Hilbert–Schmidt Operators

A Hilbert–Schmidt operator has square-summable matrix elements in an orthonormal basis. This defines a Hilbert space of operators, supplies a rigorous setting for square-integrable integral kernels, and ensures that products of two such operators are trace class. A Hilbert–Schmidt operator by itself need not have a trace.

Helpful background. Compact operators supplies singular values; positive operators supplies the meaning of $T^*T$.

The Hilbert–Schmidt norm

For a bounded operator $T$ on a separable Hilbert space $\mathcal H$, choose an orthonormal basis $(e_n)$ and define

$$\|T\|_2^2=\sum_n\|Te_n\|^2.$$

If this sum is finite, $T$ belongs to $\mathcal S_2(\mathcal H)$. The value is independent of the basis. To see this, insert another orthonormal basis $(f_m)$, use Parseval, and interchange the nonnegative sums:

$$\sum_n\|Te_n\|^2 =\sum_{m,n}|\langle f_m,Te_n\rangle|^2 =\sum_m\|T^*f_m\|^2.$$

The last expression no longer depends on $(e_n)$. Repeating the argument proves independence of both bases and $\|T^*\|_2=\|T\|_2$.

For the projections $P_N$ onto $\operatorname{span}\{e_1,\ldots,e_N\}$, the finite-rank operators $TP_N$ converge to $T$ in Hilbert–Schmidt norm. Since $\|S\|\leq\|S\|_2$, they also converge in operator norm. Thus every Hilbert–Schmidt operator is compact and

$$\|T\|_2^2=\sum_n s_n(T)^2.$$

Here and below, a subscript $2$ on an operator norm means Hilbert–Schmidt norm, not the induced Euclidean operator norm sometimes denoted by the same symbol in numerical linear algebra.

A Hilbert space of operators

With the inner product

$$\langle S,T\rangle_2=\operatorname{Tr}(S^*T),$$

the space $\mathcal S_2$ is complete. The product under the trace is trace class; this follows from the product estimate established below. Equivalently the inner product is the absolutely convergent sum $\sum_n\langle Se_n,Te_n\rangle$.

The rank-one operators $|e_j\rangle\langle e_k|$ form an orthonormal basis of this operator Hilbert space. This turns coefficient expansions into genuine norm-convergent expansions, without treating arbitrary bounded operators as square-summable vectors.

For bounded $A,B$,

$$\|ATB\|_2\leq\|A\|\,\|T\|_2\,\|B\|.$$

The left multiplication bound follows directly from the basis sum. The right bound follows by taking adjoints. Thus $\mathcal S_2$ is a two-sided ideal. These estimates allow bounded observables to act continuously on square roots of density operators.

Why two square-summable factors have a trace

If $S,T\in\mathcal S_2$, then $ST\in\mathcal S_1$ and

$$\|ST\|_1\leq\|S\|_2\|T\|_2.$$

For finite-rank operators, use trace-norm duality and Hilbert–Schmidt Cauchy–Schwarz:

$$|\operatorname{Tr}(UST)| \leq\|US\|_2\|T\|_2 \leq\|S\|_2\|T\|_2 \quad(\|U\|\leq1).$$

Taking the supremum gives the estimate. Approximate general $S,T$ by finite-rank operators in $\|\cdot\|_2$; the estimate makes their products Cauchy in trace norm, and the operator-norm limit identifies the limit as $ST$. This also proves the general statement without summing a conditionally convergent diagonal. Simon provides the full Schatten-ideal framework.

In particular, if $\rho$ is a density operator, then $\rho^{1/2}\in\mathcal S_2$ with $\|\rho^{1/2}\|_2=1$. This is why $\rho^{1/2}A\rho^{1/2}$ has a trace for every bounded observable $A$.

Square-integrable kernels

Let $(X,\mu)$ be a sigma-finite measure space and $K\in L^2(X\times X)$. The integral operator

$$(T_Kf)(x)=\int_XK(x,y)f(y)\,d\mu(y)$$

is Hilbert–Schmidt and satisfies $\|T_K\|_2=\|K\|_{L^2(X\times X)}$. Cauchy–Schwarz first shows boundedness; Parseval in the $y$ variable then identifies the norm. Kernels are equivalence classes, so their diagonal values are not automatically defined.

For example, $K(x,y)=xy$ on $[0,1]^2$ gives the rank-one operator $|f\rangle\langle f|$ with $f(x)=x$. Its Hilbert–Schmidt norm is $\|f\|^2=1/3$, agreeing with the double integral. By contrast, the diagonal operator $Te_n=n^{-1}e_n$ has $\|T\|_2^2=\pi^2/6$ but infinite positive trace. A Hilbert–Schmidt kernel therefore does not, without extra hypotheses, justify writing $\operatorname{Tr}T=\int K(x,x)\,dx$.

Exercises

1. For normalized $\psi,\phi$, compute the Hilbert–Schmidt distance between their pure-state projectors.

Solution

Expanding the square gives $\|P_\psi-P_\phi\|_2^2=2-2|\langle\psi,\phi\rangle|^2$. This relates operator-space geometry to the transition probability in Core Formalism. It is not the trace distance; the two norms agree only after the appropriate rank-dependent factor in this example.

2. Can an infinite-rank Hilbert–Schmidt operator be an isometry?

Solution

No. An isometry satisfies $\|Te_n\|=1$ for every orthonormal basis vector, so the defining square sum diverges on an infinite-dimensional domain.

References

• Holevo, Alexander S., Quantum Systems, Channels, Information: A Mathematical Introduction, 2nd revised and expanded ed., De Gruyter, 2019 — square roots of states and operator representations. Publisher.
• Simon, Barry, Trace Ideals and Their Applications, 2nd ed., Mathematical Surveys and Monographs 120, American Mathematical Society, 2005 — Hilbert–Schmidt norms, kernels, and ideal inequalities. Publisher.