pmatrix} 0.03&1/(25\sqrt2)\\ 1/(25\sqrt2)&0.56 \end{pmatrix},\\ p_1&=0.59. \end{aligned}$$

The probabilities sum to one because $M_0^\dagger M_0+M_1^\dagger M_1=I$. Dividing each branch by its own trace yields

$$\rho_{0\mid0}= \begin{pmatrix} 27/41&3\sqrt2/41\\ 3\sqrt2/41&14/41 \end{pmatrix},$$ $$\rho_{1\mid1}= \begin{pmatrix} 3/59&2\sqrt2/59\\ 2\sqrt2/59&56/59 \end{pmatrix}.$$

If the result is unread, sum before normalization:

$$\rho'=\begin{pmatrix} 0.3&1/(10\sqrt2)\\ 1/(10\sqrt2)&0.7 \end{pmatrix}.$$

Replacing $\widetilde\rho_y$ immediately by $\rho_{y\mid y}$ sets every stored trace to one. It erases $p_y$, so later code cannot reproduce outcome frequencies, likelihoods, or correctly weighted feedforward branches.

6. Change the process-matrix basis

First replace the Pauli basis by $P/\sqrt2$. Then consider a nonorthogonal basis change whose first two columns express $A'_0=I$ and $A'_1=I+X$, with $A'_2=Y$, $A'_3=Z$. Find the new process matrices for the worked Pauli channel.

Solution

Scaling every basis element by $1/\sqrt2$ requires $\chi'=2\chi$, since each product $A'_m\rho A_n'^\dagger$ supplies a factor $1/2$. For the second change write $B'=BC$, where

$$C=\begin{pmatrix}1&1&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}.$$

Then $\chi'=C^{-1}\chi(C^{-1})^\dagger$, giving

$$\chi'=\begin{pmatrix} 0.70&-0.15&0&0\\-0.15&0.15&0&0\\ 0&0&0.10&0\\0&0&0&0.20 \end{pmatrix}.$$

The cross terms cancel the unwanted $I\rho X$ and $X\rho I$ terms introduced by $(I+X)\rho(I+X)$. Substituting reproduces the original Pauli channel. An adjoint in place of $C^{-1}$ would not. Neither basis change creates the extra intervention slots of a multitime process tensor.

7. Propagate an observable backward

For the worked Pauli example channel and $A=0.2I+0.4X-0.3Y+0.5Z$, find $\Phi^\dagger(A)$. Verify forward and backward expectation values on the worked input state.

Solution

The Pauli channel is self-adjoint in the Hilbert–Schmidt inner product because each Pauli is Hermitian and the weights are real. Its adjoint has the same Pauli contractions $(0.4,0.3,0.5)$. Therefore

$$\Phi^\dagger(A)=0.2I+0.16X -0.09Y+0.25Z.$$

The output state’s Bloch vector is $\boldsymbol r'=(0.24,-0.12,0.10)$. For $A=a_0I+\boldsymbol a\cdot\boldsymbol\sigma$, the expectation is $a_0+\boldsymbol a\cdot\boldsymbol r'$, giving

$$0.2+0.4(0.24)-0.3(-0.12) +0.5(0.10)=0.382.$$

Using the original input vector $(0.6,-0.4,0.2)$ with the adjoint coefficients gives

$$0.2+0.16(0.6)-0.09(-0.4) +0.25(0.2)=0.382.$$

Hence $\operatorname{Tr}[A\Phi(\rho)]=\operatorname{Tr}[\Phi^\dagger(A)\rho]$ in the explicit fixture. Agreement checks the PTM orientation as well as the adjoint action.

8. Price a full superoperator

For an $n$-qubit channel with equal input and output spaces, compute the number of entries and complex128 storage of a dense Liouville superoperator. Evaluate the result at five and ten qubits and explain why direct Kraus action can save memory.

Solution

The Hilbert-space dimension is $d=2^n$. Liouville space has dimension $d^2$, so a square dense superoperator contains

$$d^4=(2^n)^4=2^{4n}$$

complex entries. Complex128 stores 16 bytes per entry before allocator overhead, decompositions, or scratch arrays.

At five qubits, $2^{20}$ entries require $2^{24}$ bytes, exactly 16 MiB. At ten qubits, $2^{40}$ entries require $2^{44}$ bytes, exactly 16 TiB. Increasing the register from five to ten qubits multiplies storage by $2^{20}$.

These figures exclude temporary matrices used in composition, eigendecomposition, or reshuffling. Large-system work should use matrix-free Kraus action, locality, tensor-network structure, trajectories, or another problem-specific representation. Noise Simulation explains those scalable algorithms and their numerical checks.

References

• M.-D. Choi, “Completely Positive Linear Maps on Complex Matrices,” Linear Algebra and its Applications 10, 285–290 (1975), doi:10.1016/0024-3795(75)90075-0.
• T. F. Havel, “Robust Procedures for Converting among Lindblad, Kraus and Matrix Representations of Quantum Dynamical Semigroups,” Journal of Mathematical Physics 44, 534–557 (2003), doi:10.1063/1.1518555.
• A. Jamiołkowski, “Linear Transformations Which Preserve Trace and Positive Semidefiniteness of Operators,” Reports on Mathematical Physics 3, 275–278 (1972), doi:10.1016/0034-4877(72)90011-0.
• E. Nielsen, J. K. Gamble, K. Rudinger, T. Scholten, K. Young, and R. Blume-Kohout, “Gate Set Tomography,” Quantum 5, 557 (2021), doi:10.22331/q-2021-10-05-557.
• F. A. Pollock, C. Rodríguez-Rosario, T. Frauenheim, M. Paternostro, and K. Modi, “Non-Markovian Quantum Processes: Complete Framework and Efficient Characterization,” Physical Review A 97, 012127 (2018), doi:10.1103/PhysRevA.97.012127.
• M. B. Ruskai, S. Szarek, and E. Werner, “An Analysis of Completely-Positive Trace-Preserving Maps on M2,” Linear Algebra and its Applications 347, 159–187 (2002), doi:10.1016/S0024-3795(01)00547-X.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), doi:10.1017/9781316848142.
• C. J. Wood, J. D. Biamonte, and D. G. Cory, “Tensor Networks and Graphical Calculus for Open Quantum Systems,” Quantum Information and Computation 15, 759–811 (2015), doi:10.26421/QIC15.9-10-3.