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On a Hilbert space $\mathcal{H}$, it is well known that trace class operators have a finite trace. However, there are operators which are not trace class but have a finite trace, e.g. the unilateral shift. Then, consider a (bounded) operator $A$ such that $$ c=\sum_{n=1}^{\infty}\langle x_n,A x_n\rangle<\infty, $$ for some orthonormal basis $\{x_1,x_2,\ldots\}$. I'm not sure whether this is enough for the trace of $A$ to be defined in a basis independent way. Namely, $\mathrm{tr}(A)=\sum_{n=1}^{\infty}\langle y_n,A y_n\rangle=c$ for any orthonormal basis $\{y_1,y_2,\ldots\}$.

In the classics von Neumann's book on Mathematical Foundations of Quantum Mecanics, Sec. II.11, he commented this is true (in fact, he did not consider trace class operators) using essentially the following argument: $$ c=\sum_{n=1}^{\infty}\langle x_n,A x_n\rangle=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\langle x_n,y_m\rangle \langle y_m, A x_n\rangle=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\langle y_m, A x_n\rangle\langle x_n,y_m\rangle=\sum_{m=1}^{\infty}\langle y_m,A y_m\rangle. $$ However, it is not clear to me that the double sum can be switched. Using that $|\langle u,y_m \rangle \langle y_m, v\rangle|\leq \frac12 (|\langle u,y_m \rangle|^2+|\langle v,y_m \rangle|^2)$ it is obtained that $$ \sum_{m=1}^{\infty}\langle x_n,y_n\rangle \langle y_n, A x_n\rangle $$ is absolutely convergent. Is this enough for the summation switch?

Thank you in advance.

EDIT: As Robert Israel points, $\sum_{n=1}^{\infty}\langle x_n,A x_n\rangle$ must be absolutely convergent. Is it enough?

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2 Answers 2

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If the sum $\sum_{n=1}^\infty \langle x_n, A x_n \rangle$ is only conditionally convergent, you can take the same basis $x_n$ but rearranged, and get a sum that diverges, or converges to an arbitrary value: see the Riemann rearrangement theorem.

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  • $\begingroup$ thank you!, of course you are right. However, let me ask then the same question under the condition $\sum_{n=1}^\infty|\langle x_n,Ax_n\rangle|<\infty$. It might be the case that $A$ is not trace-class (e.g. unilateral shift). However, can its trace be defined in a basis independent way? $\endgroup$ Sep 4, 2020 at 15:40
  • $\begingroup$ the answer to my last questions is also negative; with the unilateral shift is easy to prove that there are orthonormal basis $\{y_n\}$ such that $\sum_{n=1}^\infty|\langle y_n,Ay_n\rangle|=\infty$. I guess the trace class condition $\sum_{n=1}^\infty|\langle x_n,Ay_n\rangle|<\infty$ for any orthonormal sets $x_n$ and $y_n$ is the only one which ensures a proper definition of trace. $\endgroup$ Sep 5, 2020 at 0:00
  • $\begingroup$ @NessunDorma I think that if $A$ is bounded, if suffices to demand that $\sum_{n=1}^\infty \langle x_n, A x_n \rangle$ converges for some basis. Then this is true for any basis. $\endgroup$
    – Filippo
    Jun 18, 2021 at 17:51
  • $\begingroup$ @Filippo, thanks, but again the unilateral shift seems to be a counterexample to your point $\endgroup$ Jun 20, 2021 at 0:28
  • $\begingroup$ @NessunDorma Why? $\endgroup$
    – Filippo
    Jun 20, 2021 at 6:37
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Without getting too creative, stay with the unilateral shift and consider the orthonormal basis $\{y_m\}$ given by $$ \tfrac1{\sqrt2}\,(x_1+x_2),\tfrac1{\sqrt2}\,(x_1-x_2),\tfrac1{\sqrt2}\,(x_3+x_4),\tfrac1{\sqrt2}\,(x_3-x_4),\ldots $$ In this basis, the diagonal of $S$ is $$ \tfrac12,-\tfrac12,\tfrac12,-\tfrac12,\ldots $$

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  • $\begingroup$ yes, thank you! I have realized of that as commented to Robert Israel. I guess the trace class condition $\sum_{n=1}^\infty|⟨x_n,Ay_n⟩|<\infty$ for any orthonormal sets $\{x_n\}$ and $\{y_n\}$ is the only one which ensures a proper definition of trace. For a moment, I thought that a condition like $\sum_{n=1}^\infty|⟨y_n,Ay_n⟩|<\infty$ could work, but I think it only does for normal operators. Is any book addressing the issue that no non-trace class operator admits a trace? $\endgroup$ Sep 5, 2020 at 0:10

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