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No, it's not. An infinite sum like ##(\sum_{n=0}^\infty 1)## is divergent.PeterDonis said:[...] taking the trace (and seeing that it gives an infinite sum of terms each of which is ##i \hbar##) is simple.
No, it's not. An infinite sum like ##(\sum_{n=0}^\infty 1)## is divergent.PeterDonis said:[...] taking the trace (and seeing that it gives an infinite sum of terms each of which is ##i \hbar##) is simple.
Yes, I know that. I'm just saying that that infinite sum is what formally taking the trace of the infinite identity matrix gives you. I think something like that is what Ballentine intends to illustrate with the problem under discussion.strangerep said:An infinite sum like ##(\sum_{n=0}^\infty 1)## is divergent.