By basic postulates of QM, any state of a system is described by a normalised density operator. Now i fail to see why people study sub-normalized states ( e.g.: In generalised fidelity etc). I'd be glad to know any physical interpretations of the same.
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$\begingroup$ I am sorry! It seems i don't have permission to add this topic to quantum computing. I request those who have the accesses please add it. $\endgroup$– Vamsi KrishnaOct 26, 2014 at 16:09
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5$\begingroup$ Subnormalized density operators are not only very useful in studying quantum computation, but in studying quantum physics as well. $\endgroup$– Peter ShorOct 26, 2014 at 16:33
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1 Answer
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It is often mathematically very convenient to work with sub-normalized states independently of the question whether they have a direct physical meaning.
However, you can always see them as states that are logically 'AND' with a certain event. For example, consider the c-q state
$\rho_{XB} = \sum_x |x\rangle\!\langle x|_X \otimes \rho_B^x$
where $\rho_B^x$ is sub-normalized so that the whole state is normalized. Then the trace of $\rho_B^x$ just gives the probability of finding the subsystem $B$ in the renormalized, conditional state $\rho_B^x / \text{tr}(\rho_B^x)$.
For example, the probability of measuring a certain event (POVM element) $E_B$ AND x is simply given by $\text{tr}(\rho_B^x E_B)$ and we never needed to renormalize the state to write this down.
