0
$\begingroup$

Quite generally, quantum states are defined to be positive, trace-class linear maps with trace equal to one on a complex separable Hilbert space $\mathcal{H}$. If we require that these trace-class linear maps be bounded, then they are linear functionals on $\mathcal{B}(\mathcal{H})$ in particular that satisfy the properties above. However, I found an example of a trace-class linear map $A$, with domain $\mathcal{D}_{A}=\mathcal{H}$, that is unbounded, and I find this interesting. In the mathematical literature, only bounded trace-class linear operators are discussed. Is there a particular reason for that? I will provide my example as a proposition.

Proposition: There exists at least one unbounded linear map $A:\mathcal{H} \to \mathcal{H}$, which is of trace-class.

Proof: Define $A$ on the basis by $A(e_n)=n e_{n+1}$ then extend linearly. $A$ is an unbounded operator with domain $\mathcal{D}_{A}=\mathcal{H}$ and it is of trace class, since: \begin{equation} Tr(A)=\sum_{i=1}^{\text{dim}(\mathcal{H})} \langle e_i, A(e_i) \rangle=\sum_{i=1}^{\text{dim}(\mathcal{H})} \langle e_i, i e_{i+1} \rangle=\sum_{i=1}^{\text{dim}(\mathcal{H})} \langle e_i, e_{i+1} \rangle=0. \end{equation} Of course, this map is not considered to be a legit state in quantum theory, as its trace is not one. However, does there exist an unbounded linear map $B$ with $\mathcal{D}_{B}=\mathcal{H}$, which would qualify as a quantum state? What would be its physical interpretation?

$\endgroup$
5
  • 1
    $\begingroup$ You have not shown that $A$ is trace-class - that the naive expression for the trace converges in one particular basis is not enough. In fact, the usual definition of trace-class only applies to bounded operators since it involves $\lvert A\rvert$, the "square root" of $A^\dagger A$. But the $A^\dagger$ part is exactly what fails for unbounded operators - if $A^\dagger$ is also everywhere defined, then $A$ is bounded. The trace-class operators are a subset of the bounded operators more or less by definition. $\endgroup$
    – ACuriousMind
    Commented Oct 5, 2023 at 23:06
  • $\begingroup$ I am confident that the notion of being trace-class is basis independent. Could you provide a proof that the operator defined by me in a different basis has divergent expression for the trace? $\endgroup$
    – ProphetX
    Commented Oct 5, 2023 at 23:19
  • 3
    $\begingroup$ The expression for the trace is basis-independent if the operator is trace-class! But you cannot show that an operator is trace-class by showing the expression converges in one basis (why do you think the usual definition involves $\lvert A\rvert$ and not just $A$?). As for your specific operator: Consider $e'_i := 2^{-1/2}(e_i + e_{i+1})$. Then $\sum_i \langle e'_i, Ae'_i\rangle = \sum_i 2^{-1}$ diverges. $\endgroup$
    – ACuriousMind
    Commented Oct 5, 2023 at 23:39
  • $\begingroup$ Yes, you are right. Thanks. However, I did not see anywhere the definition of trace-class without boundedness. Could you perhaps write it down in full generality as an answer, then prove that trace class implies bounded, using the Hellinger-Toeplitz theorem, as you mentioned? (Also perhaps this reformulated question rather belongs to the mathematics SE instead of physics) $\endgroup$
    – ProphetX
    Commented Oct 5, 2023 at 23:53
  • $\begingroup$ Regarding your last question: No, irrespective of the trace-class business: Any positive semi-definite, densely defined operator on a complex Hilbert space is symmetric, and if the domain is the whole Hilbert space, then by Hellinger-Toeplitz it is bounded. Density matrices/quantum states are required to be positive semi-definite, for obvious reasons. $\endgroup$ Commented Oct 6, 2023 at 6:19

1 Answer 1

1
$\begingroup$
  1. A positive everywhere-defined operator on a complex Hilbert space is necessarily bounded because it is symmetric, and symmetric everywhere-defined operators are bounded by Hellinger-Toeplitz. Hence an unbounded operator, trace-class or not, cannot be a positive operator and hence is not a quantum state. The unbounded shift operator $A$ from the question is explicitly not a positive operator because for $v = e_i - e_{i+1}$ we have $$ \langle v, Av\rangle = \langle e_i - e_{i+1}, ie_{i+1} - (i+1)e_{i-2}\rangle = - i < 0.$$

  2. An operator is trace-class iff for $\lvert A\rvert$ defined as the positive operator such that $\lvert A\rvert^2 = A^\dagger A$ the trace $$ \mathrm{tr}(A) := \sum_i \langle e_i, \lvert A\rvert e_i\rangle$$ exists. But the "polar decomposition" theorems used to guarantee the existence of $\lvert A\rvert$ assume $A$ is bounded (in another phrasing: the functional calculus for $A^\dagger A$ to obtain $\lvert A\rvert = \sqrt{A^\dagger A}$ needs $A^\dagger A$ to be self-adjoint, which fails for unbounded everywhere-defined $A$), so the very definition of trace-class does not make sense for unbounded operators.

$\endgroup$
3
  • $\begingroup$ With 1, I completely agree. With 2, I haven't seen this definition, but in your framework it is true. However, in Fredric Schuller's lectures, he does not define |A|, just says the domain has to be D_{A}=H and the trace has to exist in any ONB, and has to be equal in any ONB. In this framework, the operator that I provided, seems to fit. The counterexample of divergence, which you give seems not to be a counterexample, because the basis e'_i is not orthonormal. So, the conclusion is that Schuller defines wrongly the notion of trace-class? $\endgroup$
    – ProphetX
    Commented Oct 6, 2023 at 11:16
  • $\begingroup$ Note: also on wikipedia, it is stated that the sum does not depend on the choice of ONB, not any basis. $\endgroup$
    – ProphetX
    Commented Oct 6, 2023 at 11:20
  • $\begingroup$ @ProphetX I don't know Schuller's lectures but the definition via $\lvert A\rvert$ is perfectly standard (cf. eg. section VI of Reed & Simon I or any other standard text on functional analysis). It is also a standard result that trace-class operators are compact, and hence bounded (VI.21 in R&S); your operator is unbounded and therefore cannot be trace-class, any definition of "trace-class" that contradicts this is simply and plainly wrong. And again: You have not shown that the trace exists and is equal in every ONB - just that it exists in the one ONB you chose. $\endgroup$
    – ACuriousMind
    Commented Oct 6, 2023 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.