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Trace of an operator $A$ can be written in its eigenbasis $|a\rangle$ as $${\rm Tr}(A)=\sum_{a} \langle a|A|a\rangle.$$ Since trace is base independent, does it imply that we can also write the expression for trace as $${\rm Tr}(A)=\sum_{b} \langle b|A|b\rangle$$ in some other orthonormal basis (not the eigenbasis of $A$) $|b\rangle$?

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    $\begingroup$ Exactly, as long as it is a basis (linearly independent and complete). $\endgroup$
    – TopoLynch
    Jul 24, 2023 at 14:13

2 Answers 2

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In complex Hilbert spaces, one of the various equivalent definitions of trace class operator is a bounded operator $T: H\to H$ such that, for every Hilbert basis $\{u_a\}_{a\in A}$, the sum $$\sum_{a\in A}\langle u_a|Tu_a\rangle$$ absolutely converges to a complex number and this number does not depend on the chosen Hilbert basis. That is the trace of $T$.

Hence, independence of the Hilbert basis is in the definition itself.

In finite dimension, independence is automatic and all operators are trace class.

In the infinite dimensional case that class is a subclass of compact operators. Hence it is quite small. For instance it is not possible to approximate all bounded operators with trace class operators in the natural topology of the space of bounded operators.

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  • $\begingroup$ But is this valid even for a non-orthonormal basis? $\endgroup$ Jul 25, 2023 at 9:38
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    $\begingroup$ No, it is generally false if the basis is not orthonormal... $\endgroup$ Jul 25, 2023 at 9:47
  • $\begingroup$ But in your answer you mention that the definition is valid for any Hilbert basis which would include non-orthonormal bases too, right? $\endgroup$ Jul 25, 2023 at 9:49
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    $\begingroup$ Hilbert basis means complete orthonormal system of vectors, hence they are otrhonormal by definition. It is just matter of jargon. However the orthonormality condition was included in my answer. $\endgroup$ Jul 25, 2023 at 9:54
  • $\begingroup$ Acha I didn't know this. Understood, thank you! $\endgroup$ Jul 25, 2023 at 9:59
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Let's check that the two expressions coincide; it can be done by "playing" with the closure relations with respect to each basis, as follows : $$ \begin{array}{rcl} \mathrm{Tr}(A) &=& \displaystyle \sum_a \langle a|A|a \rangle \\ &=& \displaystyle \sum_{a,b} \langle a|A|b \rangle\langle b|a \rangle \\ &=& \displaystyle \sum_{a,b} \langle b|a \rangle\langle a|A|b \rangle \\ &=& \displaystyle \sum_b \langle b|A|b \rangle \end{array} $$

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  • $\begingroup$ Is this valid even when $|b\rangle$s are not orthonormal? $\endgroup$ Jul 24, 2023 at 18:52
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    $\begingroup$ No, because the closure relation isn't satisfied by a non-orthonormal basis. In other words, the trace is defined through a scalar product, which wouldn't conserved by a non-unitary change of basis. See e.g. physics.stackexchange.com/questions/746288/… $\endgroup$
    – Abezhiko
    Jul 24, 2023 at 20:54
  • $\begingroup$ Acha understood! $\endgroup$ Jul 25, 2023 at 9:37

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