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Quantum mechanics, harmonic oscillator and wavefunction

  1. Nov 23, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    A harmonic oscillator is initially in the state [itex]\psi (x,0)=Ae^{-\frac{\alpha ^2 x^2}{2}} \alpha x (2\alpha x +i)[/itex]. Where [itex]\alpha ^2 =\frac{m \omega}{\hbar}[/itex].
    1)Find the wavefunction for all t>0.
    2)Calculate the probability to measure the values [itex]\frac{5\hbar \omega }{2}[/itex] and [itex]\frac{7\omega \hbar }{2}[/itex] for the energy.
    3)Calculate the mean value of the energy of this oscillator.
    2. Relevant equations
    [itex]\Psi (x,t)= \psi (x)e^{-\frac{iEt}{\hbar}}[/itex].
    [itex]E=(n+1)\frac{\hbar }{2}[/itex].
    Probability to measure [itex]E_n[/itex] associated to the wavefunction [itex]\psi _n[/itex] is [itex]|c_n|^2[/itex] in the expression [itex]\Psi (x,t)= \sum _i ^{\infty } c_i \psi _i (x,t)[/itex]. I'm only using memory for all of this, so I might be wrong for some things.

    3. The attempt at a solution
    Now that I think about it... can I consider that the particle is in a stationary state? I.e. that [itex]\psi (x,0)=\psi (x,t)[/itex]? Hmm I think not, that would be too easy to answer to 1).
    Now even more confused. Stationary state would imply that [itex]\Psi (x,t)=\Psi(x)[/itex] and so the equation [itex]\Psi (x,t)= \psi (x)e^{-\frac{iEt}{\hbar}}[/itex] looks wrong, which is I think impossible.
    I hope someone can shred some light on my understanding of intro to QM.
    Thank you very much.
     
  2. jcsd
  3. Nov 23, 2011 #2

    BruceW

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    I think you've misunderstood the meaning of stationary state. Maybe you should read through its definition again.

    I would personally just use the time-dependent Schrodinger equation. But I think you could answer the questions with the relevant equations you have given.

    I should also say - in usual notation, capital Psi means the actual wave function (which is a function of time and space), and if the wave function can be separated into a function of space multiplied by a function of time, then the function of space is often denoted by lower-case psi.

    Your notation seems to disobey this convention, which makes it a bit confusing.
     
  4. Nov 23, 2011 #3

    fluidistic

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    Ok thank you, heading into this. (I mean the read part)
    Indeed, and thanks for pointing this out. I didn't know about the convention, although I used it on my draft unlike here.
     
  5. Nov 23, 2011 #4

    fluidistic

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    I just read about stationary state. And I can say that I can't consider the particle to be in a stationary state. A stationary state implies a definite energy (I think with a probability 1). For the harmonic oscillator a stationary state could be for example if the particle is in the ground state (n=0 with probability 1), or the first excited state (n= 1 with probability 1).
    So I would have [itex]\Psi (x,t) = \psi _n (x) e^{-\frac{i E_n t}{\hbar}}[/itex] for some n.
    I hope I'm right on this. (Please correct me if I said something wrong)
    Ok, I'll reconsider the problem now. Will post here if I'm still stuck.
     
  6. Nov 23, 2011 #5

    BruceW

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    Yep, that's all right.
     
  7. Nov 23, 2011 #6

    fluidistic

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    Ok thank you for confirming.
    Well I'm having some hard time with the problem.
    My idea is to calculate [itex]\psi _i (x,t)[/itex]. Because [itex]\Psi (x,t)= \sum _{i=0} \psi _i (x,t)[/itex] with [itex]\sum _i |c_i|^2=1[/itex].
    Thanks to a previous problem, I know that [itex]c_n=\int _{-\infty}^{\infty} \psi _i ^* (x) \psi (x,0)dx[/itex].
    Where [itex]\psi _i (x)[/itex] are worth a somehow "nasty" expressions but which is somehow a real valued function. (Hermite polynomials with constants of normalization).
    In other words [itex]c_n=\int _{-\infty}^{\infty} \psi _i ^* (x) \psi (x,0)dx=c_n=\int _{-\infty}^{\infty} \psi _i (x) \psi (x,0)dx[/itex].
    But in a solution to another problem, I've seen that [itex]c_n=\int _{-\infty}^{\infty} \psi ^* (x,0)\psi _i (x)dx[/itex] which would here give a totally different result. So I have a doubt about what [itex]c_n[/itex] I should consider.
    There might be a quicker way to solve the problem that I'm missing. My friend told me he solved the problem within 5 minutes and the answer was something like 1/2 and 1/2 for the probabilities or something like that (I'm not sure I remember so well!). I don't think he had used Hermite polynomials though...
     
  8. Nov 24, 2011 #7

    BruceW

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    This is correct. (Where i and n are the same). So from this, you could explicitly do the integration to find the coefficients. Its an overly-complicated way of doing it though.

    You have [itex]\Psi(x,0)[/itex] and you are given the [itex]\psi_n(x)[/itex] right? So have a look at their form and think what the coefficients would be. You don't need to do integrals to find the answer.
     
  9. Nov 24, 2011 #8

    fluidistic

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    1)[itex]\Psi (x,t)= \sum _{n=0}^{\infty } c_n \psi _n (x,t)= \sum _{n=0}^{\infty } c_n \psi _n (x) e^{-\frac{iEt}{\hbar}}[/itex].
    So that [itex]\Psi (x,0)= \sum _{n=0}^{\infty} c_n \psi _n (x)=2 \alpha ^2 x^2 Ae^{-\frac{\alpha ^2 x^2}{2}}+i \alpha x A e^{-\alpha ^2 \frac{x^2}{2}}[/itex].
    So my answer for part 1 would be this multiplied by [itex]e^{-\frac{iEt}{\hbar}}[/itex].
    The first 2 Hermite polynomials are 1 and 2x. So I can already smell that the oscillator is in the superposition of the states n=0 and n=1.
    For part 2), I must look at [itex]\psi _n (y)= \left ( \frac{\beta}{\pi} \right ) ^{1/4} \frac{1}{\sqrt {2^n n!}}H_n (y)e^{-y^2/2}[/itex] where [itex]y=x \sqrt \beta[/itex] and [itex]\beta =\alpha ^2[/itex] and identify the coefficients [itex]c_n[/itex]?
    Though if I'm correct that the oscillator is only in n=0 and n=1, I could already answer "0" to part 2.
     
    Last edited: Nov 24, 2011
  10. Nov 24, 2011 #9

    vela

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    In Dirac notation, you would write
    [tex]| \psi \rangle = \sum_i | \phi_i \rangle\langle \phi_i | \psi \rangle[/tex]where [itex]c_i = \langle \phi_i | \psi \rangle = \int \phi_i^*(x)\psi(x)\,dx[/itex]. This corresponds to the expression
    [tex]\psi(x) = \sum_i c_i \phi_i(x)[/tex]In other words, you definitely want the eigenstate ϕi being the one conjugated when expanding a state ψ in terms of them.
     
    Last edited: Nov 24, 2011
  11. Nov 24, 2011 #10

    vela

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    How do you get x2 without using the n=2 state?
     
  12. Nov 24, 2011 #11

    fluidistic

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    Thanks, I'm going to spend some time understanding this (I didn't learn Dirac's notation yet, but I've sneaked into it).
    My bad, you're right, it's impossible. So there's indeed the state with [itex]E_2=\frac{3 \hbar \omega}{2}[/itex].
    By the way I edited my previous post (for answer to part 1).
     
  13. Nov 24, 2011 #12

    fluidistic

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    Part 2:
    After several tries (and errors!), I have come to the conclusion that the oscillator is in the superposition of the states n=0, n=1 and n=2. (100% sure)
    [itex]\Psi (x,0)=Ae^{-\frac{\alpha ^2 x^2}{2}} (2\alpha ^2 x^2 +i\alpha x)=c_0 \psi _0 (x)+c_1 \psi _1 (x)+c_2 \psi _2 (x)[/itex].
    Where [itex]\psi _0 (\alpha x )= \pi ^{-1/4} \sqrt { \frac{m\omega }{\hbar}} e^{-\frac{\alpha ^2 x^2}{2}}[/itex].
    [itex]\psi _1 (\alpha x)= \pi ^{-1/4} \sqrt { \frac{m \omega}{2 \hbar }} \cdot 2 \alpha x e^{-\frac{\alpha ^2 x^2}{2}}[/itex].
    [itex]\psi _2 (\alpha x)=\pi ^{-1/4} \sqrt { \frac{m \omega }{8\hbar }} (4 \alpha ^2 x^2 -2)e^{-\frac{\alpha ^2 x^2}{2}}[/itex].
    I skip the arithmetics. I reach that [itex]|c_0|^2=\frac{1}{11}[/itex], [itex]|c_1|^2=\frac{2}{11}[/itex], [itex]|c_2|^2=\frac{8}{11}[/itex].
    However it's only for t=0. Did they ask for all t's?

    So the answer for t=0, would be 0 probability for [itex]E_3=\frac{5\hbar \omega }{2}[/itex] and [itex]E_4=\frac{7\hbar \omega }{2}[/itex].

    Edit: I made an error in the energy expression (bad memory of mine!). It's [itex]E_n= \left ( n+\frac{1}{2} \right ) \hbar \omega[/itex].
    So I reach that the probability to measure [itex]E_2=\frac{5\hbar \omega }{2}[/itex] is 8/11, or around 72.7% while the probabilty to measure [itex]E_=\frac{7\hbar \omega }{2}[/itex] is 0.
     
    Last edited: Nov 24, 2011
  14. Nov 24, 2011 #13

    vela

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    Your expression for the time evolution needs a slight correction. It should be
    [tex]\Psi(x,t) = \sum_{n=0}^\infty c_n\psi_n(x)e^{-iE_nt/\hbar}[/tex]Each component evolves with frequency [itex]\omega_n = E_n/\hbar[/itex] corresponding to its energy. You left the subscript n off of the energy earlier.

    By the way, I got a slightly different expansion than you did. I found [itex]c_0 = \sqrt{2/7}[/itex], [itex]c_1 = i/\sqrt{7}[/itex], and [itex]c_2 = 2/\sqrt{7}[/itex].
     
  15. Nov 24, 2011 #14

    fluidistic

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    Ah you're right, I shouldn't have dropped the n subscript.
    Ok I post my work:
    [itex](c_0 \psi _0 +c_1 \psi _1+c_2 \psi _2)(\alpha x )=\pi ^{-1/4} e^{-\frac{\alpha ^2 x^2}{2}} \left [ c_0 \sqrt {\frac{m \omega }{\hbar}} - c_1 \sqrt {\frac{m \omega }{ 2 \hbar}} \cdot 2 \alpha x + \sqrt {\frac{m \omega }{ 8 \hbar}} c_2 (4 \alpha ^2 x^2 -2) \right ][/itex].
    But here I have the condition that the term that does not contains x must vanish, because this expression is worth [itex]Ae^{-\frac{\alpha ^2 x^2}{2}}(2 \alpha ^2 x^2 +i \alpha x)[/itex] and there's no term that doesn't contain x.
    Mathematically this implies that [itex]c_0 \sqrt {\frac{m \omega }{ \hbar}}-c_1 \sqrt {\frac{m \omega }{ 2 \hbar}} =0 \Rightarrow c_0=\frac{c_1}{\sqrt 2 }[/itex].
    Now equating the terms with [itex]x^2[/itex]: [tex]A(2\alpha ^2 x^2)=\pi ^{-1/4} c_2 \alpha ^2 x^2 \sqrt {\frac{m \omega }{ \hbar}} \Rightarrow A= \pi ^{-1/4} c_2 \sqrt {\frac{m \omega }{ 2 \hbar}}[/tex].
    On the other hand, by equating the terms with x: [tex]Ai \alpha x= \pi ^{-1/4} c_1 \sqrt {\frac{m \omega }{ 2 \hbar}} \cdot 2 \alpha x \Rightarrow A=-2i \pi ^{-1/4}c_1 \sqrt {\frac{m \omega }{ 2 \hbar}}[/tex].
    These 2 equalities with [itex]A[/itex] gives me [itex]c_2 \sqrt {\frac{m \omega }{ 2 \hbar}}=-2i c_1\sqrt {\frac{m \omega }{ 2 \hbar}} \Rightarrow c_2=-2i c_1[/itex].
    Now I use the fact that [itex]|c_0|^2+|c_1|^2+|c_2|^2=1[/itex]. So that [itex]\frac{|c_1|^2}{2}+|c_1|^2+4|c_1|^2=1 \Rightarrow |c_1|^2=\frac{2}{11} \Rightarrow |c_0|^2=\frac{1}{11}[/itex] and [itex]|c_2|^2=\frac{8}{11}[/itex].
     
  16. Nov 24, 2011 #15

    vela

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    When you looked at the constant term, you should be considering c2, not c1.
     
  17. Nov 24, 2011 #16

    fluidistic

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    Indeed, thank you. I've redone the algebra and reached the same result as you. I'll try to do part 3) now.
     
  18. Nov 24, 2011 #17

    fluidistic

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    Hmm actually I didn't find the values of [itex]c_i[/itex]'s; only their modulus to the 2nd power. And I need them for part 3). How did you find out that for instance [itex]c_1[/itex] was worth [itex]i/\sqrt 7[/itex] rather than [itex]1/\sqrt 7[/itex]?
    I'm about to try [itex]\langle E \rangle= \int _{-\infty}^{\infty} \Psi ^* (x,t) \hat H \Psi (x,t) dx[/itex].
     
  19. Nov 25, 2011 #18

    vela

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    Remember the wave function is unique only up to a complex phase. You can arbitrarily pick the phase for one of the constants. Your equations will then allow you to determine the relative phases of the other constants.
     
  20. Nov 25, 2011 #19

    Redbelly98

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    Please note: I have moved this thread to Advanced Physics.
     
  21. Nov 25, 2011 #20

    fluidistic

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    Ah I didn't know this, good to know.
    To calculate [itex]\langle E \rangle[/itex], can I just say it's equal to [itex]|c_0|^2E_0+|c_1|^2E_1+|c_2|^2E_2[/itex]?
     
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