Proof Exercises

A useful proof exercise forces a choice of topology, identifies a hypothesis that cannot be dropped, or connects results from different chapters. These four challenges use the established results in the linked articles and ask for additional deductions. Before opening a solution, write down the spaces on which every map acts.

Helpful background. Compact Operators, Hilbert Space Tensor Products, and Normal States supply the main tools.

Compactness of tensor products

Let $A\in B(\mathcal H_1)$ and $B\in B(\mathcal H_2)$ be nonzero bounded operators on complex Hilbert spaces. Prove that $A\otimes B$ is compact if and only if both $A$ and $B$ are compact. Explain why the nonzero assumption matters.

Solution

If $A$ and $B$ are compact, choose finite-rank approximants $A_n\to A$ and $B_n\to B$ in operator norm. Their tensor products have finite rank. The tensor norm identity and the decomposition

$$A\otimes B-A_n\otimes B_n =(A-A_n)\otimes B+A_n\otimes(B-B_n)$$

show norm convergence, because the norm-convergent sequence $A_n$ is bounded. Norm limits of compact operators are compact.

Conversely, since $B\ne0$, choose $u,v\in\mathcal H_2$ with $\langle v,Bu\rangle\ne0$. The bounded insertion and slice maps

$$J_u:x\mapsto x\otimes u,\qquad R_v:x\otimes y\mapsto\langle v,y\rangle x$$

satisfy

$$R_v(A\otimes B)J_u=\langle v,Bu\rangle A.$$

Compact operators form a two-sided ideal under composition with bounded maps. Thus $A$ is compact. Interchanging the two factors and using $A\ne0$ proves compactness of $B$.

If one factor is zero, the tensor product is zero and compact regardless of the other factor. The exercise also explains why a compact observable on one subsystem need not remain compact after adjoining an infinite-dimensional spectator: the other factor is then an identity that is not compact. This is a property of the enlarged Hilbert space, not a failure of the original subsystem approximation.

Uniform channel approximation on compact state families

Let $\Phi_n$ and $\Phi$ be channels on the trace-class operators of a Hilbert space, and suppose $\|\Phi_n(\rho)-\Phi(\rho)\|_1\to0$ for every density operator $\rho$. Prove uniform convergence over any trace-norm compact set $K$ of density operators.

Then explain how an energy constraint can make this result applicable when $H\geq0$ has finite-rank spectral projections $P_R=\mathbf1_{[0,R]}(H)$ for every finite $R$, with $P_R\to I$ strongly.

Solution

Fix $\varepsilon>0$. Compactness provides a finite $\varepsilon$-net $\rho_1,\ldots,\rho_m$ in $K$. Trace-norm contractivity on differences of states gives, whenever $\|\rho-\rho_j\|_1<\varepsilon$,

$$\|\Phi_n(\rho)-\Phi(\rho)\|_1 \leq2\varepsilon+ \|\Phi_n(\rho_j)-\Phi(\rho_j)\|_1.$$

For sufficiently large $n$, the last term is smaller than $\varepsilon$ for every point of the finite net. Taking the supremum over $K$ and then sending $\varepsilon$ to zero proves the assertion. Pointwise convergence alone would not supply a finite set of tests on a noncompact input family.

For $K_E=\{\rho:\operatorname{Tr}(\rho H)\leq E\}$, spectral positivity yields

$$\operatorname{Tr}\bigl(\rho(I-P_R)\bigr)\leq E/R \quad(R>0).$$

The trace-class cutoff estimates at Quantum Channels in Infinite Dimensions turn this uniform probability tail into a uniformly small trace-norm error for a channel that keeps $P_R\rho P_R$ and places the discarded probability in a fixed state in $P_R\mathcal H$. The cutoff states occupy a compact finite-dimensional state space. Thus $K_E$ is totally bounded. It is closed because the positive energy expectation is lower semicontinuous in trace norm, as follows by approximating $H$ with bounded increasing spectral truncations. Completeness of trace class now makes $K_E$ compact.

The resulting uniform approximation concerns this constrained family without an arbitrary ancillary system. It does not establish unconstrained diamond-norm convergence. It also requires finite-rank energy cutoffs: a bound on an observable with an infinitely degenerate low-energy sector need not produce compactness.

The spectrum does not control a nonnormal resolvent norm

For $M>0$, let

$$A_M=\begin{pmatrix}0&M\\0&0\end{pmatrix}.$$

Compute $\sigma(A_M)$ and show that the resolvent at $z=1$ can have arbitrarily large norm despite a fixed distance from the spectrum. Identify the hypothesis behind the familiar reciprocal-distance equality.

Solution

Since $A_M^2=0$, its spectrum is $\{0\}$. Also

$$(A_M-I)^{-1} =-(I+A_M) =\begin{pmatrix}-1&-M\\0&-1\end{pmatrix}.$$

Applying this inverse to the second unit vector gives a vector of norm $\sqrt{M^2+1}$. Hence

$$\|(A_M-I)^{-1}\|\geq\sqrt{M^2+1}, \qquad \operatorname{dist}(1,\sigma(A_M))=1.$$

For a normal operator, spectral calculus gives the norm of the resolvent as the supremum of $|\lambda-z|^{-1}$ over the spectrum. Here $A_M$ is nonnormal: $A_MA_M^\dagger=\operatorname{diag}(M^2,0)$ and $A_M^\dagger A_M=\operatorname{diag}(0,M^2)$. The spectral set alone therefore does not provide the equality. The spectral cheat sheet explicitly confines that equality to the normal setting.

This finite-dimensional example needs no domain pathology. Large response to a small forcing can arise from nonnormality even when the resolvent parameter remains a unit distance from every eigenvalue.

A sequence that escapes every finite-dimensional subspace

On $\ell^2(\mathbb N)$ let $\rho_n=|n\rangle\langle n|$. Prove that $\operatorname{Tr}(\rho_nK)\to0$ for every compact $K$. Can $\rho_n$ converge in trace norm, or converge pointwise on all of $B(\ell^2)$ to a normal state?

Solution

The basis vectors $|n\rangle$ converge weakly to zero. Compactness implies $\|K|n\rangle\|\to0$, so

$$|\operatorname{Tr}(\rho_nK)| =|\langle n|K|n\rangle| \leq\|K|n\rangle\|\longrightarrow0.$$

For $m\ne n$, $\|\rho_n-\rho_m\|_1=2$, so the sequence is not trace-norm Cauchy. If it had a pointwise limit that were a normal state $\omega(A)=\operatorname{Tr}(\rho A)$, then every fixed rank-one basis projection would have expectation zero. Positivity would imply $\operatorname{Tr}\rho=\sum_j\langle j|\rho|j\rangle=0$, contradicting normalization.

In fact this sequence has no pointwise limit on all bounded operators: the diagonal operator with entries $(-1)^n$ gives alternating expectations. Weak-star compactness of the full algebraic state space nevertheless supplies convergent subnets. Any such cluster state annihilates all compact operators but takes value one on $I$, and is therefore singular. This connects the normal-state criterion to escape of probability through infinitely many mutually orthogonal sectors; convergence against compact tests is too weak to preserve a normal density-operator description.

References

• Bratteli, Ola, and Derek W. Robinson. Operator Algebras and Quantum Statistical Mechanics 1, 2nd ed. Springer, 1987. doi:10.1007/978-3-662-02520-8. Algebraic and normal states.
• Hall, Brian C. Quantum Theory for Mathematicians. Graduate Texts in Mathematics 267, Springer, 2013. doi:10.1007/978-1-4614-7116-5. Compact operators, tensor products, and spectral calculus.
• Holevo, Alexander S. Quantum Systems, Channels, Information: A Mathematical Introduction, 2nd ed. De Gruyter, 2019. doi:10.1515/9783110642490. Trace-class channel convergence and energy constraints.