Given operator, show the Hamiltonian

[spacetimedude]

Homework Statement

Given $$\hat{P}_r\psi=-i\hbar\frac{1}{r}\frac{\partial}{\partial{r}}(r\psi)$$, show $$\hat{H}=\frac{1}{2m}(\hat{P}^2_r+\frac{\hat{L}^2}{r^2})$$

Homework Equations

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

 

[PeroK]

Homework Statement

Given $$\hat{P}_r\psi=-i\hbar\frac{1}{r}\frac{\partial}{\partial{r}}(r\psi)$$, show $$\hat{H}=\frac{1}{2m}(\hat{P}^2_r+\frac{\hat{L}^2}{r^2})$$

Homework Equations

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

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

 

[spacetimedude]

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

$\hat{P}_r\hat{P}_r\psi=-\hbar\frac{\partial}{\partial{r}}\frac{\partial}{\partial{r}}\psi$. 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

 

[PeroK]

$\hat{P}_r\hat{P}_r\psi=-\hbar\frac{\partial}{\partial{r}}\frac{\partial}{\partial{r}}\psi$. 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

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!

 

[spacetimedude]

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!

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?

 

[PeroK]

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 get what I had in the previous comment?

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

 

[spacetimedude]

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

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

 

[PeroK]

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

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.

Your question was "why?"

 

[spacetimedude]

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

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

 

[PeroK]

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

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.

 

[spacetimedude]

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.

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

 

[PeroK]

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

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

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

 

[spacetimedude]

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

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

Ah got it! Thank you very much :)