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I How to factorize the hydrogen atom Hamiltonian?

  1. Dec 8, 2017 #1
    Hello,
    The hydrogen atom Hamiltonian is
    $$H=\frac{p^2}{2m} -\frac{e^2}{r}\tag{1}$$
    with e the elementary charge,m the mass of the electron,r the radius from the nucleus and p,the momentum. Apparently we can factorize H $$H=\gamma +\frac{1}{2m}\sum_{k=1}^{3}\left(\hat p_k+i\beta\frac{\hat x_k}{r}\right)\left(\hat p_k-i\beta\frac{\hat x_k}{r}\right)\tag{2}$$
    for suitable constants β and γ that you can calculate. I assume the operator identity:
    $$\hat{A}^2+\hat{B}^2=(\hat{A}-i\hat{B})(\hat{A}+i\hat{B})-i[\hat{A},\hat{B}]$$was used.
    Can someone explain to me how we can start with formula (1) and make the position operator appear in (2)?
    Here is the source https://ocw.mit.edu/courses/physics...all-2013/lecture-notes/MIT8_05F13_Chap_09.pdf pages 33-34
    Thank you.
     
  2. jcsd
  3. Dec 8, 2017 #2
    The easiest way to verify the given factorization is to work backwards, starting with the right-hand side of the given identity and using the operator identity you gave:
    $$ \begin{eqnarray*}
    \sum_{k = 1}^3 \left( p_k + i \beta \frac{x_k}{r} \right) \left( p_k - i \beta \frac{x_k}{r} \right) &= \sum_{k = 1}^3 \left( p_k^2 + \beta^2 \frac{x_k^2}{r^2} + i \left[ p_k, \frac{\beta}{r} x_k \right] \right) \\
    &= p^2 + \beta^2 + \sum_{k = 1}^3 \beta i \left[ p_k, \frac{x_k}{r} \right]
    \end{eqnarray*}$$
    The only things left to deal with are the commutators ## \left[ p_k, \frac{x_k}{r} \right] ##. I'm sure there's a clever way of calculating this basis-free, but it's easy enough to grind it out in the position basis:
    $$ \begin{eqnarray*}
    \left[ p_k, \frac{x_k}{r} \right] \psi &= -i \hbar \frac{\partial}{\partial x_k} \left( \frac{x_k}{r} \psi \right) + i \hbar \left( \frac{x_k}{r} \right) \frac{\partial}{\partial x_k} \psi \\
    &= -i \hbar \left( \frac{1}{r} - \frac{x_k^2}{r^3} \right) \psi
    \end{eqnarray*}$$
    so we have ## \left[ p_k, \frac{x_k}{r} \right] = -i\hbar / r + i \hbar x_k^2 / r^3 ##. In particular, since ##\Sigma_{k = 1}^3 x_k^2 = r^2##, the sum I started with is just
    $$ p^2 + (\beta^2 + 3 \beta \hbar) - \frac{\beta \hbar}{r} $$
    Plugging this into the expression for the Hamiltonian allows one to easily solve for ##\beta## and ##\gamma##.

    As for how you'd actually come up with this factorization if you didn't already know it, all I can say is that after fiddling around with enough problems like this, the expression you gave seems like a plausible guess. Once you guess the correct factored form, it's a simple matter to work backwards as above to figure out what the constants must be.
     
  4. Dec 8, 2017 #3
    Thank you for this answer. I made a mistake in calculating the commutator so I didn't have the term $$-\frac{\beta\hbar}{r}$$. If I understand correctly,I am left with:
    $$-\frac{e^2}{r}=\gamma +\beta^2 +3\beta\hbar -\frac{\beta\hbar}{r}$$
    and since β and γ are constants: Do we have$$-\frac{e^2}{r}=-\frac{\beta\hbar}{r} $$ ?
     
  5. Dec 8, 2017 #4
    You're forgetting that the sum I calculated is multiplied by ##1 / (2m)## in the Hamiltonian. Otherwise, yes.
     
  6. Dec 8, 2017 #5
    Thank you very much.
     
  7. Dec 9, 2017 #6
    I think there is a mistake here
    $$\sum_{k = 1}^3 \beta i [ p_k, \frac{x_k}{r}]=\beta i(\frac {-3i\hbar}{r}+\frac {i\hbar r^2}{r^3})$$
    So $$
    \sum_{k = 1}^3 \left( p_k + i \beta \frac{x_k}{r} \right) \left( p_k - i \beta \frac{x_k}{r} \right)
    =p^2+\beta^2+\frac {2\hbar \beta}{r}$$
    but after seeing your question https://www.physicsforums.com/threads/commutator-of-p-and-x-r.933864/ the answer given makes it much simpler.
     
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