Hey there Sign in to join this conversationNew here? Join for free

Probability of the lowest eigenenergy, quantum theory

Announcements Posted on
Study Help needs new mods! 14-04-2014
We're up for a Webby! Vote TSR to help us win. 10-04-2014
    • Thread Starter
    • 9 followers
    Online

    ReputationRep:


    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
    Attached Thumbnails
    Click image for larger version. 

Name:	quantumq.png 
Views:	35 
Size:	57.3 KB 
ID:	142725  
    • 1 follower
    Offline

    ReputationRep:
    (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
    • Thread Starter
    • 9 followers
    Online

    ReputationRep:
    (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.
    • 1 follower
    Offline

    ReputationRep:
    (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.
    • Thread Starter
    • 9 followers
    Online

    ReputationRep:
    (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
    • Thread Starter
    • 9 followers
    Online

    ReputationRep:
    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?

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?

    this is what you'll be called on TSR

  2. this can't be left blank
    this email is already registered. Forgotten your password?

    never shared and never spammed

  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By completing the slider below you agree to The Student Room's terms & conditions and site rules

  2. Slide the button to the right to create your account

    Slide to join now Processing…

    You don't slide that way? No problem.

Updated: April 18, 2012
Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.