I'm working through Mattuck's "A Guide to Feynman Diagrams in the Many-Body Problem", but I'm stuck on a bit which I feel should be trivial.

In section 3.2 (p 43 in the Dover edition) he gives a Hydrogen atom as an example of a system which can be considered to have a p-dependent potential term.

In his words,

"the Hamiltonian of the center of mass motion of a Hydrogen atom is $H = p^2 / (m + m_e)$ where $m$ = proton mass and $m_e$ = electron mass.

(First question: shouldn't there be a factor 2 in the denominator here?)

This may be broken up into

$$H = \frac{p^2}{2m} - \frac{m_e}{(m_e+m)m}p^2$$

and the second term treated as if it were a perturbating potential."

Second question: I can't for the life of me figure out how he breaks it up like that, whether I include the aforementioned factor 2 or not..

Both multiplying by $\frac{m - m_e}{m - m_e}$ and ignoring terms quadratic in $m_e$, and deriving the first order taylor expansion around $m$ (in both cases inclluding the factor 2 which I believe should be there) give

$$\frac{p^2}{2m} - \frac{p^2m_e}{2m^2}$$

which looks kind of close, but not quite it.

Of course this is not a hugely significant part of the text but it's really bugging me so a bit of help would be greatly appreciated...


You are right; Hamiltonian for center of mass of hydrogen atom should be :


Where $p$ is momentum of Hydrogen atom (please check what is $p$ in your book).

Now you can also write it as :

$H=\displaystyle\frac{p^2}{2m}(\frac {m}{m+m_e})$

$=\displaystyle\frac{p^2}{2m}(\frac {(m+m_e)-m_e}{m+m_e})$

$=\displaystyle\frac{p^2}{2m}(1-\frac {m_e}{m+m_e})$

$=\displaystyle\frac{p^2}{2m}-\frac {m_e}{2m(m+m_e)}p^2$

So there must be some typos in your book.

  • $\begingroup$ ya, now I have removed it :) $\endgroup$ – user10001 Jul 11 '12 at 5:12
  • $\begingroup$ Thanks a lot, yes $p$ is the momentum of the center of mass of the atom... So there's no ignoring of terms even, and the author has forgotten the factor 2 twice. $\endgroup$ – mszep Jul 11 '12 at 8:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.