Trace-Class and Hilbert-Schmidt Operators

Trace-class and Hilbert-Schmidt operators are special bounded operators on a Hilbert space whose singular values are summable in stronger ways than ordinary boundedness requires. They are the infinite-dimensional operator classes behind traces, density operators, purities, integral kernels, and partial traces.

The practical slogan is:

Boundedness controls action on vectors; trace-class and Hilbert-Schmidt conditions control sums over infinitely many basis directions.

In finite-dimensional Hilbert spaces every operator is trace-class and Hilbert-Schmidt. The distinction becomes essential in wave mechanics, quantum statistical mechanics, continuous-variable systems, and open-system theory.

Why Quantum Mechanics Needs Them

Quantum mechanics uses traces constantly:

$$\operatorname{Tr}\rho=1, \qquad \langle A\rangle = \operatorname{Tr}(\rho A), \qquad \rho_A = \operatorname{Tr}_B\rho_{AB}.$$

In finite dimensions these formulas are ordinary matrix operations. In infinite dimensions, not every bounded operator has a finite trace. For example, the identity operator on an infinite-dimensional separable Hilbert space is bounded, but

$$\operatorname{Tr}I = \sum_{n=1}^{\infty} \langle e_n\vert I e_n\rangle = \sum_{n=1}^{\infty}1 = \infty.$$

Trace-class operators are the operators for which the trace is genuinely finite and basis-independent. Hilbert-Schmidt operators are a slightly larger class with a finite square-sum norm. They are useful for kernels, purities, compactness estimates, and operator-space Hilbert structures.

Singular Values in Infinite Dimension

Let $T$ be a compact operator on a Hilbert space $\mathcal H$. Its absolute value is

$$\lvert T\rvert = (T^\dagger T)^{1/2}.$$

The singular values of $T$ are the eigenvalues of $\lvert T\rvert$, counted with multiplicity and listed as

$$s_1(T)\ge s_2(T)\ge\cdots\ge0.$$

This is the infinite-dimensional analogue of the singular values in Singular Value Decomposition. For trace-class and Hilbert-Schmidt operators, these singular values decay fast enough to make the sums below finite.

Hilbert-Schmidt Operators

An operator $T$ is Hilbert-Schmidt if

$$\lVert T\rVert_2^2 = \operatorname{Tr}(T^\dagger T) = \sum_{n=1}^{\infty} s_n(T)^2 < \infty.$$

Equivalently, for any orthonormal basis $\{e_n\}$,

$$\lVert T\rVert_2^2 = \sum_{m,n} \lvert\langle e_m\vert T e_n\rangle\rvert^2 < \infty.$$

The value is independent of the orthonormal basis. Hilbert-Schmidt operators form a Hilbert space with inner product

$$\langle S,T\rangle_{\mathrm{HS}} = \operatorname{Tr}(S^\dagger T).$$

This is an inner product on operators, not the original state Hilbert space. It is useful when operators themselves are treated as vectors, for example in finite-temperature methods, quantum information, and numerical approximations.

Trace-Class Operators

An operator $T$ is trace-class if

$$\lVert T\rVert_1 = \operatorname{Tr}\lvert T\rvert = \sum_{n=1}^{\infty} s_n(T) < \infty.$$

The norm $\lVert T\rVert_1$ is called the trace norm or nuclear norm. Trace-class is stronger than Hilbert-Schmidt:

$$\text{trace-class} \subset \text{Hilbert-Schmidt} \subset \text{compact} \subset \text{bounded}.$$

For a trace-class operator, the trace is defined by

$$\operatorname{Tr}T = \sum_{n=1}^{\infty} \langle e_n\vert T e_n\rangle,$$

and this series is absolutely convergent in the appropriate trace-class sense and independent of the orthonormal basis.

If $B$ is bounded and $T$ is trace-class, then $BT$ and $TB$ are trace-class, and

$$\operatorname{Tr}(BT) = \operatorname{Tr}(TB).$$

This is the infinite-dimensional version of the cyclic trace property used in density-operator calculations. The hypotheses matter: cyclicity is not a license to rearrange arbitrary unbounded or non-trace-class products.

Diagonal Examples on Sequence Space

Let $\mathcal H=\ell^2(\mathbb N)$ with orthonormal basis $\{e_n\}$, and define a diagonal operator

$$T e_n = a_n e_n.$$

For this normal diagonal example, the singular values are $\lvert a_n\rvert$.

If $a_n=1$, then $T=I$ is bounded but not Hilbert-Schmidt:

$$\sum_n \lvert a_n\rvert^2 = \sum_n 1 = \infty.$$

If $a_n=1/n$, then

$$\sum_{n=1}^{\infty} \frac{1}{n^2} < \infty, \qquad \sum_{n=1}^{\infty} \frac{1}{n} = \infty.$$

So $T$ is Hilbert-Schmidt but not trace-class.

If $a_n=1/n^2$, then both sums converge, so $T$ is trace-class.

These examples show why infinite-dimensional trace formulas need more than boundedness.

Rank-One and Finite-Rank Operators

For $\phi,\psi\in\mathcal H$, the rank-one operator

$$T = \lvert\phi\rangle\langle\psi\rvert$$

acts by

$$T\eta = \phi\,\langle\psi\vert\eta\rangle.$$

It is trace-class and Hilbert-Schmidt. Its Hilbert-Schmidt norm is

$$\lVert T\rVert_2 = \lVert\phi\rVert\,\lVert\psi\rVert,$$

and its trace is

$$\operatorname{Tr}T = \langle\psi\vert\phi\rangle.$$

Finite-rank operators are finite sums of rank-one operators, so they are trace-class. Infinite-dimensional trace-class operators can be approximated in trace norm by finite-rank operators, which is one reason finite-dimensional intuition remains useful when applied carefully.

Density Operators

In an infinite-dimensional Hilbert space, a density operator is not merely a positive bounded operator with formal trace one. It is a positive trace-class operator $\rho$ satisfying

$$\rho\ge0, \qquad \operatorname{Tr}\rho=1.$$

If

$$\rho = \sum_n p_n \lvert n\rangle\langle n\rvert$$

in an orthonormal eigenbasis, then

$$p_n\ge0, \qquad \sum_n p_n=1.$$

Every density operator is Hilbert-Schmidt because

$$\operatorname{Tr}(\rho^2) = \sum_n p_n^2 \le \sum_n p_n = 1.$$

The quantity $\operatorname{Tr}(\rho^2)$ is the purity. It equals $1$ for pure states and is smaller for mixed states, when the usual finite-dimensional interpretation applies. The physics of density operators is developed in Density Operators; this page records the functional-analytic condition that makes their traces finite.

Expectation Values

If $\rho$ is trace-class and $A$ is bounded, then $\rho A$ is trace-class and the expectation value

$$\operatorname{Tr}(\rho A)$$

is well-defined. This is the mathematical condition behind the trace rule in Trace Rule for Expectation Values.

If $A$ is unbounded, extra domain and integrability conditions are required. It is not enough to write $\operatorname{Tr}(\rho A)$ formally. The spectral-measure condition is the density-operator analogue of asking whether a state has finite expectation value for an unbounded observable; see Domains of Operators and Spectral Theorem, Practical Version.

Integral Kernels

Many wave-mechanics operators are represented by kernels:

$$(T_K\psi)(x) = \int K(x,y)\psi(y)\,dy.$$

If

$$\int \int \lvert K(x,y)\rvert^2\,dx\,dy < \infty,$$

then $T_K$ is Hilbert-Schmidt and

$$\lVert T_K\rVert_2^2 = \int \int \lvert K(x,y)\rvert^2\,dx\,dy.$$

Trace-class is stronger and cannot be checked merely by looking at the square-integrability of the kernel. Under suitable additional conditions, the trace of a positive trace-class integral operator may be computed from its diagonal kernel, but this is a theorem with hypotheses, not a general rule for every formal kernel.

Products of Hilbert-Schmidt Operators

Hilbert-Schmidt operators are not always trace-class, but products of two Hilbert-Schmidt operators are trace-class. If $S$ and $T$ are Hilbert-Schmidt, then $S^\dagger T$ is trace-class and

$$\lvert\operatorname{Tr}(S^\dagger T)\rvert \le \lVert S\rVert_2\lVert T\rVert_2.$$

This is why the Hilbert-Schmidt inner product is well-defined. It is also the operator analogue of the Cauchy-Schwarz inequality.

Partial Trace Preview

For finite-dimensional composite systems, the partial trace can be computed by summing over an orthonormal basis of the discarded subsystem. In infinite dimensions, the correct input class is trace-class.

If $T_{AB}$ is trace-class on

$$\mathcal H_A\otimes\mathcal H_B,$$

then $\operatorname{Tr}_B T_{AB}$ is the unique trace-class operator on $\mathcal H_A$ satisfying

$$\operatorname{Tr}_A \bigl[ (\operatorname{Tr}_B T_{AB})M_A \bigr] = \operatorname{Tr}_{AB} \bigl[ T_{AB}(M_A\otimes I_B) \bigr]$$

for every bounded operator $M_A$ on $\mathcal H_A$.

For density operators, this says that reducing a state by partial trace preserves local expectation values and produces another trace-class density operator. The computational rules and finite-dimensional examples are developed in Partial Trace.

Common Mistakes

• Assuming every bounded operator has a finite trace.
• Treating the identity on an infinite-dimensional Hilbert space as a density operator after “normalizing by infinity.”
• Using the cyclic trace property without trace-class or boundedness hypotheses.
• Confusing the trace norm $\lVert T\rVert_1$ with the Hilbert-space norm of a vector.
• Assuming every Hilbert-Schmidt operator is trace-class.
• Computing an infinite-dimensional partial trace without checking that the joint operator is trace-class.
• Writing $\operatorname{Tr}(\rho A)$ for an unbounded $A$ without checking domains or spectral integrability.

Cross-Links

• Bounded Operators
• Adjoint Operators
• Singular Value Decomposition
• Spectral Theorem, Practical Version
• Domains of Operators
• Tensor Products
• Density Operators
• Trace Rule for Expectation Values
• Partial Trace
• Reduced Density Operators

References

• B. Simon, Trace Ideals and Their Applications, 2nd ed., American Mathematical Society, 2005.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, Academic Press, 1980.
• J. B. Conway, A Course in Functional Analysis, 2nd ed., Springer, 1990.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018.

Exercises

1. Let $T e_n=(1/n)e_n$ on $\ell^2(\mathbb N)$. Is $T$ bounded, Hilbert-Schmidt, trace-class?

Solution

The operator is bounded because $\sup_n 1/n=1$. It is Hilbert-Schmidt because

$$\sum_{n=1}^{\infty} \frac{1}{n^2} < \infty.$$

It is not trace-class because

$$\sum_{n=1}^{\infty} \frac{1}{n} = \infty.$$

2. Show that the pure-state projector $\rho_\psi=\lvert\psi\rangle\langle\psi\rvert$ is trace-class when $\lVert\psi\rVert=1$.

Solution

It is rank one, hence finite-rank, hence trace-class. Its trace is

$$\operatorname{Tr} \bigl( \lvert\psi\rangle\langle\psi\rvert \bigr) = \langle\psi\vert\psi\rangle = 1.$$

3. Let $\rho=\sum_n p_n\lvert n\rangle\langle n\rvert$ with $p_n\ge0$ and $\sum_n p_n=1$. Show that $\rho$ is Hilbert-Schmidt.

Solution

For a positive diagonal density operator, the Hilbert-Schmidt norm squared is

$$\lVert\rho\rVert_2^2 = \operatorname{Tr}(\rho^2) = \sum_n p_n^2.$$

Since $0\le p_n\le1$ and $\sum_n p_n=1$,

$$\sum_n p_n^2 \le \sum_n p_n = 1.$$

Thus $\rho$ is Hilbert-Schmidt.

4. Suppose $A$ is bounded on $\mathcal H_A$ and $B$ is trace-class on $\mathcal H_B$. Show that $\operatorname{Tr}_B(A\otimes B)=A\,\operatorname{Tr}B$.

Solution

For any bounded $M_A$,

$$\begin{aligned} \operatorname{Tr}_A \bigl[ (A\,\operatorname{Tr}B)M_A \bigr] &= \operatorname{Tr}_A(AM_A)\operatorname{Tr}_B B\\ &= \operatorname{Tr}_{AB} \bigl[ (A\otimes B)(M_A\otimes I_B) \bigr]. \end{aligned}$$

By the characterizing property of the partial trace, the reduced operator is $A\,\operatorname{Tr}B$.