Four and a Half Axioms for Finite-Dimensional Quantum Probability

Abstract

It is an old idea, lately out of fashion but now experiencing a revival, that quantum mechanics may best be understood, not as a physical theory with a problematic probabilistic interpretation, but as something closer to a probability calculus per se. However, from this angle, the rather special C ^*-algebraic apparatus of quantum probability theory stands in need of further motivation. One would like to find additional principles, having clear physical and/or probabilistic content, on the basis of which this apparatus can be reconstructed. In this paper, I explore one route to such a derivation of finite-dimensional quantum mechanics, by means of a set of strong, but probabilistically intelligible, axioms. Stated very informally, these require that systems appear completely classical as restricted to a single measurement, that different measurements, and likewise different pure states, be equivalent (up to the action of a compact group of symmetries), and that every state be the marginal of a bipartite non-signaling state perfectly correlating two measurements. This much yields a mathematical representation of (basic, discrete) measurements as orthonormal subsets of, and states, by vectors in, an ordered real Hilbert space – in the quantum case, the space of Hermitian operators, with its usual tracial inner product. One final postulate (a simple minimization principle, still in need of a clear interpretation) forces the positive cone of this space to be homogeneous and self-dual and hence, to be the state space of a formally real Jordan algebra. From here, the route to the standard framework of finite-dimensional quantum mechanics is quite short.

Appendices

Appendix A: Entropy and Sharpness

The following considerations may offer some independent motivation for Axiom 1. There are two natural ways to extend the definition of entropy to states on a test space. If $\alpha$ is a state on a locally finite, finite-dimensional test space $\mathfrak{A}$, then Minkowsky’s theorem tells us that $\alpha$ has a finite decomposition as a mixture $\alpha = \sum\nolimits_i \,{t_i}{\alpha_i}$ of pure states ${\alpha_1},...,{\alpha_n}$. Define the mixing entropy of $\alpha$, $S(\alpha )$, to be the infimum of $H({t_1},...,{t_n}) = - \sum\nolimits_i \,{t_i}log({t_i})$ over all such convex decompositions of $\alpha$. Alternatively, one can consider the local entropy ${H_E}(\alpha ) = H(\alpha {|_E}) = - \sum\nolimits_{x \in E} \,\alpha (x)log(\alpha (x))$. Define the measurement entropy of $\alpha$, $H(\alpha )$,to be the infimum value of the local measurement entropies ${H_E}$ over all tests $E$.

Suppose now that the group $G$ figuring in Axiom 2 is compact. One can then endow $\mathfrak{A}$ with the structure of a compact topological test space in the sense of [30]. Assuming that all states in $\Omega$ are continuous as functions $X \to \mathbb{R}$, it follows [36, Lemma 6] then the infimum defining $H$ is actually achieved, i.e., $H(\alpha ) = {H_E}(\alpha )$ for some test $E \in \mathfrak{A}$. An easy consequence is that $H(\alpha ) = 0$ iff $\alpha (x) = 1$ for some $x \in X(\mathfrak{A})$. One can also show [36] that $S(\alpha ) = 0$ iff $\alpha$ is a limit of pure states. Consequently, if the set of pure states is closed, we have $S(\alpha ) = 0$ iff $\alpha$ is pure.

In both classical and quantum cases, $S = H$. One might consider taking this as a general postulate:

$$ {\hbox{Postulate A: }}H(\alpha ){ \,= \,}S(\alpha ){\hbox{ for every state }}\alpha \in \Omega { }{. } $$

An immediate consequence is that, subject to the topological assumptions discussed above, a pure state (with mixing entropy $S(\alpha ) = 0$) must have local measurement entropy ${H_E}(\alpha ) = 0$ for some test E, whence, there must be some outcome $x \in E$ with $\alpha (x) = 1$. Conversely, for every $x \in X$, if $\alpha (x) = 1$, then ${H_E}(\alpha ) = 0$ for any E containing outcome x, whence, $H(\alpha ) = 0$. But then $S(\alpha ) = 0$ as well, and α is therefore pure. If $\mathfrak{A}$ is unital, meaning that every outcome has probability 1 in at least one state, then it follows that $\mathfrak{A}$ is actually sharp. Moreover, we see that every pure state has the form ε _x for some x In this case, the second half of Axiom 2 follows automatically from the first. Further discussion of Postulate A can be found in the paper [36], where theories satisfying it are termed monoentropic.

Appendix B: An Alternative Route to Homogeneity

We say that the space V is weakly self-dual iff there exists an order-isomorphism – that is, a positive, invertible linear map with positive inverse – $\phi :{V^* } \to V$. Note that such a map corresponds to a positive bilinear form $\omega :{V^* } \times {V^* } \to \mathbb{R}$ via $\omega (x,y) = \phi (x)(y)$, hence, to a non-signaling bipartite state on $\mathfrak{A}$. We call a bipartite state ω an isomorphism state iff the positive linear map $\hat{\omega }:{V^* } \to V$ given by $\hat{\omega }(x)(y) = \omega (x,y)$ is invertible. One can show [37] that any such state is pure. Note that as u belongs to the interior of $V_{+}^*$, if ω is an isomorphism state, we must have ${\omega_1} = \hat{\omega }(u)$ in the interior of ${V_{+} }$. This suggests the following alternative to Axioms 5:

$$ {\hbox{Postulate B: Every interior state is the marginal of an isomorphism state }} $$

Lemma [37]

Subject to Postulate B alone, V is weakly self-dual and homogeneous.

Proof

For there to exist an isomorphism state, V must be weakly self-dual. For homogeneity, let α and β belong to the interior of ${V_{+} }$. Then Postulate B implies that there exist isomorphism states ω and μ with $\alpha = \hat{\omega }(u)$ and $\beta = \hat{\mu }(u)$. Thus, $\beta = (\mu \circ {\omega^{ - 1}})(\alpha )$. As $\mu \circ {\omega^{ - 1}}$ is an order-automorphism of V, it follows that the cone is homogeneous.□

Postulate B is similar in flavor to Axiom 4, but seems somewhat awkward in its reference only to states in the interior of ${V_{+} }$. It would be desirable to find a single, natural principle implying both of these axioms. Further work in this direction can be found in [37]

Appendix C: An Alternative Route to Self-Duality

An alternative proof of Proposition 1 (the self-duality of ${V_{+} }$) appeals to the fact [38, Lemma 1.0] that a finite-dimensional ordered space A is self-dual w.r.t a given inner product iff every vector $a \in A$ has a unique Jordan decomposition $a = {a_{+} } - {a_{-} }$ with $\left\langle {{a_{+} },{a_{-} }} \right\rangle = 0$. We’ll need the following

Lemma

Suppose A carries a positive inner product, with respect to which every element of A has an orthogonal Jordan decomposition. Then ${A_{+} }$ is self-dual.

Proof

It suffices to show that the orthogonal Jordan decomposition is unique. Suppose ${a_{+} } - {a_{-} } = {b_{+} } - {b_{-} }$ are two orthogonal Jordan decompositions of an element $a \in A$, and that the inner product is positive. We ${a_{+} } - {b_{+} } = {a_{-} } - {b_{-} } = :x \in A$, so that

$$ 0 \le \parallel x{\parallel^2} = \left\langle {{a_{+} } - {b_{+} },{a_{-} } - {b_{-} }} \right\rangle = - (\left\langle {{b_{+} },{a_{-} }} \right\rangle + \left\langle {{a_{+} },{b_{-} }} \right\rangle ). $$

But since the inner product is positive, this last quantity is non-positive: evidently, we must have

$$ \left\langle {{a_{+} },{b_{-} }} \right\rangle = \left\langle {{a_{-} },{b_{+} }} \right\rangle = 0, $$

whence, x = 0, whence, ${a_{+} } = {b_{+} }$ and ${a_{-} } = {b_{-} }$: the decomposition is unique, as advertised. □

Let us say that a model $(\mathfrak{A},\Omega )$ is spectral iff it satisfies the conclusion of Lemma 6 — that is, if every state $\mu \in \Omega$ can be expanded as $\sum\nolimits_{x \in E} \,\mu (x){\varepsilon_x}$ where $E \in \mathfrak{A}$ and, for each $x \in E$, ${\varepsilon_x}$ is a state with ${\varepsilon_x}(x) = 1$.

Theorem A

Suppose $V(\mathfrak{A},\Omega )$ is spectral, that $\mathfrak{A}$ is 2-symmetric, and that Provisional Postulate 2 holds. Then $V(\mathfrak{A})$ is self-dual.

Proof

If $f:E \to \mathbb{R}$, where $E \in \mathfrak{A}$, let ${a_f} = \sum\nolimits_{x \in E} \,f(x)x$. Note that this gives us a positive linear mapping ${R^E} \to V{(\mathfrak{A})^* }$. That $\mathfrak{A}$ is spectral implies that every positive element of ${V^* }$ has a representation as a _f for some $f \geqslant 0$ on some $E \in \mathfrak{A}$. Notice that $u = {a_1}$ for the constant function $1:E \to \mathbb{R}$ on any test $E \in \mathfrak{A}$.

Now let ${v_x} = {q_x} + cu$, where ${q_x} = x - \left\langle {x,u} \right\rangle u = (1 - \left\langle {x,u} \right\rangle )x$, as in Lemma 2, so that ${v_x} \bot {v_y}$ for $x \ne y$ in $E$. If $f \in {\mathbb{R}^E}$, let ${v_f} = \sum\nolimits_{x \in E} \,f(x){v_x}$. Note that

$$ {v_f} = \sum\limits_{x \in E} \,f(x){v_x} = \sum\limits_{x \in E} \,f(x)(1 - \left\langle {x,u} \right\rangle + nc)x. $$

Setting $g \equiv 1 - \left\langle {x,u} \right\rangle + nc$ (noting that this is constant!), we have ${v_f} = \sum\nolimits_{x \in E} \,f(x)gx = {a_{fg}}$. In particular, ${a_g} = v = ncu$, so that $g \ne 0$. Thus, we have ${a_f} = {a_{f/gg}} = {v_{f/g}}$. Thus, if $g \ne 0$, every $a = {a_f}$ in ${V^* }$ has an orthogonal resolution with respect to an orthonormal set $\{ {v_x}|x \in E\}$ for some $E \in \mathfrak{A}$. Finally, since (by our provisional Postulate 2) we have ${v_x} \ge 0$ for every $x \in E$, every vector with an orthogonal resolution relative to the set $\{ {v_x}|x \in E\}$ has an orthogonal Jordan decomposition.□

Appendix D: Invariant Positive Inner Products on ${\mathcal{L}_h}({\mathbf{H}})$

Let H be a complex Hilbert space of dimension n, with frame manual $\mathfrak{F}$ and unit sphere X. We seek to classify the unitarily invariant inner products on ${\mathfrak{L}_\mathfrak{h}}({\mathbf{H}})$ that are positive on the positive cone of the latter, and to show that all of these are automatically minimizing.

As remarked above, Gleason’s Theorem provides an isomorphism between the space $V(\mathfrak{F})$ of signed weights on $\mathfrak{F}$, and the space $\mathop \mathfrak{L}\nolimits_h ({\mathbf{H}})$ of Hermitian operators on H: for every $\alpha \in V(\mathfrak{F})$, there is a unique ${W_\alpha } \in \mathop \mathfrak{L}\nolimits_h ({\mathbf{H}})$ with $\alpha (x) = \left\langle {{W_\alpha }x,x} \right\rangle$ for all $x \in X$. We also have a dual isomorphism ${V^* }(\mathfrak{F}) \simeq \mathop \mathfrak{L}\nolimits_h ({\mathbf{H}})$, sending each $a \in {V^* }(\mathfrak{F})$ to an Hermitian operator A _a with ${\hbox{Tr}}({A_a}{W_\alpha }) = a(\alpha )$ for all $\alpha \in V(\mathfrak{F})$. Note that in this representation, the order unit is represented by the identity operator 1 on H. If U is a unitary operator on H, understood as acting on X, then the natural action on $V(\mathfrak{F})$ is given by $U(\alpha )(x) = \alpha ({U^{ - 1}}x)$ for all $\alpha \in V(\mathfrak{F})$ and all $X \in X$. Thus, we have $\left\langle {{W_{U\alpha }}x,x} \right\rangle = \left\langle {{W_\alpha }{U^{ - 1}}x,{U^{ - 1}}x} \right\rangle$, whence, ${W_{U\alpha }} = U{W_\alpha }{U^* }$ for all states α. In other words, the natural representation of U(H) on $V(\mathfrak{F}({\mathbf{H}})) \simeq \mathop \mathfrak{L}\nolimits_h ({\mathbf{H}})$ is exactly its usual adjoint action. It follows that the dual action of U(H) on ${V^* }(\mathfrak{F})$ is again the adjoint action $A \mapsto {U^* }AU$. Noting that 1 and 1^⊥, the space of trace-0 Hermitian operators, are both invariant under this action, it follows that the two are orthogonal with respect to any unitarily invariant inner product on ${V^* }(\mathfrak{F})$. Also, since the adjoint representation of U(H) on 1^⊥ is irreducible [39, p. 20], it follows from Schur’s Lemma that up to normalization, there is only one unitarily invariant inner product on the latter – in other words, any invariant inner product on 1^⊥ has the form $\left\langle {a,b} \right\rangle = \tfrac{\lambda }{n}{\hbox{Tr}}(ab)$ for some λ > 0, with λ > 1 corresponding to the normalized trace inner product. Hence, an invariant inner product on ${\mathbf{V}} = \left\langle 1 \right\rangle \oplus {1^\bot }$ is entirely determined by the normalization of 1 and the choice of λ. Taking ||1|| = 1, we have that, for any $a = s1 + {a_o}$ and $b = t1 + {b_o}$, where ${a_o},{b_o} \in {1^\bot }$ and $s,t \in \mathbb{R}$, we have

$$ \left\langle {s1 + {a_o},t1 + {b_o}} \right\rangle = st + \frac{\lambda }{n}{\hbox{Tr}}({a_o}{b_o}). $$

We require that $\left\langle {a,b} \right\rangle \ge 0$ for all positive $a,b \in {V^* }$. The spectral theorem tells us that this is equivalent to requiring that $\left\langle {{p_x},{p_y}} \right\rangle \ge 0$ for all rank-one projections${P_x},{P_y}\,(x,y, \in X)$. Writing ${P_x} = \tfrac{1}{n}1 + ({P_x} - \tfrac{1}{n}1)$, and similarly for ${P_y}$, we have

$$ \begin{array}{lll}\left\langle {{P_x},{P_y}} \right\rangle \,\,= & \frac{1}{{{n^2}}} + {\rm {\lambda Tr}}\left( {\left( {{P_x} - \frac{1}{n}1} \right)\left( {{P_y} - \frac{1}{n}1} \right)} \right) \\ \quad\quad\quad\quad= & \frac{1}{{{n^2}}} + \frac{{\rm{\lambda }}}{n}{\hbox{Tr}}\left( {{P_x}{P_y} - \frac{{{P_x} + {P_y}}}{n} + \frac{1}{{{n^2}}}1} \right) \\ \quad\quad\quad\quad=& \frac{1}{{{n^2}}} + \frac{{\rm{\lambda }}}{n}\left( {{\hbox{Tr}}({P_x}{P_y}) - \frac{2}{n} + \frac{1}{n}} \right) \\ \quad\quad\quad\quad=& \frac{1}{{{n^2}}} + \frac{{\rm{\lambda }}}{n}\left( {|{{\left\langle {x,y} \right\rangle }_o}{|^2} - \frac{1}{n}} \right) \\ \quad\quad\quad\quad=& \frac{{1 - {\rm{\lambda }}}}{{{n^2}}} + \frac{{\rm{\lambda }}}{n}|{\left\langle {x,y} \right\rangle_o}{|^2}, \\\end{array} $$

where $\left\langle, \right\rangle$ is the inner product on H. This will be non-negative for all choices of unit vectors x and y (in particular, for x and y orthogonal) iff $0 < \lambda \le 1$ – in which case, the minimum value of $\left\langle {{P_x},{P_y}} \right\rangle$ occurs exactly when $x \bot y$, so such an inner product is automatically minimizing.