Quantum Information for Black-Hole Physics

Black-hole information arguments combine ordinary quantum mechanics with claims about correlations, inaccessible subsystems, noisy evolution, and recovery. Those claims can be tested in finite-dimensional models before any horizon physics is added. This module supplies that bounded test: it is not a complete quantum-information course, and its toy transfer problem is not a model of black-hole evaporation.

Required background. Quantum States, Observables, and Measurement develops the required quantum-mechanics capability; use the Quantum Mechanics and Field Theory Diagnostic to check your understanding of states, operators, and unitary evolution. Calculus and Linear Algebra for Relativity supplies eigenvalues, traces, and tensor products.

Helpful background. States, Processes, and the Laws of Thermodynamics develops the recommended thermodynamics capability; the Thermodynamics and Statistical Physics Bridge connects entropy to equilibrium statistical mechanics.

You should leave able to compute and interpret density operators, entanglement entropy, quantum channels, and elementary recovery conditions, then apply those constraints to a deliberately limited black-hole-information analogue.

Quantum-information diagnostic

1. Let $\rho=(I+r\sigma_z)/2$. For which real $r$ is $\rho$ a density operator? Find its von Neumann entropy.
2. For $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$, compute the reduced state $\rho_A$ and the entanglement entropy across $A|B$.
3. A dephasing channel has Kraus operators $K_0=\sqrt{1-p}\,I$ and $K_1=\sqrt p\,Z$, with $0\leq p\leq1$. Verify trace preservation and determine what happens to the off-diagonal entries of a qubit state.
4. Suppose a unitary copier and a fixed blank state obey $U|\psi\rangle|0\rangle=|\psi\rangle|\psi\rangle$ for every pure state. Use preservation of inner products to show why this cannot hold for two arbitrary nonorthogonal states.
5. The three-qubit repetition code stores $\alpha|0\rangle+\beta|1\rangle$ as $\alpha|000\rangle+\beta|111\rangle$. If the second qubit flips, what are the outcomes of the stabilizers $Z_1Z_2$ and $Z_2Z_3$, and how is the state recovered?

Hints and solutions

Hints for 1

Diagonalize in the $\sigma_z$ basis. Positivity constrains both eigenvalues, and the entropy is $-\operatorname{tr}(\rho\ln\rho)$.

Solution to 1

The eigenvalues are

$$\lambda_\pm=\frac{1\pm r}{2}.$$

Hermiticity and unit trace already hold. Positivity requires $\lambda_\pm\geq0$, hence $|r|\leq1$. With the convention $0\ln0=0$,

$$S(\rho) =-\frac{1+r}{2}\ln\frac{1+r}{2} -\frac{1-r}{2}\ln\frac{1-r}{2}.$$

The state is pure at $|r|=1$ and maximally mixed at $r=0$, where $S=\ln2$.

Hints for 2

Expand $|\Phi^+\rangle\langle\Phi^+|$ and trace the second factor using $\operatorname{tr}(|i\rangle\langle j|)=\delta_{ij}$.

Solution to 2

The cross terms contain $|0\rangle_B\langle1|$ or $|1\rangle_B\langle0|$ and vanish under the partial trace. Therefore

$$\rho_A=\operatorname{tr}_B|\Phi^+\rangle\langle\Phi^+| =\frac12\bigl(|0\rangle\langle0|+|1\rangle\langle1|\bigr) =\frac I2.$$

Its entropy is $S(\rho_A)=\ln2$. The joint state is pure, so this nonzero reduced entropy measures entanglement rather than classical ignorance about a mixed joint preparation.

Hints for 3

Check $\sum_aK_a^\dagger K_a=I$. Write $\rho$ as a $2\times2$ matrix and use $Z=\operatorname{diag}(1,-1)$.

Solution to 3

The completeness relation is

$$K_0^\dagger K_0+K_1^\dagger K_1=(1-p)I+pI=I,$$

so the channel preserves trace. For

$$\rho=\begin{pmatrix}a&b\\ b^*&1-a\end{pmatrix},$$

one finds

$$\mathcal D_p(\rho)=(1-p)\rho+pZ\rho Z =\begin{pmatrix}a&(1-2p)b\\(1-2p)b^*&1-a\end{pmatrix}.$$

Populations are unchanged while coherences are multiplied by $1-2p$. At $p=1/2$ they vanish; at $p=1$ the channel is the unitary $Z$ operation, not maximal decoherence.

Hints for 4

Take the inner product of the proposed input-output relation for states $|\psi\rangle$ and $|\phi\rangle$.

Solution to 4

Unitarity preserves the input overlap:

$$\langle\psi|\phi\rangle =\bigl(\langle\psi|\langle0|\bigr) \bigl(|\phi\rangle|0\rangle\bigr).$$

The copied outputs would instead have overlap

$$\bigl(\langle\psi|\langle\psi|\bigr) \bigl(|\phi\rangle|\phi\rangle\bigr) =\langle\psi|\phi\rangle^2.$$

Thus $z=\langle\psi|\phi\rangle$ would have to satisfy $z=z^2$, which permits only $z=0$ or $z=1$. A single unitary can copy a mutually orthogonal set, but not arbitrary unknown nonorthogonal states.

Hints for 5

$X_2$ anticommutes with each stabilizer containing $Z_2$. The syndrome identifies the location without measuring the logical amplitudes.

Solution to 5

The encoded state has stabilizer eigenvalues $(+1,+1)$. After the error it is

$$X_2\bigl(\alpha|000\rangle+\beta|111\rangle\bigr) =\alpha|010\rangle+\beta|101\rangle.$$

Because $X_2$ anticommutes with both $Z_1Z_2$ and $Z_2Z_3$, the syndrome is $(-1,-1)$. Applying $X_2$ again restores the encoded state. The stabilizer measurements reveal the error sector but not $\alpha$ or $\beta$; recovery would fail if the noise lay outside the code’s correctable set.

Interpret the result

• Secure: you distinguish a joint state from its reductions, verify a channel rather than merely naming it, and state what information a recovery operation can and cannot restore.
• Uncertain: the physical interpretation is sound but a partial trace, Kraus check, or syndrome calculation needs one hint. Repeat that item with a changed parameter or error location.
• Not yet: negative eigenvalues are accepted as probabilities, reduced entropy is confused with entropy of the pure joint state, every completely positive map is assumed reversible, or no-cloning is treated as a ban on copying known orthogonal data.

The result is low-stakes and capability-specific. One weak item calls for focused review, not a verdict on readiness for the whole subject.

Focused review

States, reductions, and entropy

A density operator is positive, has unit trace, and acts on a declared Hilbert space. Composite systems use tensor products; discarding a subsystem means a partial trace, not deleting indices by inspection. Review Density Operators, Entanglement, and Reduced States and recompute one two-qubit reduction before continuing.

Channels and distinguishability

A quantum channel is linear, completely positive, and trace preserving. A Kraus representation $\mathcal E(\rho)=\sum_aK_a\rho K_a^\dagger$ is trace preserving exactly when $\sum_aK_a^\dagger K_a=I$. Noise cannot increase the operational distinguishability of two inputs; any proposed recovery must respect that data-processing constraint.

Recovery has hypotheses

Recovery is always relative to a code subspace and an error set. A successful decoder on selected states does not invert the channel on arbitrary inputs. State the encoding, accessible output, noise model, and success criterion before turning a recovery calculation into an information claim.

Statistical and thermodynamic entropy

Von Neumann entropy $S(\rho)=-\operatorname{tr}(\rho\ln\rho)$ depends on the state and the chosen subsystem. Its use in equilibrium ensembles is developed in Quantum Statistical Mechanics and Density Operators; horizon entropy adds physical and gravitational structure not supplied by the finite-dimensional definition alone.

Transfer task: a flagged erasure analogue

Let a reference qubit $R$ and a system qubit $B$ begin in $|\Phi^+\rangle_{RB}$. A flagged erasure channel sends $B$ unchanged with probability $1-\varepsilon$ and replaces it by a state $|e\rangle$, orthogonal to the qubit output space, with probability $\varepsilon$. Find $S(R)$, $S(B')$, $S(RB')$, and the mutual information $I(R:B')$. What does the result permit you to say about recovery?

Hint 1

The output is block diagonal:

$$\rho_{RB'}=(1-\varepsilon)|\Phi^+\rangle\langle\Phi^+| +\varepsilon\frac{I_R}{2}\otimes|e\rangle\langle e|.$$

For orthogonal blocks, include the binary mixing entropy $H_2(\varepsilon)$.

Hint 2

Compute each block’s internal entropy, then use $I(R:B')=S(R)+S(B')-S(RB')$.

Solution

The channel acts only on $B$, so $\rho_R=I_R/2$ and

$$S(R)=\ln2.$$

The $B'$ marginal has an erasure eigenvalue $\varepsilon$ and two qubit eigenvalues $(1-\varepsilon)/2$, hence

$$S(B')=H_2(\varepsilon)+(1-\varepsilon)\ln2.$$

The unerased joint block is pure, whereas the erased block contains the maximally mixed reference. Therefore

$$S(RB')=H_2(\varepsilon)+\varepsilon\ln2.$$

It follows that

$$I(R:B')=2(1-\varepsilon)\ln2.$$

The flag tells a decoder when loss occurred, and the unerased branch retains the original entanglement. In the erased branch, however, $B'$ carries no system correlated with $R$; an operation on $B'$ alone cannot recreate that lost entanglement. This calculation illustrates channel and recovery constraints only. Identifying $B$ with a black-hole interior or $B'$ with Hawking radiation would require a dynamical model, a factorization prescription, and a statement of which observables are accessible.

Common errors

• Calling every mixed reduced state a classically uncertain ensemble, even when it comes from a pure entangled state.
• Using entropy as though it were attached to a region without first specifying the state, algebra, regulator, and subsystem.
• Inferring reversibility from trace preservation.
• Claiming information recovery without naming the code space, error set, accessible degrees of freedom, and fidelity criterion.

Return to your route

After the changed transfer problem, return to Black-Hole Information, String and Holographic Black Holes, or AdS Black Holes and Holography.