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Homework Help: Find wavefunction of harmonic oscillator

  1. Jan 16, 2014 #1
    1. The problem statement, all variables and given/known data
    We want to prepare a particle in state ##\psi ## under following conditions:
    1. Let energy be ##E=\frac{5}{4}\hbar \omega ##
    2. Probability, that we will measure energy greater than ##2\hbar \omega## is ##0##
    3. ##<x>=0##


    2. Relevant equations



    3. The attempt at a solution

    I have no idea, yet i tried to do some magic.

    ##E=\frac{5}{4}\hbar \omega ## than I guess that this is actually a sum of all states? or is it not?

    So... ##\frac{5}{4}\hbar \omega=\sum (n+1/2)\hbar \omega ##

    The second condition:

    if ##P(E>2\hbar \omega)=\sum\left | C_n \right |^2E_n=\sum\left | C_n \right |^2\hbar\omega (n+1/2)=0## than

    ##P(E<2\hbar \omega)=\sum\left | C_n \right |^2E_n=\sum\left | C_n \right |^2\hbar\omega (n+1/2)=1##

    Now I don't know how to use third condition.. :/ I don't even know if this makes any sense?
     
  2. jcsd
  3. Jan 17, 2014 #2

    vanhees71

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    Since [itex]5 \hbar \omega/4[/itex] is not an eigenstate of the Hamiltonian,
    [tex]\hat{H}=\frac{\hat{p}^2}{2m}+\frac{m \omega^2}{2} \hat{x}^2,[/itex]
    there is no eigenstate with that value of the energy. Do you mean, the average energy is [itex]5/4 \hbar \omega[/itex]? But then the state is not uniquely defined.
     
  4. Jan 17, 2014 #3
    Well it only says that the energy is ##\frac{5}{4}\hbar \omega ##. I also think there is no eigenstate with that value of energy, therefore I assume that this is meant for the average energy.
     
  5. Jan 17, 2014 #4

    DrClaude

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    That should be ##P(E \le 2\hbar \omega)##.

    What is the value of ##\langle x\rangle## for an eigenstate of the h.o.? What happens if you sum two states of the same parity? What happens if you sum two states of opposite parity?
     
  6. Jan 17, 2014 #5
    I though the value of ##<x>## for any eigenstate of the harmonic oscillator is ##0##. But now, when i came across this problem I have to admit that I am not really sure about that.
    Thinking in classical physics the average ##x## is ##0##.

    If i sum the two states of the opposite parity you get nothing. (straight line)
     
  7. Jan 17, 2014 #6

    DrClaude

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    The fact that an eigenstate is either symmetric or anti-symmetric with respect to ##x=0## should clue you in.

    Really? Try ploting ##\psi_0 + \psi_1##.
     
  8. Jan 17, 2014 #7
  9. Jan 17, 2014 #8

    DrClaude

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    No. Look carefully at ##\psi_0##. What do you expect ##\langle x \rangle## to be?

    We're interested in the sum of wave functions. Here are ##\psi_0##, ##\psi_1##, and ##(\psi_0 + \psi_1)/\sqrt{2}##. What can you say about ##\langle x \rangle## in each case?
     

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  10. Jan 17, 2014 #9
    Aaaha.

    ##\left | \psi \right |^2## is the probability to find the particle at x. Clearely, for ##\psi _0## the ##<x>=0##. So, for all symmetrical ##\psi ## the ##<x>## is ##0##. Can we say anything about ##<x>## of anti-symmetrical functions, except that ##<x>\neq 0##?
     
  11. Jan 17, 2014 #10

    DrClaude

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    Sorry, but you're going at it wrong.
    $$
    \begin{align}
    \langle x \rangle &= \int_{-\infty}^{\infty} x | \psi(x) |^2 \; \mathrm{d}x \\
    &= \int_{-\infty}^{0} x | \psi(x) |^2 \; \mathrm{d}x + \int_{0}^{\infty} x | \psi(x) |^2 \; \mathrm{d}x \\
    &= - \int_{0}^{\infty} (-x) | \psi(-x) |^2 \; \mathrm{d}(-x) + \int_{0}^{\infty} x | \psi(x) |^2 \; \mathrm{d}x \\
    &= - \int_{0}^{\infty} x | \psi(x) |^2 \; \mathrm{d}x + \int_{0}^{\infty} x | \psi(x) |^2 \; \mathrm{d}x \\
    &= 0
    \end{align}
    $$
    where I made use of ##| \psi(-x) |^2 = | \psi(x) |^2## for ##\psi## either symmetric or anti-symmetric.

    If ##\psi## is neither symmetric or anti-symmetric, then ##\langle x \rangle## is most often not zero. So either you make your ##\psi## symmetric or anti-symmetric or (hint, hint) use a clever superposition of symmetric and anti-symmetric wave functions such that ##| \psi(x) |^2## itself is symmetric [STRIKE]or anti-symmetric[/STRIKE].

    [Edit: I corrected the silliness just above. ##| \psi(x) |^2## obviously can't be anti-symmetric.]
     
    Last edited: Jan 17, 2014
  12. Jan 17, 2014 #11
    o_O This kinda of frightens me. I guess I don't know a thing about harmonic oscillators or wavefunctions in general.

    Since ##<x>=0## in my case, than ##\psi ## is a combination of symmetric and anti-symmetric wavefunctions. BUT, how can I be sure that we are talking about ##\psi _0## and ##\psi _1## here and not about ##\psi _{3432434}## and ##\psi _{23}## (random functions)?
     
  13. Jan 17, 2014 #12

    DrClaude

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    But you're learning, aren't you?

    I'll repeat: ##\psi## will either be
    • a superposition of symmetric wave functions
    • a superposition of anti-symmetric wave functions
    • a clever superposition of symmetric and anti-symmetric wave functions, such that ##| \psi |^2## is symmetric

    Remember that you are considering here the third criterion for ##\psi##. You have the first two criteria that also play a role. I used ##\psi_0## and ##\psi_1## only as an example.
     
  14. Jan 17, 2014 #13
    I am, but still... I should have known that. It's frustrating me now. -.- Nevermind.

    First condition: ##E=\frac{5}{4}\hbar \omega##
    Second condition: ##P(E\leq 2\hbar \omega)=\sum\left | C_n \right |^2E_n=\sum\left | C_n \right |^2\hbar\omega (n+1/2)=1##
    and the third is written by you.

    If the first condition is not the average energy, than ##E=\frac{5}{4}\hbar \omega=\hbar \omega (n+1/2)##. Inserting this in second condition gives me ##\sum\left | C_n \right |^2=\frac{4}{5\hbar \omega}## where ##C_n## can be complex.

    Now what? To me this kinda feels like ##\psi ## will be a superposition of symmetric and anti-symmetric wave functions.
     
  15. Jan 17, 2014 #14

    DrClaude

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    You have an ##\hbar \omega## too much.

    What does criterion 2 tell you?
     
  16. Jan 17, 2014 #15

    BruceW

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    yes. sorry to barge in. But as DrClaude is implying, your current equation for condition 2 is not correct. condition 2 is telling you something about the possible outcome of an energy measurement.
     
  17. Jan 17, 2014 #16

    DrClaude

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    Thanks Bruce. I read too fast and missed that.
     
  18. Jan 17, 2014 #17

    vanhees71

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    I think, here a lot is totally confused now. So let's first repeat the basics.

    (1) A (pure) state in quantum theory can be described by a wave function [itex]\psi(x)[/itex] that is square integrable, i.e., the integral
    [tex]\int_{\mathbb{R}} \mathrm{d} x |\psi(x)|^2[/tex]
    exists.

    (2) The probability distribution to find the particle at place [itex]x[/itex] is given by [itex]|\psi(x)|^2[/itex].

    (3) Any observable is represented by a self-adjoint operator on the Hilbert space of wave functions.

    (4) A possible value of the observable is given by a (generalized) eigenvalue of the self-adjoint operator representing it.

    For the harmonic oscillator the Hamiltonian (operator representing the total energy) is given by
    [tex]\hat{H}=\frac{\hat{p}^2}{2m}+\frac{m \omega^2}{2}.[/tex]
    The momentum operator is given by [itex]-\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} x}[/itex]. The eigenvalues of [itex]\hat{H}[/itex] are
    [tex]E_n=\frac{\hbar \omega}{2}(2n+1), \quad n \in \{0,1,2,\ldots \}=\mathbb{N}_0.[/tex]

    The normalized eigenfunctions of the Hamiltonian [itex]u_n(x)[/itex] build a complete set of orthonormal functions in the Hilbert space, i.e., any state is given by a superposition of these eigenfunctions:
    [tex]\psi(x)=\sum_{n=0}^{\infty} C_n u_n(x), \quad C_n=\langle u_n |\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \; u_n^*(x) \psi(x).[/tex]
    In the following let [itex]\psi[/itex] be normalized, i.e.,
    [tex]\int_{\mathbb{R}} \mathrm{d} x |\psi(x)|^2=\sum_{n=0}^{\infty} |C_n|^2=1.[/tex]

    As stated earlier the constraint 1. is impossible to fulfill, because the energy cannot be [itex]5 \hbar \omega/4[/itex]. So I guess the statement should be that the mean enery,
    [tex]\langle E \rangle = \sum_{n=0}^{\infty} |C_n|^2 E_n=\frac{5 \hbar \omega}{4}.[/tex]

    Now you should be able to put everything together. The trick is to express all the constraints (1)-(3) in terms of the coefficients [itex]C_n[/itex], with (1) corrected as said above (because otherwise the question doesn't make any sense or the answer is trivially that there is no state that fulfills constraint (1)).
     
  19. Jan 17, 2014 #18
    First condition:

    ##\sum_{n=0}^{\infty}\left | C_n \right |^2E_n=\frac{5}{4}\hbar \omega ##

    Second should say that... The energy ## E## of ##\psi ## which I am looking is for sure ##E\leq 2 \hbar \omega ## so ##(n+1/2)\leq 2 ## and finally

    ##n\leq \frac{3}{2}##

    So my ##\psi ## can only be a combination of ##\psi _0## and ##\psi _1##

    or not?

    First and second condition together now give me

    ##\left | C_0 \right |^2E_0+\left | C_1 \right |^2E_1=\frac{5}{4}\hbar \omega##

    ##\left | C_0 \right |^2+3\left | C_1 \right |^2=\frac{5}{2}##

    Now I know that ##\psi = C_0 \psi _0 + C_1\psi _1##

    Third condition, if I understood all of you right, says that

    ##<x>=\int_{-\infty}^{\infty}\psi ^*x\psi dx=\int_{-\infty}^{\infty}(\left | C_0 \right |^2\left | \psi_0 \right |^2x+\left | C_1 \right |^2\left | \psi_1 \right |^2x)dx=0##

    I hope this mens that ##\left | C_0 \right |^2+\left | C_1 \right |^2=0##

    Now I have a system of two equations with two constants

    ##\left | C_0 \right |^2+\left | C_1 \right |^2=0##

    ##\left | C_0 \right |^2+3\left | C_1 \right |^2=\frac{5}{2}##

    Which give me ##\left | C_1 \right |^2=\frac{5}{4}## and ##\left | C_0 \right |^2=-\frac{5}{4}## which is a bit confusing since I am working with absolute values here...

    What did I do wrong this time?
     
  20. Jan 17, 2014 #19

    DrClaude

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    Correct.

    You're fine up to here. Now you should normalize ##\psi## before continuing further.

    That is not correct. You can't transform ##\psi^* \psi## into ##\left | C_0 \right |^2\left | \psi_0 \right |^2 +\left | C_1 \right |^2\left | \psi_1 \right |^2##.

    There are many ways to proceed here. I suggest the following:
    Write that integral by correctly substituting ##\psi^* \psi##, with coefficients ##C_0## and ##C_1##. Then rewrite the ##C##'s as ##a + i b##, where ##a## and ##b## are real. What you will be looking for inside the integral is a sum of symmetric and anti-symmetric terms. What you want to avoid are terms that have no defined symmetry, as they are the ones that lead to ##\langle x \rangle \neq 0##.
     
  21. Jan 17, 2014 #20

    vanhees71

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    Ok, the average-energy condition is correct, giving indeed
    [tex]|C_0|^2+3 |C_1|^2=\frac{5}{2}.[/tex]
    Then what's the condition for the correct normalization of the wave function?

    Further, your wave function is
    [tex]\psi_x=C_0 u_0(x)+C_1 u_1(x).[/tex]
    The average position is
    [tex]\langle x \rangle = \int_{\mathbb{R}} \mathrm{d} x \psi^*(x) x \psi(x)=\int_{\mathbb{R}} \mathrm{d} x x |\psi(x)|^2.[/tex]
    Now carefully (!!!) calculate [itex]|\psi(x)|^2[/itex] and plug this into the equation for the average. What do you get then?

    To finally solve the problem you only need to know that you can choose the overall phase of the wave function arbitrarily, i.e., you can assume, e.g., [itex]C_0 \geq 0[/itex].
     
  22. Jan 17, 2014 #21
    ##|C_0|^2+3 |C_1|^2=\frac{5}{2}##

    from normalization also

    ##|C_0|^2+ |C_1|^2=1##


    Carefully ( =) ) calculated ##\left | \psi (x) \right |^2=\left | C_0 \right |^2\left | \psi _0 \right |^2+C_0^*C_1\psi _0^*\psi _1+C_1^*C_0\psi _1^*\psi _0+\left | C_1 \right |^2\left | \psi _1 \right |^2##
     
  23. Jan 17, 2014 #22

    DrClaude

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    Good. Now, to get ##\langle x \rangle = 0##, you need all terms to have a definite parity, which is not the case for ##C_0^*C_1\psi _0^*\psi _1## and ##C_1^*C_0\psi _1^*\psi _0##, so you need to get rid of them.
     
  24. Jan 17, 2014 #23

    BruceW

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    why don't they have definite parity? (playing devil's advocate)

    edit: if skrat has notes written down somewhere with the form of ##\psi_0## and ##\psi_1## then it would be possible to explicitly see what ##\langle x \rangle## is, and then work out what ##C_0## and ##C_1## need to be.
     
    Last edited: Jan 17, 2014
  25. Jan 17, 2014 #24
    Since I am having some troubles translating what "definite parity" would mean, I am going with the second option.

    ##\psi _n=\frac{1}{\sqrt{a}\pi ^{1/4}}\sqrt{\frac{1}{2^{n}n!}}H_n(\frac{x}{a})e^{-\frac{x^2}{2a^2}}##

    so

    ##\left \langle x \right \rangle=\int_{-\infty }^{\infty }\psi _1x\psi _0\mathrm{d}x=\frac{1}{a}\frac{1}{\sqrt{\pi }}\int_{-\infty }^{\infty }\frac{1}{\sqrt{2}}(2\frac{x}{a})xe^{-\frac{x^2}{a^2}}=\frac{2}{a^2\sqrt{2\pi }}2\int_{0}^{\infty }x^2e^{-\frac{x^2}{a^2}}=\frac{a}{\sqrt{2}}##

    where ##\sqrt{\frac{\hbar}{m\omega }}##

    I hope everything is ok..
     
  26. Jan 17, 2014 #25

    DrClaude

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    I'm sorry if this confuses you. Bruce's method might be easier for you. Just to clarify: I meant that the terms ##\psi_1^* x \psi_0## and ##\psi_0^* x \psi_1## are neither symmetric nor anti-symmetric, and therefore will always contribute something to ##\langle x \rangle##, which you figured out from the integral.

    So, you need ##\langle x \rangle = 0##. What does that set as a condition on ##C_0## and ##C_1##?
     
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