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Infinite Square well

  1. May 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider an infinite square well defined by the potential energy function
    U=0 for 0<x<a and U = ∞ otherwise
    Consider a superposed state represented by the wave function ## \Psi(x,t)## given at time t=0 by
    $$\Psi(x,0) = N \{(-\psi_1(x) + (1+ i)\psi_2(x)\}$$

    1. Assume that the constant N is real and positive. Fix N such that Ψ(x, 0) is normalised to one particle.
    2. Find Ψ(x, t) for all times t. Is your result correctly normalised? Is this state stationary?
    3. In terms of {En}, what is the average energy <E> of the superposed state? Does it depend on t?
    4. Calculate <x> as a function of time t.
    5. Given ##E_n=(\frac{\hbar^{2}}{2m})( \frac{n\pi}{a})^{2}## , find the average time for an electron to move back and forth once in a well 1 nm wide. Compare this with the classical result for an electron with energy equal to <E>.




    2. Relevant equations


    3. The attempt at a solution
    I am really struggling with what the lecturer is asking here. I have an answer to 1. ##N=\frac{1}{\sqrt{3}}## then I am not sure where to go.
    for 2. I think I get from the TDSE

    $$ \Psi(x,0)=\frac{1}{\sqrt{3}} \{(-\psi_1(x)e^{-iE_1t/\hbar} + (1+ i)\psi_2(x) e^{-iE_2t/\hbar}\} $$

    I am not sure if it is correctly normalised but It is not stationary as E1≠E2


    Could someone please point me in the right direction. I have both Griffiths and Serway and have read through but cannot find anything that helps me understand what is being asked. If someone could either talk me through it or point to relevant sections in the text I would be very grateful.

    Thanks
     
  2. jcsd
  3. May 8, 2015 #2
    Your always doing the same problems as me at the same time - we must be in the same class! As far as the problem goes:
    1. I got the same N
    2. Im a little stuck here too, but i suppose
    Ψ(x,t)=∑cn(t)Ψn(t)
    We have Ψ1,Ψ2 so we need to find c1,c2 where
    cn(t)=cn(0)exp(-i*En*t/ħ)
     
  4. May 9, 2015 #3

    blue_leaf77

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    Then to be sure if your result is correctly normalized, you need to calculate the inner product with itself and see if it yields unity. 'Non-stationaryness' should not affect the result of any inner product due to the unitarity property of the time evolution operator you have applied to your wavefunction.
     
  5. May 9, 2015 #4

    vela

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    Well how did you arrive at this answer? Presumably you required ##\int \psi^*(x)\psi(x)\,dx = 1## because that's what normalized means.

    That's right.

    Again, what does it mean for wave function to be normalized?
     
  6. May 10, 2015 #5
    I think I am getting somewhere...(thanks)

    for a function to be normalized I need ##\int \psi^*(x)\psi(x)\,dx = 1## so it is normalised and not stationary

    for 3
    Here $$\langle E\rangle = \int \Psi^*(x)E\Psi(x)\,dx$$

    spits out an answer that doesn't depend on t

    Here i need ## \langle x\rangle =\int \langle x\rangle\Psi^*(x)\Psi(x)\,dx##

    The integrals involving cross terms ψ1ψ1= 0 and the ψn2= 1 leaving
    $$ \langle x\rangle =\frac{1}{3} \langle x\rangle+\frac{2}{3} \langle x\rangle$$

    I think I should have a value for x somewhere to give me an actual answer but am not sure what that should be...
     
  7. May 10, 2015 #6

    blue_leaf77

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    That's what one should expect when the Hamiltonian is time independent.
    Since the question asked you to calculate the time dependency of <x>, you should have used the wavefunctions at time t to do the calculation.
     
  8. May 10, 2015 #7
    I thought I had by calculating <x> using $$ \Psi(x,0)=\frac{1}{\sqrt{3}} \{(-\psi_1(x)e^{-iE_1t/\hbar} + (1+ i)\psi_2(x) e^{-iE_2t/\hbar}\} $$
     
  9. May 10, 2015 #8

    blue_leaf77

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    Alright then, just wondered why t didn't appear in your expression.
     
  10. May 10, 2015 #9
    From ##\psi_1^*\psi_1## I get ##-\psi_1^*(x)e^{+iE_1t/\hbar} \psi_1(x) e^{-iE_2t/\hbar}## which gives ##\psi_1^*(x)\psi_1(x) ## and as they are normalized to one particle ##\int\psi_1^*(x)\psi_1(x) = 1 ## leaving me with just the coefficients.

    Yes??
     
  11. May 10, 2015 #10

    BvU

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    Yes
    Note that your post #7 is a bit confusing: You write ##\Psi(x,0)=##, followed by ##\Psi(x,t)## but you don't state what you do to calculate ##<x>## ...

    And I forget the important part: what about the cross terms ? Check carefully if they cancel ...

    [edit 3] No ! Sorry:

    The Yes above is for the normalization, but it doesn't hold for ##\int \psi_1^* x \psi_1## !
     
    Last edited: May 10, 2015
  12. May 10, 2015 #11

    BvU

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    Could you oblige me and elucidate on post #3: your "spits out an answer that doesn' t depend on t " sounds as if it's obvious to you. What exactly do you do to calculate ##< E > ## ?
     
  13. May 10, 2015 #12
    To calculate ##\langle x \rangle## I used should have been $$<x>=\int \langle x\rangle\Psi^*(x)\Psi(x)dx$$

    And I didn't check if the cross terms canceled I just 'presumed' they did as they were normalised and therefore ##\int\psi_1^*(x)e^{+iE_1t/\hbar} \psi_2(x) e^{-iE_2t/\hbar} =0## I think you are saying I presumed wrong.
     
  14. May 10, 2015 #13

    BvU

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    Can't do that. x is an operator (a simple multiplication), ## <x> ## is a number. You can pull the number in front of the ##\int##, but not the unadorned x !
     
  15. May 10, 2015 #14
    So it should read ##<x>=\int x\Psi^*(x)\Psi(x)dx## and I and doing the correct thing?

    (thanks for helping by the way...)
     
  16. May 10, 2015 #15

    BvU

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    Better to write $$
    <x>=\int \Psi^*(x)\; x \;\Psi(x)dx$$although for this simple multiplication factor case it doesn't matter. (But for the momentum, for example it does matter!)
     
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