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Expecting the possible event of zero probability

  1. Nov 8, 2014 #1
    Consider a potential well in 1 dimension defined by
    $$
    V(x)=
    \begin{cases}
    +\infty &\text{if}& x<0 \text{ and } x>L\\
    0 &\text{if} &0\leq x\leq L
    \end{cases}
    $$

    The probability to find the particle at any particular point [tex] x [/tex] is zero.

    $$P(\{x\}) = \int_S \rho(x)\mathrm{d}x=0 ;\forall\; x \in \mathbb{R}$$.

    $$S = \{x\}$$ is a null set w.r.t. to the usual integration measure and therefore for each point the assigned probability is zero.

    Let's suppose that the energy level is $$E_2$$ so the wavefunction is given by $$\psi_2(x)=(\frac {2}{L})^{\frac {1}{2}} \sin(\frac {2 \pi x}{L})=\psi_{2}^*(x)$$.

    Now calculate the "expectation value" of position operator $$\hat{x}$$

    $$<x>_2=<E_2|\hat{x}|E_2>=\frac{2}{L}\int_0^L\psi_{2}^*(x)x\psi_2(x)dx$$

    So $$<x>_2=\frac{2}{L}\int_0^L x\sin^2(\frac {2 \pi x}{L})dx=\frac {L}{2}$$

    Now you say that you "will observe, on average, the particle" at $$x=\frac {L}{2}$$

    Can we say that here we indeed "expect" to measure or observe this "possible" event which has "zero" probability, i.e. occurs "almost never"?

    Can we say that, respecting the uncertainty principle, if you localize the particle at any particular point, that is with zero uncertainty, then its momentum tends to infinity and thus will not be in this potential well.

    Do we need to redefine our common meaning for the words, expectation, possible, and probable?

    Also we see that $$\psi_2(\frac {L}{2})=0 \to |\psi_2(\frac {L}{2})|^2=0$$

    But what is or how we should interpret the value of $$|\psi_2(\frac {L}{2})|^2$$

    Is $$x=L/2$$ any different from other points given this particular wave function?
     
    Last edited: Nov 8, 2014
  2. jcsd
  3. Nov 8, 2014 #2

    ShayanJ

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    Gold Member

    You based what you said on a wrong assumption. Its not that the probability for observing the particle is zero for all x. Its only zero for the places where the potential is infinite. For the zero potential region, you should find the wave function and that will give you the probability at each point.
     
  4. Nov 8, 2014 #3
    [tex] \rho [/tex] is a probability density and must be integrated over some subset of [tex]\mathbb R[/tex] to actually gain a probability.

    But $$S = \{x\}$$ is a null set w.r.t. to the usual integration measure and therefore for each point the assigned probability is zero.
     
  5. Nov 8, 2014 #4

    ShayanJ

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    Gold Member

    I understand it now. But that's not how things work in QM. It just doesn't make sense physically to integrate from x to x and get zero and conclude that its impossible for the particle to be at x! I mean, it wasn't in the mind of the people who were developing QM so they didn't mean to get something from this so I think you can't get something out of it too.
    Also, all things you did are mathematical. Mathematical tricks can't produce physical phenomena!
     
  6. Nov 8, 2014 #5
    I didn't conclude that it is impossible to find the particle in a particular, neither mathematically nor physically. However, said, by math, that it should be assigned zero probability. That is precisely the point here, impossible here does not relate or is equivalent to improbable. In QM formalism, an event could be improbable, that is to occur almost never, yet it is a possible event. Ordinary language is, of course, not helpful here.
     
  7. Nov 11, 2014 #6

    Jano L.

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    Gold Member

    Yes, events with zero probability are possible events. Zero probability never implies the event is impossible. The deduction is correct only in the opposite direction; impossible event necessarily has zero probability (or is not even considered to have probability). This has nothing to do with wave functions - it is probability theory.

    You can say it, but a lot of people will not agree. High momentum does not mean the particle is outside potential well. For the infinite potential well, it is possible to have positive probability for any finite interval of momenta.

    No, I do not think so.



    Density of probability of position at ##x=L/2## corresponding to ##\psi_2##.
     
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