Necessary conditions for the uncertainty principle

[andresB]

In (1D) quantum mechanics we deal with wavefunctions $\psi(x)$ belonging to $L^{2}(\mathbb{R})$. We then have the position and momentum operators $$\hat{x}=x, \quad\hat{p}=-i\hbar\frac{\partial}{\partial x},$$
with the canonical commutation relation $[\hat{x},\hat{p}]=i\hbar$. The textbooks then use the variance to prove the Uncertainty principle $\sigma_{x}\sigma_{p}\geq\frac{\hbar}{2}.$

So far, so good. However, consider the position-momentum wavefunctions $\psi(x,p)$ belonging to $L^{2}(\mathbb{R}^{2})$. We can define the prequantization operators $$\hat{x}_{pre}=i\hbar\frac{\partial}{\partial p}+x, \quad\hat{p}_{pre}=-i\hbar\frac{\partial}{\partial x},$$
and they also obey the canonical commutation relation $[\hat{x}_{pre},\hat{p}_{pre}]=i\hbar.$ The thing is, there is no position-momentum uncertainty principle at this level, and $\psi(x,p)$ is allowed to be as narrow in phase space as we want.

Why? What necessary condition for the uncertainty principle is missing at the prequantization level?

 

[PeterDonis]

We can define the prequantization operators

Where is this coming from? Do you have a reference?

 

[renormalize]

Where is this coming from? Do you have a reference?

This seems to be relevant: https://en.wikipedia.org/wiki/Phase-space_wavefunctions
Phase-space representation of quantum state vectors is a formulation of quantum mechanics elaborating the phase-space formulation with a Hilbert space. It "is obtained within the framework of the relative-state formulation. For this purpose, the Hilbert space of a quantum system is enlarged by introducing an auxiliary quantum system. Relative-position state and relative-momentum state are defined in the extended Hilbert space of the composite quantum system and expressions of basic operators such as canonical position and momentum operators, acting on these states, are obtained." Thus, it is possible to assign a meaning to the wave function in phase space, ψ(x,p,t), as a quasiamplitude, associated to a quasiprobability distribution.

 

[PeterDonis]

This seems to be relevant

This is about the phase space representation of quantum state vectors.

There's nothing at all about "prequantization" operators, which is what the OP was making claims about.

 

[andresB]

Where is this coming from? Do you have a reference?

Ah, sorry. It's from geometric quantization https://arxiv.org/pdf/1801.02307

At the level of prequantization, wavefunction dynamics is entirely equivalent to classical (statistical) mechanics, where the density $\rho(q,p)=\left|\psi(q,p)\right|^{2}$ follows the Liouville equation.

 

[PeterDonis]

It's from geometric quantization

Ok. Then the issue you describe would only be an issue for that particular approach to QM.

 

[Demystifier]

Nice question and apparent paradox, let me demystify it.

To simplify the math I will work in units $\hbar=1$, so that position and momentum can be measured in the same units. And to reduce the apparent mystery to something well understood, I will change the notation by writing $p\equiv y$. With this notation the wave function $\psi(x,p)$ is written as $\psi(x,y)$, and the mysterious pre-quantum operators become
$$x_{pre}=i\partial_y+x=-p_y+x$$
$$p_{pre}=-i\partial_x=p_x$$
Now everything should be clear, because all this can be understood by standard QM, with $y$ interpreted as another position variable independent of $x$. The wave function $\psi(x,y)$ can simultaneously be arbitrarily narrow in both $x$ and $y$, because $x$ and $y$ commute. The non-commuting operators $x_{pre}$ and $p_{pre}$ obey
$$[x_{pre},p_{pre}]=i$$
so we have the usual uncertainty relation
$$\Delta x_{pre} \Delta p_{pre} \geq \frac{1}{2}$$
At the same time we have
$$\Delta x \Delta y \geq 0$$
There is no paradox and no mystery. The catch is that $p=y$ is very different from the momentum operator $p_{pre}=p_x$, the apparent paradox stems from misidentification of those two different operators.

Or to answer the question by OP, no additional condition is missing. What OP misses is to realize that $p$ is not $p_{pre}$.

 

[Demystifier]

At the level of prequantization, wavefunction dynamics is entirely equivalent to classical (statistical) mechanics, where the density $\rho(q,p)=\left|\psi(q,p)\right|^{2}$ follows the Liouville equation.

Which, by writing $q=x$, $p=y$, becomes the usual quantum mechanical continuity equation.

So pre-quantization, at this level, is equivalent to quantization in a doubled number of space dimensions.