$\DeclareMathOperator\tr{tr}$One begins with a quantum mechanical system, i.e. a unital $C^*$-algebra $A$.

It is common to begin the discussion with embedding $A$ into the algebra of bounded operators $\mathcal{B}$ on some Hilbert space $H$.

A state is defined as a positive linear functional $\varphi: A\rightarrow \mathbb{C}$ taking $1$ to $1$. Since neither $A$ nor $\mathcal{B}$ is a Hilbert space, we can't use Riesz's Representation Theorem directly.

In Physics texts that I have encountered, however, a state is often cast as a trace $1$ operator $\alpha$ in $\mathcal{B}$, and its action on an observable $\rho$ is by $\tr(\rho\alpha)$.

Recall that observables are self dual, and so $\tr(\rho\alpha)=\tr(\rho^* \alpha)$, which is highly suggestive of the subalgebra of $\mathcal{B}$ given by the Hilbert–Schmidt operators, which is a Hilbert space with the inner product $\alpha\cdot\beta=\tr(\alpha^*\beta)$.

Indeed if you restrict $\varphi$ to the algebra of Hilbert–Schmidt operators, then $\varphi$ can be associated with a trace $1$ operator via Riesz's Representation Theorem.


  1. What is going on here? Are we saying that states (defined as positive linear functionals taking $1$ to $1$) are completely determined by their restriction to the subalgebra of Hilbert–Schmidt operators in $A$? Or are there two competing definitions of "states" here? And if so, what is the merit of having these two different definitions?
  2. In https://math.stackexchange.com/questions/77820/a-question-about-pure-state they appear to suggest that not all pure states are represented by projections to one dimensional subspaces of $H$. This confuses me, because I thought those are exactly the pure states. Is this somehow an issue with diverging definitions, related perhaps to my first question?
  • $\begingroup$ Isn't Q1 answered in the Wikipedia definition of a state: a positive linear operator of unit trace, and hence by definition a Hilbert-Schmidt operator. The restriction to unit trace is a matter of normalization, in the same sense that a probability distribution is normalized to unity. Pure states satisfy the additional requirement that the operator squares to itself. $\endgroup$ Feb 1, 2021 at 20:37
  • $\begingroup$ No. It is clear that a positive operator of unit trace is Hilbert Schmidt, and that it defines a positive linear functional on $A$ taking $1$ to $1$. It is not clear why any positive linear function on $A$ is determined by its restriction to the sub-algebra of $A$ made up of only the Hilbert Schmidt operators. The reason to normalize, i.e. look at unit trace operators, is irrelevant to the question. $\endgroup$
    – Andrew NC
    Feb 1, 2021 at 20:42
  • 1
    $\begingroup$ Are you perhaps confusing two different Riesz representation theorems (not saying you are, just a hunch)? If the $C^*$ algebra is commutative, so $A=C(K)$, then the Riesz representation theorem (the one about measures being the dual of $C(K)$) says that the states are exactly the prob measures. What we are dealing with here is a non-commutative version of this, so perhaps it's natural to use the same words. $\endgroup$ Feb 1, 2021 at 21:33
  • $\begingroup$ No, I'm not concerned with the interpretation of states as probability measures, which I am aware of and understand perfectly. What I am concerned with is that a state is a priori a positive funcitonal taking 1 to 1 defined on the $C^*$ algebra, which is a Banach space, but potentially not a Hilbert space. Then the question is: why are physics textbooks treating it as if the $C^*$ algebra is a Hilbert space? This seems to make sense only if you restrict to the Hilbert Schmidt operators among the $C^*$ algebra. $\endgroup$
    – Andrew NC
    Feb 1, 2021 at 21:59
  • $\begingroup$ Why do you think that the collection of all positive linear functions on $A$ is the actual object of interest? $\endgroup$
    – Buzz
    Feb 1, 2021 at 22:07

2 Answers 2


Okay, there is a lot of confusion in this question.

First, I'm not sure why you say ``it is common to begin the discussion with embedding $A$'' into $B(H)$. The point of the C${}^*$-algebra approach to quantum mechanics is doing things in a representation-independent manner, so I would say it's unusual to begin the discussion this way.

You are right that a state on a C${}^*$-algebra is a unital positive linear functional. This isn't really related to the Riesz representation theorem. You seem to want to apply that theorem to the Hilbert-Schmidt operators, which constitute a Hilbert space, but they do not constitute a C*-algebra, and many C${}^*$-algebras contain no Hilbert-Schmidt operators. (So the answer to question 1 is an emphatic ``no''.)

Part of the confusion may have to do with the distinction between pure and mixed states. In the Hilbert space approach to quantum mechanics, pure states are represented by unit vectors in the Hilbert space and mixed states are represented by positive, norm one, trace-class operators. If you're working with C${}^*$-algebras then the mixed states are the states defined above, and the pure states are the extreme points of the set of mixed states. You can use the GNS representation to put a C${}^*$-algebra into a Hilbert space in such a way that the states on $A$ (in the C${}^*$-algebra sense) extend to states on $B(H)$ (in the Hilbert space sense).

Finally, the issue in the question you linked to arises because someone is applying the C${}^*$-algebra definition of states to $B(H)$. This is a little more subtle because we can turn C${}^*$-algebra states into Hilbert space states by embedding the C${}^*$-algebra into some $B(H)$, but if you start with $B(H)$ you might have to embed it in a larger $B(K)$.

(Maybe I should add that looking at states on $B(H)$ in the C${}^*$-algebra sense is something a C${}^*$-algebraist might do, but it's not something mathematical physicists typically do.)

My answer to this question goes into further detail about why one would bother with the C${}^*$-algebra approach to quantum mechanics, etc.

  • $\begingroup$ Let's zoom in on "You can use the GNS representation to put a $C^*$-algebra into a Hilbert space in such a way that the states on 𝐴 (in the $C^*$-algebra sense) extend to states on $\mathcal{B}(H)$ (in the Hilbert space sense)." Why exactly is this true? This is precisely saying that states (as functionals) are represented (with the trace inner product). Does the GNS construction embed the entire $C^*$ algebra into Hilbert Schmidt operators, is that the point? In other words, the GNS construction is special in that way? $\endgroup$
    – Andrew NC
    Feb 1, 2021 at 23:00
  • 1
    $\begingroup$ "Why exactly is this true" --- look at any source on the GNS representation and you'll get a proof. This is the fundamental property of the GNS construction. In fact this construction extends an arbitrary state on a C${}^*$-algebra to a *pure* (vector) state on the $B(H)$ in which it embeds. But this has little or nothing to do with Hilbert-Schmidt operators. $\endgroup$
    – Nik Weaver
    Feb 1, 2021 at 23:10
  • $\begingroup$ Aha! I think I get it. In other words, if $\varphi$ is a state and $\rho$ is an observable in the $C^*$ algebra, then in the GNS construction $\varphi(\rho)=tr(\rho\alpha)$ for some $\alpha$, despite $\rho$ not being a Hilbert Schmidt operator. So in other words, while this is reminiscent of the inner product for Hilbert Schmidt operators, that's the extent of it, there's no full relationship there. And, you say, furthermore these $\alpha$'s are all in fact pure in the Hilbert sense in that they are projections to one dimensional spaces. $\endgroup$
    – Andrew NC
    Feb 1, 2021 at 23:20
  • 2
    $\begingroup$ Okay, yes, this gets a little complicated. Basically, any state can be made pure by putting all the action in a large enough Hilbert space. Another comment is that in the von Neumann algebra approach to quantum mechanics we are more interested in normal states, and $B(H)$ is a von Neumann algebra, and the usual (vector and trace-class) states indeed correspond exactly to the pure and mixed normal states in the von Neumann algebra sense. $\endgroup$
    – Nik Weaver
    Feb 1, 2021 at 23:35
  • 2
    $\begingroup$ Also do take a look at Pedro Lauridsen Ribeiro's answer, it adds some good points. $\endgroup$
    – Nik Weaver
    Feb 1, 2021 at 23:36

$\DeclareMathOperator\Ann{Ann}\DeclareMathOperator\Tr{Tr}$My answer is somewhat complementary to Nik Weaver's, and admitedly more focused on Question 2 since I have nothing more to add to the latter regarding Question 1.

When you deal with a not necessarily commutative C${}^*\!$-algebra $\mathfrak{A}$, the Riesz representation theorem is no longer the tool one uses to represent states. What is usually done is the so-called Gel'fand–Naimark–Segal (GNS) construction of a ${}^*\!$-representation $\pi_\omega:\mathfrak{A}\rightarrow\mathfrak{B}(\mathscr{H}_\omega)$ of $\mathfrak{A}$ in a Hilbert space $\mathscr{H}_\omega$ starting from a state $\omega$ on $\mathfrak{A}$.

Let us recall the GNS construction for convenience. We assume for now that $\mathfrak{A}$ has a unit $\mathbb{1}$. Recall that a state on $\mathfrak{A}$ is a linear map $\omega:\mathfrak{A}\rightarrow\mathbb{C}$ such that $\omega(a^*a)\geq 0$ for all $a\in\mathfrak{A}$ and $\omega(\mathbb{1})=1$. This implies that $$\mathfrak{A}\times\mathfrak{A}\ni(a,b)\mapsto\omega(a^*b)$$ is a positive semidefinite Hermitian sesquilinear form on $\mathfrak{A}$, and therefore satisfies the Cauchy–Schwarz inequality $$\lvert\omega(a^*b)\rvert^2\leq\omega(a^*a)\omega(b^*b)\leq\|a\|^2\omega(b^*b)\ ,\quad a,b\in\mathfrak{A}\ $$ (the latter inequality comes from the fact that $b=\|a\|^2\mathbb{1}-a^*a$ is a positive element of $\mathfrak{A}$, i.e. it has the form $b=c^*c$ for some $c\in\mathfrak{A}$) This means that the so-called annihilator $\Ann \omega$ of $\omega$ $$\Ann \omega=\{a\in\mathfrak{A} \mathrel\vert \omega(a^*a)=0\}$$ is a left ideal (hence a vector subspace) of $\mathfrak{A}$, consisting of the zero-seminorm elements of $\mathfrak{A}$ with respect to the seminorm $\|\cdot\|_\omega$ on $\mathfrak{A}$ induced by this sesquilinear form: $$\|a\|_\omega=\sqrt{\omega(a^* a)}\ ,\quad a\in\mathfrak{A}\ .$$ Hence, the latter induces a complex scalar product on $\mathfrak{A}/\Ann \omega$ - if $[a],[b]$ are the respective equivalence classes of $a,b\in\mathfrak{A}$ modulo $\Ann \omega$, we write $$\langle[a],[b]\rangle=\omega(a^*b)\ .$$ Moreover, since $\Ann \omega$ is a left ideal, $\mathfrak{A}$ has a natural left action on $\mathfrak{A}/\Ann \omega$ as $$\pi_\omega(a)[b]=[ab]\ .$$ This defines a ${}^*\!$-representation $\pi_\omega$ on $\mathfrak{A}/\Ann \omega$ which satisfies $$\|\pi_\omega(a)[b]\|\leq\|a\|\,\|[b]\|=\|a\|\,\|b\|_\omega\ ,\quad a,b\in\mathfrak{A}\ ,$$ thanks to the Cauchy–Schwarz inequality. This means that $\pi_\omega$ extends uniquely to a ${}^*\!$-representation of $\mathfrak{A}$ in the Hilbert space $\mathscr{H}_\omega=\overline{\mathfrak{A}/\Ann \omega}$ by bounded linear operators therein. The state $\omega$ is then represented in $\mathscr{H}_\omega$ by the unit-norm element $\Omega_\omega=[\mathbb{1}]$, for $\omega(a)=\langle\Omega_\omega,\pi_\omega(a)\Omega_\omega\rangle$ for all $a\in\mathfrak{A}$.

As you can see, the GNS construction "kind of" identifies $\mathfrak{A}$ with (a dense subspace of) the Hilbert space $\mathscr{H}_\omega$ - that is, modulo $\Ann \omega$. If the state is faithful, i.e. $\Ann \omega=\{0\}$, then $\mathfrak{A}$ is indeed identified with (a dense subspace of) $\mathscr{H}_\omega$ - this is equivalent to $\pi_\omega$ being injective, i.e. faithful, and to $\pi_\omega$ being isometric. This happens regardless of $\mathfrak{A}$ having Hilbert-Schmidt elements or not. However, even though any nonzero C${}^*\!$-algebra possesses a good deal of faithful states (this is a consequence of the Hahn-Banach theorem and leads to the important Gel'fand-Naimark theorem identifying abstract C${}^*\!$-algebras with closed ${}^*\!$-subalgebras of bounded linear operators in a Hilbert space), not all states of $\mathfrak{A}$ are faithful.

Depending on which kind of C${}^*\!$-algebra $\mathfrak{A}$ and reference state $\omega$ you have, another given state $\eta$ on $\mathfrak{A}$ may or may not be representable as trace-class operators $\rho_\eta$ in $\mathscr{H}_\omega$ with unit trace, that is, given a linear map $\eta:\mathfrak{A}\rightarrow\mathbb{C}$ such that $\eta(a^*a)\geq 0$ for all $a\in\mathfrak{A}$ and $\eta(\mathbb{1})=1$ there may or may not be a $0\leq\rho_\eta\in\mathfrak{B}(\mathscr{H}_\omega)$ with $\Tr(\rho_\eta)=1$ such that $\eta(a)=\Tr(\rho_\eta\pi_\omega(a))$ for all $a\in\mathfrak{A}$.

A situation where this is true regardless of which $\omega$ you choose is when $\mathfrak{A}$ is finite dimensional (i.e. a ${}^*\!$-algebra of matrices). This remains true for $\mathfrak{A}=\mathfrak{B}(\mathscr{H})$ with a separable Hilbert space $\mathscr{H}$ - i.e. a type-I factor (as mentioned at the end of Nik Weaver's answer), which is the case of the algebra of observables for physical systems with finitely many degrees of freedom. On the other hand, for the kind of C${}^*\!$-algebras that appear as algebras of observables for physical systems with infinitely many degrees of freedom (e.g. thermodynamic limits of quantum statistical systems and quantum field theory), this is usually false — as examples, one may cite thermal equilibrium states at different temperatures (in the thermodynamic limit), distinct pure thermodynamic phases of a non-pure thermal equilibrium state, different superselection sectors in quantum field theory, etc.

In physical terms, infinitely many degrees of freedom usually is a manifestation of locality. More precisely, one usually considers the self-adjoint elements of $\mathfrak{A}$ as local observables measured within certain space(-time) regions (or limits of Cauchy sequences thereof). This means that states $\eta$ of the form $\eta(a)=\Tr(\rho_\eta\pi_\omega(a))$ may be seen as states "accessible" from $\omega$ through "physically allowed local operations", at least to an arbitrary degree of accuracy. Any other state is seen as "disjoint" from $\omega$. In infinitely extended space(-time) regions, there are usually plenty of mutually disjoint states on $\mathfrak{A}$.

  • 4
    $\begingroup$ TeX note: \DeclareMathOperator (or its one-shot version \operatorname) takes care of spacing, so you can write just $\operatorname{Ann} \omega$ (\operatorname{Ann} \omega or $\DeclareMathOperator\Ann{Ann}$ … $\Ann \omega$) instead of $\mathrm{Ann}\,\omega$ (\mathrm{Ann}\,\omega). $\endgroup$
    – LSpice
    Feb 1, 2021 at 23:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.