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advanced32 min studyLesson 367 in Physics

Schmidt decomposition and bipartite purity

Use singular values to expose entanglement in a pure bipartite state and calculate its subsystem purity independently of local bases.

Before you begin
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The question for today

What you’ll understand

  • Find Schmidt coefficients from a coefficient matrix or reduced-state spectrum.
  • Use Schmidt rank to distinguish product and entangled pure states.
  • Calculate local purity and explain the limits of this pure-state criterion.

Entanglement in a pure state can be hidden by an awkward product basis. The Schmidt decomposition chooses local orthonormal bases that reveal its invariant weights. Those weights are simultaneously the nonzero eigenvalues of either reduced density operator. They make a direct connection between a global pure state's inability to factor and the mixedness seen locally, while clarifying why the same test cannot diagnose arbitrary mixed composite states without further information.

Singular values identify the useful local bases

Write the amplitudes of a normalized bipartite pure state as a rectangular matrix C. Singular-value decomposition separates that matrix into local unitary factors and nonnegative singular values. The resulting state is a sum of paired local orthonormal vectors, with one nonnegative coefficient per pair. Unlike the raw coefficients in an arbitrary product basis, these Schmidt coefficients are invariant under local unitary changes of coordinates or local physical unitary operations.

Coefficient conventions matter when amplitudes are complex. If a numerical SVD is written as C=UΣV†C=U\Sigma V^\dagger, the B-side Schmidt kets have the complex conjugates of V's columns in the original B basis. This conjugation converts the matrix's dagger contraction into the tensor-product coefficient expansion. In real examples it is invisible, which makes it an easy mistake to carry into complex calculations. Degenerate Schmidt values also allow paired local-basis rotations, so the coefficients can be unique while the chosen vectors are not.

∣Ψ⟩=∑r=1Rsr∣ur⟩A∣vr⟩B,sr>0,∑rsr2=1,ρA=∑rsr2∣ur⟩⟨ur∣|\Psi\rangle=\sum_{r=1}^{R}s_r|u_r\rangle_A|v_r\rangle_B,\qquad s_r>0,\quad\sum_rs_r^2=1,\qquad \rho_A=\sum_rs_r^2|u_r\rangle\langle u_r|
The local Schmidt vectors are orthonormal, the dimensionless sr are the positive singular values of the normalized coefficient matrix, and R is Schmidt rank. The formula applies to a pure bipartite state with the stated tensor factorization. Zero coefficients can be omitted.↔ Scroll wide equations horizontally.

Local mixedness has a precise pure-state interpretation

Tracing either side removes cross terms between orthogonal partner vectors. Both reduced operators therefore have the same nonzero eigenvalues, the squared Schmidt coefficients, even when their Hilbert spaces have different dimensions. A single nonzero coefficient gives a product state with pure reductions. More than one gives an entangled pure state with mixed reductions. Counting nonzero amplitudes in an arbitrary basis is not a substitute: a product state can have many nonzero product-basis components.

Local purity is the sum of the fourth powers of the Schmidt coefficients. Equal weights over R terms give the smallest purity permitted at that rank, namely one divided by R. The von Neumann entropy of either reduction provides another pure-state entanglement measure, expressed in bits when logarithms are base two. Equal local entropies here reflect the pure global assumption; a generic mixed composite state need not have equal marginal entropies.

tr⁡(ρA2)=∑rsr4,S(ρA)=−∑rsr2log⁡2(sr2),R=1 ⟺ ∣Ψ⟩ is a product state\operatorname{tr}(\rho_A^2)=\sum_rs_r^4,\qquad S(\rho_A)=-\sum_rs_r^2\log_2(s_r^2),\qquad R=1\ \Longleftrightarrow\ |\Psi\rangle\ \text{is a product state}
Purity is dimensionless and S is in bits. The product-state equivalence and interpretation of S as entanglement concern normalized pure states of two specified subsystems. Schmidt rank is at most the smaller subsystem dimension.↔ Scroll wide equations horizontally.

Which features depend on the assumptions?

A classical mixture of |00⟩ and |11⟩ can have the same maximally mixed reductions as a Bell state while having no entanglement. Local purity alone cannot distinguish them if global purity is not established. The Schmidt decomposition is a theorem about a vector, not a unique decomposition of an arbitrary mixed operator into paired vectors. Testing entanglement in mixed states needs additional criteria beyond the present calculation.

The decomposition also does not extend in the same simple paired form to every pure state of three or more subsystems. Grouping several subsystems into one side restores a bipartition, but the resulting Schmidt data describe that selected cut. In practice, first declare the tensor factorization and normalization, then compute a reduced spectrum or SVD. Finally check for degeneracies and zeros before interpreting numerical coefficients as distinct physical correlation modes.

Work it through

A Bell structure in a rotated local basis

Find a Schmidt decomposition of ∣Ψ⟩=(∣00⟩+∣01⟩+∣10⟩−∣11⟩)/2|\Psi\rangle=(|00\rangle+|01\rangle+|10\rangle-|11\rangle)/2. Compute its reduced purity and entanglement entropy in bits.

  1. The coefficient matrix is C=12(111−1)C=\tfrac12\begin{pmatrix}1&1\\1&-1\end{pmatrix}. Multiplying gives CC†=I/2CC^\dagger=I/2, so its two singular values are both 1/21/\sqrt2. This immediately fixes both reduced spectra.

  2. Group terms by the A label: ∣Ψ⟩=(∣0⟩∣+⟩+∣1⟩∣−⟩)/2|\Psi\rangle=(|0\rangle|+\rangle+|1\rangle|-\rangle)/\sqrt2. The computational A vectors and X-basis B vectors are orthonormal, making this a Schmidt decomposition with rank two.

  3. Purity is 2(1/2)4=1/22(1/\sqrt2)^4=1/2, and entropy is −2(1/2)log⁡2(1/2)=1-2(1/2)\log_2(1/2)=1 bit. All four original product-basis amplitudes were nonzero, but only two Schmidt terms are needed.

The result

The Schmidt weights are 1/2 and 1/2, with rank two, local purity 1/2, and entanglement entropy one bit. The basis rotation on B exposes the structure without changing those invariant weights.

The living notebook

Explore the model

Interactive lab

Interactive lab · matrix

Normalize singular values into Schmidt coefficients

Reset and set both controls to a=1 and b=0. The selected matrix has singular values 1 and 1; after dividing by their Euclidean norm, predict Schmidt weights 1/2 and 1/2. For a fresh comparison, Reset and set both controls to a=2 and b=0. Singular values 2 and 1 give weights 4/5 and 1/5 and local purity 17/25. These quantum quantities are calculated from the readouts, not reported by the instrument.

Normalize singular values into Schmidt coefficientsThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. 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 forSingular values become Schmidt coefficients only after state normalization. The triangular family is always invertible, so it cannot display an exact rank-one product coefficient matrix or serve as a general entanglement solver.

Pause & reason

Check your understanding

For a normalized pure state of two specified subsystems, what proves that it is a product state?

Your turn

Try it yourself

A two-qubit pure state has normalized coefficient matrix C=110(1212)C=\frac1{\sqrt{10}}\begin{pmatrix}1&2\\1&2\end{pmatrix}. Determine its Schmidt rank, factor it if possible, and find local purity.

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

Give me a hint

The rows are identical, so the matrix has rank one. Look for a normalized row-direction vector and a normalized column-direction vector.

Compare with the solution

The state factors as [(∣0⟩+∣1⟩)/2]⊗[(∣0⟩+2∣1⟩)/5][(|0\rangle+|1\rangle)/\sqrt2]\otimes[(|0\rangle+2|1\rangle)/\sqrt5]. Thus its only positive singular value is one, its Schmidt rank is one, and both reduced purities equal one. All four coefficients are nonzero, demonstrating that amplitude counting in a fixed product basis is not an entanglement criterion. The total squared matrix norm is (1+4+1+4)/10=1, consistent with the factorization.

Keep the idea

Take it with you

  • Schmidt coefficients are normalized singular values of a pure-state coefficient matrix.
  • For pure bipartite states, Schmidt rank above one is equivalent to mixed reduced states.
  • Local mixedness alone does not certify entanglement when the global state may be mixed.

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