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Infinite Square Well for E<0

  1. Jul 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that E<0 solutions to the infinite square well potential are not applicable (precisely show that boundary conditions are not satisfied when E<0)

    2. Relevant equations

    Time independent Schrodinger equation

    V(x) = 0 , 0<x<a
    V(x) = inf otherwise

    3. The attempt at a solution

    The heart of the problem lies in defining [tex]k = \sqrt{2mE}/\hbar[/tex] or [tex]K = \sqrt{-2mE}/\hbar[/tex]

    if I use k such that E>0 then I can find the boundary conditions and I can find the energy eigenvalues which are also real and positive. When E<0 I can do the exact same thing, I can satisfy the boundary conditions and find the energy eigenvalues which are positive, now this causes a contradiction (E<0 but E>0 in our solution) so that we can say there are no solutions for E<0. But I cannot see how E<0 affects my boundary conditions.

    If I use K, then we get the real exponential solutions to the Schrodinger equation and it does not matter whether E<0 or E>0 we cannot satisfy the boundary conditions. Where we should be able to for E>0.

    I think I am missing something mathwise here, do I have to (and I mean "have to" as in "must") define a real variable (k,K) when simplifying the Schrodinger Equation I do not think there would be a restriction on that because I may not know how the potential looks like for some stuff. I thought maybe I was being sloppy on complex numbers when I was squaring them but powers of imaginary numbers are defined just as the real numbers.

    You see I am confused, why the way I define a variable makes the problem unsolveable (in the case of K).

    Thanks in advance
     
    Last edited: Jul 13, 2009
  2. jcsd
  3. Jul 14, 2009 #2
    Lets go through this step by step. The Time Independent Schrodinger Equation inside the well reduces to

    [tex]-\frac{\hbar^2}{2m}\frac{d^2 \psi}{dx^2} = E\psi[/tex]

    or

    [tex]\frac{d^2\psi}{dx^2} + \frac{2mE}{\hbar^2}\psi = 0[/tex]

    Define

    [tex]k^2 = \frac{2mE}{\hbar^2}[/tex]

    The "general" solution can be written as

    [tex]\psi(x) = A e^{ikx} + B e^{-ikx}[/tex]

    Note that I have not made any statement about the sign of E or equivalently, about the real or complex nature of k. (If you will, [itex]k = \sqrt{2mE/\hbar^2}[/itex] and the two solutions correspond to +k and -k)

    Now, lets impose the boundary condition at x = 0 and x = a. At both these points, the wavefunction must vanish. This yields

    [tex]A + B = 0[/tex]
    [tex]A e^{ika} + Be^{-ika} = 0[/tex]

    With B = -A, the second equation reduces to

    [tex]A(e^{ika}-e^{-ika}) = 0[/tex]

    For a nontrivial solution, [itex]A\neq 0[/itex]. This means that

    [tex]e^{ika}-e^{-ika} = 0[/tex]

    or equivalently

    [tex]sin(ka) = 0[/tex]

    Now, if the energy E were to be negative, k would be purely imaginary, of the form

    [tex]k = i\kappa[/tex]

    so

    [tex]sinh(\kappa a) = 0[/tex]

    the only solution of which is [itex]\kappa = 0[/itex] or equivalently, [itex]E = 0[/itex]. This shows that there is no nontrivial (nonzero energy) solution of negative energy. So, we return to our old friend [itex]sin(ka) = 0[/itex] for real values of k and positive energies. This equation gives you the allowed energies that the well can support.

    Its always better to reason ab-initio before you can get a feel for the solutions. That will come after you analyze the solutions to potentials which can be solved exactly. As you can see, this systematic approach of defining [itex]k^2[/itex] instead of [itex]k[/itex] directly and considering the two possibilities separately is better than losing track of the k's and K's.

    Physically, a simple exponentially decaying (or increasing) solution cannot possibly vanish at both walls, which is what your wavefunction must do in this problem. So, at the outset, the possibility of real exponentials is ruled out in this problem. (In contrast with a "finite" square well -- where V doesn't shoot off to infinity at the walls -- where you will encounter both kinds of solutions.)
     
  4. Jul 14, 2009 #3

    Ygggdrasil

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    Without going into so much math, you can also think of the physical quantity that E represents and make an argument that way. (hint: in an infinite square well there is no potential energy so all energy is...)
     
  5. Jul 14, 2009 #4
    Thank you for this lengthy response, I understand all these but through your solution right before we jump to

    [tex]sinh(\kappa a) = 0[/tex]

    We can still go on in this manner

    [tex]sin(ka) = 0[/tex]

    [tex]k = i\frac{\sqrt{2m\left|E\right|}}{\hbar} = \frac{n \pi}{a}[/tex]

    Then

    [tex]\left|E\right| = -\frac{(n\pi\hbar)^2}{2ma^2}[/tex]

    Which taking into account the absolute value leads to a contradiction again, in terms of the Energy.

    What I am trying to say is that one can show E<0 solutions are not applicable only by reaching a contradiction.

    That is to say "not satisfying the boundary conditions" only depends on your way of solving the equation, if you define k sloppily you will end up with the real exponential solutions where you cannot satisfy the boundary conditions, it has nothing to do with E>0 or E<0.

    As in:

    I am not putting a restriction on E right now.

    [tex]Define: k = \frac{\sqrt{-2mE}}{\hbar}[/tex]

    [tex]\frac{d^2\psi}{dx^2} =k^2\psi[/tex]

    Here I end up with the real exponential solutions, and I realize that only for E<0 the real exponentials can be regarded as complex exponentials and go on from there. Thus I can say that E<0 solutions are not applicable because they do not satisfy the boundary conditions. Or I can still go on assuming k has imaginary factors and reach a contradiction for E<0.

    You see my question is not looking for an answer, what I am struggling through is that in order to show E<0 solutions are not applicable by showing that they do not satisfy the boundary conditions I must be sloppy about defining k and be aware of it. Whereas showing E<0 solutions are not applicable by contradiction is physically and mathematically more satisfying because it does not depend on my definition of k.

    What I am asking is if you are seeing anything that is wrong with this argument.

    Thanks in advance
     
  6. Jul 14, 2009 #5

    cepheid

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    Start with the Schrödinger equation (h means hbar in what follows):

    ψʺ(x) = -(2mE/h2)ψ(x)​

    Note that the negative sign here comes from the equation itself. Define k as follows (I don't think there has been any loss of generality so far. We aren't assuming anything about the sign of E yet):

    k2 = (2mE/h2) ​

    Then Schrödinger's equation becomes:

    ψʺ(x) = -k2ψ(x)​

    If E > 0, then k is real, and this equation:

    ψʺ(x) = -k2ψ(x)​

    is the differential equation for SHM (because of the negative sign). Its solutions are therefore sinusoids (or you can write complex exponentials if you prefer):

    ψ(x) = Asin(kx) + Bcos(kx) ​

    which can be made to satisfy the boundary conditions.

    IF E < 0, then k is imaginary. I.e.:

    k = iκ​

    where κ is a positive real number defined by:

    κ = (2m|E|/h2)½

    Schrödinger's equation becomes:

    ψʺ(x) = -k2ψ(x) = -i2κ2 ψ(x) = κ2 ψ(x) ​

    This is NOT a differential equation for SHM, because the second derivative is proportional to the function itself, not the negative of the function itself. This equation's general solutions are therefore not sinusoids, but rather hyperbolic sine and cosine (or you can use real exponentials if you prefer):

    ψ(x) = Asinh(κx) + Bcosh(κx) ​

    which cannot be made to satisfy the boundary conditions.

    In the above, I picked a consistent definition for k in both cases, and assumed nothing about the sign of E. What problem do you have with this argument?
     
    Last edited: Jul 14, 2009
  7. Jul 14, 2009 #6

    cepheid

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    *SIGH*, okay, I actually read the thread this time. HERE is the error in your reasoning:

    NO, you CAN'T "still go on in that manner", because the fact that sine is a solution COMES from the fact that k is real. If k is imaginary, then sine is NOT a solution, so OF COURSE pretending that it is will lead to a contradiction.

    By the way, the contradiction is already apparent in the very last line of yours that I quoted above, because you have:

    some imaginary number = some real number

    You didn't even need to carry on to the next line.



    This exercise has nothing to do with "reaching a contradiction." You reached contradictions because you did stupid things. This problem can be solved by showing that E < 0 solutions don't satisfy the boundary conditions, as requested.


    Defining k this way, you have actually introduced TWO sign changes that cancel each other out. Therefore, you get exactly the same differential equations and solutions for E > 0 and E < 0 respectively, and your answer will be the same.
     
  8. Jul 15, 2009 #7
    I agree on your second post about some imaginary number = some real number argument. But when solving for someother thing you may not be aware of that. For example we may have an imaginary mass for this particular example (not physically correct, but in general you may come across a quantity which you are not aware if it is imaginary or not)

    Also for a differential equation of this form

    [tex]\frac{d^2}{dx^2}\psi = \left(k\right)^2\psi[/tex]

    both real exponential and complex exponential solutions (if you assume k is imaginary) are applicable, it corrects itself anyway after you plug in the actual quantities, I personally do not feel comfortable with "here is SHM DE" and "here is the positively signed DE".


    But in general I agree with your argument, for this particular problem given that we actually do know the sign of E, m and have an idea of k being imaginary or not so one can as well proceed in the usual way and show that boundary conditions are not satisfied.

    Thanks.
     
  9. Jul 15, 2009 #8

    cepheid

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    Hey guguma,

    First of all, sorry for being a jerk. You raise an interesting point which is that it seems like:

    ψ(x) = Aekx + Be-kx

    is a solution to

    ψʺ(x) = k2ψ(x) ​

    For ANY k [itex] \in \mathbb{C} [/itex]. So we could take this more general approach, and not label the DE at all. The functional form of the solutions would not be determined until we substituted the appropriate k (whatever it was in the case being considered) into the exponentials. The two cases were are interested in are k being real or imaginary (leading to k2 being positive or negative).

    However, I stand by my statement that you ran into contradictions only because you used the wrong functional form for your solution (wrong given the case you were considering: real or imaginary). I also stand by my statement that no matter which of the two ways k is defined, you get consistent answers for E < 0 and E > 0.
     
  10. Jul 15, 2009 #9
    First of all no need to feel sorry I was not offended.

    I completely agree with what you are saying. And I never thought differently regarding your last statement about consistent answers.

    To be honest I am doing a review of QM right now, I never gave much thought about the teeny weeny details of calculations before and now I am trying to do that, I believe a have learned a bit more from this discussion even though it was a simple problem and for that I am grateful.

    Thanks.
     
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