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Is it possible to construct a well defined inner-product (and therefore orthonormality) within the set of self-adjoint trace-class linear operators? In the affirmative case, dynamics could be analyzed in Hilbert space, which seem way more simple that Banach spaces.

Which is the fundamental reason why this is not possible?

Qmechanic
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Alejandro D. Somoza
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1 Answers1

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It is possible indeed !! It is called Hilbert Schmidt scalar product, it is defined in a Hilbert space of bounded compact operators including trace class operators.

$$\langle A|B\rangle := tr(A^\dagger B)\:.$$

The space of Hilbert Schmidt operators is made of all bounded operators $A$ in the considered Hilbert space, such that $A^\dagger A$ is trace class. It is in fact possible to reformulate all QM using that notion.

MariusMatutiae
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Valter Moretti
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  • I knew this product is used to define the norm in trace-class bounded operators, but I thought there was some kind of fundamental problem, id.est that this norm is not induced by this Hilbert Schimdt inner product. Anyway, when studying the theory of dynamical maps, why is it so convenient to formulate the theory in terms of Banach spaces instead of Hilbert spaces with the Hilbert Schimdt product? Thanks for your answer, any good reference about this? – Alejandro D. Somoza Aug 15 '14 at 09:59
  • There are at least three natural notions of norm actually in this context. The trace class Banach space has another norm, which is not Hilbert (have a look here http://physics.stackexchange.com/questions/121337/density-matrix-formalism/121367#121367). Sorry, I think I am not familiar with the theory of dynamical maps. What is a dynamical map? (Is it the same as quantum operation?) – Valter Moretti Aug 15 '14 at 10:07
  • The Hibert-Schmidt ideal has Hilbert space structure, while the others ideals (like the trace class one) have not. They are all subsets of the compact operators, and satisfy inequalities similar to the $L^p$ functions (e.g. Hölder). A very complete reference is the book of Barry Simon Trace ideals and their applications. – yuggib Aug 15 '14 at 10:12
  • there is also the classic book by R.Schatten. Norm ideals of completely continuous operators. Springer, Berlin (1960) and several results can be found in Dunford Schwartz books. – Valter Moretti Aug 15 '14 at 10:18
  • States are self-adjoint positive trace class operators. If they are pure (i.e. rank-one projectors), they are also Hilbert-Schmidt, and the two norms are equivalent (there is an argument by R.Seiringer about that, but I don't know the reference). However for general mixed states, it is not assured that they are Hilbert-Schmidt, but only trace class... – yuggib Aug 15 '14 at 10:20
  • Nevertheless usually the dynamics is defined as a unitary operator on the Hilbert space. Only for open systems, there is a formulation (see the Lindbladian) in terms of evolution of semigroups on the trace operators; at least in my knowledge of QM. – yuggib Aug 15 '14 at 10:21
  • Well, trace class $\Rightarrow$ Hilbert Schmidt, since trace class is a $^$- two-sided-ideal...So all generally mixed quantum states are* Hilbert Schmidt. – Valter Moretti Aug 15 '14 at 10:23
  • Thank guys for the references. A dynamical map is simply a superoperator on the reduced system: $\epsilon_{(t,t_0): \rho_{\rm S}(t_0)\rightarrow \rho_{\rm S}(t_1)}$ A quantum operation is the so-called Universal Dynamical Map which is independent of the state it acts upon (the associated Krauss operators do not depend on $\rho_{\rm S}$. I'm following this reference: http://arxiv.org/abs/1104.5242 – Alejandro D. Somoza Aug 15 '14 at 10:25