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

Question

The set of trace-class operators $\mathcal{B_1(H)}$ on the Hilbert space $\mathcal{H}$ is like the Banach space $l^1$, while the set of bounded operators $\mathcal{B_\infty(H)}$ is like the Banach space $l^\infty$. It is known that $\mathcal{B_\infty(H)}$ is the dual of $\mathcal{B_1(H)}$, and thus we have the state-effect duality: $p(E|\rho) = \operatorname{tr}[\rho E]$ where the state $\rho \in \mathcal{B_1(H)}$, the effect $E \in \mathcal{B_\infty(H)}$, and $p(E|\rho)$ is the probability of the state $\rho$ having the effect $E$.

However, if we restrict ourselves to pure states and projection measurements, then we can consider solely $\mathcal{H} \cong l^2$, so that the dual of $\mathcal{H}$ is itself, as much as $l^2$ is the dual of itself.

Now my question is, why don't we identify the state $\rho$ and the effects $E$ as the Hilbert-Schidt operators $\mathcal{B_2(H)}$, so that the states and the effects are naturally isomorphic, just as in the pure state case?

Answer 1

I think Tobias Fünke 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 finite dimensional case we require $\operatorname{tr}(\rho) = 1$).

In particlar we want the trace-class and not Hilbert-Schmidt because there are Hilbert-Schmidt operators which look like \begin{align} A = \sum_j \frac{1}{j} |j\rangle\langle j|, \end{align} because $\sum_j j^{-2} = \frac{\pi^2}{6}$, now if you try to do computations with this thing as if its a state you're going to get infinite probabilities, because $\sum_j j^{-1} = \infty$.

For example given a quantum state one thing you can always do is apply a "trivial measurement", which just returns the answer $1$ whatever input state you give. Oviously this thing should have expctation-value 1, but if you try to apply it to the $A$ I gave above and use linearity to get the answer you get expectation value infinity.

Answer 2

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 states can be “defined” as HS operators.

What can be done is to identify every unit-trace trace class positive operator $\rho$ with its square root $T_\rho := \sqrt{\rho}$ which is, in fact, Hilbert Schmidt. With this definition, $$tr(\rho E) = \langle T_\rho| E T_\rho\rangle_{HS}$$ which resembles the formula of expectation values for pure states.

Actually, we can also change $T\to TU$ where $U$ is a partial isometry without changing the identity above. Every HS defines that way a state, even if the correspondence is not one-to-one. All HS operators $T$ such that $$\rho :=TT^* $$ has unit trace represent the state $\rho$. I think that, if $\rho$ is given, these are all of the form $\sqrt{\rho}U$ where $U$ is any partial isometry.