Given operator, show the Hamiltonian

1. Feb 25, 2017

spacetimedude

1. The problem statement, all variables and given/known data
Given $$\hat{P}_r\psi=-i\hbar\frac{1}{r}\frac{\partial}{\partial{r}}(r\psi)$$, show $$\hat{H}=\frac{1}{2m}(\hat{P}^2_r+\frac{\hat{L}^2}{r^2})$$

2. Relevant equations

3. The attempt at a solution
The solution starts out with $$\hat{P}^2\psi=-\hbar^2\frac{1}{r}\frac{\partial}{\partial{r}}\frac{\partial}{\partial{r}}(r\psi)$$ but I am not sure why it's 1/r isntead of 1/r^2 when I square P?

Thank you

2. Feb 25, 2017

PeroK

Write $\hat{P_r}^2 \psi = \hat{P_r} \hat{P_r}\psi$ out in full.

3. Feb 25, 2017

spacetimedude

$\hat{P}_r\hat{P}_r\psi=-\hbar\frac{\partial}{\partial{r}}\frac{\partial}{\partial{r}}\psi$. Then do we just multiply in r/r into the right hand side?
EDIT: Probably not because can't just multiply something in the derivative

4. Feb 25, 2017

PeroK

Yes, was just going to say. You need to focus on how $P$ is defined (literally) and calculate $P^2$. No short cuts, no presumptions.

For example, where did the $i$'s go? You presumed those away (with no ill effects), but you wrongly presumed other things away!

5. Feb 25, 2017

spacetimedude

Is it right to assume, in the question, $\hat{P}_r\psi=-i\hbar\frac{\partial}{\partial{r}}\psi$? Then if we disregard the $\psi$, then it's just multiplying two momentum operators and $(-i\hbar)(-i\hbar)\frac{\partial}{\partial{r}}\frac{\partial}{\partial{r}}$ and put the $\psi$ to both sides and get what I had in the previous comment?

6. Feb 25, 2017

PeroK

Given $$\hat{P}_r\psi=-i\hbar\frac{1}{r}\frac{\partial}{\partial{r}}(r\psi)$$

7. Feb 25, 2017

spacetimedude

Hmm, what would $\hat{P}_r$ be in our case?

8. Feb 25, 2017

PeroK

You're given the $\hat{P_r}$ above. What is the square of that operator?

In fact, you are given $\hat{P_r}^2$ as well.

9. Feb 25, 2017

spacetimedude

If I square the right side, wouldn't it square $\psi$ as well? I have $$\hat{P}^2_r\psi=-\hbar^2\frac{1}{r^2}(\frac{\partial}{\partial{r}}(r\psi))^2$$

10. Feb 25, 2017

PeroK

Look, to get there you have done about 10 algebraic steps in your head. Some correctly, some wrongly. You need to focus on what you have, write it out in full, no presumptions, one step at a time and see what you get.

11. Feb 25, 2017

spacetimedude

$$\hat{P}_r\psi=-i\hbar\frac{1}{r}\frac{\partial}{\partial{r}}(r\psi)$$
$$\hat{P}_r^2\psi=(-i\hbar)^2\frac{1}{r}\frac{\partial}{\partial{r}}(r\psi)\frac{1}{r}\frac{\partial}{\partial{r}}(r\psi)$$
$$\hat{P}^2_r\psi=-\hbar^2(\frac{1}{r}\frac{\partial}{\partial{r}}(r\psi))^2$$
$$\hat{P}^2_r\psi=-\hbar^2\frac{1}{r^2}(\frac{\partial}{\partial{r}}(r\psi))^2$$
This is what I have on my paper. I don't think I am assuming anything?

12. Feb 25, 2017

PeroK

Okay, your mistake was more fundamental. The product of operators is composition:

$\hat{P}_r^2\psi = \hat{P}_r(\hat{P}_r \psi) = -i \hbar \frac{1}{r}\frac{\partial}{\partial{r}} (r \hat{P}_r \psi) = -i \hbar \frac{1}{r}\frac{\partial}{\partial{r}} (r[-i \hbar \frac{1}{r}\frac{\partial}{\partial{r}} (r \psi)])$

13. Feb 25, 2017

spacetimedude

Ah got it! Thank you very much :)