The Topos of Star-Rings is Quantum

I've noticed that the classifying topos of star-rings behaves in a "quantum" fashion.

I mean this in a precise sense. I will describe a simple question in the internal logic, whose solution quantitatively reproduces behavior seen in the quantum harmonic oscillator.

Topos

Allow me to outline the salient features of this topos.

• It contains a generic star-ring, \(C\). This has interesting subsets: \(R = {x : C | x = x^*}\), the set of "reals", and \(R^{\geq 0}\), the set of "positive numbers", consisting of finite sums of elements of the form \(x x^*\).
• It is a monoidal topos. For any type \(X\), there are adjoint functors \(X * -\) and \(X \wand -\) from the topos to itself. The monoidal structure captures the concept of commutativity: \(C * C = \{(a,b) : C \times C \mid xy = yx\}\).
• Functions from \(R\) are polynomials. The choice between \(\to\) and \(\wand\) determines whether or not the coefficients commute with the variable. Typically, we want the coefficients to commute, so \(\wand\) is preferred.
• It is a presheaf topos. Objects are functors from finitely-presented star-rings to sets; morphisms are natural transformations.

Setup

We will focus on the following object. \[A = \{a \in C | a a^* - a^* a = 1\}\] How does it behave?

Well, it has no elements in the global context. The relevant classifying topos is a presheaf topos, in which the interpretation of \(A\) is the functor \(R \mapsto \{a \in R \mid a a^* - a^* a = 1\}\). This has no global sections, since that set is empty for some values of \(R\).

But even though we can't pin down elements of \(A\), we can describe plenty of probability distributions over it. Linear maps \((A \wand C) \wand C\), which map real functions to reals, positive functions to positive numbers, and the constant \(1\) to \(1\).

Note that the construction of probability distributions is functorial. If \(f : X \wand Y\), we get a map \(f : \mathrm{Prob}(X) \wand \mathrm{Prob}(Y)\). So if we have a probability distribution on \(A\), we can ask for the probability distribution of its real part, or its squared magnitude.

So I pose the following question.

Find an element of \(\mathrm{Prob}(A)\), defined in the global context, that in some reasonable sense minimizes squared magnitude. Mapping into \(\mathrm{Prob}(R)\) by taking the first coordinate, what does the distribution look like?

Global Elements

Let's start by understanding global elements.

The interpretation of \(C\) in the classifying topos is the forgetful functor, from finitely-presented star-rings to sets. A global element of this presheaf is given by an element of the initial star-ring, \(\mathbb{Z}\). So every globally-defined number is an integer.

What about probability distributions?

Consider first the set \((R \wand C) \wand C\). Its global elements correspond to elements of \(C\) in context \(R \wand C\). And every function \(R \wand C\) is given uniquely as a polynomial in one variable. So a global element of \((R \wand C) \wand C\) interprets as a natural transformation, assigning to each finitely-presented star-ring \(R\) a function \(R[x] \to R\).

When our map \((R \wand C) \wand C\) is linear, the same will be true of the functions making up the natural transformation. Such functions are uniquely determined by their values at the monomials. So the natural transformation might as well consist of functions \(\mathbb{N} \to R\) instead. But such things are simply infinite lists of global elements of \(R\).

In short, the linear maps \((R \wand C) \wand C\) available in global context are those of the form \((a \mapsto \sum_n c_n a^n) \mapsto \sum_n c_n k_n\), where the \(k_n\) are fixed integers. The global elements of \(\mathrm{Prob}(R)\) are therefore a subset of these.

By similar logic, every element of \(\mathrm{Prob}(A)\) has the form \((a \mapsto \sum_{m,n} c_{m,n} (a^*)^m a^n) \mapsto \sum_{m,n} c_{m,n} k_{m,n}\), for fixed integer values of \(k_{m,n}\).

Solution

With that understood, let's tackle the question.

Let \(E : \mathrm{Prob}(A)\) be an arbitrary probability distribution, defined in the global context. By the above discussion, it must take the form \(E(a \mapsto \sum_{m,n} c_{m,n} (a^*)^m a^n) = \sum_{m,n} c_{m,n} k_{m,n}\), for some fixed integers \(k_{m,n}\).

The right notion of "minimizing expected magnitude" seems to be to minimize \(E((a^* a)^n)\) for all \(n \in \mathbb{N}\). They will always be positive, and \(E(1)\) will always equal one, but we are able to push them down to those bounds.

This constraint, that \(E((a^* a)^n) = 0\) for all \(n \geq 1\), is equivalent the statement that \(k_{n,n} = 0\) for all \(n \neq 0\). Given this, we observe that \(k_{m,m} + k_{n,n} \pm 2k_{m,n}\) is always positive. This implies \(k_{m,n} = 0\) unless \(m=n=0\). So there's only one possible solution.

And it is indeed a solution. \[\begin{align*} &E\left(\left(\sum_{m,n} c_{m,n} (a^*)^ma^n\right)^* \left(\sum_{m,n} c_{m,n} (a^*)^ma^n\right)\right) \\ &= \sum_{m,n,m',n'} E((a^*)^n a^m c_{m,n}^* c_{m',n'} (a^*)^{m'} a^{n'}) \\ &= \sum_{m} E((a^*)^0 a^m c_{m,0}^* c_{m,0} (a^*)^m a^0) \\ &= \sum_{m} E((c_{m,0} (a^*)^m)^* (c_{m,0} (a^*)^m)) \\ & \text{is positive} \end{align*}\]

So we have found the distribution. It is written most simply as \(E(a \mapsto \sum_{m,n} \sum_{m,n} c_{m,n} (a^*)^m a^n) = c_{0,0}\).

Now, what is the distribution of the real part? We actually can't answer that: the real part is only defined if \(2\) is invertible! So let's compute the distribution of twice the real part.

\[\begin{align*} E' &: (R \wand C) \wand C \\ E' &= f \mapsto E(f \circ 2\Re) \\ &= f \mapsto E(a \mapsto f(a + a^*)) \\ &= \left(x \mapsto \sum_n c_n x^n\right) \mapsto \sum_n c_n E(a \mapsto (a+a^*)^n) \end{align*}\]

We need the constant term of \((a+a^*)^n\), when it is reexpressed with the \(a^*\)s preceding the \(a\)s. This equals the number of ways of putting an \(a\) or an \(a^*\) in each of \(n\) slots, then matching each \(a\) to a later \(a^*\). This is simply the number of matchings of \(n\) items, which is \((n-1)!!\) when \(n\) is even, and zero when \(n\) is odd.

We conclude: \[E'\left(x \mapsto \sum_n c_n x^n\right) = \sum_n (2n-1)!! c_{2n}\]

One final touch. Notice that \(\int_{-\infty}^\infty \frac{e^{-x^2}}{\sqrt{2\pi}} \sum_n c_n x^n = \sum_n (2n-1)!! c_{2n}\). So our solution seems to be describing a standard normal distribution.

The Quantum Connection

What does all this have to do with quantum mechanics?

Our set \(A\) essentially captures the state space of the quantum harmonic oscillator, expressed in terms of the ladder operators. There are no global elements, corresponding to the fact that we can't simultaneously determine position and momentum. Probability distributions over \(A\) act like mixed states of the quantum system.

In our problem, we minimized the expected Hamiltonian, then computed the probability distribution of \(a + a^\dagger = \sqrt{2}x\). This yielded a standard normal distribution for \(\sqrt{2}x\). And indeed, traditional quantum mechanical theory says that distribution should be \(\frac{1}{\sqrt{2}}\psi_0^2\left(\frac{x}{\sqrt{2}}\right) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\), which is also a standard normal.

In other words, we have reproduced the ground state of the quantum harmonic oscillator. I expect to be able to reproduce more of quantum mechanics in this framework, and I'm particularly curious to see what happens in systems with an infinite number of degrees of freedom.