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About wavepacket unit and probability

  1. Jul 19, 2015 #1

    KFC

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    Hi all,
    In quantum mechanics, we consider the squared modulo of a wave function has meaning of probability, so does it mean a wave packet should be unitless? I am reading some materials online about the Gaussian wave packet http://www.colorado.edu/physics/phys2170/phys2170_sp07/downloads/Gaussian.pdf, there at eq. 7.10, we have

    ##|\psi|^2 = \frac{1}{\sqrt{\pi}\sigma}\exp(-x^2/\sigma)##

    I think ##\sigma## has the unit of meter so whatever inside the ##\exp## has no unit. But what about the coefficient? It has unit of 1/meter. When I am reading it, it looks like that ##|\psi|^2## is density of probability but not probability.
     
  2. jcsd
  3. Jul 19, 2015 #2

    jtbell

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    Indeed, ##|\psi|^2## is a probability density. In order to get an actual probability (which is unitless), you have to integrate over some interval in x: $$P = \int_a^b {|\psi|^2 dx}$$ gives the probability that the particle is in the range ##a \le x \le b##, provided that ##\psi## is normalized so that $$\int_{-\infty}^{+\infty} {|\psi|^2 dx} = 1$$
     
  4. Jul 20, 2015 #3
    |ψ|2 needs to have units of inverse meters because probability is dimensionless. remember that the dx in the integral has units of meters as well so
    |ψ|2dx has no units
     
  5. Jul 20, 2015 #4

    bhobba

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    That's wrong.

    Its dimensionless. Remember its the expansion of a state in position eigenvectors and since such expansions are arbitrary it must be dimensionless.

    ie |u> = ∫|x><x|u> dx. By definition <x|u> is the wave-function and by construction obviously dimensionless as |x> and |u> are both a quantum state and have the same dimension. Quantum states are dimensionless - but that is just by the by.

    Thanks
    Bill
     
  6. Jul 20, 2015 #5
    If you want to view it as a probability density then you would require it have units of inverse vol/area/length depending on the dimensionality of the problem.
     
  7. Jul 20, 2015 #6

    bhobba

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    The probability density comes from the Born rule which is independent of units.

    It does not matter how you twist or turn |u> = ∫|x><x|u> dx. implies <x|u> has no units since |x> and |u> are the same thing. There is simply no way it can be anything else.

    A probability density does not have the units of what its a density in - in fact its dimensionless.

    Thanks
    Bill
     
  8. Jul 20, 2015 #7
    I disagree. Units depend on the basis you project onto. If you project a position vector onto the x unit vector you get a length. If you project it onto the azimuthal unit vector you will get an angle. Are you implying that integrating a dimensionless quantity over an interval of length will still leave you dimensionless?dx is not a dimensionless measure.
     
  9. Jul 20, 2015 #8

    bhobba

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    I am saying exactly what I said. Look at the equation |u> = ∫|x><x|u> dx. All you are doing in integrating is multiplying something by a number and summing. Now if you obtain the same thing as what you multiply and sum ie a state what you multiply must be dimensionless.

    But if you want to look at it as a probability density its the same thing. For example the 1/2 probability you assign to the heads and tails of a one dollar coin does not give it the dimensions of dollars. Probabilities are, by definition from the Kolmogorov axioms, dimensionless. They are a number assigned to an event.

    Thanks
    Bill
     
  10. Jul 20, 2015 #9
    mod psi squared is not a probability....for a given x, mod psi squared gives you a finite number, but I'm sure you know the actual probability of any given x is 0. You are correct, probabilities are dimensionless, but mod psi is NOT a probability. <x|u> has dimension inverse length, the measure dx has units length and returns the sum to dimensionless.
     
  11. Jul 20, 2015 #10

    bhobba

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    Ok - probability density - it changes nothing. Its still dimensionless.

    What's the density of the Dirac measure? Its what's used to represent discreet distributions by densities.

    Thanks
    Bill
     
    Last edited: Jul 20, 2015
  12. Jul 20, 2015 #11
    It's not dimensionless. you liked jtbell's response which agrees with me. Clearly stated it's not dimensionless until you integrate it.
     
  13. Jul 20, 2015 #12

    bhobba

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    That's not what he said.

    Thanks
    Bill
     
  14. Jul 20, 2015 #13

    stevendaryl

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    This is one of those cases where there is a distinction between [itex]\sum_n |n\rangle \langle n|[/itex] and [itex]\int |x\rangle \langle x| dx[/itex]. For a discrete basis, [itex]|n\rangle[/itex] is dimensionless, but for a continuous basis, [itex]|x\rangle[/itex] has dimensions of [itex]L^{\frac{-d}{2}}[/itex] so that [itex]|x\rangle \langle x|\ dx^d[/itex] is dimensionless.
     
  15. Jul 20, 2015 #14

    bhobba

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    You have a function whose integral gives a probability. You have agreed probability is dimensionless. Differentiate a dimensionless quantity to give the original density and its still dimensionless.

    Thanks
    Bill
     
  16. Jul 20, 2015 #15

    bhobba

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    The dimensions of |x><x| is not the issue.

    Thanks
    Bill
     
  17. Jul 20, 2015 #16

    stevendaryl

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    Oh? Then I misunderstood what was being discussed.
     
  18. Jul 20, 2015 #17
    The units are in the integration measure and the function, not the basis vectors. ∫dx has units of length...
     
  19. Jul 20, 2015 #18

    bhobba

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    We are discussing the dimensions of wave-functions. By definition a wave function is <x|u> in the following:
    |u> = ∫|x><x|u> dx

    This is shorthand for the limit of a sum of |xi> where the |xi> goes to a continuum. Since that is exactly the same thing as |u> the weights in that sum must be dimensionless.

    Thanks
    Bill
     
  20. Jul 20, 2015 #19

    bhobba

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    The dx there has units of probability. Its a probability density and applies to events - not lengths. It is the EVENT you get an observation in the infinitesimal interval dx.

    Thanks
    Bill
     
  21. Jul 20, 2015 #20
    <x'|x> is the delta distribution, delta has units inverse length
     
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