Probability of the lowest eigenenergy, quantum theory

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  1. wanderlust.xx's Avatar
    1 18-04-2012  19:33
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    Probability of the lowest eigenenergy, quantum theory


    So I have my solutions of part b, but I'm not sure how to go about c.

    What eigenfunction am I using, exactly?

    For the region |x|<a I have the eigenfunction \phi_n(x) = A cosnx + B sinnx. The lowest eigenenergy should be at n=0 I think, but doing this leaves just the constant A. Do we work this constant out using a boundary condition?

    Edit: Oh of course I'm using c_n = <\phi_n , \psi> and P(E=E_n)=|c_n|^2
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    Last edited by wanderlust.xx; 18-04-2012 at 19:37.
  2. ben-smith's Avatar
    2 18-04-2012  20:15
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    Re: Probability of the lowest eigenenergy, quantum theory
    (Original post by wanderlust.xx)


    So I have my solutions of part b, but I'm not sure how to go about c.

    What eigenfunction am I using, exactly?

    For the region |x|<a I have the eigenfunction \phi_n(x) = A cosnx + B sinnx. The lowest eigenenergy should be at n=0 I think, but doing this leaves just the constant A. Do we work this constant out using a boundary condition?

    Edit: Oh of course I'm using c_n = <\phi_n , \psi> and P(E=E_n)=|c_n|^2
    Normalising your wavefunction should yield A and, yes, n=0 is the lowest energy. This can be seen by the fact that n clearly determines the frequency of the eigenstates and E=h*f
    Last edited by ben-smith; 18-04-2012 at 20:17.
  3. wanderlust.xx's Avatar
    3 18-04-2012  20:30
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    Re: Probability of the lowest eigenenergy, quantum theory
    (Original post by ben-smith)
    Normalising your wavefunction should yield A and, yes, n=0 is the lowest energy. This can be seen by the fact that n clearly determines the frequency of the eigenstates and E=h*f
    Oh yeah.

    So after normalising I got |A|^2 = \frac{1}{2a}.

    Thus

    \dfrac{1}{a\sqrt{2}} \displaystyle \int_{-a}^{a} sin(\dfrac{\pi x}{a}) dx = \dfrac{\sqrt{2}}{\pi}

    Hence

    |c_0|^2 = P(E_n = E_0) = \dfrac{4}{\pi^2}

    Does this look right? Seems fine to me, but I'd appreciate a correction if needed!

    Thanks for the help.
  4. ben-smith's Avatar
    4 18-04-2012  20:58
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    Re: Probability of the lowest eigenenergy, quantum theory
    (Original post by wanderlust.xx)
    Oh yeah.

    So after normalising I got |A|^2 = \frac{1}{2a}.

    Thus

    \dfrac{1}{a\sqrt{2}} \displaystyle \int_{-a}^{a} sin(\dfrac{\pi x}{a}) dx = \dfrac{\sqrt{2}}{\pi}

    Hence

    |c_0|^2 = P(E_n = E_0) = \dfrac{4}{\pi^2}

    Does this look right? Seems fine to me, but I'd appreciate a correction if needed!

    Thanks for the help.
    Actually, having done a bit of reading, I have feeling n=0 is not actually the lowest energy eigenstate as it makes the wave function discontinuous at the sides of the box. sorry about that.
  5. wanderlust.xx's Avatar
    5 18-04-2012  21:36
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    Re: Probability of the lowest eigenenergy, quantum theory
    (Original post by ben-smith)
    Actually, having done a bit of reading, I have feeling n=0 is not actually the lowest energy eigenstate as it makes the wave function discontinuous at the sides of the box. sorry about that.
    The problem now is that normalising becomes a pain in the arse, since we'd have \phi_1(x) = Acosx+Bsinx, \ |x|<a
    Last edited by wanderlust.xx; 18-04-2012 at 21:43.
  6. wanderlust.xx's Avatar
    6 18-04-2012  21:51
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    Re: Probability of the lowest eigenenergy, quantum theory
    Okay, here's another idea. Since the potential is symmetric, I can split our solutions into even and odd states, yielding

    Even:
    \psi_n^{even}(x)=\dfrac{1}{\sqrt {a}}cos\dfrac{(n-\frac{1}{2})\pix}{a}, \ n= 1,2,... \ |x| < a

    E_n^{even} = \dfrac{\hbar^2 \pi^2}{2ma^2}(n-\dfrac{1}{2})^2

    Odd:

    \psi_n^{odd}(x)=\dfrac{1}{\sqrt{ a}}sin\dfrac{n \pi x}{a}, \ n= 1,2,... \ |x| < a


    E_n^{odd} = \dfrac{n^2 \hbar^2 \pi^2}{2ma^2}

    Hence our lowest eigenenergies for Even and Odd parity states respectively will be

    E_1^{even} = \dfrac{\hbar^2 \pi^2}{8ma^2}

    E_1^{odd} = \dfrac{\hbar^2 \pi^2}{2ma^2}

    but now clearly we have E_1^{even} < E_1^{odd}, so we use the even eigenfunction \psi_1^{even}(x)=\dfrac{1}{\sqrt {a}}cos\dfrac{\pi x}{2a}

    in

    c_1 = <\psi_1^{even}, \psi>

    Does this work? If not, why not?
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