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Quantum mechanics, harmonic oscillator

  1. Oct 23, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Consider a classical particle in an unidimensional harmonic potential. Let A be the amplitude of the oscillation of the particle at a given energy. Show that the probability to find the particule between [itex]x[/itex] and [itex]x+dx[/itex] is given by [itex]P(x)dx=\frac{dx}{\pi \sqrt {A^2-x^2}}[/itex].
    1)Graph P(x) and compare it with the probability to find a quantum particle between x and dx for the first state of the unidimensional harmonic quantum oscillator.
    2)What is the probability to find the classical and quantum particles in some position between 0 and A/2?
    3)Graph the density of probability for the state n=10 of the harmonic quantum oscillator and compare it with the classical density of probability of the particle for this same energy.

    2. Relevant equations
    Details in next section.
    [itex]T=\frac{2 \pi}{\omega}[/itex].

    3. The attempt at a solution
    So my strategy to show the mathematical relation of P(x) is to write find what percentage of its time the particle passes from x to dx compared to half its period (the particle "runs" a distance of A in that time).
    So I start with [itex]x(t)=A\cos (\omega t)[/itex]. I want to get dt in function of dx. So I find t in function of x first. This gives me [itex]t= \left [ \arccos \left ( \frac{x}{A} \right ) \right ] \frac{1}{\omega}[/itex].
    Now I take the differential [itex]d/dx[/itex] in both sides of this equation:
    [itex]\frac{dt}{dx}= \left ( \frac{1}{\omega} \right ) \frac{d}{dx}\left [ \arccos \left ( \frac{x}{A} \right ) \right ]=-\frac{1}{\omega} \left [ \frac{1}{\sqrt {A^2-x^2}} \right ]=-\frac{1}{\omega} \left ( \frac{1}{\sqrt {A^2-x^2}} \right )=-\frac{T}{2\pi} \left ( \frac{1}{\sqrt {A^2-x^2}}\right ) \Rightarrow dt =- \frac{T}{2 \pi} \left ( \frac{1}{\sqrt {A^2-x^2}} \right ) dx[/itex].
    But I know that if [itex]t_1-t_0=T/2[/itex], [itex]x_0=0[/itex] and [itex]x_1=A[/itex] can be associated to [itex]t_0[/itex] and [itex]t_1[/itex] respectively. So that I can write [itex]\frac{T}{2}=-\frac{T}{2\pi} \int _0 ^A \frac{dx}{\sqrt {A^2-x^2}} \Rightarrow 1=-\frac{1}{\pi} \cdot \frac{\pi}{2}=-\frac{1}{2}[/itex] which is obvisouly wrong.
    I don't where I went wrong. I hope someone can help me.
     
  2. jcsd
  3. Oct 23, 2011 #2

    ehild

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    You set t0=0 and t1=T/2. What is cos(wt) at t=0 and and at t=T/2?

    ehild
     
  4. Oct 23, 2011 #3

    fluidistic

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    Oh nice... Now I see my error. I get 1 for both. So rather than t=T/2 I must pick t=T/4.
    Now I reach 1=-1 though.
    [itex]T/4=-\frac{T}{2\pi} \int _0 ^A \frac{dx}{\sqrt {A^2-x^2}} \Rightarrow 1=-1[/itex].

    Edit: Probably I should inverse the limits of integration... but still I don't really understand why it's not the same.
     
    Last edited: Oct 23, 2011
  5. Oct 23, 2011 #4

    fluidistic

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    I'm somehow at a loss.
    I know that in a time t=T/4, the particle runs a distance A.
    In a time t=dt the particle runs a distance in terms of an [arctan (fraction involving a square root ) ] and differential x. I'm not sure about this though.
    I've been 2 more hours on this problem, trying to reach the result but I don't get it.
    I tried dt/(T/4) to get the probability for the particle to be between x and x+dx but this didn't work well. My answer differs from a factor of -1/2 from the answer if I do this.
    I tried other stuff but senless I think. I'm lost.
     
  6. Oct 23, 2011 #5

    ehild

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    You integrated the time from t=0 to t=T/2.
    At t=0, x = Acos(w*0)=A, as cos(0)=1
    At t=T/2, x==Acos(w*T/2) Acos(pi)=-A, as cos(pi)=-1.
    You have to integrate from x=A to x=-A.

    By the way, you can choose x=Asin(wt) and then you do not have problems with the minuses.

    ehild
     
  7. Oct 23, 2011 #6

    fluidistic

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    Yes I realized this. What I did is integrating dt from 0 to T/4 and the expression with dx from A to 0. Now I get 1=1 as it should.
    But I'm totally lost after this.
    I thought that the probability would be given by [itex]dt/(T/4)=\frac{4dt}{T}=-\frac{2}{\pi} \frac{dx}{\sqrt {A^2-x^2}}[/itex], which is as I said in my previous post, different from the answer by a factor -1/2.
     
  8. Oct 23, 2011 #7

    ehild

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    The probability of finding the particle between x and x+dx is the length of the time interval during the particle is in the desired x interval, divided by the time period. Draw a cosine curve from t=0 to T. How long is the particle between x and x+dx? Is it only one interval? Do the relative signs of dx and dt count?

    ehild
     
  9. Oct 23, 2011 #8

    fluidistic

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    Oh thank you, that's right.
    I drew the sketch and the particle passes twice at every position (but one), the sign doesn't matter and I reach the desired result.
    Now trying to solve 1, 2 and 3.
    For the quantum oscillator, I think that [itex]\Psi (x)=(km)^{1/8} \hbar ^{-1/4}\pi ^{-1/2} e^{-\frac{(km)^{1/2}}{2\hbar}x^2}[/itex]. I reached this by solving part 1) of https://www.physicsforums.com/showthread.php?t=442195. (that was 1 year ago but I redid all the algebra now).
    So that [itex]\int _{x}^{x+dx}|\Psi (x)|^2dx= (km)^{1/4}\hbar ^{-1/2} \pi ^{-1} e^{-\frac{\sqrt {km}}{\hbar }x^2}dx[/itex].
    What I expect to see in the graphs is that P(x) will probably be slightly bigger on the edges of the potential for the quantum oscillator compared to the classical one.
    I'll try to see this on a graph but I'm not even 100% sure what I expect is right.


    Edit: I attach the graph of both functions for arbritrary constants. I had to play for several minutes with the coefficients to get such a graph. I am quite surprised especially for the quantum oscillator graph (blueish line). The reddish line is the classical oscillador.
     

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    Last edited: Oct 23, 2011
  10. Oct 23, 2011 #9

    ehild

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    Are you sure? It is so easy to Google it... Does the first state mean n=0 or n=1?

    You need to compare the same oscillators with the same energy. Choose the constants accordingly. Write the probabilities in terms of ω, and choose the amplitude of the classical oscillator so its energy is [itex]\frac{ωh}{2\pi }(n+1/2)[/itex].


    ehild
     
  11. Oct 26, 2011 #10

    fluidistic

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    Ok I've progressed on this subject. First state means n=0.
    For the quantum oscillator, I know that [itex]E= \left ( n+\frac{1}{2} \right ) \hbar \omega[/itex].
     
    Last edited: Oct 26, 2011
  12. Oct 27, 2011 #11

    ehild

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    And what is the energy of the classical oscillator?
    n=0 means no real oscillation. Te particle is nearly at its equilibrium position. If you choose a classical particle with mass much larger than that of an atom, it is confined in a very small interval around x=0.

    ehild
     
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