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Prediction of observables from wavefunction

  1. Dec 15, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    The wavefunction of a particle is given as $$u(r,\theta,\phi) = AR(r)f(\theta)\cos(2\phi),$$ where ##f## is an unknown function of ##\theta##. What can be predicted about the results of measuring
    a) The z component of angular momentum
    b)The square of the angular momentum

    2. Relevant equations
    Operators representing ##\hat{L}_z = -i\hbar \frac{\partial}{\partial \phi}## and ##\hat{L}^2 = -\hbar^2 \left(\frac{1}{\sin\theta} \frac{\partial}{\partial \theta} \left(\sin\theta\frac{\partial}{\partial \theta}\right) + \frac{1}{\sin^2\theta} \frac{\partial^2}{\partial \phi^2}\right)##.

    3. The attempt at a solution
    To obtain these predictions, I thought I could act with the two operators above in turn on ##u##. If ##u## was an eigenfunction of the operators, then I would know for sure the possible outcomes of the measurement. $$a) \hat{L}_z u(r,\theta,\phi) = -i\hbar A R(r)f(\theta) \frac{\partial}{\partial \phi} \left(\frac{e^{i2\phi} + e^{-i2\phi}}{2}\right) = 2\hbar AR(r)f(\theta) \left(\frac{e^{-2i\phi} - e^{2i\phi}}{2}\right) $$and so ##u## is not an eigenfunction of the operator. How do I obtain the possible outcomes then?
     
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  3. Dec 15, 2013 #2

    vela

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    You want to express u in terms of the eigenfunctions of Lz, not apply Lz to u. So what are the eigenfunctions of Lz?
     
  4. Dec 15, 2013 #3

    CAF123

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    Hi vela,
    Do you mean what are a possible set of eigenfunctions of Lz? I would go with the spherical harmonics here since we are also interested in L2. In that case, I think $$u = \frac{1}{2}AR(r)f(\theta) \frac{8 \pi}{3} \frac{1}{\sin^2 \theta}\left( (Y_1^{-1})^2 + (Y_1^1)^2\right)$$ is a possible linear combination of the spherical harmonics equal to u. Is this what you had in mind?
     
  5. Dec 15, 2013 #4

    vela

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    Kinda. The problem with what you did is that you assumed ##l=1##, but there are other spherical harmonics that are proportional to ##e^{\pm i\phi}## as well as ##e^{\pm 2i\phi}##. Remember you want a linear combination, so you don't want to be squaring ##Y_{lm}##. Restrict yourself to just considering ##\hat{L}_z## for now.
     
  6. Dec 15, 2013 #5

    CAF123

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    Perhaps consider the more general form of the eigenfunctions of ##\hat{L}_z##. They could be of the form ##R(r)f(\theta)\exp(\pm i2\phi)## in which case $$u = \frac{1}{2}A\left(R(r)f(\theta)\exp(i2\phi) + R(r)f(\theta)\exp(-i2\phi)\right)$$ Acting with ##\hat{L}_z##, I do not get the form of an eigenvalue equation.

    Should I write ##u## in terms of the spherical harmonics with ##m=2## and leave ##\ell## arbritary?
     
  7. Dec 16, 2013 #6

    vela

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    But you have a superposition of eigenstates, right? So what are the possible outcomes of the measurement and how do you calculate the probabilities when you have a superposition?
     
  8. Dec 17, 2013 #7

    CAF123

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    Okay, I see. If I let the eigenfunctions of ##\hat{L}_z## be ##u_{1,2} = \frac{1}{2}AR(r)f(\theta)\exp(\pm i2\phi)## then I obtain the following; $$\hat{L}_z u = 2\hbar u_1 -2\hbar u_2$$ so the possible outcomes are ##\pm 2 \hbar##.

    Here is my analysis; If system is state ##u_1(u_2)##, then measure ##L_z = +(-)2\hbar## with probability 1. If system in state ##u## then the probability of each value is found by projecting the eigenstate onto the wavefunction. Consider the probability of obtaining ##+2\hbar##; $$P(L_z = 2 \hbar) = |\langle u_1 | u \rangle |^2 = \left|\int d^3 r u_1^* [u_1 + u_2]\right|^2 = 1.$$ But this result does not make sense because it implies there is a zero probability of obtaining ##-2\hbar##, which is not the case.

    Can you see where I went wrong here?
     
  9. Dec 17, 2013 #8

    vela

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    You're not using a normalized wave function.
     
  10. Dec 17, 2013 #9

    CAF123

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    Because I do not have the functional forms of R or f, is it correct to say that it is not possible to obtain the normalization constant?
     
  11. Dec 18, 2013 #10

    CAF123

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    I think I can argue that the probability of obtaining ##+2\hbar## or ##-2\hbar## is equal and thus equal to 1/2 since these are the only possible outcomes. I said this was the case because the only difference in the eigenstates is in the sign of the complex exponential and so it is an argument based on symmetry. However, since there is no explicit form of R or f, how would I be able to obtain the normalizable constant?

    For ##\hat{L}^2##, the most obvious choice of eigenfunctions I see at the moment are the spherical harmonics. Since we are interested in linear combinations, then ##\ell \geq 2## otherwise we would end up with what I had in an earlier post with the eigenfunctions squared.

    If we now assume a particular value of ##\ell## and write ##u## in terms of the appropriate spherical harmonics, since ##\hat{L}^2## is dependent on ##\theta##, we cannot know anymore about the possible values of ##L^2## without knowing the explicit functional form ##f(\theta)##.

    Is this reasonable?
     
  12. Dec 19, 2013 #11

    vela

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    You can solve for A in terms of integrals involving R and f.


    Right. Without knowing f, the best you can say is that you won't get results corresponding to l=0 or l=1.
     
  13. Dec 19, 2013 #12

    CAF123

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    This gives the following equation for A2 for which I can then find A; $$A^2 = \frac{2}{\pi} \frac{1}{\int_0^\infty R(r)^* R(r) r^2 dr} \cdot \frac{1}{\int_0^\pi f(\theta)^* f(\theta) \sin \theta d\theta}$$
    I still don't think I can compute the probability because we do not know what R or f is.

    I think the exact same analysis(i.e same two outcomes (##\pm 2\hbar##) and condition on ##\ell## for ##\hat{L}^2##) applies to the wavefunction ##u = AR(r)f(\theta)\cos^2\phi##. Is that so?
    Reasoning: Reexpress cos∅ in terms of exponential and when it is squared, we get the same form as the previous u except for a cross term which is just a constant so when Lz (and L2) acts it vanishes anyway.
     
  14. Dec 19, 2013 #13

    vela

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    Have you tried?


    Measurements have nothing to do with applying operators.
     
  15. Dec 20, 2013 #14

    CAF123

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    Yes, if I look for the same probability as in an earlier post, then the integral is $$\int_V d^3\,r \left(\frac{1}{2}AR(r)^*f(\theta)^* \exp(-i2\phi)\right)\left[\frac{1}{2}A\left(R(r)f(\theta)\exp(i2\phi) + R(r)f(\theta)\exp(-i2\phi)\right)\right],$$which after simplification gives $$\frac{1}{4}A^2 \int_V d^3 r R(r)^* f(\theta)^* \left[R(r)f(\theta) + \exp(-i4\phi)\right]$$ and I cannot continue because I do not know R or f.

    I am not sure I understand - by expressing u as a L.C of eigenstates acting on u with the right operator corresponding to the observable,I obtain the possible outcomes. For ##u = AR(r)f(\theta)\cos^2\phi = \frac{1}{4}AR(r)f(\theta)\left(e^{i2\phi} + e^{-2i\phi} + 2\right)## I can write u as a L.C of three distinct eigenstates of ##L_z##: ##u_{1,2} = \frac{1}{4}AR(r)f(\theta)\exp(\pm i2\phi)## and ##u_3 = \frac{1}{4}AR(r)f(\theta)\cdot 2## which happens to be a trivial eigenstate. So now that I have u expressed as a L.c of eigenstates, I follow the same method previously, and obtain the exact same outcomes (##\pm 2 \hbar##) since ##\hat{L}_z u_3 = 0##.
     
  16. Dec 20, 2013 #15

    vela

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    You made an algebra mistake in getting to your last expression. It should be
    $$\frac{1}{4}A^2 \int_V d^3 r R(r)^* f(\theta)^* R(r)f(\theta) \left[1 + \exp(-i4\phi)\right]$$ which you can write as
    $$\frac{1}{4}A^2 \int dr\,r^2 R(r)^*R(r) \int d\theta\, \sin\theta\ f(\theta)^* f(\theta) \int d\phi\,\left[1 + \exp(-i4\phi)\right].$$

    Fine up to here, though I wouldn't call ##u_3## a trivial eigenstate, whatever that is.

    You don't apply ##\hat{L}_z## to determine the outcomes. Think about it. If your method were correct, you'd never find a particle in an ##m=0## state.

    The proper method is to express the state as a linear combination of the eigenstates of ##\hat{L}_z## and then examine the coefficients to determine the probabilities.
     
  17. Dec 20, 2013 #16

    CAF123

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    Going back to the first case for u = (1/2)AR(r)f(θ)(exp(+i2∅)+exp-(i2∅)), the eigenstates I chose were u1,2 = (1/2)AR(r)f(θ)exp(±i2∅). I said before that u = u1+u2 but this is not true. If instead I choose the eigenstates to be those functions above but without the factors of 1/2, then u = (1/2)[u1+u2] and now I see that the probabilities are 1/2, by inspecting the coefficients. Now, is it correct to say that to get the outcomes (of L_z) I then act with Lz on each of u1 and u2?

    The error in expressing u has carried along here. The correct value of A2 has instead a factor 1/π not 2/π. If I now compute $$\frac{1}{2}\int_V d^3 r AR(r)^*f(\theta)^*\exp(-i4\phi)[AR(r)f(\theta)(\exp(i4\phi) + \exp(-i4\phi)) = \langle u_1 | u \rangle$$ I get an answer of 1, and I should get 1/2.
    I get 1/2 when instead I computed ##\int d^3r u_1^*(\frac{1}{2}(u_1+u_2))##, but this did not take into consideration any normalization constant.
    Many thanks.
     
  18. Dec 20, 2013 #17

    vela

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    The coefficient isn't the probability; it's equal to the amplitude if the wave function is normalized.
     
  19. Dec 20, 2013 #18

    CAF123

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    I managed to sort out everything in my last post, thanks. But is it correct to say that to obtain the possible outcomes of ##{L}_z##, I need to act with ##\hat{L}_z##only onto the eigenfunctions of ##\hat{L}_z##?
    For example, normalized u looks like ##u = \frac{1}{\sqrt{2}}\left(u_1 + u_2\right)##. To obtain the possible values of L2, I act with the operator on the eigenfunctions to obtain ##\pm 2\hbar##?

    If this is correct, then going back to the second example of u (=AR(r)f(θ)cos2∅), is it correct to say that the possible values of Lz are ##\pm 2 \hbar## and ##0## each with probability 1/3.

    Similarly, for L2, as it depends on θ, I think the analysis is the same as for the first form of u.
     
    Last edited: Dec 20, 2013
  20. Dec 20, 2013 #19

    vela

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    Yeah, you can do that, but it's usually not necessary. When you expand a wave function in terms of the eigenfunctions, you typically have already found the eigenfunctions, and in the process of doing that, you had to find the eigenvalues first. For example, in the case of ##\hat{L}_z##, you know that the eigenfunctions are of the form ##N e^{im\phi}## with eigenvalue ##m\hbar##. So you can just look at the function and identify what values of ##m## are relevant.

    The possible outcomes are correct, but you didn't get the probabilities right. You want to express the normalized wave function in terms of the normalized eigenfunctions. Then the coefficients will be the probability amplitudes.
     
  21. Dec 21, 2013 #20

    CAF123

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    This is what I tried, here is my workings; Reexpress ##u = AR(r)f(\theta)\cos^2\phi = \frac{1}{4}\left(AR(r)f(\theta) \cdot 2 + AR(r)f(\theta)e^{-i2\phi} + AR(r)f(\theta)e^{i2\phi}\right) = \frac{1}{4}\left(u_1 + u_2 + u_3\right).## This is the non-normalized form of u. To normalize, introduce a constant C and find C such that ##\int_V d^3 r u^* u = 1 \Rightarrow C = 4/\sqrt{3}## This means the normalized u is ##\frac{1}{\sqrt{3}}\left(u_1+u_2+u_3\right)## and the square of the amplitudes add up to 1, where ##|c_1|^2 = |c_2|^2 = |c_3|^2 = 1/3##.
     
    Last edited: Dec 21, 2013
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