Talk:Quantum operation

Comment on pure v. impure states

In usual textbooks of QM one distinguishes pure and impure states (at the moment I do not have access to any text on QM, I'm writing from memory, so I cannot quote sources). I would write:

An impure state of a quantum system is represented on a Hilbert space $\scriptstyle {\mathcal {H}}$ by a non-negative definite trace class operator on $\scriptstyle {\mathcal {H}}$ with trace equal to one. Such operators are called density operators.

After projective measurement the state has become pure, i.e., $\scriptstyle \rho '={\frac {P_{i}\rho P_{i}}{{\rm {tr}}(P_{i}\rho P_{i})}}$, describe a density that is a delta-function with as peak the pure state ψ_i. --Paul Wormer 20:36, 11 April 2009 (UTC)

Beste Paul, I'm not familiar with the terminology 'impure state', so I'm not sure how to respond. I guess this is what some people would call a superposition state? Or could it be synonymous with mixed state? Hendra I. Nurdin 01:21, 12 April 2009 (UTC)

Beste Hendra, apparently you know quite well what a pure state is, as I noticed when I read your article mixed state (which I hadn't seen before you linked to it in your answer). So, yes, an impure state is the same as a mixed state. I believe "impure" is the term used by Messiah. Groetjes, --Paul Wormer 16:30, 12 April 2009 (UTC)

Typo?

Second paragraph:

for all $\scriptstyle \rho \,\in \,{\mathcal {D}}({\mathcal {K}})$,

Should this not be:

for all $\scriptstyle \rho \,\in \,{\mathcal {D}}({\mathcal {H}})$?

Or do I miss something completely? --Paul Wormer 20:40, 11 April 2009 (UTC)

Oops, typo! Thanks for spotting that one. It's been corrected now. Hendra I. Nurdin 01:21, 12 April 2009 (UTC)