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Infinite Square Well Probability of getting Ground State Energy

  1. Dec 2, 2007 #1
    1. The problem statement, all variables and given/known data
    Particle is in a tube with infinitely strong walls at x=-L/2 and x=L/2/ Suppose at t = 0 the electron known not to be in the left half of the tube, but you have no informations about where it might be in the right half---it is equally likely to be anywhere on the right side.

    (a) Using this information, determine the initial wave function Psi(x,t=0) for this electron.

    (b) If you were to measure the energy of the lectron at t=0, find the probability of getting E_1, the ground state energy for this tube.


    2. Relevant equations



    3. The attempt at a solution

    Well for part (a) I think I have a solution.

    Psi(x,t=0) = Psi(x) = A Sin(n pi x/L)

    Then found the normalization constant A....since we know that the electron is initially known to be in the right half of the well A= 2/Sqrt(L)....I think this is right..


    For part (b) I'm a little confused ... I know that

    E_n= (n^2 pi^2 hbar^2) / (2 m L^2) and I think that

    Prob(E=E_1) = |<E_1|Psi>|^2 = |<E_1|x|E_n>|^2

    Is that right? And what if I want the probability density where t>0...I know I have to just include the term Exp[-iE_n t/ hbar]...but is that it? Please help me ASAP...lol
     
    Last edited: Dec 2, 2007
  2. jcsd
  3. Dec 2, 2007 #2

    Gokul43201

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    No, this is not correct. You only know that initially, the electron is equally likely to be found at any spot in the right half. What probability density function will you associate with this information? What wavefunction does that give you?
     
  4. Dec 2, 2007 #3
    ok so if you say that then i think that

    Psi(x,t=0) = Sqrt(2/L) Sin(n pi x /L) for 0<=x<=L/2
    and 0 for -L/2<=x<=0

    but i dont know how to find the ground state energy now? I know that the probability density unction now would just be

    |Psi(x,t)|^2 = (2/L) Sin(n pi x/ L)^2

    am i right? what about the ground state energy?
     
  5. Dec 2, 2007 #4

    nrqed

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    No, you want the probability to be the same for all points on the right half. What psi(x,t=0) gives this?
     
  6. Dec 2, 2007 #5

    Gokul43201

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    Aminh, forget that the particle is in an infinite well for now. If someone told you that you are equally likely to find a particle anywhere from x=0 to x=L, what would be the probability density function describing this situation? Can you describe its shape?
     
  7. Dec 2, 2007 #6
    Gokul- I may be wrong- but just because we don't know where an electron is over an interval doesn't mean that the electron has a wave-function which corresponds to our ignorance. For all we know- the electron has a sharply peaked wave-function at x=0.7.

    I know this doesn't help the OP, so sorry for confusing things.
     
  8. Dec 3, 2007 #7

    nrqed

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    You are right...but then the question says it is equally likely to be found anywhere in that half which now specifies the wavefunction.
     
  9. Dec 3, 2007 #8
    I agree, but the fact that it is equally likely to be anywhere could equally well be taken as a measure of our ignorance. I suppose what the question really means to say that the probability distribution associated with the wave-function is flat over that region.
     
  10. Dec 3, 2007 #9

    then in that case my origniaal post should have the right Psi(x,t=0) right?

    now how do i foudn the probability of getting the ground state energy?
     
  11. Dec 3, 2007 #10
    by the way guys this is the first time i'm using this site....thanx a lot in advance....the thing is i'm the only student in my quantum mechanics class so I have noone else to work with...

    and i think the probability density function when t>0 should just look like a sin[n pi x/l)^2 graph..
     
  12. Dec 3, 2007 #11

    Gokul43201

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    Does this graph have the same height everywhere? You want the probability to be equal everywhere in the right side, so the PDF will be rectangular (uniform distribution). Now can you describe this as a function and normalize it?

    christian: Only one of those two interpretations allows us to solve the problem, so we choose that one. :smile:
     
    Last edited: Dec 3, 2007
  13. Dec 3, 2007 #12

    we are told the the INITIALLY the electron is found in the right side of the infinite potential well...however as t>0 this does not have to be true...the electron can then be anywhere

    i'm confused by what u are saying....still i would like some help on finding the ground state energy...thanx in advance
     
  14. Dec 3, 2007 #13
    Aminh: Gokul is a really smart person- and it would be in your best interest to follow his directions. In particular, try and answer his question first:

    "Does this graph have the same height everywhere?"

    Forget about t>0 for now. Concentrate at what happens when t=0.
     
  15. Dec 3, 2007 #14
    i think i get it now...i need to first find a function Psi(x,t=0) whose modulus squared
    |Psi(x,t=0)|^2 gives me zero for the left side and a constant for the right side....ok got that...i will try this onw....how about the probability of finding the electron in the ground state energy though...how do i find that?

    do i take

    Prob(E=E_1) = |<E_1|Psi>|^2 with the Psi(x,t=0) that satisfies the setup? I think I have to do that....but i'm getting a lil confused of how the algebra should look?
     
  16. Dec 3, 2007 #15
    ok so i think i understand the problem but i can't find an apporpriate Psi(x,t=0) that will give me a constant PDF on the right side and zero PDF on the left? i tried looking at the website that Gokul mentioned but i'm confused by what they are trying to say..
     
  17. Dec 3, 2007 #16

    Gokul43201

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    aminh, if this is the first time you are seeing a problem like this, it is going to take a bit of work. There are several steps involved in this problem, that you must tackle one at a time. Let's first find the initial wavefuntion. See if this makes sense to you:

    [tex]\psi(x) = 0~,~~x ~\epsilon~ [-L/2,0) [/tex]

    [tex]\psi(x) = A~,~~x ~\epsilon~ [0,L/2] [/tex]

    where A is some real constant. To determine A, you must normalize the wavefunction. Since the particle is definitely somewhere in the right half, we want

    [tex]\int_0^{L/2}dx |\psi(x)|^2 = 1 [/tex]

    Plug A into the above and solve. What do you get for the value of A?

    That gives you the complete initial wavefunction. This is the end of step one.

    Now the system is allowed to evolve from this state. The particle now becomes "aware" of its boundaries and starts to act like a particle in an infinite well. The next step is then to see if we can express [itex]\psi(x,t=0)[/itex] as a linear combination of the energy eigenstates of the infinite well. Fortunately for us, the eigenstates of the infinite well are exactly the basis functions of the Fourier expansion of an odd function: [itex]sin(n \pi x/L)[/itex].

    Have you performed a Fourier expansion before? If not, you will need to learn how to do that first.

    http://mathworld.wolfram.com/FourierSeries.html
     
    Last edited: Dec 3, 2007
  18. Dec 3, 2007 #17
    An aside to Gokul: psi(x)=Aexp[i f(x)] is a more general solution in the region x<L/2. Wouldn't that change the integral |<psi|0>|^2 ?
     
  19. Dec 3, 2007 #18

    Avodyne

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    Reply to christianjb: yes and yes. The problem is ill posed. However, Gokul has undoubtedly written down what the student is expected to write down by whoever wrote the problem.
     
  20. Dec 3, 2007 #19
    OK: I don't want to derail this thread into discussing various niceties of QM, but it strikes me that this problem is so ill-posed as to be ultimately more damaging than beneficial.

    This question gives a false impression that the phase of the wave-function is unimportant. A wave-function must be specified by magnitude and phase at each point (in the x basis). As I already mentioned- the question also leads to the false impression that our uncertainty of the wave-function is reflected by the inherent probability density of the wave-function.

    (I may have overlooked something here- it wouldn't be the first time. My intuition about QM is quite shaky.)
     
  21. Dec 4, 2007 #20

    Avodyne

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    No, this is the initial state, which can in principle be any normalizable state. The phase is allowed.
     
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