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The question is: A particle of mass m is attached to a fixed point in space by a massless rigid rod of length a and can freely rotate about this point. Find the quantum energy levels of the system. What is the degeneracy of each energy level?

I used rotational kinetic energy:

$E=\frac{1}{2}I\omega^2=\frac{L^2}{2I}$

and then substituted $I=ma^2$ and $L=\hbar \sqrt{l(l+1)}$ to get:

$E_l=\frac{\hbar^2l(l+1)}{2ma^2}$.

So the energies are quantized as expected. But what is the degeneracy of each level? Plugging in a bunch of values for $l$ doesn't show any $l$s with similar energy so far. Is it correct that the degeneracy of each level is $0$?

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  • $\begingroup$ How can the mass be attached to a fixed point in space and at the same time rotate freely about this point? $\endgroup$ – descheleschilder Apr 8 at 18:52
  • $\begingroup$ @descheleschilder Do you have an issue with the word "freely" or the word "rotate"? $\endgroup$ – probably_someone Apr 8 at 18:56
  • $\begingroup$ I'm having trouble to discriminate between the fixed point in space and the point about which the rod can freely rotate. Is it meant that a mass fixed on a massless rod rotates freely around the opposite side of the rod (with no mass attached to it) which is fixed in space? $\endgroup$ – descheleschilder Apr 8 at 19:05
  • $\begingroup$ @descheleschilder One end of the rod is fixed to a point in space. The rod can assume any orientation as long as the end is at that particular point. It might be easier to think of the equivalent problem: a particle is constrained to move on a fixed sphere, with no other forces besides the constraint (which is what is meant by "free" in this context). $\endgroup$ – probably_someone Apr 8 at 19:10
  • $\begingroup$ That makes it clear! $\endgroup$ – descheleschilder Apr 8 at 19:12
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For each $l$ there exists $2l+1$ possible values of $m$. Since $m$ must be an integer, and $-l\leq m\leq l$, expanding out the associated Legendre function:

$P_l^m(x)\equiv (-1)^m(1-x^2)^{m/2}(\frac{d}{dx})^mP_l(x)$

where $P_l(x)$ is the $l$th Legendre polynomial in $x$, will show that there is $2l+1$ degeneracies. The solutions to the theta dependence of the angular equation due to separation of variables of the spherical Schrodinger equation are the Legendre polynomials in $\cos(\theta)$.

For example:

$P_0^0=1$, $P_1^1=-\sin(\theta)$, $P_1^0=\cos(\theta)$, etc...

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    $\begingroup$ Basically, the eigenfunctions of the Hamiltonian are the spherical harmonics, and the eigen-energies are parameterized by the total angular momentum quantum number $l$, yielding a degeneracy of $2l+1$ for each level because there are $2l+1$ different possible $z$-projection quantum numbers $m$ for each $l$ ($-l\leq m \leq l$ and the $m$'s are integers). $\endgroup$ – march Apr 8 at 22:57
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As @probably_someone wrote in his comment:

It might be easier to think of the equivalent problem: a particle is constrained to move on a fixed sphere, with no other forces besides the constraint (which is what is meant by "free" in this context).

Now I guess that the non-degenerate energy levels you wrote down are correct if we confine the rotation to one plane. But there is an infinity of planes in which the rod can rotate, which means there is an infinite degeneracy for all energy levels.

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  • $\begingroup$ I don't think this is right. In a hydrogen atom, the electron is allowed to "rotate" about any axis, and we describe this arbitrary-axis rotation using the spherical harmonics (so $l$'s and $m$'s are relevant). In that case, you don't get infinite degeneracy at each $l$-level. Rather, you get the $2l+1$-fold degeneracy as explained in the other answer. $\endgroup$ – march Apr 8 at 21:37
  • $\begingroup$ This isn't an equivalent situation as the hydrogen atom. $\endgroup$ – descheleschilder Apr 8 at 22:16
  • $\begingroup$ Sure, but the particle is still bound: It's fixed to a rigid rod and therefore allowed to move on the surface of a sphere. For this reason, the eigenfunctions of the Hamiltonian are spherical harmonics, and so the levels are parameterized by the angular momentum quantum number $l$, and the degeneracy of each level is given by $2l+1$ (accounting for all allowed $z$-"projections" of the angular momentum onto the $z$-axis). I brought up the hydrogen atom as an analogous problem, but one where there is actually more freedom of movement than this one. $\endgroup$ – march Apr 8 at 22:54
  • $\begingroup$ I understand. But what is the potential energy in this case? Can this situation be compared with the $l=0$ situation in a hydrogen atom? And ain't it so that the faster the rod rotates, the higher the energy is (unlike the energy in a hydrogen atom that keeps the same energy in a state with fixed $l$)? If the mass had an electric charge what would happen if it was hit by a photon? $\endgroup$ – descheleschilder Apr 9 at 10:17
  • $\begingroup$ I guess you could imagine a very deep delta-function potential $\delta(r-R)$, where $R$ is the radius of the sphere on which the particle moves so that the radial part of the wave function is highly localized around $r=R$. And as for the other question, since the moment of inertia is constant (fixed mass at fixed radius), the faster the rod rotates, the larger the angular momentum (semi-classically speaking), and so the energy increases with $l$ exactly as the kinetic energy increases. $\endgroup$ – march Apr 9 at 16:18

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