Schmidt Decomposition Caveats

The Schmidt theorem concerns a normalizable pure vector and a specified bipartition. Its coefficients are countable even when coordinates are continuous, and its consequences do not transfer unchanged to mixed states or to three separate parties. This page extends the scope discussion in the canonical Schmidt Decomposition article.

Helpful background. Schmidt Decomposition in Hilbert Space proves the countable theorem and truncation bounds. Purification in Infinite Dimensions explains the support-isometry boundary.

Continuous coordinates do not make Schmidt labels continuous

For $\Psi\in L^2(\mathbb R^2,dx\,dy)$, the tensor identification with $L^2(\mathbb R)\otimes L^2(\mathbb R)$ gives

$$\Psi(x,y)=\sum_ns_nu_n(x)v_n(y)$$

in $L^2$ norm. The associated integral map is Hilbert–Schmidt because $\int|\Psi(x,y)|^2\,dx\,dy<\infty$. Its nonzero singular values form a discrete countable list tending to zero.

The equality need not hold pointwise for arbitrary chosen representatives, and termwise differentiation is not justified merely by norm convergence. A kinetic-energy calculation requires additional Sobolev or operator-domain control.

By contrast, the formal perfectly correlated kernel $\delta(x-y)$ is not in $L^2(\mathbb R^2)$. It represents the identity integral kernel on a suitable test space, rather than a normalizable bipartite vector. The identity on infinite-dimensional $L^2$ is not Hilbert–Schmidt. Thus a formal expression with a continuous family of equally weighted exact-position pairs lies outside the theorem’s hypotheses.

This is a normalization obstruction, not a failure of the Schmidt theorem for legitimate continuous-variable wave packets.

Nonzero coefficients do not record the whole ambient space

The Schmidt sum describes the supports of the two reduced states. Basis vectors in the kernels do not appear, but those kernel dimensions matter when asking for a unitary on a prescribed whole Hilbert space.

In finite dimensions, matching Schmidt coefficients with the fixed ambient dimensions determines local-unitary equivalence. In infinite dimensions, matching the nonzero list alone need not do so: one reduced state can have full support while another with the same positive eigenvalues has a nontrivial kernel.

A unitary must preserve that kernel dimension. With matching nonzero coefficients and matching support-complement dimensions on both sides, the paired support maps can be extended to whole-space unitaries. Without those complement conditions, the natural conclusion concerns isometries or partial isometries on the occupied supports. The explicit counterexample and support-map argument are given in Purification in Infinite Dimensions.

Degenerate nonzero coefficients allow paired basis rotations, but every such eigenspace has finite dimension for a normalizable vector. The zero coefficient is different: its multiplicity may be infinite.

An operator Schmidt expansion is not a separable ensemble

A density operator is trace class and hence Hilbert–Schmidt. It can therefore be treated as a vector in an operator Hilbert space and expanded in orthonormal Hilbert–Schmidt operator bases:

$$\rho=\sum_k t_k A_k\otimes B_k.$$

Here orthogonality uses $\operatorname{Tr}(A_j^*A_k)$, and convergence is initially Hilbert–Schmidt convergence. The factors need not be positive or trace one. Even when the original operator is a density operator, this is not a convex decomposition into physical product states.

For example,

$$\rho=\tfrac12|00\rangle\langle00| +\tfrac12|11\rangle\langle11|$$

has operator Schmidt rank two, since the two local rank-one projectors are orthonormal in Hilbert–Schmidt inner product. Yet the displayed expression is an explicit separable mixture. Operator Schmidt rank greater than one diagnoses nonproduct structure, not mixed-state entanglement.

Conversely, diagonalizing a mixed state’s density operator into entangled eigenvectors does not rule out a different product ensemble. The separability question is a convex-set question, developed in Separable and Entangled Mixed States.

Three parties need a declared cut

The theorem applies to $A:(BC)$, $(AB):C$, or another chosen bipartition. It does not generally produce a sum

$$\Psi=\sum_ns_n a_n\otimes b_n\otimes c_n$$

with all three local families orthonormal.

If such a form existed, all three single-party reduced states would have the same nonzero eigenvalues $s_n^2$. Take instead a Bell pair on $AB$ tensored with a pure state of $C$. Its $A$ and $B$ marginals have rank two, whereas its $C$ marginal has rank one. The proposed three-family form is impossible.

Nevertheless this state has an ordinary Schmidt decomposition across every bipartition: rank two across $A:(BC)$, rank two across $B:(AC)$, and rank one across $(AB):C$. There is no contradiction because the factors being paired have changed.

Rank accuracy is different from entropy accuracy

An infinite Schmidt rank can be well approximated in Hilbert norm by finite rank; the omitted squared weights give the exact best error. But a small omitted probability need not imply a small entropy contribution, since it may be spread over extremely many modes.

Consequently a truncation certificate for bounded measurement predictions is not automatically a certificate for entanglement entropy. The relevant additional tail or energy hypotheses belong to Entanglement Entropy Caveats. A numerical rank determined by a threshold is an approximation parameter, not an exact invariant of the state.

Exercises

1. Can a normalizable vector have infinitely many Schmidt coefficients all equal to a fixed positive number?

Solution

No: the sum of their squares diverges. A finite rank-$r$ maximally entangled state has coefficients $r^{-1/2}$, which change with $r$. Sending $r$ to infinity does not produce a normalized state with constant nonzero coefficients.

2. Why does convergence of wavefunctions in $L^2(\mathbb R^2)$ not guarantee convergence of their kinetic energies?

Solution

Kinetic energy is unbounded and depends on derivatives. A component with vanishing norm can occupy momenta whose squared magnitude grows fast enough to retain or increase its energy contribution. Control in the kinetic-energy form norm, rather than $L^2$ norm alone, is needed for an energy conclusion.

References

• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013 — Hilbert-space tensors and operator-domain distinctions. Publisher.
• Holevo, Alexander S., Quantum Systems, Channels, Information: A Mathematical Introduction, second revised and expanded edition, De Gruyter, 2019 — infinite-dimensional states, purification, and information quantities. Publisher.
• Watrous, John, The Theory of Quantum Information, Cambridge University Press, 2018 — vector and operator decompositions and the convex structure of entanglement. Author’s text.