0
$\begingroup$

I'm new to this site and I would like some help about a tiny matter.

We know that the Heisenberg Uncertainty Principle can be either focused on the Displacement/Momentum formula and the Energy/Time formula:

$$\Delta{x}\Delta{p} \ge \hbar/2$$

$$\Delta{E}\Delta{t} \ge \hbar/2$$

Also if $$E=0$$ then $$\Delta{p} = 0$$ and would violate the uncertainty principle since energy can never be zero.

In the first case, we know that if the displacement x is definite, then the momentum is unknown, same with energy and time. If time changes then energy is definite and vice versa in both cases since the wave function is invariant.

So my question is, if energy is definite, what relation does it have to momentum and wavelength? Is it true that definite energy has two momentum for both max/min points of the wave function, one positive and one negative and therefore can never have definite momentum?

I'm still new to Quantum Physics so forgive me if I have gotten certain concepts wrong.

Improve this question
$\endgroup$
4
  • $\begingroup$ Where did you get the idea that $E=0 \implies \Delta p=0$? $\endgroup$
    – DanielSank
    Commented May 7, 2018 at 22:44
  • 2
    $\begingroup$ Your taking a very classical view of quantum physics. This always leads to problems. Also it appears you're thinking $\Delta$ means difference. It does not. It means the standard deviation of the measured value. $\endgroup$
    – JEB
    Commented May 7, 2018 at 23:31
  • $\begingroup$ DanielSank, inside a potential well, doesn't momentum go hand to hand with energy? If Energy is definite isn't momentum definite too? Or is it too classical still? $\endgroup$
    – EBE
    Commented May 8, 2018 at 19:25
  • $\begingroup$ JEB, well a deviation is a type of difference, but if I understood it correctly, this deviation occurs because of the equipment used not being able to exactly measure these quantum observables right? $\endgroup$
    – EBE
    Commented May 8, 2018 at 19:27

1 Answer 1

Reset to default
1
$\begingroup$

The measurements of any quantum observable (momentum, energy, position, etc.) follow a statistical distribution which will have some standard deviation/error. As such you can never have $\Delta E=0$, for all quantum observables are susceptible to random error. It's akin to how its impossible to make a perfect measurement.

Simply because energy is "definite" (I assume you really mean quantized), does not mean its expectation value with no error bars.

Improve this answer
$\endgroup$
1
  • $\begingroup$ That's one of my issues when it comes to try to understand this to be honest, I don't get an exact explanation about what "definite" means really in these cases, certainly not in my book... $\endgroup$
    – EBE
    Commented May 8, 2018 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.