How to factorize the hydrogen atom Hamiltonian?

[yamata1]

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.

 

[VKint]

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.

 

[yamata1]

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$$-\frac{e^2}{r}=-\frac{\beta\hbar}{r} $$ ?

 

[VKint]

You're forgetting that the sum I calculated is multiplied by $1 / (2m)$ in the Hamiltonian. Otherwise, yes.

 

[yamata1]

You're forgetting that the sum I calculated is multiplied by $1 / (2m)$ in the Hamiltonian. Otherwise, yes.

Thank you very much.

 

[yamata1]

I think there is a mistake here

the sum I started with is just$$p^2 + (\beta^2 + 3 \beta \hbar) - \frac{\beta \hbar}{r}$$

$$\sum_{k = 1}^3 \beta i [ p_k, \frac{x_k}{r}]=\beta i(\frac {-3i\hbar}{r}+\frac {i\hbar r^2}{r^3})$$
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.