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Understanding proof for Heisenberg uncertainty

  1. Jan 17, 2016 #1
    I've uploaded a proof of the Heisenberg uncertainty principle from Konishi's QM. I just don't quite understand one part: what is the significance of the discriminant being less than or equal to 0? Wouldn't this just result in ## \alpha = R \pm iZ ##? Why would this be desired in this proof?
     

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  3. Jan 17, 2016 #2

    Samy_A

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    He finds an inequality that must hold for all real ##\alpha##.
    Simplified, it says: ##{\alpha}^{2}+b\alpha +c \geq 0##
    The discriminant of the equation ##{\alpha}^{2}+b\alpha +c = 0## is ##D=b²-4c##.
    If ##D>0##, the equation will have two different real roots, ##r_1## and ##r_2##, so you get ##{\alpha}^{2}+b\alpha +c=(\alpha-r_1)(\alpha-r_2) \geq 0## (still for all real ##\alpha##).
    But this can't be true for values of ##\alpha## lying between the two roots. Therefore ##D \leq 0##.
     
    Last edited: Jan 17, 2016
  4. Jan 17, 2016 #3

    fresh_42

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    If I made no mistake just set ##α = ħ / (2<(P-P_0)^2>)##.
     
  5. Jan 17, 2016 #4

    stevendaryl

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    I'm not sure about the discriminate, but the equation he derives, true for any [itex]\alpha[/itex], is:

    [itex]A - B \alpha + C \alpha^2 \geq 0[/itex]

    where [itex]A = \langle (Q - Q_0)^2 \rangle[/itex], [itex]B = \hbar[/itex], and [itex]C = \langle (P- P_0)^2 \rangle[/itex]

    So if it's true for every [itex]\alpha[/itex], then in particular, it's true when [itex]\alpha = \frac{B}{2C}[/itex]. Plugging this into the inequality gives:
    [itex]A - \frac{B^2}{2C} + \frac{B^2}{4C} \geq 0[/itex]

    Which implies [itex]AC - \frac{B^2}{4} \geq 0[/itex], or [itex]\sqrt{A}\sqrt{C} \geq \frac{B}{2}[/itex]

    Going back to the definitions of [itex]A[/itex], [itex]B[/itex] and [itex]C[/itex] gives us the uncertainty principle:

    [itex]\Delta Q \Delta P \geq \frac{\hbar}{2}[/itex]

    where [itex]\Delta Q = \sqrt{\langle (Q - Q_0)^2 \rangle}[/itex] and [itex]\Delta P = \sqrt{\langle (P - P_0)^2 \rangle}[/itex]
     
  6. Jan 17, 2016 #5
    Okay, that makes sense since if it's between the two roots then the inequality is not satisfied. But if the discriminant is less than 0, then isn't ## \alpha## now complex and thus not real? Therefore the initial assumption that ## \alpha## is real for the inequality doesn't stand?
     
  7. Jan 17, 2016 #6

    Samy_A

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    The inequality holds for all real ##\alpha##. That leads to the condition that the discriminant must be 0 or less. He doesn't care about non-real roots of the equation. Sure they will exist if the discriminant is less than 0, but the inequality has been derived for real ##\alpha##.
     
    Last edited: Jan 17, 2016
  8. Jan 17, 2016 #7

    stevendaryl

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    No, in the original derivation, [itex]\alpha[/itex] is just declared to be an arbitrary real number. It's not the solution to the equation [itex]\alpha^2 + b \alpha + c = 0[/itex] (which would be complex if [itex]b^2 - 4c < 0[/itex]).
     
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