For finite matrices the trace is invariant under similarity transformations, and the trace of a commutator is zero. This is not necessarily true for infinite matrices. But the trace of H in your example is anyway infinite, so why should we be bothered by an extra infinite imaginary term?

I think it would be more interesting to consider a trace-class operator like exp(-t H), with your H and t > 0, and check to which extent the trace is invariant under similarity transformations. I suspect one can come up with examples of T where invariance is still violated.

@Kare@Christian

Thaks for giving time for my question. Please give references showing :: trace variant in

infinite matrix.

Biswanath

Biswanath> Please give references

The Wikipedia article is a good start point.

Googling "trace class" provides more (and reveals that the concept is also used in completely different contexts). As I interpret the definition, it is sufficiently restrictive to assure that the trace of trace-class operators is invariant under similarity transformations (i.e., change of basis) -- otherwise it is by definition not a proper change of basis... So it is unclear if anything interesting can result from such a restrictive approach.

https://en.wikipedia.org/wiki/Trace_class

@Kare

Thanks for reference . Please clarify the statement

• (1) if H is infinie dimensional  then T can not have a bounded inverse .
• Does it mean eigenvalues of T^-1 never exist ?
• Biswanath