Local Tomography and the Jordan Structure of Quantum Theory

Abstract

Using a result of H. Hanche-Olsen, we show that (subject to fairly natural constraints on what constitutes a system, and on what constitutes a composite system), orthodox finite-dimensional complex quantum mechanics with superselection rules is the only non-signaling probabilistic theory in which (i) individual systems are Jordan algebras (equivalently, their cones of unnormalized states are homogeneous and self-dual), (ii) composites are locally tomographic (meaning that states are determined by the joint probabilities they assign to measurement outcomes on the component systems) and (iii) at least one system has the structure of a qubit. Using this result, we also characterize finite dimensional quantum theory among probabilistic theories having the structure of a dagger-monoidal category.

Appendix: The Koecher-Vinberg Theorem

This appendix contains a detailed sketch of the proof of the Koecher-Vinberg Theorem. This is almost entirely extracted from Faraut and Koranyi [13], to whom we refer for many of the details, but with a few minor modifications in order to obtain the precise form of the theorem (our Theorem 2.3) that we required above.

In what follows, $\mathbf{E}$ is a finite-dimensional order-unit space with a self-dual positive cone $\mathbf{E}_+$. By this we mean that there exists a self-dualizing inner product on $\mathbf{E}$. Let $\text{ Aut }(\mathbf{E})$ denote the group of order-automorphisms of $\mathbf{E}$. This is a Lie group, as is any closed subgroup $G \le \text{ Aut }(\mathbf{E})$. When $G$ is connected and acts homogeneously on $\mathbf{E}_+$ (that is, transitively on the interior of $\mathbf{E}_+$), we can use this action to construct a Jordan product on $\mathbf{E}$, as per Theorem 2.3, which, for convenience, we now restate:

Theorem 2.3

Let $G$ be a closed, connected subgroup of $\text{ Aut }(\mathbf{E})$ and let ${\mathfrak g}_u$ denote the Lie algebra of $G_u$, the stabilizer of $u$ in $G$. Then

• (a) It is possible to choose a self-dualizing inner product on $\mathbf{E}_+$ in such a way that $G_u = G \cap O(\mathbf{E})$, where $O(\mathbf{E})$ is the orthogonal group with respect to the chosen inner product;

• (b) If $G = G^{\dagger }$ with respect to this inner product, then

  $$\begin{aligned} {\mathfrak g}_u = \{ X \in {\mathfrak g}| X^{\dagger } = -X\} = \{ X \in {\mathfrak g}| Xu = 0\}, \end{aligned}$$

  and ${\mathfrak g}= {\mathfrak p}\oplus {\mathfrak g}_u$, where ${\mathfrak p}= \{ X \in {\mathfrak g}| X^{\dagger } = X\}$.

• (c) In this case the mapping ${\mathfrak p}\rightarrow \mathbf{E}$, given by $X \mapsto Xu$, is an isomorphism of linear spaces. Letting $L_a$ denote the unique element $X \in {\mathfrak p}$ with $X u = a$, define

  $$\begin{aligned} a \bullet b = L_{a} b \end{aligned}$$

  for all $a, b \in \mathbf{E}$. Then $\bullet$ makes $\mathbf{E}$ a formally real Jordan algebra, with identity element $u$.

We break the proof into a series of Lemmas. Throughout, $G$ is a connected, closed subgroup of $\text{ Aut }(\mathbf{E})$, acting homogeneously on the interior, $\mathbf{E}^{\circ }_{+}$, of the cone $\mathbf{E}_+$.

Lemma A

If $g \in \text{ Aut }(\mathbf{E})$, then $g^{*} \in \text{ Aut }(\mathbf{E})$, where $g^{*}$ is the adjoint with respect to any self-dualizing inner product.

Proof

If $g \in \text{ Aut }(\mathbf{E})$ preserves $\mathbf{E}_+$, then $g^{\dagger }$ preserves $\mathbf{E}^{+} = \mathbf{E}_+$. $\square$

Lemma B

Any compact subgroup of $\text{ Aut }(\mathbf{E})$ fixes some point $a$ in the interior of $\mathbf{E}_+$. In particular, a maximal compact subgroup is a stabilizer, and vice versa. Thus, all maximal compact subgroups of $\text{ Aut }(\mathbf{E})$ are conjugate.

For a proof, see [13], Proposition I.1.8ff.

Lemma C

For a suitable choice of self-dualizing inner product, $O(\mathbf{E}) \cap G \le G_u$, where $O(\mathbf{E})$ is the orthogonal group relative to the chosen inner product.

Proof

If $\langle , \rangle$ is any inner product on $\mathbf{E}$ with respect to which $\mathbf{E}_{+} = \mathbf{E}^{+}$, one can show that $O(\mathbf{E}) \cap \text{ Aut }(\mathbf{E})$ is the stabilizer of some $a \in \mathbf{E}_{+}^{\circ }$ ([13], Proposition I.1.9). It follows that $O(\mathbf{E}) \cap G \le G_{a}$. Since $G$ acts transitively on $\mathbf{E}_{+}^{\circ }$, we can find some $g \in G$ with $ga = u$; replacing $\langle , \rangle$, if necessary, by the inner product $\langle x, y \rangle _{g} : = \langle g x, g y \rangle$—which is also self-dualizing, by [13], Proposition I.1.7—we can assume that $a = u$, whence, that $O(\mathbf{E}) \cap G \le G_u$. $\square$

This gives us part (a) of the Theorem. Now let $K = G \cap O(\mathbf{E})$, and let ${\mathfrak k}$ denote $K$’s Lie algebra. Notice that ${\mathfrak k}= {\mathfrak g}\cap \mathfrak {o}(\mathbf{E}) = \{ X \in {\mathfrak g}| X^{\dagger } = -X\}$.

Corollary 1

Let the inner product on $\mathbf{E}$ be chosen as per Lemma C. Let $K = G \cap O(\mathbf{E})$, and let ${\mathfrak k}$ be the Lie algebra of $K$, and let ${\mathfrak g}_u$ denote the Lie algebra of $G_u \le G$. Then

• (a) $G_{u}$ is connected;

• (b) $G_{u} = K$;

• (c) ${\mathfrak g}_u = {\mathfrak k}$;

Proof

(a) $G/G_u$ is homeomorphic to the simply connected space $\mathbf{E}^{\circ }_{+}$; hence, as $G$ is connected, so is $G_{u}$ (see, e.g., [21], Proposition 1.94). It follows that if $G_u$ and $K$ have the same Lieq algebra, they coincide—in other words, (b) follows from (c). To prove (c), note that since $K \le G_u$, by the choice of inner product, we have ${\mathfrak k}\le {\mathfrak g}_u$. For every $X \in {\mathfrak g}_u \le {\mathfrak g}$, we have the decomposition $X = X_1 + X_2$ with $X_1$ self-adjoint and $X_2$, skew-adjoint. Since $X_2 \in {\mathfrak k}\subseteq {\mathfrak g}_{u}$, it follows that $X_1 = X - X_2 \in {\mathfrak g}_u$ as well, whence, $e^{tX_1} \in G_u$ for all $t$. However, $G_u$ is compact, so this implies that $e^{tX_1}$ is bounded as a function of $t$. Since $X_1$ is self-adjoint, this is possible only if $X_1 = 0$. Hence, $X = X_2 \in {\mathfrak k}$, and we have ${\mathfrak g}_u \le {\mathfrak k}$. $\square$

Proof of Part (b) of Theorem 2.3: Suppose that $G$ is self-adjoint with respect to the self-dualizing inner product of Lemma C. Then, for every $X \in {\mathfrak g}$, $X^{*} \in {\mathfrak g}$. To see this, let $X = \gamma '(0)$ where $\gamma : {\mathbb R} \rightarrow G$ is a smooth path with $\gamma (0) = \text{1 }$, and note that $\gamma ^{*} : t \mapsto \gamma (t)^{*}$ is another such path, with ${\gamma ^{*}}'(0) = X^{*}$. Thus, for every $X \in {\mathfrak g}$, $X_1 := \left( X + X^{\dagger }\right) /2$ and $X_2 = \left( X - X^{\dagger }\right) /2$ also lie in ${\mathfrak g}$; $X = X_1 + X_2$, so ${\mathfrak g}$ decomposes as the direct sum ${\mathfrak g}= {\mathfrak p}+ {\mathfrak k}$, where ${\mathfrak p}$ and ${\mathfrak k}$ are respectively the spaces of self-adjoint and skew-adjoint elements of ${\mathfrak g}$.$\square$

Remark

Note also, for later reference, that if $X, Y \in {\mathfrak p}$, we also have $[X,Y] \in {\mathfrak g}$ and $[X,Y]^{\dagger } = [Y,X] = -[X,Y]$, i.e., $[X,Y] \in {\mathfrak k}$.

The interior, $\mathbf{E}^{\circ }_{+}$, of the cone $\mathbf{E}_+$ is a smooth manifold, on which $G$ acts smoothly. Thus, we have a canonical smooth mapping $\phi : G \rightarrow \mathbf{E}^{\circ }_{+}$ given by $g \mapsto gu$. Differentiating this, we obtain a linear mapping $d\phi (\text{1 }) : {\mathfrak g}\rightarrow T_{u}\left( \mathbf{E}^{\circ }_{+}\right) = \mathbf{E}$. Explicitly, if $X = \gamma '(0) \in {\mathfrak g}$, where $\gamma$ is a smooth path in $G$ with $\gamma (0) = \text{1 }$, then

$$\begin{aligned} d\phi (\text{1 })(X) = \frac{d}{dt} \phi (\gamma (t))_{t = 0} = \gamma '(0)u = Xu. \end{aligned}$$

(3)

Lemma D

Let $X \in {\mathfrak g}$. Then $X \in {\mathfrak k}$ iff $Xu = 0$.

Proof

Suppose $d\phi (\text{1 })(X) = Xu = 0$, where $X = \gamma '(0)$. The vector-valued function $v(t) = e^{tX}u$ then satisfies

$$\begin{aligned} v' = Xv = Xe^{tX}u = e^{tX}Xu = 0. \end{aligned}$$

It follows that $v$ is constant, i.e, that $e^{tX}u = u$ for all $t$. But then $e^{tX} \in G_u$, so that $X = \frac{d}{dt} e^{tX}|_{t =0} \in {\mathfrak k}$. Conversely, if $X \in {\mathfrak k}$, then $X = \gamma '(0)$ where $\gamma (t) \in K$ and $\gamma (0) = \text{1 }$, so that $Xu = \gamma '(0) u = \left[ \frac{d}{dt}(\gamma (t)u)\right] _{t = 0} - \left[ \gamma (t) \frac{d}{dt} u\right] _{t=0} = \left[ \frac{d}{dt}(\gamma (t)u)\right] _{t = 0} = \frac{d}{dt} u |_{t = 0} = 0$ (the last equality uses $\gamma (t) \in K$ and $Ku=u$). $\square$

Corollary 2

$d\phi (\text{1 }) : {\mathfrak p}\simeq \text{ ran }(d\phi (\text{1 })) \le \mathbf{E}$.

Proof

By Lemma D and (0), ${\mathfrak k}$ is the kernel of $d\phi (\text{1 })$; as established above (in the proof of part (b) of Lemma C, ${\mathfrak g}= {\mathfrak p}\oplus {\mathfrak k}$. $\square$

Lemma E

$\text{ ran }(d\phi (\text{1 }))$ is all of $\mathbf{E}$, i.e., $d\phi (\text{1 }) : {\mathfrak p}\simeq \mathbf{E}$.

Proof

Note, first, that $T_{g}(G) = gT_{1}(G) = g{\mathfrak g}$ for any $g \in G$. We also have

$$\begin{aligned} (d\phi (g))(gX) = gXu \end{aligned}$$

for all $X \in {\mathfrak g}$. Hence, $\text{ ran }(d\phi (g)) = g \text{ ran }(d\phi (1))$. If the latter is not all of $\mathbf{E}$, then every $g$ is a critical point of $\phi$, whence, every point $\phi (g) = gu \in \mathbf{E}$ is a critical value. Sard’s Theorem now tells us that $Gu = \mathbf{E}^{\circ }_{+}$ has measure zero, a contradiction. $\square$

Construction of the Jordan product Now define $L_x \in {\mathfrak p}$ to be the unique self-adjoint element of ${\mathfrak g}$ with $L_x u = x$. Set $x \bullet y = L_x y$ for all $x, y \in \mathbf{E}$. This is evidently bilinear. A series of computations (see [13], pp. 49-50) shows that it makes $\mathbf{E}$ a formally real Jordan algebra with identity element $u$. Specifically,

• (1) $x \bullet y = y \bullet x$ since $x \bullet y - y \bullet x = x \bullet (y \bullet u)u - y \bullet (x \bullet u) = [L_x,L_y]u = 0$ (as remarked above following the proof of Corollary 1, $X, Y \in {\mathfrak p}\Rightarrow [X,Y] \in {\mathfrak k}$, and, by Lemma D, ${\mathfrak k}u = 0$);

• (2) $x \bullet u = L_x u = x$ by definition of $L_x$, so $u$ serves as the identity;

• (3) The product satisfies the Jacobi identity. This is proved exactly as in [13]. Note that the argument uses the fact that the inner product is associative, which follows from $L_x$ being self-adjoint. This also then gives us that the Jordan algebra $\mathbf{E}$ is formally real.

This completes the proof of the Koecher-Vinberg Theorem. $\square$