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Homework Help: Calculating Potential of a groundstate Hydrogen Atom

  1. May 1, 2009 #1


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    1. The problem statement, all variables and given/known data

    The groundstate energy of the hydrogen atom can be calculated using the variational
    principle. The normalised groundstate wavefunction is:

    [tex] \psi_{100} = R(r)_{10} \cdot Y_{00} [/tex]

    with [tex] R_{10} = 2Ae^{-3/2}e^(r/a) [/tex] and [tex] Y_{00} = \frac{1}{\sqrt{4z\pi}} [/tex]

    A is the so called variational parameter.


    Calculate the potential energy of the groundstate as a function of A.


    Calculate the kinetic energy of the groundstate as a function of A.

    2. Relevant equations

    Schrödinger Equation:

    [tex] -\frac{\hbar^2}{2m_e}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial R}{\partial r}\right) +\frac{\hbar^2l(l+1)}{2m_er^2}R - \frac{Zr^2}{(4\pi\epsilon_0)r}R = ER [/tex]

    3. The attempt at a solution

    I am not quite sure, but would I be ruight in saying that if a solve the LHS, and then divide through by R, this would give:

    [tex] \frac{1}{R}\left( -\frac{\hbar^2}{2m_e}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial R}{\partial r}\right) +\frac{\hbar^2l(l+1)}{2m_er^2}R - \frac{Zr^2}{(4\pi\epsilon_0)r}R\right) = E [/tex]

    Would this be the potential energy, or is it the total energy?

    Many thanks,

  2. jcsd
  3. May 1, 2009 #2


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    Homework Helper

    The hydrogen nucleus is to be treated as a point charge. What, then, is potential seen by an electron due to this point charge is the electron is at position [itex]\bold r[/itex]?
  4. May 1, 2009 #3
    I'm sorry to say you've very messed up on this problem. You definitely cannot divide the R out the way you've shown in your attempted solution. What they're doing here is giving you a possible solution of the general form A*exp(-r/a) where A is chosen to properly normalize the function and a (little a) gives you the spatial confinement of the wave function. The idea of the problem is that instead of actually solving the Schroedinger equation from scratch, if you assume a solution of the given form (which happens to be the correct form) you can actually get the exact value for the decay rate by just minimizing the energy. To help you out they give you the Schroedinger equation in polar coordinates, and to confuse you they throw in a term for the angular energy (?) which is zero for this case anyways.

    They want you to apply the given calculus to the target function "R" and see if you can minimuze it by playing with the value of "a". It's tricky because you've got to keep the normalization correct. Beyond that, I think you have to understand about multiplying bra-ket states because if I'm not mistaken you have to operate on R with all that calculus and then multiply again by R, which only makes sense if you know about bras and kets.
  5. May 2, 2009 #4


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    I see, so the Schrödinger equation is:

    [tex] -\frac{\hbar^2}{2m_e}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial R}{\partial r}\right) +\frac{\hbar^2l(l+1)}{2m_er^2}R - \frac{Zr^2}{(4\pi\epsilon_0)r}R = ER [/tex]

    And we know that:

    [tex] R_{10} = 2Ae^{-3/2}e^{(r/a)} [/tex]

    So should I first insert this value of R into SE, like so:

    [tex] -\frac{\hbar^2}{2m_e}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial 2Ae^{-3/2}e^{(r/a)}}{\partial r}\right) +\frac{\hbar^2l(l+1)}{2m_er^2}2Ae^{-3/2}e^{(r/a)} - \frac{Zr^2}{(4\pi\epsilon_0)r}2Ae^{-3/2}e^{(r/a)} = E2Ae^{-3/2}e^{(r/a)} [/tex]

    and then solve the differentiation on the very first term?
  6. May 2, 2009 #5
    Yes, that's how it starts for sure. I'm also pretty sure the component in (l)(l+1) is just a red herring that you can throw away.

    Once you've done the differentiation I'm not sure what comes next. The normalization factor which they seem to give you should help but I don't understand why it's in terms of z and not a. I also don't really get the relation between big A and little a in your trial function...I would really want big A defined as a function of little a so the overall integral of R*R comes out to unity as you vary a. But I don't see how they give you this in the setting up of the problem.
  7. May 2, 2009 #6


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    okay, so with the first differential, we can lump together the constahnts, like so:

    [tex]Be^{r/a} [/tex]


    [tex] B = 2Ae^{-3/2} [/tex]

    If we differentiate this it gives:

    [tex] \frac{B}{a}e^{r/a} [/tex]

    so we now multiply this by r^2, to give:

    [tex] r^2\frac{B}{a}e^{r/a} [/tex]

    which we now differentiate again, using the product rule, to give:

    [tex] r^2 \frac{B}{a^2}e^{r/a} + 2r\frac{B}{a}e^{r/a} [/tex]

    so thus we now have:

    [tex] -\frac{\hbar^2}{2m_e}\frac{1}{r^2}\left(r^2 \frac{B}{a^2}e^{r/a} + 2r\frac{B}{a}e^{r/a}\right) +\frac{\hbar^2l(l+1)}{2m_er^2}2Ae^{-3/2}e^{(r/a)} - \frac{Zr^2}{(4\pi\epsilon_0)r}2Ae^{-3/2}e^{(r/a)} = E2Ae^{-3/2}e^{(r/a)} [/tex]

    which we can cancel down slightly like so:

    [tex] -\frac{\hbar^2}{2m_e}\left(\frac{B}{a^2}e^{r/a} + \frac{2}{r}\frac{B}{a}e^{r/a}\right) +\frac{\hbar^2l(l+1)}{2m_er^2}2Ae^{-3/2}e^{(r/a)} - \frac{Zr^2}{(4\pi\epsilon_0)r}2Ae^{-3/2}e^{(r/a)} = E2Ae^{-3/2}e^{(r/a)} [/tex]

    Also, z isn't anything to do with the normalization, it is the number of protons (atomic number)

    So what would come next to find the potential?
    Last edited: May 2, 2009
  8. May 2, 2009 #7
    OK, that helps about z. So we don't need to worry about it. But I thought some more about the normalization and I think I figured out how it is supposed to work: you can think of "little a" as a confinement parameter that tells you the approximate radius of confinement for the electron. If you make "a" twice as big, the electron is spread over 8 times the volume. So to normalize the value of R^2, you need to divide R by a factor of root 8, or in other words 1/a^(3/2). I'm pretty sure that's what is meant where you show e to this same exponent...that must be a mistake.

    So back to the problem: you are basically taking R to be of the form


    which gives you one single parameter to play with. Applying the Hamiltonian with your del-squared operator in cylindrical coordinates you get:

    a^(-3/2)*exp(-r/a)[ -1/a^2 + 2/ar - 1/r ] = E*a^(-3/2)*exp(-r/a)

    (I've thrown out the stuff in (l)(l+1) because there is no angular dependence).
    Now there are two directions you can go. The traditional method for solving the Schroedinger equation has you looking for the RHS to be a multiple of the LHS. And the only way to get that is for a=2 which gets rid of everything in 1/r. And that's basically the answer.

    But it seems they're asking you to do it a different way. Ignore the RHS and just focus on the LHS as the total energy. Well, it's not quite the total energy yet: first you have to multiply by R-conjugate (which is just R in this case) and integrate over all space, which is in other words the integrand times r-squared-dr*4pi. It's not a TERRIBLY hard integral...it's going to be a-cubed times a bunch of stuff; there are some additional normalization factors but they won't really make a difference. Having done the integral you are then supposed to look at how it depends on the value of little-a, and MINIMIZE it. I haven't worked it through yet but that's how it should go.
  9. May 4, 2009 #8


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    But when you say it will give the total energy, will this be the total potential energy, because the first part asks for total potential energy, and then the second asks for total kinetic energy.
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