advanced27 min studyLesson 423 in Physics

Von Neumann entropy and purification

Compute entropy from spectra and use an explicit tensor factor and partial trace to distinguish local uncertainty from global mixedness.

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What you’ll understand

  • Calculate von Neumann entropy in bits from eigenvalues.
  • Construct a finite-dimensional purification and trace the correct factor.
  • Explain why a mixed marginal need not imply a mixed global state.

A random measurement outcome does not by itself mean that a state is mixed. A known pure qubit can produce a fair bit in an incompatible measurement basis. Von Neumann entropy instead uses the density operator's eigenvalues, so it is unchanged by a unitary change of basis. Purification adds a second perspective: uncertainty for a restricted observer can coexist with a completely pure description of a larger system. Keeping the observer's accessible tensor factor explicit resolves the apparent conflict.

Entropy belongs to a spectrum

Let the nonnegative eigenvalues of a normalized finite-dimensional density operator be lambda values summing to one. Define entropy by the negative sum of each eigenvalue times its base-two logarithm, with zero times log zero assigned its limiting value zero. The result is measured in bits. It vanishes precisely for a pure state and reaches log base two of dimension for the maximally mixed state. Multiplication by Boltzmann's constant and conversion to natural logarithms gives a thermodynamic entropy convention, which we do not use here.

A projective measurement in the eigenbasis has outcome Shannon entropy equal to this spectral entropy. Other complete rank-one projective bases can produce larger outcome entropy. For example, a pure plus state has zero von Neumann entropy but yields one bit in the zero–one basis. This distinction prevents a common mistake: inserting measurement frequencies from a single arbitrary basis into the spectral formula does not reconstruct the entropy of an unknown quantum state.

S(ρ)=−Tr⁡(ρlog⁡2ρ)=−∑iλilog⁡2λi,0≤S(ρ)≤log⁡2dS(\rho)=-\operatorname{Tr}(\rho\log_2\rho)=-\sum_i\lambda_i\log_2\lambda_i,\qquad 0\le S(\rho)\le\log_2 d
The dimension d is finite, eigenvalues are dimensionless, and 0 log 0 is defined by continuity. S is information in bits, without a Boltzmann constant.↔ Scroll wide equations horizontally.

Introduce a reference and trace it out

Write a spectral decomposition on system A and introduce orthonormal reference states on factor R. Weight each product eigenvector by the square root of the corresponding eigenvalue. The resulting normalized joint vector purifies the original density operator. In the outer product, cross terms carry reference operators with different labels. Taking their reference trace gives zero because the reference basis is orthonormal, leaving exactly the original weighted projectors on A.

The partial trace is defined by preserving every expectation of an A-only observable. In components, sum matrix elements whose two reference indices agree; do not merely select the upper block of the joint matrix. A reference dimension at least the rank of the state is sufficient. Purifications are not unique: a unitary on the reference leaves the A marginal unchanged. More generally the corresponding statement uses an isometry when reference spaces have different dimensions.

∣Ψ⟩AR=∑iλi∣i⟩A∣i⟩R,ρA=Tr⁡R∣Ψ⟩⟨Ψ∣,(ρA)ij=∑r(ρAR)ir,jr|\Psi\rangle_{AR}=\sum_i\sqrt{\lambda_i}|i\rangle_A|i\rangle_R,\qquad \rho_A=\operatorname{Tr}_R|\Psi\rangle\langle\Psi|,\qquad (\rho_A)_{ij}=\sum_r(\rho_{AR})_{ir,jr}
A and R name independent tensor factors. The component identity holds in a product basis, and the joint purification has entropy zero.↔ Scroll wide equations horizontally.

Entanglement and ordinary mixing differ

For a pure bipartite state, both reduced states share the same nonzero Schmidt eigenvalues. Their equal entropy quantifies pure-state entanglement across that specified partition. A product pure state gives zero; a maximally entangled pair of qubits gives one bit locally while the joint entropy remains zero. Entropy here measures inaccessible correlations with the other factor rather than uncertainty about which global pure vector was prepared.

For a mixed joint state, the local entropy alone is not an entanglement measure. A fair classical mixture of the product states zero-zero and one-one has the same one-bit local entropies as a Bell pair, but its joint entropy is one bit and it is separable. Even an uncorrelated product of maximally mixed qubits has those local entropies. Additional global information is therefore necessary to distinguish entanglement, classical correlation, and independent noise. The Hilbert-space partition and allowed operations are part of any entanglement claim.

Purify a biased mixed qubit

Purify ρA=diag⁡(3/4,1/4)\rho_A=\operatorname{diag}(3/4,1/4) using one reference qubit. Compute the two local entropies and the joint entropy.

  1. Choose ∣Ψ⟩AR=3∣00⟩/2+∣11⟩/2|\Psi\rangle_{AR}=\sqrt3|00\rangle/2+|11\rangle/2. The squared coefficients sum to one, so this is a normalized vector on the ordered factors A and R.

  2. In the outer product, tracing R removes ∣0⟩R⟨1∣|0\rangle_R\langle1| and its adjoint, while the diagonal reference projectors each have trace one. Thus the A marginal is the required diagonal matrix, and the R marginal has the same eigenvalues.

  3. Compute −(3/4)log⁡2(3/4)−(1/4)log⁡2(1/4)=2−(3/4)log⁡23≈0.811278-(3/4)\log_2(3/4)-(1/4)\log_2(1/4)=2-(3/4)\log_2 3\approx0.811278 bits. The joint operator is a rank-one projector with eigenvalues one and zeros, so its entropy is zero.

The result

Both marginals have entropy about 0.811278 bits; the joint purification has entropy zero and is entangled across A:R.

Explore the model

Interactive lab

Interactive lab · probability

Classical spectral weights

For each fresh comparison, Reset and set both controls. Reset and set success probability p to 0.25 and trials n to 1. Predict bar weights 0.75 and 0.25; use these as the two spectral weights in the worked entropy calculation. Repeat a fresh run with both controls p=0.5 and n=1 to obtain equal weights and one bit of binary entropy by hand. These exact binomial bars are classical probabilities; the lab displays no density matrix, entropy, purification, or entanglement.

Classical spectral weightsThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA classical probability spectrum can support an entropy calculation without representing the coherences of a purification.

Check your understanding

A joint pure state has a reduced qubit state I/2. What follows across that bipartition?

Try it yourself

Compare the Bell state (∣00⟩+∣11⟩)/2(|00\rangle+|11\rangle)/\sqrt2 with the mixture (∣00⟩⟨00∣+∣11⟩⟨11∣)/2(|00\rangle\langle00|+|11\rangle\langle11|)/2. Find S(A), S(B), and S(AB) for each.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Their marginals agree, but their joint ranks and spectra differ.

Compare with the solution

Both have marginals I/2, so S(A)=S(B)=1 bit. The Bell projector has joint spectrum (1,0,0,0), giving S(AB)=0. The mixture has spectrum (1/2,1/2,0,0), giving S(AB)=1 bit. It is explicitly a convex mixture of product projectors and is separable. Matching all single-subsystem statistics therefore does not determine whether the joint state is entangled.

Take it with you

  • Von Neumann entropy uses eigenvalues and a declared logarithm base.
  • A purification lives on a larger tensor product and reproduces the state by partial trace.
  • Local mixedness quantifies entanglement for a globally pure state, with no such conclusion for a general mixed state.

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