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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.

Lawton
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  • 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. – Kurt G. Apr 05 '22 at 06:52
  • @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? – Lawton Apr 06 '22 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. – Kurt G. Apr 06 '22 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. – Kurt G. Apr 06 '22 at 13:24
  • That should have been squared uncertainty $\sigma_A^2$. – Kurt G. Apr 06 '22 at 13:30
  • @KurtG. So you're saying that I need to already know the uncertainty of one value to find the uncertainty of the other value? – Lawton Apr 06 '22 at 13:38
  • No. I just said that, unless the observables commute, you will always have a minimal uncertainty in at least one of them. I cannot answer your question how you determine it exactly. – Kurt G. Apr 06 '22 at 13:48

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