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Quantum Mechanics - quick integral question.

  1. Dec 26, 2013 #1
    ψnlm, ψnlm', ψnlm'' are energy eigenfunctions of hydrogen, and a function of r, θ and Ø.

    (they all have the same n and l, but different m's)

    I am given ψ(r,t=0)=ψnlm+ψnlm'+ψnlm'' .

    I am after the probability density as a function of θ, and so need to integrate over r and Ø.

    I have been introduced to ψnlm=Rnl(r)Θlm(θ) ϕm(Ø)

    I am checking my understanding of integration tecniques

    First of all, energy eigenfunctions are orthogonal when integrated over all space in r/θ/Ø.

    So I integrate over Ø and attain |RnlΘlm|^2+|RnlΘlm'|^2+|RnlΘl m''|^2

    Which is fine, so now I need to integrate over r.


    ∫ ( |RnlΘlm|^2+|RnlΘlm'|^2+|RnlΘl m''|^2 ) r^2 dr. *

    But I know that Rnl(r),Θlm(θ), ϕm (Ø) are separately normalised.

    So ∫ (Rnl)^2 r^2 dr =1 ( Here Rnl is real ! )

    So from * I atttain :

    |Θlm|^2+|Θlm'|^2+|Θlm''|^2

    which gives me the wrong answer.

    If anyone could help shed some light on this, that would be greatly appreciated !

    The correct solution uses the orthogonal tecnique, and then continues explicitly , as a pose to using the normalization technique, so I suspect this may be were I am going wrong...
     
  2. jcsd
  3. Dec 26, 2013 #2

    Simon Bridge

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    Aside: you can place LaTeX expressions between double-dollar signs
    ##\renewcommand{\F}{\mathrm{\Phi}}
    \renewcommand{\T}{\mathrm{\Theta}}
    \renewcommand{\R}{\mathrm{R}}
    \renewcommand{\dr}{\mathrm{\;\mathrm{d}r}}
    \renewcommand{\dth}{\mathrm{\;\mathrm{d}\theta}}
    \renewcommand{\dph}{\mathrm{\;\mathrm{d}\phi}}##
    $$\psi_{nlm}=\R_{nl}(r)\T_{lm}(\theta)\F_m(\phi)$$

    You appear to have done:
    $$p(\theta)=\int_0^\infty \int_0^\pi |\Psi(\vec r, t=0)|^2 r^2\dph\dr$$ ... the question that immediately springs up is: where does the ##r^2## come from? The volume element has an ##r^2 \sin\theta## in it... I suspect you just need to be more rigorous with your definitions.
     
  4. Dec 27, 2013 #3
    But I only want to integrate over r and ϕ as I am after a p(θ)?
     
  5. Dec 27, 2013 #4

    Simon Bridge

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    ##\renewcommand{\F}{\mathrm{\Phi}}
    \renewcommand{\T}{\mathrm{\Theta}}
    \renewcommand{\R}{\mathrm{R}}
    \renewcommand{\d}{\mathrm{\;\mathrm{d}}}
    \renewcommand{\dr}{\mathrm{\;\mathrm{d}r}}
    \renewcommand{\dth}{\mathrm{\;\mathrm{d}\theta}}
    \renewcommand{\dph}{\mathrm{\;\mathrm{d}\phi}}##
    That's right.

    Normally $$P_{vol} = \iiint_{vol}|\Psi(\vec r)|\d\tau$$ .... which is to say: the probability of finding the particle in a specified volume is the integral over that volume of the square-modulus of the position-wavefunction.

    The volume is indicated by ##\tau## because ##V## is already used in context to represent a potential.
    ##\d\tau = r^2\sin\theta\dr\dth\dph## would be the volume element in spherical coordinates.
    (Where does that ##r^2## come from? Can you derive this volume element?)

    You want to find ##p(\theta)##, which is not the same thing.
    Can you express ##p(\theta)## in words: what does this bunch of symbols mean?

    You seem to be thinking that you should do the above volume integral except for the ##\theta## part ... but (a) that is not what you did above and (b) does that make sense? But I cannot actually tell what you are thinking ... if you don't answer questions, I cannot help you.
     
  6. Dec 29, 2013 #5
    a) That is what I was thinking, and I am struggling to see how this is not what I did.
    b) I think I see now, this doesnt make any sense at all as r^2 is not a volume element.

    I thought this part wouldn't be relevant to my issue, but due to the volume element I now see it was - the question asked to find the probablity density functin as a function of μ, where μ=cosθ, st θ terms no longer appear:

    |dμ|=sinθ

    D(μ)dμ = [itex]\int\int[/itex](ψnlm+ψnlm'+ψnlm'')*(ψnlm+ψnlm'+ψnlm'') r^2 dr dμ dø
    => D(μ) = [itex]^{2∏}_{0}[/itex][itex]\int[/itex][itex]^{∞}_{0}[/itex][itex]\int[/itex](ψnlm+ψnlm'+ψnlm'')*(ψnlm+ψnlm'+ψnlm'') r^2 dr dø.

    which was my starting point.
     
  7. Dec 29, 2013 #6

    vela

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    ##r^2\,dr## has units of volume.

    You can also write the volume element as ##d\tau = r^2\,dr\,d(\cos\theta)\,d\phi## where the limits for ##\cos\theta## run from -1 to 1.

    You mean ##d\mu = \sin\theta\,d\theta##. You can't have a lone differential.

    You said your answer was incorrect. How do you know? What are you supposed to get?
     
  8. Dec 30, 2013 #7
    r[itex]^{2}[/itex] dr is not the relevant volume element - here derived from the Jacobian for spherical coordinates.

    The solution I am trying to follow uses the method in post 5, where sin/cos [itex]\theta[/itex] and d[itex]\theta[/itex] terms no longer appear.

    As in post 1 I thought it was a later step where I went wrong - the orthogonal or normalisation assumptions/knowledge application.

    The solution attains a number as the answer, whereas I attain a function of [itex]\theta[/itex]. (the wave functions are all specified).
     
  9. Dec 30, 2013 #8

    vela

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    No one said it was. You seemed to have implied that ##r^2## wasn't correct because it didn't have units of volume. I was pointing out the volume element includes ##dr## so that the units work out.

    Did you provide us with the complete problem statement? I don't see how you can get just a number if ##l## isn't specified and you're not integrating over ##\theta##.
     
  10. Dec 31, 2013 #9
    I saw that, and the reply was merely aimed to be a correction in recognition of your point.

    All n,l,m are specified with the wave functions.

    Here's the complete problem statement:

    ψ[itex]_{210}[/itex]=[itex]\sqrt{\frac{1}{2a^{3}\pi}}[/itex][itex]\frac{r}{4a}[/itex]exp[itex]^{\frac{-r}{2a}}[/itex]cos[itex]\theta[/itex]

    ψ[itex]_{211}[/itex]=[itex]\sqrt{\frac{-1}{a^{3}\pi}}[/itex][itex]\frac{r}{8a}[/itex]exp[itex]^{\frac{-r}{2a}}[/itex]sin[itex]\theta[/itex]exp[itex]^{i\phi}[/itex]

    ψ[itex]_{21-1}[/itex]=ψ[itex]^{*}[/itex][itex]_{211}[/itex]

    ψ[itex]_{(\underline{r},t=0)}[/itex]=[itex]\frac{1}{\sqrt{3}}[/itex](ψ[itex]_{210}[/itex]+ψ[itex]_{211}[/itex]+ψ[itex]_{21-1}[/itex])

    The question then is: what is the probability density distribution as a function of μ = cos[itex]\theta[/itex]
     
  11. Dec 31, 2013 #10

    vela

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    OK, that makes more sense. That combination of spherical harmonics is spherically symmetric, so you should get a uniform distribution as a function of ##\cos \theta##. It seems to me the problem may be that you didn't break up the original wave functions properly into ##R_{nl}(r)##, ##\Theta_{lm}(\theta)##, and ##\Phi_m(\phi)##.
     
  12. Jan 1, 2014 #11
    The only obvious mistake in breaking up the wave function that I can see is where the normalization constant should go.
    But regardless of how I do this, I attain a function of θ, rather than 1/2.
     
  13. Jan 1, 2014 #12

    vela

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    Well, we can't tell what you're doing wrong if you don't show us your work.
     
  14. Jan 2, 2014 #13
    D([itex]\mu[/itex])d[itex]\mu[/itex]=[itex]^{∞}_{0}[/itex][itex]\int[/itex][itex]^{2\pi}_{0}[/itex][itex]\int[/itex]([itex]\psi[/itex][itex]_{210}[/itex]+[itex]\psi[/itex][itex]_{211}[/itex]+[itex]\psi[/itex][itex]_{21-1}[/itex])*([itex]\psi[/itex][itex]_{210}[/itex]+[itex]\psi[/itex][itex]_{211}[/itex]+[itex]\psi[/itex][itex]_{21-1}[/itex]) r[itex]^{2}[/itex]drd[itex]\mu[/itex]d[itex]\phi[/itex]

    => D([itex]\mu[/itex])= ([itex]\psi[/itex][itex]_{210}[/itex]+[itex]\psi[/itex][itex]_{211}[/itex]+[itex]\psi[/itex][itex]_{21-1}[/itex])*([itex]\psi[/itex][itex]_{210}[/itex]+[itex]\psi[/itex][itex]_{211}[/itex]+[itex]\psi[/itex][itex]_{21-1}[/itex]) r[itex]^{2}[/itex]drd[itex]\phi[/itex]

    Integrate over [itex]\phi[/itex]:

    [itex]^{2\pi}_{0}[/itex][itex]\int[/itex]([itex]\psi[/itex][itex]_{210}[/itex]+[itex]\psi[/itex][itex]_{211}[/itex]+[itex]\psi[/itex][itex]_{21-1}[/itex])*([itex]\psi[/itex][itex]_{210}[/itex]+[itex]\psi[/itex][itex]_{211}[/itex]+[itex]\psi[/itex][itex]_{21-1}[/itex]) r[itex]^{2}[/itex]d[itex]\phi[/itex] = |R[itex]_{21}[/itex][itex]\Theta[/itex][itex]_{10}[/itex]|[itex]^{2}[/itex] + |R[itex]_{21}[/itex][itex]\Theta[/itex][itex]_{11}[/itex]|[itex]^{2}[/itex] + |R[itex]_{21}[/itex][itex]\Theta[/itex][itex]_{1-1}[/itex]|[itex]^{2}[/itex]


    Integrate over r:

    [itex]^{∞}_{0}[/itex][itex]\int[/itex] = |R[itex]_{21}[/itex][itex]\Theta[/itex][itex]_{10}[/itex]|[itex]^{2}[/itex] + |R[itex]_{21}[/itex][itex]\Theta[/itex][itex]_{11}[/itex]|[itex]^{2}[/itex] + |R[itex]_{21}[/itex][itex]\Theta[/itex][itex]_{1-1}[/itex]|[itex]^{2}[/itex] dr = [itex]^{∞}_{0}[/itex][itex]\int[/itex] |R[itex]_{21}[/itex]|[itex]^{2}[/itex]|[itex]\Theta[/itex][itex]_{10}[/itex]|[itex]^{2}[/itex] + |R[itex]_{21}[/itex]|[itex]^{2}[/itex]|[itex]\Theta[/itex][itex]_{11}[/itex]|[itex]^{2}[/itex] + |R[itex]_{21}[/itex]|[itex]^{2}[/itex]|[itex]\Theta[/itex][itex]_{1-1}[/itex]|[itex]^{2}[/itex] dr

    = [itex]^{∞}_{0}[/itex][itex]\int[/itex] |[itex]\Theta[/itex][itex]_{10}[/itex]|[itex]^{2}[/itex]+|[itex]\Theta[/itex][itex]_{11}[/itex]|[itex]^{2}[/itex]+|[itex]\Theta[/itex][itex]_{1-1}[/itex]|[itex]^{2}[/itex]

    (by normalisation of Rnl's)

    = cos[itex]^{2}[/itex][itex]\theta[/itex] + sin[itex]^{2}[/itex][itex]\theta[/itex] + sin[itex]^{2}[/itex][itex]\theta[/itex]
     
  15. Jan 2, 2014 #14

    vela

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    You're not getting the normalization of the ##\Theta_{lm}##'s correct.

    Take ##\psi_{210}##. The constant at the front of
    $$\psi_{210} = \sqrt{\frac{1}{8\pi a_0^3}}\left(\frac{r}{2a_0}\right)e^{-r/2a_0}\cos\theta.$$ is a combination of the normalization constants for ##R_{21}##, ##\Theta_{10}##, and ##\Phi_{0}##. You need to separate them out correctly.
     
  16. Jan 5, 2014 #15
    Thanks for your help, I can see were this is going.
     
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