Hilbert Space Tensor Products

A tensor product of bounded operators extends uniquely from product vectors to the completed composite Hilbert space. An unbounded local observable requires a domain that includes more than finite sums of product vectors. These two extension problems must be distinguished. The canonical Tensor Products of Hilbert Spaces article supplies the composition rule and coordinate conventions; this page supplies the operator estimates behind that rule.

Helpful background. Algebraic Tensor Products and Completion constructs the space. Bounded Operators and The Unbounded Spectral Theorem supply the two operator settings.

Bounded tensor operators extend continuously

Let $A:\mathcal H_A\to\mathcal K_A$ and $B:\mathcal H_B\to\mathcal K_B$ be bounded. On algebraic tensors define

$$(A\otimes B)\sum_jx_j\otimes y_j =\sum_jAx_j\otimes By_j.$$

Bilinearity makes this well-defined. To prove boundedness, first take an orthonormal expansion in the unchanged factor. A finite algebraic tensor can be written $z=\sum_j e_j\otimes y_j$ with orthonormal $e_j$, giving

$$\|(I\otimes B)z\|^2 =\sum_j\|By_j\|^2 \leq\|B\|^2\sum_j\|y_j\|^2.$$

The analogous estimate holds for $A\otimes I$. Their composition yields

$$\|(A\otimes B)z\|\leq\|A\|\|B\|\|z\|.$$

Density therefore gives a unique bounded extension to the Hilbert completion. Testing unit simple tensors whose factors approach the respective norm suprema proves the exact identity

$$\|A\otimes B\|=\|A\|\,\|B\|.$$

No norm-attaining vector is needed: approximating sequences suffice.

Adjoint, composition, and limiting operations

Testing inner products on simple tensors and extending by continuity gives

$$(A\otimes B)^*=A^*\otimes B^*.$$

Compatible bounded compositions satisfy $(A\otimes B)(C\otimes D)=AC\otimes BD$. Hence local bounded operators $A\otimes I$ and $I\otimes B$ commute on the whole space. If the local factors are unitaries or orthogonal projections, their tensor product has the same property.

Strong convergence also passes through tensor products under the appropriate uniform bounds. Suppose $A_n\to A$ and $B_n\to B$ strongly. The uniform boundedness principle gives bounded norm sequences. On a simple tensor,

$$(A_n\otimes B_n-A\otimes B)(x\otimes y) =(A_n-A)x\otimes B_ny +Ax\otimes(B_n-B)y,$$

which tends to zero. Density and the uniform operator-norm bounds extend convergence to every composite vector. The same argument without uniform bounds would not justify that last step.

In particular, increasing finite-rank projections $P_n\to I$ and $Q_n\to I$ give $P_n\otimes Q_n\to I$ strongly. They approximate each state vector, although their operator-norm distance from the identity remains one whenever a nonzero subspace is omitted.

A local unbounded observable has a larger domain

Let $A$ be self-adjoint on $\mathcal H_A$ and assume the second factor is separable with basis $(f_j)$. Write

$$\Psi=\sum_j\psi_j\otimes f_j, \qquad \sum_j\|\psi_j\|^2<\infty.$$

The self-adjoint realization of $A\otimes I$ has domain

$$D(A\otimes I)= \left\{\Psi: \psi_j\in D(A)\ \forall j,\quad \sum_j\|A\psi_j\|^2<\infty \right\},$$

and acts by $\sum_jA\psi_j\otimes f_j$. This is the closure of the operator initially defined on $D(A)\odot\mathcal H_B$. Truncating the second-factor expansion approximates both the vector and its image in norm, proving that algebraic domain is a core.

Equivalently its spectral projections are $E_A(\Delta)\otimes I$. The spectral-integrability description proves that the domain does not depend on the chosen second-factor basis.

Two self-adjoint observables on different factors strongly commute because their spectral projections commute. This statement remains valid even if neither algebraic product is defined on every vector. The joint calculus supplies their self-adjoint sums and functions with the corresponding maximal domains.

Spin components make the domain concrete

For a spin-$1/2$ particle, $\mathcal H=L^2(\mathbb R)\otimes\mathbb C^2$ is the space of two-component wavefunctions $(\psi_\uparrow,\psi_\downarrow)$. Position acts on both components:

$$D(Q\otimes I_2)= \{(\psi_\uparrow,\psi_\downarrow): x\psi_\uparrow,\ x\psi_\downarrow\in L^2\}.$$

Spin is bounded and has no additional domain restriction. This distinction is physical: adding a finite internal factor does not make spatial position bounded.

Adding an infinite spectator can also destroy compactness. If $K\neq0$ is compact and $(f_j)$ is an infinite orthonormal family, choose $x$ with $Kx\neq0$. The images $Kx\otimes f_j$ of the bounded sequence $x\otimes f_j$ have no convergent subsequence. Thus $K\otimes I$ is not compact. A compact subsystem operator need not be compact on the total system.

Exercises

1. If both $K$ and $L$ are compact, show that $K\otimes L$ is compact.

Solution

Choose finite-rank norm approximations $K_n\to K$ and $L_n\to L$. Their tensor products are finite rank, and the tensor norm identity gives $\|K_n\otimes L_n-K\otimes L\| \leq\|K_n-K\|\|L_n\|+\|K\|\|L_n-L\|\to0$. Compact operators are norm closed.

2. For $Ae_n=ne_n$ on $\ell^2$, does $\Psi=C\sum_{n\geq1}n^{-1}e_n\otimes f_n$ belong to $D(A\otimes I)$?

Solution

It is a Hilbert-space vector because $\sum n^{-2}<\infty$. Its image would have coefficients $C$, so the proposed image has infinite squared norm. It is outside the operator domain. Normalization alone does not guarantee a finite local second moment.

References

• Dereziński, Jan, and Christian Gérard, Mathematics of Quantization and Quantum Fields, Cambridge University Press, 2013, Chapter 3 — tensor products of spaces and operators. Publisher.
• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013 — tensor products and unbounded observables. Publisher.
• Reed, Michael, and Barry Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, revised and enlarged edition, Academic Press, 1980 — closed operators, tensor products, and spectral domains.