Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof of Heisenberg's Uncertainty Principle

  1. Mar 8, 2005 #1
    Is there any mathematical or physical proof or derivation of Heisenberg's Uncertainty principle out there? Can someone send me a link to one or provide a proof if it isn't too complicated? I know that in quantum mechanics if two operators don't commute then we can't measure both of these simulataneously. Why does this correlation exist?
  2. jcsd
  3. Mar 8, 2005 #2
  4. Mar 8, 2005 #3

    Because Mother Nature has chosen to behave like this

  5. Mar 8, 2005 #4
    Besides, check out the original manuscript in my last entry of my journal

  6. Mar 8, 2005 #5
    I don't really follow this derivation. Why does he have to show that (deltaA')^2=(deltaA')^2?

    And how do we know |<A'psi,B'psi>-<B'psi,A'psi>|=2|Im<A'psi,B'psi>|
  7. Mar 8, 2005 #6
  8. Mar 8, 2005 #7


    User Avatar
    Science Advisor
    Homework Helper

    The nonmathematized version (the one involving wave-mechanics formalism) is in the beginning of Davydov's book,IIRC.Anyway,every book on QM has the proof for the generalized version:

    [tex] \Delta \mathcal{A}\cdot\Delta \mathcal{B}\geq\frac{1}{2} |\langle[\hat{A},\hat{B}]_{-}\rangle _{|\psi\rangle}| [/tex]

    Last edited: Mar 8, 2005
  9. Mar 8, 2005 #8
    Daniel, why is there a minus sign as the subscript of the commutators? I never understood that.
  10. Mar 8, 2005 #9


    User Avatar
    Science Advisor
    Homework Helper

    Because it is the commutator...?
    [tex] [\hat{A},\hat{B}]_{-}=:\hat{A}\hat{B}-\hat{B}\hat{A} [/tex]

    and that's how it's elegantly specified the fact that one speaks about commutators of (linear) operators...

  11. Mar 8, 2005 #10
    Oh... OK. I thought the commutator automatically implied a negative sign, and we have the anti-commutator for the version with the plus sign. Ok never mind about that.

    I have another question though. If we reverse the order of the commutator, i.e.

    [tex][\hat{A},\hat{B}] = -[\hat{B}, \hat{A}][/tex], we get a minus sign in the uncertainty. But is that of any significance if the product of the variances is negative as opposed to positive?
  12. Mar 8, 2005 #11


    User Avatar
    Science Advisor
    Homework Helper

    If you haven't seen so far,there's a modulus after performing the average of the commutator on the (pure) quantum state [itex] |\psi\rangle [/itex]

  13. Mar 8, 2005 #12
    Oh yes, of course. Sorry, it's 04:30AM here in England, and my mind's not thinking straight!
  14. Mar 9, 2005 #13

    If I can understand this, I understand the derivation of the Uncertainty Principle. But I can't make sense of this one line in the original link provided by Marlon.
  15. Mar 9, 2005 #14


    User Avatar
    Science Advisor
    Homework Helper


    [tex]\langle \psi|\phi \rangle = \langle \phi|\psi \rangle^*[/tex]
    where the * denotes complex conjugation.


    [tex]\langle A\psi|B \psi \rangle=\langle B\psi|A \psi \rangle^*[/tex]

    For any complex number [itex]z[/itex] we have [itex]z-z^*=2i\Im(z)[/itex].
  16. Mar 9, 2005 #15


    User Avatar
    Science Advisor
    Homework Helper

    This is very simple.

    [tex] \langle \psi|\hat{A}\hat{B}|\psi\rangle =:u\in \mathbb{C} [/tex] (1)

    Then,using the property:

    [tex] \langle \psi|\hat{A}\hat{B}|\psi\rangle = \langle \psi|\hat{B}\hat{A}|\psi\rangle ^{*} [/tex] (2)

    ,we can write the LHS of the equality u wish to prove as:

    [tex] |u-u^{*}| [/tex] (3)

    The RHS of the equality you want to prove is

    [tex] 2|Im \ u | [/tex] (4)

    Take the generic algebraic for "u"

    [tex] u=:a+ib [/tex] (5) [tex] \Rightarrow u^{*}=a-ib [/tex] (6)

    and then

    [tex] |u-u^{*}|=|2ib|=2|b|=2|Im \ u| [/tex] (7)


    Last edited: Mar 9, 2005
  17. Mar 9, 2005 #16

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    He's not showing that. He's showing that [itex]( \Delta A )^2=< \psi , A'^2 \psi >[/itex]. The fact that he ended up with [itex]( \Delta A )^2=( \Delta A )^2[/itex] simply means that he completed the proof.
  18. Mar 9, 2005 #17
    Thanks a lot everyone. Much appreciated.
  19. Mar 16, 2005 #18
    The very simplest answer is this.

    If you want to determine the frequency of a signal by counting
    pulses, your frequency determination gets more accurate the longer
    you count. In QM, frequency determies energy so the less time you
    have to count the frequency, the less certain you will be of the
    exact energy.
  20. Mar 16, 2005 #19


    User Avatar
    Science Advisor
    Homework Helper

    Physicists are not really fond of descriptive explanations.They need maths.

  21. Mar 16, 2005 #20
    I don't use maths because some of the people on this board are in dire need of descriptive
    explanations. I'll leave the maths to the Physicists amoung you.

    Edit: ..and let's not forget that there are many who could understand the
    basics of much of physics but are confounded by the equations. It's partly
    for them that descriptives are healty. The other group who needs it are the
    naive budding physicists who have been made to beleive that descriptive
    explanations are at odds with good science. There are times when only
    the math will do and descriptives aren't helpful, like the nature of electronic
    spin. But there are other times (like the uncertainty principle) when
    whole philosopies get built on the opacity of the equations. Quantum Uncertainty
    is simple to comprehend visually for simple configurations and it is healthy, instructive
    and right to do so.
    Last edited: Mar 16, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook