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

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?

• Anyway, note that not all Hilbert Schmidt operators are trace-class. But if you want to have a normalized probability distribution, you need that $\rho$ is trace-normalized. Commented Jun 30 at 7:51
• @TobiasFünke : As much as I know, I am referring to the bounded operators, although many operators in quantum mechanics such as position and momentum operators are unbounded. The effects are bounded though, and this is sort of a standart formulation of quantum measurements. See Quantum Measurement by Paul Busch et. al. Commented Jun 30 at 7:52
• @TobiasFünke : could you elaborate more on $\rho$ being trace-normalized? Does it entail requiring $\rho$ to be a trace-class operator rather than a Hilbert-Schmidt operator? A ket in $\mathcal{H}$ can be normalized and it is $l^2$, not $l^1$, though I see there is no statistical mixture in a pure state. Commented Jun 30 at 7:57
• @TobiasFünke : The set of bounded operators is the dual of the set of trace class operators, but in this, the set of compact operators are the predual of the set of trace class operators. Commented Jun 30 at 7:59
• @TobiasFünke : I'm sorry for the confusion. I use $B_\infty$ to draw an analogy to $l^\infty$. In this book, the set of bounded operators is denoted by $\mathcal{L(H)}$. See section 9.1 and 9.2 for details on states and effects. Commented Jun 30 at 8:04

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.

• This makes total sense. Though I still feel very upset about the states and the effects not being naturally isomorphic like the bras and kets are. I wonder if this would lead to some interesting consequences. Commented Jun 30 at 9:40

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.

• Why can't this procedure be applied to effects? Commented Jun 30 at 10:55
• It can be applied indeed, and it gives rise to Kraus operators. But the Hilbert Schmidt product needs to apply the procedure to $\rho$... Commented Jun 30 at 11:39