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5 votes
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Why we use trace-class operators and bounded operators in quantum mechanics?

I think Tobias F√ľnke has essentially answerered this question already, but to be as explicit as possible: we need the states to be trace-class so we can obtain normalized probabilities (just as in the ...
ors's user avatar
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4 votes
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Jensen's inequality on (super)operator exponential

Substituting your definition you are asking whether for all Lindbladians $\mathcal L$, all times $t\geq 0$, and all reference states $\rho$ it holds that $$ {\rm tr}(\rho e^{t\mathcal L}(\rho))\geq e^{...
Frederik vom Ende's user avatar
2 votes
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The meaning of a representation in one-dimensional quantum mechanics

In abstract terms the meaning of representation in these cases is the ordinary meaning of representation in the sense of representation theory. Concretely, you are representing the Heisenberg algebra ...
ACuriousMind's user avatar
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2 votes

Why we use trace-class operators and bounded operators in quantum mechanics?

The effects are not Hilbert Schmidt in general, so to identify them with HS operators would be too restrictive. However, even if density operators are just a part of the Hilbert-Schmidt operators, the ...
Valter Moretti's user avatar
1 vote
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Non-uniqueness of Glauber-Sudarshan $P$-function

As was previously mentioned in the comments, the decomposition formula $$\rho = \int \mathrm d^2\alpha P_\rho(\alpha) |\alpha\rangle\!\langle\alpha|,$$ uniquely determines the structure of $P_\rho(\...
glS's user avatar
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