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Heisenberger principle

  1. Apr 23, 2012 #1
    Please explain if the product of uncertainities related to to the measurement of position and momentum of particle is independent or depends on velocity of particle.
     
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  3. Apr 23, 2012 #2

    Fredrik

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    The uncertainties depend on the state.

    If the particle has a velocity, i.e. if the state is such that there's a vector v such that a velocity measurement will certainly have the result v, then the momentum uncertainty is zero, no matter what v is.

    It's "Heisenberg" by the way.
     
  4. Apr 23, 2012 #3

    phinds

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    I must be misunderstanding what you have said, since what I am hearing you say is that if you know the velocity precisely, then you know the momentum precisely.

    Isn't that just the opposite of what Heisenberg said?
     
  5. Apr 23, 2012 #4
    No, Heisenberg says that if you know the momentum precisely, you cannot know the position precisely. Since momentum is directly proportional to velocity for non-relativistic motion, knowing the momentum and velocity precisely at the same time should be straightforward.
     
  6. Apr 23, 2012 #5

    phinds

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    OK, thanks. I mistook velocity for position. Not bright.
     
  7. Apr 23, 2012 #6
    I think it is related to the mass of particle.Because : λ=h/mv.The wavelength can tell us if the particle must be treated as wave.
     
  8. Apr 23, 2012 #7
    In what quantum state?

    EDIT:

    Also, velocity in Quantum Mechanics is defined through:
    [tex]
    \hat{\mathbf{v}} = \frac{d\hat{\mathbf{r}}}{d t} = \frac{i}{\hbar} \, \left[ \hat{H}, \hat{\mathbf{r}} \right]
    [/tex]
    For a single particle in an external vector potential, you would get:
    [tex]
    \hat{\mathbf{v}} = \frac{1}{m} \, \left( \hat{\mathbf{p}} - q \, \mathbf{A}(\hat{\mathbf{r}}, t) \right)
    [/tex]
    Taking the commutator of [itex]\hat{v}_i[/itex] with the components [itex]\hat{x}_k[/itex], and [itex]\hat{p}_k[/itex], one obtains:
    [tex]
    \left[ \hat{v}_i, \hat{x}_k \right]= -i \, \frac{\hbar \, \hat{p}_i}{m},
    [/tex]
    and
    [tex]
    \left[ \hat{v}_i, \hat{p}_k \right] = -i \, \frac{\hbar \, q}{m} \, \frac{\partial A_i}{\partial x_k}(\hat{\mathbf{r}}, t)
    [/tex]

    Thus, in a state where [itex]\mathbf{v}[/itex] is defined, the position is not defined, and, if there is a magnetic field, neither is the momentum of the particle.
     
    Last edited: Apr 23, 2012
  9. Apr 23, 2012 #8
    Remember, pretty much nothing can be precisely known in QM, so the velocity isn't well defined if there's any uncertainty in momentum (use mv or the relativistic formula), so your question is meaningless.
     
  10. Apr 23, 2012 #9
    what i s uncertainty principle mean ?? what is it ?
     
  11. Apr 23, 2012 #10

    phinds

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    Google is your friend
     
  12. Apr 23, 2012 #11
    thank you i now what it mean ? but how is the Equation refer to that
    ΔxΔp≥ h\2
    how it' s describe if we measure the position we can't find momentum
     
  13. Apr 23, 2012 #12
    Could you please rephrase that? I don't understand the question.
     
  14. Apr 23, 2012 #13

    Fredrik

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    It's a theorem in quantum mechanics that tells us that if you perform a long series measurements of either position or momentum on identically prepared particles, then regardless of the preparation, you will get a wide range of results for either the position measurements or the momentum measurements. (The product of the widths of the ranges of measurement results will never be smaller than ħ/2). In other words, it tells us that there's no way to prepare a particle such that results of position measurements will be close to the average, and results of momentum measurements will be close to the average.
     
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