# The Pauli objection

This objection is about the asymmetry between configuration and momentum variables in the interpretation. It was part of Pauli's rejection of Bohm's causal interpretation. In Pauli's words, "the artificial asymmetry introduced in the treatment of the two variables of a canonically conjugated pair characterizes this form of theory as artificial metaphysics" (Pauli 1953).

Given that the minimal realist interpretation also prefers the configuration space, this objection is relevant and has to be answered.

## Physics itself does not show such a symmetry

But, first of all, there are important parts of physics which do not show this symmetry at all. So, in particular, the Hamilton operator has the form $$\hat{H} = \hat{p}^2 + V(\hat{q})$$, thus, a quite different dependence on both variables.

Then, as I have shown in (Schmelzer 2009) even if we restrict ourselves to Hamilton operators which have the form $$\hat{H} = \hat{p}^2 + V(\hat{q})$$, one can, for a given $$\hat{H}$$, find different pairs of conjugate operators $$\hat{p},\hat{q}$$ so that the same $$\hat{H}$$ has the same form $$\hat{H} = \hat{p}^2 + V(\hat{q})$$ but with different potentials $$V(\hat{q}$$, and, as a consequence, with different physical predictions. So, the physical predictions of the theory depend on the choice of the operators $$\hat{p},\hat{q}$$.

In principle, this is already sufficient to reject the Pauli objection.

## The different character of configuration and momentum variables already in classical theory

### Configuration variables define a system completely, momentum not

First of all, if we know the configuration space trajectory, we know everything about a particular system. The trajectory $$q(t)\in Q$$ is a complete description of reality. For the momentum trajectory, this is wrong. In the simplest case of a particle, we have $$p(t)=m\dot{q}(t)$$. But this defines the trajectory only modulo a constant shift, $$q(t) \to q(t) + q_0$$.

In a situation where one variable is sufficient to define the reality of the system completely, and the other not, they are certainly not on equal foot.

### Momentum variables depend on other parts of the universe

Given the usual formula $$p=m\dot{q}$$, one tends to think that the momentum is essentially nothing but the velocity.

But in fact the very definition of the momentum makes the momentum definition depending on the Lagrangian: $p_i = \frac{\partial L}{\partial q^i}$

So, this gives even in the simplest case already a difference - a dependence on the mass. One may think the difference is not important, given that it is a constant. Nonetheless, in general even the physical properties of the system are not sufficent to define the momentum variable. The very definition of the momentum depends, via the Lagrangian, on other parts of the universe. So, for a charged particle, we have $L = \frac12 mv^2 + \frac{q}{c} \vec{v}⋅\vec{A}(x,t)−q \phi(x,t)$ so that $\vec{p} = m\vec{v}(t) + \frac{q}{c} \vec{A}(x,t),$ that means, the value of the momentum depends even on the state of other fields, in particular of the electromagnetic field.

That means even if the trajectory $$q(t)$$ is given and fixed, the momentum is yet unknown, its value depends on the environment. If the environment is different, say, if we use a different EM field to measure it, we obtain different values for the momentum.

So, momentum measurement is contextual already in classical theory.

### In general, there is no equivalence between the Lagrange and the Hamilton formalism

Let's also not forget that in the general case there is no equivalence between the Lagrange and the Hamilton formalism. We have to assume that for every $$p$$ and $$q$$ there exist at most one $$v$$ such that $$p = \frac{\partial}{\partial v} L(q,v)$$. It follows that the Lagrangian has to be a convex function of $$v$$. Else, for the Lagrangian there exists no equivalent Hamilton formalism.

The same problem holds in the reverse direction too. For each $$\dot{q}$$ and $$q$$ there should exist at most one solution $$p$$ of the Hamilton equation $$\dot{q} = \frac{\partial}{\partial p} H(p,q)$$. So, the Hamilton function should be a convex function of $$p$$ for every $$q$$ too.

## The Lagrange formalism is more fundamental

So, once there is, in general, no equivalence between the Lagrange and the Hamilton formalism, the question which formalism is the more fundamental.

What can be used to find an answer? I think the points mentioned above already give good enough hints that it is the Lagrange formalism which is more fundamental. The independent variable of the Lagrange formalism completely describes the system under consideration, and nothing else. Instead, the physics as defined by the rest of world is defined by a single function, the Lagrangian.

Once we recognize that already on the classical level the Lagrange formalism is more fundamental, this gives even more certainty that the configuration space, which is what the Lagrange formalism is using, is a more fundamental object than the phase space used in the Hamilton formalism.