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Basic QM technicalities

  1. Nov 22, 2006 #1
    Im working through some basic QM problem for school, but sadly im still in high school and none of my teachers have studied physics :frown: . So just curious about a few things:

    Often for certain situations there are multiple (atleast in mathematic form) solutions to the schrödinger Eq. Are these all the same? eg.
    Famous infinite potential well. 2 solutions are Psi=Asin(kx) but also powers of e are possible. When entirely normalized, etc, are they essentially the same probability distribution?

    With tunneling, the wave function is shown to have a value in the V>E area. But this would mean that there is a possibilty of finding it here, which is nonsense. If this probability would be disregarded, wouldnt this disturb the normalizing of the wavefunction? (the V>E would still be part of the normalization :S) This would the same with a finite potential well, as there are results for the wavefunction in V>E areas. How does normalizing work in these situations?

    And one last question. Are wavefunctions are purely attained through solving things like the schrödinger equations, or can they also attained through other means?

    It would help my understanding a lot of you help my out with this curiosities.
     
  2. jcsd
  3. Nov 22, 2006 #2
    There's a different solution for each eigenvalue of the Schrödinger equation. The eigenvalues in this case are the allowed energy levels for the particle in the well. And since the Schrödinger equation is a linear equation, any sum (superposition) of these solutions is also a solution. Any good QM book will do a better job of explaining this than I can.

    It sure seems like nonsense, but that's the way the world works.

    They can be computed numerically, as long as you also maintain the requirement that the integral of the squared modulus of the wave function converges. For an example of this, see the book Basic Concepts of Physics by Chalmers Sherwin. A reasonably large university library should have this book.
     
  4. Nov 22, 2006 #3
    Rather than leave it with that flip response, here's the Wikipaedia entry on quantum tunneling, where you can see that there are technologies, like the scanning tunneling microscope, that depend on the reality of quantum tunneling.
     
  5. Nov 23, 2006 #4
    i get the tunnelung part, as thats beyond the V>E part where it turns E>V again, but if there was a chance of finding the particle in the "hill" as its usually called, that would kinda violate any conservation of energy. That what im wondering about.

    and as my math isnt what it should be, what is an eigenvalue exactly?
     
    Last edited: Nov 23, 2006
  6. Nov 23, 2006 #5

    vanesch

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    Don't worry too much about that. As long as the particle is in an energy eigenstate, it does have indeed a finite amplitude to be in "the forbidden zone". However, this is not a probability. In order to turn this into a probability, you need to do a measurement. Now, we are touching here upon the delicate and ununderstood part of quantum theory: the "measurement problem". There are different views on this, all have their problems and advantages. But let us not go into that for the moment.
    What needs to be done to turn this amplitude into an observation "the particle is in this forbidden zone" is in any case an interaction with a measurement apparatus, and the energy will come from this interaction with the measurement apparatus.


    For a linear operator O, if you find a vector v such that:
    O(v) = lambda . v
    (most of the time that's not so!), then lambda is an eigenvalue of the operator O.
     
  7. Nov 23, 2006 #6
    thanks, that helps a lot!

    So for the operator of energy for example. After you "apply" it you just have to algebra the wavefunction lose again and the value that it is multiplied with is then the function for the property of that operator, if i understand correctly how that works.

    Then im curious about the position operator, as i belive that is x times the wavefunction. I thought that you found the position by doing the whole absolute square integral thing. What is the different between these functions? What is it that they predict?
     
    Last edited: Nov 23, 2006
  8. Nov 24, 2006 #7

    vanesch

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    Yes. A strange way of stating it, but yes.
    Now, notice that most functions do not have this property! (for a given operator). These are then functions that describe states that do not have a well-defined value for the quantity the operator stands for. It's the entire mystery of quantum theory, in a nutshell: the fact that there exist (many) states that do not have well-defined values for measureable quantities (and which are NOT statistical ensembles, no matter what one can sometimes read).

    In fact, the entire formalism of QM has been explicitly constructed that way: that if you have a state, S1, that gives you a certain value a1 for a given quantity, say, A, and a state S2 that gives you another value a2 for A, that you can also have states (the superpositions of states S1 and S2) such that we do not have a well-defined value for A, and that upon measuring A, we have a certain probability to observe a1, and another probability to observe a2. But, again, the "superposed state" is NOT a statistical ensemble.

    In my example, the state (the wavefunction f1 describing the state) S1 would then be an eigenfunction of the operator A, with eigenvalue a1 ; and the state (the wavefunction f2 describing the state) S2 would then be an eigenfunction of A with the eigenvalue a2.
    In other words: A f1 = a1.f1 and A f2 = a2. f2

    But clearly, a general state f = u.f1+ v.f2 (with u and v complex numbers), which is a superposition of the states S1 and S2, is not an eigenfunction of A:

    A f = A(u.f1 + v.f2) = u.a1.f1 + v.a2.f2 is NOT some number times f.

    It turns out (and that's the big mystery), that such a state, when we measure the quantity A (and only then), gives us, with probability |u|^2 a result equal to a1, and with probability |v|^2 a result equal to a2. Afterwards (once we have the result), if we had a1, we can now say that the state is well-described by the function f1 (and hence the state S1) (or S2 if we had a2 as a result).

    But it is an error to think that the state f was a statistical mixture of the states S1 and S2, because we can do other measurements on that state which make this come out wrong.

    It is exactly the same !
    The position operator has as eigenfunction for the state "particle is in position x0" the wavefunction delta(x-x0) (Dirac delta).

    X delta(x-x0) = x0 delta(x-x0).


    Now, your general wavefunction psi(x) = integral psi(x0) delta(x-x0) dx0.

    So you can see this psi(x) as a superposition of delta(x-x0), delta(x-x1)...
    with coefficient (as was the u and v above) psi(x0), psi(x1) ...

    If you now measure the position (and only if you do!), then you will observe one of the possible eigenvalues x0, with probability, the coefficient of the eigenfunction, squared, which is here |psi(x0)|^2.
     
  9. Dec 9, 2006 #8
    just came across another strange something. In this book im working out of, they start introducing the 3D inf-V well, but the formula they show is En=(h^2)/(8mL^2) x (nx^2 + ny^2 + nz^2). I was wondering what L is, as there are 3 spacial dimensions!?!? Would love an explanation :D.
     
  10. Dec 9, 2006 #9

    jtbell

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    That formula is for a cubical box, with all sides the same length. For a rectangular box with unequal sides, I think it's

    [tex]E_n = \frac{h^2}{8m} \left( \frac{n_x^2}{L_x^2} + \frac{n_y^2}{L_y^2} + \frac{n_z^2}{L_z^2} \right) [/tex]
     
  11. Dec 12, 2006 #10
    thats makes a bit more sense, thanks
     
  12. Dec 12, 2006 #11
    one more thing which has been puzzling me constantly, is what those accents and strange symbols on top of most of the symbols (like E, p etc) are? If Im not mistaken they are unneccesary...
     
  13. Dec 12, 2006 #12

    dextercioby

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    No, they're not. They're used to distinguish operators from scalars.

    Daniel.
     
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