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Quantum: energy equation

  1. Feb 28, 2014 #1
    1. The problem statement, all variables and given/known data

    2yosmmf.png

    Part (a): By writing L2 in einstein notation, show that p2 can be written as:

    Part(b): Show ##\vec{p}.\hat {\vec{r}} - \hat {\vec{r}}.\vec{p} = -2i\hbar\frac{1}{r}##

    Part(c): Show ##\vec{r}.\vec{p} = rp_r + i\hbar##

    Part (d): Show ##p^2 = p_r^2 + \frac{L}{r^2}##

    I missed out terms in part (a), I couldn't get parts (b) and (c) of this question.

    2. Relevant equations



    3. The attempt at a solution

    Part (a)

    [tex]L^2 = \epsilon_{ijk}x_j p_k \epsilon_{ilm} x_l p_m[/tex]
    [tex] = \epsilon_{ijk}\epsilon{ilm} x_j p_k x_l p_m[/tex]
    [tex] = \left(\delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}\right)x_jp_kx_lp_m[/tex]

    [tex] = x_jp_kx_jp_k - x_jp_kx_kp_j [/tex]
    [tex] = x_j^2p_k^2 - x_jp_jp_kx_k[/tex]
    [tex] = x_j^2p_k^2 - x_jp_j[/tex]
    [tex] = r^2p^2 - (\vec{r} . \vec {p})[/tex]


    Part(b)

    [tex]\vec{p}.\hat {\vec{r}} - \hat {\vec{r}}.\vec{p} = -i\hbar\left(\nabla . (\frac{\vec{r}}{r})\right) + i\hbar\left(\frac{\vec{r}}{r}.\nabla\right)[/tex]

    Now using product rule:

    [tex] = -i\hbar\left[ \vec{r}.(\nabla \frac{1}{r}) + \frac{1}{r}\nabla . \vec{r}\right] + i\hbar\left[\frac{1}{r} \vec{r}.\nabla\right][/tex]

    Now, ##\nabla \frac{1}{r} = -\frac{1}{r^3}\vec{r}## and ##\nabla . \vec{r} = 3##. Using these results, we obtain:

    [tex] = \frac{i\hbar}{r}[/tex]

    Part (c)

    Using result of part (b):

    [tex]\vec{r}.\vec{p} = 2i\hbar + r(\vec{p}.\hat{\vec{r}}) [/tex]
    [tex] = 2i\hbar + \vec{p}.\vec{r}[/tex]
    [tex] = 2i\hbar -i\hbar\nabla . \vec{r}[/tex]
    [tex] = 2i\hbar - 3i\hbar[/tex]
    [tex] = -i\hbar[/tex]

    Part(d)

    Chuck parts (b) and (c) into the equation and you get the answer.

    The physical interpretation is:

    Total Energy = Radial Kinetic Energy + Rotational Energy
     
    Last edited: Feb 28, 2014
  2. jcsd
  3. Feb 28, 2014 #2
    For part (a), I think you made a mistake on the 5th line. You aren't allowed to just swap the order of x's and p's.
     
  4. Mar 1, 2014 #3

    ChrisVer

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    Gold Member

    for part b, try acting with the difference on a function....and avoid playing algebraically just with the operators alone...
    For example after using the chain rule, you would cancel out terms with having a function being acted on...
     
  5. Mar 1, 2014 #4
    OK here's what I got:

    [tex]x_jp_kx_jp_k - x_jp_kx_kp_j[/tex]

    Now i change ##p_k## in first and second term to ##-i\hbar\partial_k##:

    [tex]-i\hbar x_j\partial_k(x_jp_k) +i\hbar x_j\partial_k(x_kp_j)[/tex]

    Using product rule: ##\partial_kx_j = \delta_{kj}##

    [tex]i\hbar \left[-x_jp_k\delta_{kj} - x_j^2\partial_kp_k + x_jp_j + x_jx_k\partial_kp_j\right][/tex]

    Converting ##\partial_k = \frac{i}{\hbar}p_k##:

    [tex]i\hbar \left[ -x_jp_j - \frac{i}{\hbar} x_j^2p_k^2 + x_jp_j +\frac{i}{\hbar}x_jx_kp_kp_j\right][/tex]
     
    Last edited: Mar 1, 2014
  6. Mar 1, 2014 #5

    ChrisVer

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    Gold Member

    you should use the commutation relations for exchanging the x and p's....So in general:
    [itex] p_{i} x_{j}= -i \hbar δ_{ij} + x_{j} p_{i} [/itex]

    it's still not trivial to write the momentum as the derivative and making it act on x alone... you are still having operators and as such they act on functions .... I'll repeat, avoid treating them as functions alone and work algebraically so "free".

    For example take the commutator of p and x:
    [itex] [p_{i},x_{j}] = p_{i}x_{j} - x_{j} p_{i} = -i\hbar δ_{ij} + x_{j} i\hbar \frac{d}{dx_{i}}[/itex]
    is not correct

    Instead you take:

    [itex] [p_{i},x_{j}] f= p_{i}x_{j}f - x_{j} p_{i}f= -i\hbar \frac{d}{dx_{i}}(x_{j}f)+ i\hbar x_{j} \frac{d}{dx_{i}}f = -i\hbar δ_{ij} f[/itex]
    so you take
    [itex] [p_{i},x_{j}]= p_{i}x_{j} - x_{j} p_{i} =-i\hbar δ_{ij}[/itex]
     
    Last edited: Mar 1, 2014
  7. Mar 1, 2014 #6
    That's exactly the relation I was looking for. thanks alot
     
  8. Mar 1, 2014 #7

    ChrisVer

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    Gold Member

    :smile: it's basic (before the central potential you are trying to work with)
     
  9. Mar 1, 2014 #8

    We need to show this: ##x_j^2p_k^2 - (x_jp_j)(x_kp_k) +i\hbar(x_jp_j)##

    Starting:

    [tex]x_jp_kx_jp_k - x_jp_kx_kp_j[/tex]

    Using ##[x_i,p_j] = x_ip_j - p_jx_i = i\hbar \delta_{ij}##:

    [tex]x_j(x_jp_k - i\hbar \delta_{jk})p_k - x_j(x_kp_k - i\hbar)p_j[/tex]
    [tex] x_j^2p_k^2 - i\hbar x_jp_j - x_j(x_kp_k)p_j + i\hbar x_jp_j[/tex]

    Using ##[x_kp_k,p_j] = x_kp_kp_j - p_jx_kp_k = i\hbar \delta_{jk}p_k##

    [tex] = x_j^2p_k^2 - i\hbar x_jp_j - x_j(i\hbar\delta_{jk}p_k + p_jx_kp_k) + i\hbar x_jp_j[/tex]

    [tex] = x_j^2p_k^2 - i\hbar x_jp_j - x_jp_jx_kp_k[/tex]
    [tex] = r^2p^2 - i\hbar (\vec{r}.\vec{p}) - (\vec{r}.\vec{p})(\vec{r}.\vec{p})[/tex]

    which is different from the answer..
     
    Last edited: Mar 1, 2014
  10. Mar 2, 2014 #9
  11. Mar 11, 2014 #10
  12. Mar 18, 2014 #11
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