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Hydrogen atom eigenstate: proving orthogonality of states

  1. Nov 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Hey dudes

    So here's the question:
    Consider the first excited Hydrogen atom eigenstate eigenstate [itex]\psi_{2,1,1}=R_{2,1}(r)Y_{11}(\theta, \phi)[/itex] with [itex]Y_{11}≈e^{i\phi}sin(\theta)[/itex]. You may assume that [itex]Y_{11}[/itex] is correctly normalized.

    (a)Show that [itex]\psi_{2,1,1}[/itex] is orthogonal to the eigenstates [itex]\psi_{2,1,0}=R_{2,1}(r)Y_{1,0}(\theta,\phi)[/itex] and [itex]\psi_{2,1,-1}=R_{2,1}(r)Y_{1,-1}(\theta,\phi)[/itex] with [itex]Y_{1,0}≈cos(\theta)[/itex] and [itex]Y_{1,-1}≈e^{-i\phi}sin(\theta)[/itex].

    2. Relevant equations

    I dont think there is any...

    3. The attempt at a solution

    I'm completely dumbfounded here. So i have no idea....i know that orthogonality can be tested by applying the same operator on two eigenstates..for example, if we have two states [itex]\psi_{i}, \psi_{j}[/itex] that correspond to two different eigenvalues [itex]a_{i}, a_{j}[/itex] of an operator [itex]A[/itex], then [itex]A\psi_{i}=a_{i}\psi_{i}[/itex] and [itex]A\psi_{j}=a_{j}\psi_{j}[/itex]...so then [itex](a_{i}-a_{j})<\psi_{i}|\psi_{j}>=0[/itex]...right? but I dont know how to apply that here...
     
  2. jcsd
  3. Nov 27, 2013 #2

    vela

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    In that proof, what was the point of showing ##(a_i - a_j)\langle \psi_i \vert \psi_j \rangle = 0##?
     
  4. Nov 27, 2013 #3
    Okay so I think ive solved that part. Basically you have to do the integral of [itex]\psi_{2,1,1}^{*}\psi_{2,1,0}[/itex] and you should get 0, which is what I get for both.

    But in the next question, i am required to find the normalization constant N, expectation value of r and uncertainty of r, given [itex]R_{2,1}(r)=Nre^{\frac{-r}{2a}}[/itex]

    I just need to make sure I'm doing it right. So:

    - For the normalization constant: do [itex]\int_{0}^{∞}r^{2}(Nre^{\frac{-r}{2a}})^{2}=1[/itex]

    - for the expectation value [itex]<r>[/itex], i do [itex]\int_{0}^{∞}r^{2}(Nre^{\frac{-r}{2a}})r(Nre^{\frac{-r}{2a}})[/itex]

    - for the expectation value [itex]<r^{2}>[/itex] i do [itex]\int_{0}^{∞}r^{2}(Nre^{\frac{-r}{2a}})r^{2}(Nre^{\frac{-r}{2a}})[/itex]

    Is that correct, or am I putting in the [itex]r^{2}[/itex] for no reason? i think its meant to be there...i just get crazy powers in the integral which makes it look a bit weird...
     
  5. Nov 27, 2013 #4
    Btw doing what I said above gives me [itex]N=\frac{1}{\sqrt{24a^{5}}}[/itex], and [itex]\Delta r = \sqrt{5}a[/itex]...does that look right? (I used [itex]\int_{0}^{\infty}r^{n}e^{-r/c}=n!c^{n+1}[/itex]).
     
  6. Nov 27, 2013 #5

    vela

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    That's right. You're calculating
    $$\langle r^n \rangle = \int \psi^*(\vec{r})r^n\psi(\vec{r})\,d^3\vec{r}.$$ In spherical coordinates, this becomes
    $$\langle r^n \rangle = \int_0^\infty \int_0^\pi \int_0^{2\pi} \psi^*(\vec{r})r^n\psi(\vec{r})\,r^2\sin\theta\,d\phi\,d\theta\,dr.$$ The factor of ##r^2## is part of the volume element in spherical coordinates. Now since ##\psi_{nlm} = NR_{nl}(r)Y_{lm}(\theta,\phi)##, you get
    $$\langle r^n \rangle = \int_0^\infty [NR_{nl}(r)]^* r^n [NR_{nl}(r)]\,r^2\,dr \int_0^\pi \int_0^{2\pi} Y_{lm}^*(\theta,\phi) Y_{lm}(\theta,\phi)\,\sin\theta\,d\phi\,d\theta.$$ The angular integrals evaluate to 1 because of the way ##Y_{lm}## is normalized, which leaves you with the integrals you had.
     
  7. Nov 27, 2013 #6

    vela

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    Yes, that's what I got.
     
  8. Nov 27, 2013 #7
    Thanks Vela...now just one more thing....there's a so-called "Bonus" question, which asks me to evaluate the expectation value of [itex]V(r)=-\frac{e^{2}}{4\pi \epsilon_{0}r}[/itex]. I think I know how to do this, you basically just do [itex]\int_{0}^{\infty}r^{2}R_{2,1}V(r)R_{2,1}[/itex], right?

    And lastly, the final question asks me to find the expectation value of kinetic energy.....how do I do that one? do I just use [itex]T=\frac{1}{2}mv^{2}[/itex]? or is it relativistic...?
     
  9. Nov 27, 2013 #8

    vela

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    Right.

    The wave functions you're working with are the solutions to the Schrodinger equation, which is non-relativistic. There are two ways to find the expectation value of kinetic energy that come to mind. You can express the kinetic energy in terms of the momentum operator and find the expectation value of the resulting operator, or you can deduce the result using the expectation value of V and the fact that you're working with an energy eigenstate.
     
  10. Nov 27, 2013 #9
    So should i use [itex]T=1/2 \hat{\vec{p}}v[/itex]? Im not sure how to do that...because I would need the momentum operator in spherical coordinates right? could you explain more about the second option? because I know that [itex]E_{1}=V(r)+T[/itex]...so can i use that maybe?
     
  11. Nov 27, 2013 #10

    vela

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    The kinetic energy is given by ##\hat{T} = \frac{\hat{p}^2}{2m}##. In the coordinate basis, this becomes
    $$\hat{T} = \frac{1}{2m}\left(\frac{\hbar}i \nabla\right)^2 = -\frac{\hbar^2}{2m}\nabla^2.$$ You should recognize that as the first term in the Schrodinger equation.

    So you'd need to get the expression for ##\nabla^2## in spherical coordinates, which you can easily google or find in a textbook.

    Yes, that's what you want to use, though I think you want ##E_2##, right? Take the expectation value of both sides.
     
  12. Nov 27, 2013 #11
    I found [itex]<T>=-\frac{E_{1}}{12}[/itex]...does that look good?
     
  13. Nov 27, 2013 #12

    vela

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    What does E1 represent?
     
  14. Nov 28, 2013 #13
    [itex]E_{1}=-\frac{m_{e}e^{4}}{32\pi^{2}\epsilon_{0}^{2}\hbar^{2}}[/itex]...and I also got [itex]<V(r)>=\frac{E_{1}}{2}[/itex]
     
    Last edited: Nov 28, 2013
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