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I Infinite Sq. Well sinusoidal gen. soln question

  1. Mar 3, 2016 #1
    I'm reading Griffiths' section on the infinite square well defined as having zero potential between 0 and a on the x-axis and being infinite everywhere else, and am confused about the following part when discussing the general solution inside the well. The bolded part is what confuses me, the rest is just context.

    Continuity of [tex]\psi(x)[/tex] requires that [tex] \psi(0)=\psi(a)=0,[/tex]so as to join onto the solution outside the well. What does this tell us about A and B? Well, [tex] \psi(0)=A\sin(0)+B\cos(0)=B,[/tex]so B=0, and hence [tex]\psi(x)=A\sin(kx).[/tex]Then [tex]\psi(a)=Asin(ka),[/tex]so either A=0 in which case we're left with the trivial-non-normalizable--solution [tex]\psi(x)=0,[/tex] or else [tex]\sin(ka)=0,[/tex] which means that[tex] ka=0\pm\pi\pm 2 \pi\pm 3\pi, ...[/tex]But k=0 is no good-again, that would imply [tex] \psi(x)=0[/tex] and the negative solutions give nothing new, since [tex]\sin(-\theta)=-\sin(\theta)[/tex] and we can absord the minus sign into A.

    My question here is, why do the negative solutions even matter? There are restrictions on k (E>0, mass is always for this case going to be nonzero, hbar is obviously nonzero and positive), and the general solution for the well that is used above is only applicable for the portion of the x-axis where x is always positive or zero at one of the boundaries. The restrictions on kx above make for a product of two positive real numbers, giving you another positive real number. I'm tempted to swear like a sailor because the argument will *never* be negative for that sine function, right? How does it matter?

    PS: Is there any way to put LaTeX inline on PF rather than just having it come out in separated chunks? I feel like the equations jumping every line feels like Christopher Walken reading a Physics textbook.
     
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  3. Mar 3, 2016 #2

    DrClaude

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    If you choose k < 0, then the argument will be negative. But you get the same solution as for positive k, up to a complex phase factor, so it does not represent a distinct physical solution (in other words, ##| \psi(x) |^2## is the same.

    Yes. Use ##\psi## or [itex] \psi [/itex].
     
  4. Mar 3, 2016 #3

    vanhees71

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    Easy things first: To have inline formulae enclose the expression in double hashes.

    To answer your question. You look for eigenvalues and eigenfunctions of the self-adjoing operator (setting ##\hbar=1##)
    $$\hat{H}=-\frac{\partial_x^2}{2m}$$
    in the Hilbert space ##L^2([0,a])## of square-integrable functions with "rigid boundary conditions" ##\psi(0)=\psi(a)=0##. You look for linearly independent eigenvectors, i.e., if two eigenvectors, e.g., differ only in a factor they count just as one such eigenvector. The point is that for a self-adjoint operator you can always find a complete set of orthonormal eigenvectors.

    Indeed sine and cosine functions make up these solutions since
    $$-\frac{1}{2m} u_E''(x)=E u_E(x)$$
    has solutions of the form
    $$u_n^{(+)}(x)=A \cos(k x)$$
    with ##k a=(2n+1) \pi/2## with ##n \in \mathbb{N}_0=\{0,1,2,\ldots \}## and
    $$u_n^{(-)}(x)=A \sin(k x)$$
    with ##k a=n \pi## with ##n \in \mathbb{N}=\{1,2,\ldots \}##. Here ##n=0## is no solution, because then the resulting function would be identically 0, and that's by definition not an eigenvector. The corresponding negative integers are also solutions, but they lead to the same functions (up to a sign) and thus give no new eigenvectors.

    Of course in both cases the energy eigenvalues are
    $$E=\frac{k^2}{2m}$$
    for each ##k## listed above.
     
  5. Mar 3, 2016 #4
    I guess this is just wording/choice of how to convey an idea, earlier he explicitly defined k as [itex]k \equiv \frac{\sqrt{2mE}}{\hbar} [/itex] so I'd like to assume all the following discussion unless otherwise indicated is based on that. What/when would it be sensible for k to be negative?

    vanhees71: You just lost me a little due to no fault of your own, I haven't learned about the formalism/Hilbert Space at all yet and don't know about self-adjoint(ness?), etc. I know that for my purposes right now eigenvectors are basically eigenfunctions which are in this case eigenstates but that there are subtle differences between the first and second. I get the idea of linear independence between eigen(functions) and that the ODE above is just an unsimplified general solution inside the well where k has been left as is and hbar has been set to one, but I don't follow why the cosine solution still exists since due to the boundary conditions I expect its coefficient to be zero (oh, nevermind, saw you said that soln doesn't make sense/is trivial).


    I get that since the sine function is odd, no new solutions are added, but my argument is that it wouldn't matter for the region of the well if it was odd or not because it's over a region where x is positive and k is by definition positive and thus you're never going to get [itex]\sin(-\theta)[/itex]

    I get how the distinct solutions are defined but it seems like an irrelevant or moot point to mention that the sine solution is odd because its argument in this instance is always positive anyways.
     
  6. Mar 3, 2016 #5

    vanhees71

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    I don't know, how you can understand anything about quantum mechanics, if you don't learn the important basics first. The self-adjointness of the operators that represent observables is crucial! I don't know that textbook by Griffiths, but the more postings I read about it in this forum, the more I come to the conclusion that it is confusing more than explaining :-((.

    For a good introduction to QM, see e.g. Sakurai, Modern Quantum Mechanics.
     
  7. Mar 3, 2016 #6
    It's the chapter after this, (Hilbert Space, etc) but I think my point still stands as phrased. And I'd like to buy J. J. Sakurai's books if I could afford them right now, but even the used ones on Amazon are out of my price range for just the moment :^)

    And yes, it does make for quite a bit of confusion that there's all these different teaching approaches, Griffiths avoids using Dirac notation as much as possible whereas others start from that point, and trying to learn the fundamentals while finding one source in one language and one in another is pretty frustrating, heh
     
  8. Mar 3, 2016 #7

    vanhees71

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    Too bad :-(.

    So forget about the self-adjointedness for a moment, and just solve the differential equation. It's a socalled boundary-value problem. You look for solutions of the equation
    $$\psi''(x)=-k^2 \psi(x) \quad \text{with} \quad \psi(0)=\psi(a)=0.$$
    For simplicity I've defined ##E=k^2/(2m)##.

    Now for each ##k## there are two linearly independent solutions of the differential equation, namely
    $$\psi_1(x)=A \cos (kx), \quad \psi_2=B \sin(k x)$$
    with integration constants ##A## and ##B##.

    Now to have two zeroes at ##x=0## and ##x=a## you must have ##k \in \mathbb{R}##, which implies that ##E \geq 0##. Now the boundary conditions give the ##k##s, I've derived for you in my previous posting, and of course for each ##k## also ##-k## gives a solution, but it's not really different from the ones given, because ##\psi_1## stays simply the same and ##\psi_2## just flips the sign. This doesn't give wave functions that are really different.
     
  9. Mar 3, 2016 #8
    I get how that mechanism works, I just don't understand how, in the framework of this situation, you could get a -k when you've defined it to be positive, or all of its constituents to be positive or zero so their product will be positive. 2 is positive. m is positive. E is positive or zero. hbar is positive. When would it make sense in this problem to have a k that is negative? I get that it's possible in different situations, but it isn't here if the energy is greater than or equal to zero.

    I don't disagree that dual solutions turn up that are equal because of the sine function being odd, I just don't get how it makes sense to bring it up because in this case where k is always positive (unless we have something that doesn't make sense like negative mass) and x is always positive. Regardless of whether the sine functions were odd, even, or purple with blue polka dots, the argument kx for the general solution is always going to be positive for a potential well which is zero from x=0 to x=a and infinity elsewhere.

    The function argument was always going to be positive, whether or not we were dealing with an odd function with which the (-1) could be factored out, because the parts of the function argument (k, and k's constituents, and x) are all positive in the region.


    I feel like this argument would make a bunch of sense if the well region was split across the x-axis on both negative and positive sides where you could have the argument of sin(kx) being negative (and not in a silly way like deliberately factoring in a negative sign from the constant coefficient), but with the well in the region 0<x<a, as this is defined, triple-bar equivalent, it would be true either way.
     
  10. Mar 3, 2016 #9

    vanhees71

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    The point is that you want to find all these functions.
     
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