# Deriving a non trivial pair of observables whose product of variances is EQUAL to position and momentum?

I'm operating in the schrodinger model, where my position space is $$\mathbb{R}^1$$.

I'm looking for a pair of operators $$O_1, O_2$$ such that

$$\sigma_{O_1}^2 \sigma_{O_2}^2 = \sigma_x^2 \sigma_p^2$$

Always holds for any wave function $$\Psi$$ but $$O_1$$ is not the position or momentum operator.

## A Direct Derivation Strategy:

We can define $$\mu_{O_1}, \mu_{O_2}$$ in the canonical way as the expectation of their respective operators. So we are looking for an $$O_1$$, $$O_2$$ where $$O_1$$ isn't the position or momentum operator and the following holds:

$$\int_{-\infty}^{\infty} \psi^*(x) (O_1 - \mu_{O_1})^2 \psi(x) dx \int_{-\infty}^{\infty} \psi^*(x) (O_2 - \mu_{O_2})^2 \psi(x) dx = \\ \int_{-\infty}^{\infty} \psi^*(x) (x - \mu_{x})^2 \psi(x) dx \int_{-\infty}^{\infty} \psi^*(x) \left( i\hbar\frac{\partial}{\partial x} - \mu_{p} \right)^2 \psi(x) dx$$

And of course that $$\Psi$$ obeys the schrodinger equation for some choice of potential $$V$$.

It's possible to write

$$O_1 =$$

$$\begin{matrix} a_{0,0} & + a_{0, 1} x^1 & + a_{0,2}x^2 & \dots & + \\ a_{1,0} \frac{\partial}{\partial x} & + a_{1,1}x \frac{\partial}{\partial x} & + a_{1,2}x^2 \frac{\partial}{\partial x} & \dots & + \\ a_{2,0} \frac{\partial^2}{\partial x^2} & + a_{2,1} x \frac{\partial^2}{\partial x^2} & + a_{2,2} x^2 \frac{\partial^2}{\partial x^2} & \dots & + \\ \vdots & \vdots & \vdots & \ddots \end{matrix}$$

And it similarly we have that:

$$O_2 =$$

$$\begin{matrix} b_{0,0} & + b_{0, 1} x^1 & + b_{0,2}x^2 & \dots & + \\ b_{1,0} \frac{\partial}{\partial x} & + b_{1,1}x \frac{\partial}{\partial x} & + b_{1,2}x^2 \frac{\partial}{\partial x} & \dots & + \\ b_{2,0} \frac{\partial^2}{\partial x^2} & + b_{2,1} x \frac{\partial^2}{\partial x^2} & + b_{2,2} x^2 \frac{\partial^2}{\partial x^2} & \dots & + \\ \vdots & \vdots & \vdots & \ddots \end{matrix}$$

Then the operator $$O_1^2$$ which is literally $$O_1 \circ O_1$$ can be defined and similarly for $$O_2$$ so that we can now equate

$$(E[O_1 \circ O_1] - E[O_1])(E[O_2 \circ O_1] - E[O_2]) = (E[x^2] - E[x]^2)(-\hbar^2 E[\frac{\partial^2}{\partial x^2} - i\hbar E[\frac{\partial}{\partial x}])$$

And this gives you an infinite variable, infinite equation system of polynomial equations over $$a_{i,j}, b_{r,k}$$ which might contain non trivial solutions.

But this approach ABSOLUTELY sucks in terms of work required.

## A Weaker Derivation Strategy:

We could try to find a set of operators $$O_1, O_2$$ that meet our constraints such that their commutator has units of action and is equal to $$i\hbar$$.

With some blind luck it might be the case that the variance of these operators always happens to be equal to the product of variances of position and momentum.

The third bullet point here: https://en.wikipedia.org/wiki/Uncertainty_principle#Examples

Gives a way to generate candidates but at least upon initially inspecting i haven't been able to find a solution.