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Delta function potential

  1. Sep 19, 2004 #1
    So I read that the delta function potential well has one and only one bound state. This seems to give a precise momentum and position as the bound state has a definite energy and the particle must be in the well. This seems to be a violation of the HUP. Is the physical impossibility of creating a true delta function potential the savior here?

    Ahh I forgot about the exponential fall off outside the well. So I suppose there is still some uncertainty in position, though is it enough to counter a zero uncertainty in momentum? For that matter the regular finite potential well has bound states that have exact energies (and therefore momenta) and yet the uncertainty in position seems to not be big enough to satisfy the HUP?

    Are these finite potential well bound states (stationary states) not realizable by particles. Can particles only be described by wave packets?
    Last edited: Sep 19, 2004
  2. jcsd
  3. Sep 19, 2004 #2


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    The uncertainty in the position of the particle is the amount of 'spread' in the probability distribution of the position. Likewise for the momentum.
    Even if you know the particle is inside the well, there is still some uncertainty in its position. You don't know 'where' you'll find the particle in the well, when you measure its position.

    A particle in a bound stationary state has a definite energy, but that doesn't mean it also has a definite momentum.

    The uncertainty prinicple is always satisfied. Try it with the infinite square well, or any other potential.
    Quantitatively it says:
    [tex]\sigma_x \sigma_p \geq \frac{\hbar}{2}[/tex]
    Where [itex]\sigma_x, \sigma_p[/itex] are the standard deviations of the position and momentum distributions respectively.
  4. Sep 19, 2004 #3


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    There appears to be a common error that permeates throughout your posting here. Somehow there is an impression that if a state has a definite energy, then the momentum must have the same degree of definiteness. What this implies is that the hamiltonian H always commutes with the momentum operator p. This isn't true in general. [H,p] is not always zero. This is because H typically has a term containing p and another term containing "x", or the position operator, in the potential part. And we already know p and x do not commute. So H and p do not necessarily commute.

    The only situation [H,p] = 0 is for a free particle, where V(x)=0. So just because one has a well-defined energy state, it doesn't mean one also has a well-defined momentum distribution.

  5. Sep 19, 2004 #4
    common error

    Yes thats it. I was equating states with exact energy with states that have exact momenta and this can only be done when they commute. Thanks Galileo and Zapper.
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