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Heisenberg's Uncertainty

  1. Aug 15, 2015 #1
    1. The problem statement, all variables and given/known data
    I don't get how you can use x and p instead of delta x and p in Heisenberg's principle.
    Also why is it just h bar? Isn't it supposed to be h bar/2?
    Problem 2-iv

    2. Relevant equations
    delta x* delta p >= h bar/2

    3. The attempt at a solution
    I tried to deal with this problem but I couldn't find a way, when I looked at the solution it use the heisenberg's principle without the deltas, I can't understand how this could be done

    Attached Files:

  2. jcsd
  3. Aug 15, 2015 #2


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    There's quite a few versions of Heisenberg's Uncertainty Principle. ##\Delta x \Delta p \geq h \ \text{or}\ \geq \hbar \ \text{or}\ \geq \frac{\hbar}{2} ## are the main 3 that I'm aware of, at least for position momentum uncertainty. The last reported (most accurate) that I remember was hbar over 2, but they may use a different version, depending on how old the documents are.

    I also don't see where you're talking about. I just skimmed through the solutions, but everywhere that I noticed them speaking of uncertainty they either used ##\delta x \ \text{&}\ \delta p \ \text{or} \ \Delta x \ \text{&}\ \Delta p ##
    Which problem are you referring to?

    Edit** Somehow I missed 2.iv -.-

    They're not talking about uncertainty. They're talking about the total energy of the ground state. The ground state should never be lower than 1 quantum.

    So I believe that the measured values of x and p in the ground state should be of the order of the uncertainty principle. There may be more to the story, if so, someone else would have to jump in and inform us. I'm not a huge quantum guy.
    Last edited: Aug 15, 2015
  4. Aug 16, 2015 #3
    What's 1 quantum? is it the planck constant?
  5. Aug 16, 2015 #4


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    Basically, yea. It's the increment that your energy scale progresses by.
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