Reference, Problems, and Notebooks

Use this chapter to recall a formula, choose an entanglement test, or practice a calculation. The worked problems develop the same ideas from several representations; the computational examples help reveal errors that are easy to miss by looking only at eigenvalues or normalization.

Helpful background. Tensor products fix the basis order, and reduced states explain what a partial trace must reproduce.

Find a formula or a problem

Your task	Useful page
Recall identities and conventions	Formula sheet
Recognize a standard state family	Common composite states
Choose a test for a given state class	Entanglement diagnostic table
Practice basis ordering and local operators	Tensor-product exercises
Practice reductions and conditional predictions	Partial-trace exercises
Work with exchange symmetry and occupations	Identical-particle exercises and Fock-space exercises
Implement and compare numerical calculations	Computational notebooks

If a formula’s assumptions are unfamiliar, follow its link to the derivation before substituting numbers. For practice, try a problem before opening the solution, then identify the first incorrect assumption or algebraic step rather than checking only the final answer.

Choose a test that can answer the question

A pure bipartite state is entangled exactly when its Schmidt rank exceeds one. For a mixed state, several nonzero eigenvalues of a reduced state are not enough: even a product of two thermal states has mixed marginals.

For arbitrary two-qubit or qubit–qutrit states, a positive partial transpose (PPT) is necessary and sufficient for separability. In larger bipartite dimensions, a negative partial transpose still proves entanglement, but a positive one can be inconclusive. An entanglement witness can be useful with limited measurements; failure to violate its bound does not prove separability. For a two-mode Gaussian state with one mode per party, the covariance version of PPT is complete, provided the quadrature convention and Gaussian assumption are specified. The diagnostic table compares these choices.

A complex example catches a hidden transpose

For finite product bases, write

$$|\psi\rangle=\sum_{i,j}C_{ij}|i\rangle_A|j\rangle_B, \qquad \sum_{i,j}|C_{ij}|^2=1.$$

The partial traces give

$$\rho_A=CC^\dagger,\qquad \rho_B=C^{\mathsf T}C^*=(C^\dagger C)^{\mathsf T}.$$

The second transpose follows directly from $(\rho_B)_{jk}=\sum_i C_{ij}C_{ik}^*$. The nonzero eigenvalues of both reduced states are the squared singular values of $C$, but the matrix $C^\dagger C$ itself need not be $\rho_B$ in the stated basis. For example,

$$|\psi\rangle=|0\rangle_A\frac{|0\rangle_B+i|1\rangle_B}{\sqrt2}, \qquad C=\frac1{\sqrt2}\begin{pmatrix}1&i\\0&0\end{pmatrix},$$

gives

$$\rho_B=\frac12\begin{pmatrix}1&-i\\i&1\end{pmatrix}, \qquad C^\dagger C=\frac12\begin{pmatrix}1&i\\-i&1\end{pmatrix}.$$

Both matrices are positive, have trace one, and have eigenvalues $1,0$. Yet their predictions for $\sigma_y$ are $+1$ and $-1$. A test using only real states or eigenvalues would miss the error. Schmidt decomposition explains the conjugation of the right singular vectors in the corresponding state vectors.

Check the physical predictions of a calculation

For a numerical density matrix, inspect its Hermiticity, trace, and smallest eigenvalue, using tolerances appropriate to the numerical precision and matrix scale. Record the residuals before making corrections: silently clipping a substantial negative eigenvalue can disguise an incorrect calculation.

A partial trace should also reproduce local expectations:

$$\operatorname{Tr}(\rho_A O_A) =\operatorname{Tr}[\rho_{AB}(O_A\otimes I_B)].$$

For a qubit, testing $I,\sigma_x,\sigma_y,\sigma_z$ checks the entire local operator space. Use a nonsymmetric state as well as a Bell state: the latter’s identical marginals cannot reveal that the wrong subsystem was discarded. For large systems, selected local observables supply useful checks without establishing correctness on every possible input.

Singular values are more stable comparison targets than individual singular vectors. When values are degenerate, rotations within the corresponding subspace change the vectors without changing the state decomposition’s content. A computational comparison should allow these rotations and arbitrary vector phases. Golub and Van Loan discuss the underlying matrix methods; the notebooks apply them to composite states.

Exercises

Positivity is an independent requirement

For real $\varepsilon$, consider

$$\rho_\varepsilon=\begin{pmatrix}0.6&0.2+i\varepsilon\\0.2-i\varepsilon&0.4\end{pmatrix}.$$

It is Hermitian and has trace one. Find the values of $\varepsilon$ for which it is a density matrix.

Solution

Its diagonal entries are positive and its determinant is $0.24-(0.04+\varepsilon^2)=0.20-\varepsilon^2$. A Hermitian $2\times2$ matrix with these diagonal entries is positive semidefinite exactly when this determinant is nonnegative. Thus $|\varepsilon|\leq\sqrt{0.20}$; the endpoint is allowed and gives a rank-one state.

Why three Pauli expectations suffice

Show that two normalized Hermitian qubit matrices with equal expectations of $\sigma_x,\sigma_y,\sigma_z$ are the same matrix. Why must the complex example above include the $y$ test?

Solution

Every normalized Hermitian qubit matrix has the expansion $\rho=(I+r_x\sigma_x+r_y\sigma_y+r_z\sigma_z)/2$, where $r_j=\operatorname{Tr}(\rho\sigma_j)$. Equality of all three components therefore fixes $\rho$. In the example, the two matrices have equal $x$ and $z$ components but opposite $y$ components; the imaginary coherence is precisely what those real-observable tests fail to distinguish.

A test that distinguishes the two subsystems

Use $|0\rangle_A|+y\rangle_B$, where $|+y\rangle=(|0\rangle+i|1\rangle)/\sqrt2$, as input to a routine intended to discard $B$. What local expectations would reveal that it instead returned $\rho_B$?

Solution

The correct output $|0\rangle\langle0|$ has $\langle\sigma_z\rangle=1$ and $\langle\sigma_y\rangle=0$. The wrong marginal $|+y\rangle\langle+y|$ instead gives $0$ and $1$. Both are pure normalized positive matrices, so those structural checks alone cannot identify the exchanged subsystem.

References

• G. H. Golub and C. F. Van Loan, Matrix Computations, 4th ed., Johns Hopkins University Press, 2013.
• O. Gühne and G. Tóth, “Entanglement detection,” Physics Reports 474, 1–75 (2009). DOI.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010.
• R. Simon, “Peres–Horodecki separability criterion for continuous variable systems,” Physical Review Letters 84, 2726–2729 (2000). DOI.