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Quantum harmonic oscillator most likely position

  1. Mar 2, 2016 #1
    1. The problem statement, all variables and given/known data

    Part d) of the question below.
    29xxk6a.jpg

    2. Relevant equations

    We are told NOT to use the ladder technique to find the position operator as that's not covered until our Advanced Quantum Mechanics module next year (I don't even know this technique anyway). I emailed my tutor and he said it's about basic understanding of probability density which, according to his lecture notes, is given by

    P = |φ(x)|2

    I'm also using this given information, also from his lectures:

    282pa39.jpg

    3. The attempt at a solution

    For parts a, b and c I was fine. For a) I took the derivative of the potential function and used that to find the minimum point, x0, where the particle is in stable equilibrium. For b) I expanded the potential function as a Taylor series and applied the relevant parts to the equation of motion for a harmonic oscillator about the point x0. For part c) I simply found E = E1 - E0 and used that with hc/λ to find the wavelength.

    But part d)....

    To be honest I'm totally stuck here and more than a little annoyed at how vague my lecturer has been in his correspondence to me. I don't understand how to use and apply "P = |φ(x)|2" to come to any useful answer.

    I'm pretty sure I'm supposed to be taking φ(x) = φn(x) = φ1(x) as detailed in the above screenshot, in order to be working with the correct eigenstate for this part of the problem. But from there I really don't know.

    As far as "P = |φ(x)|2" goes, I'm sure there's a need to normalise it too. In fact I'm sure there's a lot more to it than that, involving some form of P = (x |φ(x)|)2", integrating over infinity, etc...

    My brain is totally fried right now :headbang:.

    Any help would be very much appreciated.

    Thanks!
     
  2. jcsd
  3. Mar 2, 2016 #2

    TeethWhitener

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    You might be overthinking this. I'll provide an analogy here. If your lecturer puts up a bell curve (probability distribution) of the grades on the last exam, how do you know what the most likely grade in the class is?
    By analogy, ##P(x)=|\varphi(x)|^2## is the probability distribution of the positions of the particle. How do you find what the most likely position is?
     
  4. Mar 2, 2016 #3
    I would take the function and find the position value of its maximum point, where dP/dx = 0 . It's the first thing I tried.

    The problem is that when I do this I get a position of ±(√2)x0, which seems to go against the intuitive answer which is that it should plainly be x0 since that is the equilibrium position of an oscillator. When looking at more complex explanations of this problem online, it seems to be that the equilibrium point is supposed to be x0 so I don't know what I've done wrong.

    When I plot ##P(x)=|\varphi(x)|^2## on mathematica I get a curve much like the n=1 curve on the right hand side of the below image. But that suggests two most-likely positions. :oldconfused:

    V13RX.png
     
  5. Mar 2, 2016 #4

    TeethWhitener

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    Haha, well that was much easier than I thought it was going to be. Congratulations, you got the right answer. In its first excited state, there are two equally most-likely positions for the particle.

    EDIT: There's actually an intuitive reason behind this. If you go to higher and higher excited states, the most likely position approaches the oscillator's classical turning points asymptotically, an example of the correspondence principle. If you do the same exercise for a classical oscillator, you find that, in fact, the most likely positions of the oscillator are near its turning points. This is because the oscillator is slowing near its turning points and therefore spending more time in those regions.
     
  6. Mar 2, 2016 #5
    Ah but I'm still so confused and unsure about this.

    I've just found this PDF - http://www.phys.hawaii.edu/~solsen/p274_hw4_sol.pdf - Q7 (page 2) says that <x> is the central position for all energy levels.

    The function used in that example is hugely similar to the one I was working with, of the form exp(ax2).

    Why does that example come to 0 whereas my very similar problem comes to √2 ?

    EDIT wait I know why my problem came to root 2. It's because I was working was Ψ2. If I do xΨ2 I get 0 too. (Although surely it should be x0 in my case since that's my equilibrium point?!).

    Brain is still fried, ha.
     
  7. Mar 2, 2016 #6

    TeethWhitener

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    It sounds from your edit like you're on the verge of understanding most of this.
    1. ##\langle x \rangle = \int x|\varphi(x)|^2 dx = 0## but ##max(P(x)) \neq 0##. This is equivalent to saying that the mode of the probability distribution ##max(P(x))## is not necessarily equal to the mean ##\langle x\rangle##.
    2. Eq point not working out...Maybe an algebra error somewhere? For your Hermite polynomial, did you use ##2x## or ##2(\frac{x}{x_0})##?
     
  8. Mar 2, 2016 #7
    Ahh, riiiiiight, it was more than an algebra error, it was a 'reading the lecture notes' error.

    I saw Hn(x/x0) And thought it meant Hn(x)(x/x0), wherein Hn alone was the hermite polynomial.

    With that altered, I now run through the very first suggestion you made to find the maximum point of the function and get a result of ±x0. Presumably I discard the negative answer because it doesn't make sense in this case.

    I understand now what you mean about modes and means, too. The most likely position is not the same as the average position.

    Got it. Thanks!!!
     
    Last edited: Mar 2, 2016
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