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?

  • 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. 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.

  • $\begingroup$ 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

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