1. Quick check §
Find the Schmidt probabilities of √0.9|00⟩+√0.1|11⟩.
Hint 1
Probabilities are squared amplitudes.
Worked solution
They are (0.9,0.1); their sum is one.
Quantum Information Theory · Chapter 04
Entanglement as a Resource
Which entangled states can separated laboratories convert into one another?
In this chapter · 3 sections
Reading guide · preparation and approach
Starting point: Chapters 1–3; Schmidt decomposition and reduced states.
Anchor example: Converting a Bell pair into a pair with Schmidt probabilities (0.8,0.2).
Method: Compare cumulative Schmidt probabilities and track allowed operations.
Watch for: One-copy exact conversion and asymptotic approximate conversion have different criteria.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
4.1 The allowed operations
4.1.1 Local control with classical messages
Alice and Bob share a state but cannot send quantum systems to one another. They may append local ancillas, perform local instruments, exchange classical messages, and choose later operations based on the transcript. These are local operations and classical communication, abbreviated LOCC. Shared randomness is allowed; shared entanglement is a resource to be counted. A joint entangling gate between their laboratories is not free.
A bipartite pure state has a Schmidt decomposition \(|\psi\rangle=\sum_i\sqrt{\lambda_i}|i\rangle_A|i\rangle_B\), where the nonnegative probabilities sum to one. Local basis changes do not change the \(\lambda_i\). The reduced-state entropy \(H(\lambda)\) quantifies pure-state entanglement in the asymptotic setting, but the full vector controls exact single-copy conversions.
For \(\sqrt{0.8}|00\rangle+\sqrt{0.2}|11\rangle\), the local entropy is approximately 0.721928 bits. Compare the separable mixture \(0.8|00\rangle\langle00|+0.2|11\rangle\langle11|\): its local entropy is the same, yet its entanglement is zero because Alice can sample a classical label and tell Bob which local state to prepare. The global coherence distinguishes the resources. Reduced-state entropy alone therefore fails as an entanglement measure for general mixed states.
Worked check 4.1. Why can LOCC prepare the correlated mixture above but not a Bell pair from a product input?
Reasoned solution
The mixture is obtained by shared classical randomness and local preparations. Each conditioned LOCC branch maps a product input to a product state, and forgetting branches gives a separable mixture. A Bell pair is not separable.
4.2 Majorization
4.2.1 The direction of deterministic conversion
Sort the Schmidt probabilities in descending order. Write \(\lambda\prec\mu\) if \(\sum_{i=1}^k\lambda_i\le\sum_{i=1}^k\mu_i\) for every proper prefix, with equal totals. Pad shorter vectors with zeros. Nielsen’s theorem states that a bipartite pure state with vector \(\lambda\) converts exactly and deterministically by LOCC to one with vector \(\mu\) if and only if \(\lambda\prec\mu\). We use the theorem; its full constructive proof is outside this chapter.
The direction is easy to reverse accidentally. A Bell vector \((0.5,0.5)\) is majorized by \((0.8,0.2)\), so the Bell pair can be degraded into that target. The reverse deterministic transformation is impossible. Entropy decreases, consistent with the operational restriction, but entropy alone is not sufficient to compare higher-dimensional vectors.
The two-dimensional conversion can be shown explicitly. On her half of a Bell pair, Alice measures \(A_0=\operatorname{diag}(\sqrt{0.8},\sqrt{0.2})\) and \(A_1=\operatorname{diag}(\sqrt{0.2},\sqrt{0.8})\). Their squared products sum to identity. Each outcome has probability one half. The first produces the target; the second produces its swapped version. Alice sends the outcome, and both parties apply X for the second outcome. Every branch then has the desired state.
Worked check 4.2. Are vectors (0.5,0.25,0.25) and (0.45,0.4,0.15) comparable by majorization?
Reasoned solution
No. The first prefix is 0.5 versus 0.45, but the second is 0.75 versus 0.85. The inequalities change direction, so neither exact deterministic LOCC conversion is possible.
4.3 Many copies and mixed states
4.3.1 What an entanglement rate promises
A probabilistic filter can concentrate one pair. For the state with probabilities \((0.8,0.2)\), Alice’s successful Kraus operator \(\operatorname{diag}(1/2,1)\) makes the two surviving amplitudes equal. The unnormalized state has amplitudes \(\sqrt{0.2}\) on both terms, so success probability is 0.4 and the normalized result is a Bell pair. A second Kraus operator completes the instrument. Failure must be reported; ignoring it would falsely turn a conditional success into a deterministic protocol.
With many independent copies of a fixed pure state, concentration and dilution become asymptotically reversible: the Bell-pair rate is \(H(\lambda)\), with vanishing error and unrestricted classical communication. This is a coding theorem, stated rather than proved here. The typical-subspace reasoning in the next chapter explains why entropy appears. It does not imply exact conversion of one copy into a fractional Bell pair.
Mixed states are harder. Distillable entanglement measures the asymptotic Bell-pair yield, while entanglement cost measures the Bell pairs needed to prepare the state. These need not agree. A useful entanglement witness is the partial transpose: every separable state has positive partial transpose, since transposition preserves each local factor’s positivity. A negative eigenvalue certifies entanglement. Positive partial transpose alone does not certify separability in general dimensions. This chapter establishes the test’s necessity directly, without asserting a general converse.
Worked check 4.3. Find a negative partial-transpose eigenvalue for a Bell state.
Reasoned solution
The partial transpose is swap/2. On the antisymmetric vector it has eigenvalue −1/2, proving entanglement. This repeats the same matrix calculation that ruled out transposition as a physical channel.
Summary
- LOCC defines which operations are free; a conversion statement must identify its resources.
- Pure-state deterministic conversion requires \(\lambda_{\rm input}\prec\lambda_{\rm output}\).
- Exact one-copy conversion, probabilistic filtering, and asymptotic rates are different tasks.
Questions and worked problems
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
2. Quick check §
Can (0.6,0.4) convert deterministically to (0.9,0.1)?
Hint 1
Check the first prefix.
Worked solution
Yes. 0.6≤0.9 and totals agree. The initial state is more entangled.
3. Quick check §
Can a product state convert deterministically to a Bell pair by LOCC?
Hint 1
Compare (1,0) with (0.5,0.5).
Worked solution
No: 1≤0.5 fails. The allowed operations cannot create entanglement from a product input.
4. Worked problem §
Find the success probability of the same equalizing filter on probabilities (0.9,0.1).
Hint 1
Scale the larger amplitude down to the smaller.
Worked solution
Use \(A=\operatorname{diag}(1/3,1)\). Each surviving squared amplitude is 0.1, so success probability is 0.2.
5. Worked problem §
Complete the filter diag(1/2,1) into a two-outcome instrument.
Hint 1
The squared operators must sum to I.
Worked solution
The failure operator can be \(\operatorname{diag}(\sqrt3/2,0)\). Its squared product fills the missing diagonal (3/4,0).
6. Worked problem §
Compute pure-state entanglement for equal Schmidt probabilities on four levels.
Hint 1
Use the entropy of the spectrum.
Worked solution
The entropy is log₂4=2 bits, corresponding asymptotically to two Bell pairs per copy.
7. Worked problem §
Prove that partial transpose preserves positivity for a separable state.
Hint 1
Transpose Bob’s factor term by term.
Worked solution
If \(\rho=\sum_jp_j\alpha_j\otimes\beta_j\), then \(\rho^{T_B}=\sum_jp_j\alpha_j\otimes\beta_j^T\). Each term is positive and the weights are nonnegative.
8. Misconception §
Does equal reduced-state entropy imply two mixed states have equal entanglement?
Hint 1
Compare a Bell pair and a shared classical bit.
Worked solution
No. Both have maximally mixed marginals, while the shared-bit mixture is separable and the Bell pair is entangled.
9. Misconception §
Does a concentration rate of 0.72 mean one copy yields exactly 0.72 Bell pairs?
Hint 1
Identify the asymptotic quantity.
Worked solution
No. It is the limiting ratio of output Bell pairs to input copies for large block protocols with vanishing error.
10. Extension §
Give two incomparable Schmidt vectors and check both prefixes.
Hint 1
Use three probabilities so that prefixes can cross.
Worked solution
For (0.5,0.25,0.25) and (0.45,0.4,0.15), the prefixes are 0.5>0.45 and 0.75<0.85. Neither vector majorizes the other.