1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Given operator, show the Hamiltonian

  1. Feb 25, 2017 #1
    1. The problem statement, all variables and given/known data
    Given [tex] \hat{P}_r\psi=-i\hbar\frac{1}{r}\frac{\partial}{\partial{r}}(r\psi)[/tex], show [tex]\hat{H}=\frac{1}{2m}(\hat{P}^2_r+\frac{\hat{L}^2}{r^2})[/tex]

    2. Relevant equations


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

    Thank you
     
  2. jcsd
  3. Feb 25, 2017 #2

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Write ##\hat{P_r}^2 \psi = \hat{P_r} \hat{P_r}\psi## out in full.
     
  4. Feb 25, 2017 #3
    [itex]\hat{P}_r\hat{P}_r\psi=-\hbar\frac{\partial}{\partial{r}}\frac{\partial}{\partial{r}}\psi[/itex]. 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
     
  5. Feb 25, 2017 #4

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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!
     
  6. Feb 25, 2017 #5
    Is it right to assume, in the question, [itex]\hat{P}_r\psi=-i\hbar\frac{\partial}{\partial{r}}\psi[/itex]? Then if we disregard the ##\psi##, then it's just multiplying two momentum operators and [itex](-i\hbar)(-i\hbar)\frac{\partial}{\partial{r}}\frac{\partial}{\partial{r}}[/itex] and put the ##\psi## to both sides and get what I had in the previous comment?
     
  7. Feb 25, 2017 #6

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Given [tex] \hat{P}_r\psi=-i\hbar\frac{1}{r}\frac{\partial}{\partial{r}}(r\psi)[/tex]
     
  8. Feb 25, 2017 #7
    Hmm, what would ##\hat{P}_r## be in our case?
     
  9. Feb 25, 2017 #8

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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?"
     
  10. Feb 25, 2017 #9
    If I square the right side, wouldn't it square ##\psi## as well? I have [tex]\hat{P}^2_r\psi=-\hbar^2\frac{1}{r^2}(\frac{\partial}{\partial{r}}(r\psi))^2[/tex]
     
  11. Feb 25, 2017 #10

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  12. Feb 25, 2017 #11
    [tex]\hat{P}_r\psi=-i\hbar\frac{1}{r}\frac{\partial}{\partial{r}}(r\psi)[/tex]
    [tex]\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)[/tex]
    [tex]\hat{P}^2_r\psi=-\hbar^2(\frac{1}{r}\frac{\partial}{\partial{r}}(r\psi))^2[/tex]
    [tex]\hat{P}^2_r\psi=-\hbar^2\frac{1}{r^2}(\frac{\partial}{\partial{r}}(r\psi))^2[/tex]
    This is what I have on my paper. I don't think I am assuming anything?
     
  13. Feb 25, 2017 #12

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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)])##
     
  14. Feb 25, 2017 #13
    Ah got it! Thank you very much :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Given operator, show the Hamiltonian
  1. Hamiltonian Operator (Replies: 1)

  2. Hamiltonian operator (Replies: 2)

  3. Hamiltonian operator (Replies: 4)

Loading...