How Does Quantum Mechanics Describe Particle Dynamics in Spherical Coordinates?

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  • #1
unscientific
1,734
13

Homework Statement



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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.

Homework Equations





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
 
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  • #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.
 
  • #3
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...
 
  • #4
MisterX said:
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.

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]
 
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  • #5
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]
 
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  • #6
ChrisVer said:
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]
(I may forget a - in front of [itex]i \hbar[/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"

That's exactly the relation I was looking for. thanks alot
 
  • #7
unscientific said:
That's exactly the relation I was looking for. thanks alot

:smile: it's basic (before the central potential you are trying to work with)
 
  • #8
ChrisVer said:
:smile: it's basic (before the central potential you are trying to work with)
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..
 
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  • #9
bumpp
 
  • #10
bumppp
 
  • #11
bumpp
 

What is the energy equation in quantum mechanics?

The energy equation in quantum mechanics is known as the Schrödinger equation, which describes the time evolution of a quantum system. It is written as HΨ = EΨ, where H represents the Hamiltonian operator, Ψ represents the wave function, and E represents the energy of the system.

How is the energy equation derived in quantum mechanics?

The energy equation is derived using the principles of wave-particle duality and the uncertainty principle. It combines the classical energy equation with the de Broglie wavelength equation to account for the particle-like and wave-like behavior of quantum particles.

What is the significance of the energy equation in quantum mechanics?

The energy equation allows us to predict the behavior of quantum systems and calculate their energy levels. It is a fundamental equation in quantum mechanics and is used to understand various phenomena, such as atomic and molecular structures, chemical reactions, and nuclear reactions.

How does the energy equation relate to the Heisenberg uncertainty principle?

The energy equation is related to the Heisenberg uncertainty principle, which states that the more precisely we know the energy of a system, the less precisely we can know its position, and vice versa. The energy equation takes into account the uncertainty in energy levels of a quantum system.

Are there any limitations to the energy equation in quantum mechanics?

The energy equation is limited to non-relativistic systems, meaning it cannot accurately describe particles moving at speeds close to the speed of light. It also does not account for the effects of gravity and is limited to describing systems at the microscopic level.

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