How do I find the uncertainty range of experimental data given this constraint on the standard deviation?

We have an inequality relating the standard deviation of position, $$σ_x$$, and the standard deviation of momentum, $$σ_p$$: $$σ_x × σ_p ≥ \,\frac{h}{4π}$$

Where $$\frac{h}{4π} = 5.27285909 × 10^{-35} \text{ J s}$$.

Say we record a position value $$x = 0.4 \text{ cm}$$ and a momentum value $$p = 1.25 × 10^{-27} \frac{\text{kg m}}{\text{s}}$$. How do I use these values for $$x$$ and $$p$$ and the relation $$σ_x × σ_p ≥ \,\frac{h}{4π}$$ to find the uncertainty in the measurements? For instance, would the uncertainty in position be on the order of nanometers? Micrometers? Picometers?

Where this question comes from:

I am reading up about Heisenberg's Uncertainty Principle, but I'm struggling with some of the math. It's been too long since my high-school physics and statistics classes.

The principle states that there is "a fundamental limit to the accuracy with which the values for certain pairs of physical quantities of a particle, such as position, $$x$$, and momentum, $$p$$, can be predicted from initial conditions". I want to know the relative scale of the uncertainty, but I don't know enough about the math and haven't been able to figure it out or find the answer.

• The proof of Heisenberg's uncertainty principle is well-known. You may find the answer on Physics stackexchange, if not, you can post a question there. If you record a position, say, $x=0.4$ cm and momentum $p= 1.25\cdot 10^{-27}\text{kg m}/\text{s}$ Heisenbergs relation says that either $x$ or $p$ (or both) are not the true postion/momentum. They carry uncertainty. How much ? It depends but must follow Heisenbergs constraint. Commented Apr 5, 2022 at 6:52
• @KurtG. I understand what you've said, but your last sentence is the point of my question: HOW do I use Heisenberg's constraint to find the uncertainty? Commented Apr 6, 2022 at 13:14
• You can't. The nature of your experiment determines the uncertainty of, say, the position measurement. Then Heisenberg says that you will have a minimal momentum uncertainty that you cannot avoid whatever experiment you perform. Think in terms of eigen states. Heisenberg says that the system cannot be simultaneously in a position and in a momentum eigenstate. Commented Apr 6, 2022 at 13:17
• I hope what I am saying now will not be too confusing: formally the uncertainty of an observable $A$ is defined as $\sigma_A:=\langle\psi|(A-\langle A\rangle)^2|\psi\rangle$ where $\psi$ is the state the system is in. You are probably asking how you are going to measure this. I don't know. Probably worth to edit your question to attract the attention of a few better physicists than I am one. Commented Apr 6, 2022 at 13:24
• That should have been squared uncertainty $\sigma_A^2$. Commented Apr 6, 2022 at 13:30