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Uncertainty of an electron in a core

  1. Jul 27, 2013 #1
    Lets say we put an electron in a box (spherical) which has a radius ##r##. I want to know if position uncertainty ##\Delta x = r## or is it ##\Delta x = 2r##?

    I need to know this to corectly calculate 1st the momentum uncertainty ##\Delta p=\frac{\hbar}{2\Delta x}## and 2nd that kinetic energy uncertainty ##\Delta E_k## using the Lorentz invariant like this:

    \begin{align}
    \Delta E^2 &= \Delta p^2 c^2 + {E_0}^2\\
    \Delta E &= \sqrt{\Delta p^2 c^2 + {E_0}^2}\\
    \Delta E_k + E_0 &= \sqrt{\Delta p^2 c^2 + {E_0}^2}\\
    \Delta E_k &= \sqrt{\Delta p^2 c^2 + {E_0}^2} - E_0\\
    \end{align}

    Furthermore. Can anyone elce confirm that this is the proper way to get the kinetic energy uncertainty ##\Delta E_k##. Is it ok that uncertainty for full energy ##\Delta E## all comes from the uncertainty in kinetic energy ##\Delta E_k## and none from rest energy (infront of which i wrote no deltas) ##E_0##?
     
  2. jcsd
  3. Jul 27, 2013 #2

    mfb

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    Is this homework? You can use both, it does not matter for a rough approximation. A proper calculation would give the correct prefactor, but that is way more complicated.

    ΔE is the minimal kinetic energy (again, neglecting small prefactors), the rest energy is constant and cannot have an uncertainty.
     
  4. Jul 27, 2013 #3
    I have been solving a homework yes, but i didn't know if i have to take ##\Delta x = r## or ##\Delta x = 2r##. But the question is theoretical. How do you mean i can use both? I still don't understand...
     
  5. Jul 27, 2013 #4

    BruceW

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    mfb means that if you were only asked to calculate the 'order of magnitude' of the uncertainty, then the teacher will say you are correct if you write [itex]\Delta x=r[/itex] or [itex]\Delta x = 2r[/itex]. (This is quite common for these types of questions). But if your teacher did not say 'order of magnitude', then assume he/she will care about which one you use.

    Now, assuming you are not meant to do any Fourier transforms, then you are not supposed to calculate the exact uncertainty in momentum. So, as you say, you have the choice between [itex]\Delta x=r[/itex] or [itex]\Delta x = 2r[/itex]. Without doing any extra calculations, can you guarantee that [itex]\Delta x<2r[/itex]? And keep in mind that you are trying to come up with a lower bound of the uncertainty. This should determine which one you choose.

    edit: Actually, I don't know if you are looking for a lower bound on uncertainty in momentum (see post 7)
     
    Last edited: Jul 27, 2013
  6. Jul 27, 2013 #5
    So i can't go wrong if i choose ##\Delta x = r## or ##\Delta x = 2r## unless my professor states which one to use. This is a good news indeed. I only used Fourier transform when dealing with Gauss to derive ##\Delta x \Delta p = \frac{\hbar}{2}##. Well i got ##\Delta x \Delta p = \hbar## which i couldn't explain after putting a week of effort in it... What a waste...
     
  7. Jul 27, 2013 #6

    BruceW

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    ah, be careful though. I would say ONLY if your professor stated 'order-of-magnitude', then you can't go wrong with either. If your professor did not say this, then I would say most likely you can go wrong, depending on which one you choose.

    Ah, I know that heartbreaking feeling, when you get to the end of a long calculation and it is off by some factor.
     
  8. Jul 27, 2013 #7

    BruceW

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    Actually, uh... This depends was the problem to find a lower bound for the uncertainty in momentum, or just to find the uncertainty in momentum. Because if it was just 'find uncertainty in momentum', then yeah most likely you can't go wrong if you choose r or 2r as uncertainty in x.
     
  9. Jul 27, 2013 #8
    Well this are the two pages that i have written to derive the uncertainty but missed it for a factor of ##1/2## (it is in the attachment). Maybee you could spot my mistake =)
     

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