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Uncertainly principle

  1. Mar 30, 2014 #1
    For the ground state of harmonic oscillator i have

    [tex] \Delta p \Delta x =\frac{\hbar}{2}[/tex]

    why if i do

    [tex]\frac{1}{F} \Delta p \Delta x \cdot F =\frac{\hbar}{2}[/tex]

    [tex] \Delta t \Delta E = \frac{\hbar}{2} [/tex]

    [tex] \Delta E = \frac{\hbar}{2} \frac{1}{\Delta t} [/tex]

    [tex] \Delta E = \frac{\hbar}{2} f [/tex]

    but my answer should be [tex] \Delta E = \frac{\hbar}{2} \omega [/tex] ???
     
  2. jcsd
  3. Mar 30, 2014 #2

    Simon Bridge

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    The delta-terms are measurements of uncertainty.
    i.e. ##\Delta E## represents the precision to which energy is measured.
     
  4. Mar 30, 2014 #3
    ok, but ##<\Delta E^2>= <E^2>-<E>^2= (<p^2>/(2m) + 0.5 m \omega^2 <x^2> )^2- ( <p>^2/(2m)+ 0.5 m \omega^2 <x>^2 ) ^2##

    ##=E_0^2-0^2=E_0 ^2 \to \Delta E = E_0##

    then, my aswer should be ##\Delta E= \hbar \omega /2##, and not ##\Delta E = \hbar f/2##
     
    Last edited: Mar 30, 2014
  5. Mar 30, 2014 #4

    dlgoff

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    hosc18.gif

    http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc4.html
     
  6. Mar 30, 2014 #5
    ok, but my question is about the value of ##\Delta E##, i don't ask about ##E_0##

    ##\Delta E## is ##\hbar \omega /2## or ##\hbar f /2## ???
     
  7. Mar 30, 2014 #6
    The mean deviation of energy in an energy eigenstate is obviously zero.
    The manipulations you made in the original post are not correct, the deltas are not differences or infinitesimals
     
  8. Mar 30, 2014 #7
    Last edited: Mar 30, 2014
  9. Mar 30, 2014 #8

    Simon Bridge

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    I think you are still very confused.

    ##\langle (\Delta E)^2\rangle## reads: "expectation of the variance of E", and you put it equal to the variance, which is an odd thing to say ... but I think you meant to write:

    ##(\Delta E)^2 = \langle E^2\rangle-\langle E\rangle^2## would be the definition of variance.

    For an energy eigenstate ##\langle H\rangle = E_n##
    You seem to want to work on the ground state:
    Who says that ##\Delta E = \hbar f/2## is correct?
    Where are you getting these ideas from?

    Checking:
    For the ground state:

    ##(\Delta E)^2 = \langle E^2\rangle-\langle E\rangle^2 = \langle E^2\rangle-\langle E\rangle^2 = E_0^2-E_0^2=0\implies \Delta E = 0##

    ... which means that the energy eigenstates of the harmonic oscillator are infinitely thin
    ... the energy of the lowest eigenstate is exactly ##E_0## and so on.

    (I think you missed out a term in your subtraction.)

    Since ##\Delta t \propto 1/\Delta E##, this means that the lifetime of the state is infinitely uncertain. i.e. the state lasts forever.

    It may be easier of you write the relations out as: $$\sigma_x\sigma_p\geq \frac{\hbar}{2},\; \sigma_E\sigma_t\geq \frac{\hbar}{2}$$

    Or it may be that you are referring to something else - I can't really tell.
    Please tell us where these ideas are coming from.

    Meantime, the statistics of the harmonic oscillator are detailed in:
    http://academic.reed.edu/physics/courses/P342.S10/Physics342/page1/files/Lecture.9.pdf
     
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