Suppose there is a system of particles interacting with each other. Is it the angular momentum of each particles which would be quantized as $\sqrt{n(n+1)} \hbar$ or is the angular momentum of the entires system.

I was of the opinion that it should be true for each particle but then I came across the following problem in Quantum Mechanics Griffiths (Problem 4.29, 2nd Indian Edition):

Prob: 2 particles of mass $m$ are attached to the ends of a massless rigid rod of length $a$. The system is free to rotate in three dimensions about the centre (but the centre itself is fixed). Show that the allowed energies of rigid rotor are $E_n = \frac{\hbar^2 n(n+1)}{ma^2}$.

Now if energy is expressed as $E=\frac{L^2}{2I}$ , ($0$ potential energy and entire kinetic energy is due to rotation about COM), the above is true since $L^2$ is quantized as $n(n+1) \hbar^2$ . However I can express the above as $E = \frac{L_1^2}{2I_1}+\frac{L_2^2}{2I_2} = 2\frac{L_1^2}{2I_1}$ , (here $L_1^2$ and $L_2^2$ are the squared magnitudes of angular momentum about origin). The above is the sum of KE of first and second particle ,which are equal. Thus the total KE is $2\frac{n(n+1)\hbar^2}{2. m (\frac{a}{2})^2} = \frac{4 n(n+1) \hbar^2}{ma^2}$

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    $\begingroup$ For a $2$ particle system, the angular momentum quantum number takes all values from $l_1+l_2$ to $|l_1-l_2|$ in steps of 1 where $l_1$ and $l_2$ are the respective quantum numbers. The probabilties of these combinations are given by the Clebsch Gordon coefficients. It's quite more involved in the case of $3$ or more particles. $\endgroup$ Jul 14, 2019 at 10:17

2 Answers 2


You can quantize it anyway you want, but some ways are better than others.

Lets say you have two non-interacting particles. Then the position of each particle relative to another one should not matter. Therefore you expect the Hamiltonian of the whole system to commute with rotation operators of both particle 1 and 2.

Lets us put it into maths. You have two particles with two wavefunctions $|j_1,m_1\rangle|j_2,m_2\rangle$, where $j$ is the angular momentum and $m$ is the projection on z-axis. You also have angular momentum operators $\mathbf{\hat{J}}_1$ and $\mathbf{\hat{J}}_1$. Rotation of particle 1 by $\theta$ around axis $\mathbf{\hat{n}}$ is given by:

$\exp\left(i\theta \mathbf{\hat{J}}_1.\mathbf{\hat{n}}\right)|j_1,m_1\rangle|j_2,m_2\rangle=\sum_{m'_1} c_{m_1 m'_1}|j_1,m'_1\rangle|j_2,m_2\rangle$

You then have a Hamiltonian $\hat{H}$ that is not changed by the rotation of particle 1. So

$\exp\left(i\theta \mathbf{\hat{J}}_1.\mathbf{\hat{n}}\right)\hat{H}\exp\left(-i\theta \mathbf{\hat{J}}_1.\mathbf{\hat{n}}\right)=\hat{H}$,

which implies $\left[\hat{H},\,\mathbf{\hat{J}}_1\right]=\mathbf{0}$ so eigenstates of the operator $\mathbf{\hat{J}}_1$ are also eigenstates of the Hamiltonian. Which means that these eigenstates do not mix as the system evolves in time. Therefore if your particle 1 is in eigenstate with values $j_1,m_1$ at one moment in time, it will stay in this state indefinitely - hence $j_1,m_1$ are 'good' quantum numbers.

Now let us look at interacting particles, now the system's energy depends on the relative positions of the two particles, hence rotating one particle whilst keeping the other one fixed will change the energy, hence $\left[\hat{H},\,\mathbf{\hat{J}}_1\right]\neq\mathbf{0}$. You can still label your particle 1 with quantum numbers $j_1,\,m_1$, but you will find that these labels will only be correct for a short period of time. You will find that your particle 1 will stay in eigenstates with $j=j_1$, but $m$ will now change in time. More precisely, particle 1 will be in time-dependent superposition of different eignestates with the same $j$ but different $m$.

What to do? Well we are still in isotropic space so full rotation of the system cannot change the energy, i.e.:

$\exp\left(i\theta \mathbf{\hat{J}}_2.\mathbf{\hat{n}}\right)\exp\left(i\theta \mathbf{\hat{J}}_1.\mathbf{\hat{n}}\right)\hat{H}\exp\left(-i\theta \mathbf{\hat{J}}_1.\mathbf{\hat{n}}\right)\exp\left(-i\theta \mathbf{\hat{J}}_2.\mathbf{\hat{n}}\right)=\hat{H}$,



So if you re-express the $|j_1,m_1\rangle|j_2,m_2\rangle$ in terms of eigenstates of the combined operator ($\mathbf{\hat{J}}=\mathbf{\hat{J}_1}+\mathbf{\hat{J}_2}$), i.e. $|j,j_1,j_2,m\rangle$, you will find that these eigenstates do not mix as the system evolves in time, hence the 'good' quantum numbers are $j,j_1,j_2,m$

A specific answer to your question is this. If you want an expression for the energy of the system in terms of eigenvalues of the eignestate it is in, then you have to work with energy eigenstates. As I showed above energy eigenstates of a system of interacting particles are eignestates of the full momentum operator ($\left[\mathbf{\hat{J}_1}+\mathbf{\hat{J}_2},\hat{H}\right]=\mathbf{0}$ but $\left[\mathbf{\hat{J}_1},\hat{H}\right]\neq\mathbf{0}$), hence you have to quantize the full angular momentum


The answer already given is comprehensive and correct. Concerning your example problem however, using your approach you must put r=a/2 for L=mvr; so that squaring gives the factor of 1/4 on the top. Otherwise, for interacting particles, the point of the exercise is presumably that in quantising such a system of particles, one calculates I=mr^2 using r=a to determine the L angular momentum. (Don't forget where the formulae come from, as Faraday said to Maxwell, more-or-less).

As you probably know, the form of the expression for L arises from that of the solutions to Laplace's equation in spherical harmonics; the eigenfunctions of L^2. In that case, what is being described is the effect that transitions in L for the system, here L=J, occur in increments of h-bar for variations in m, v and r=a.


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