I looked for the reason, but it seems to be taken for granted that it must be $J(J+1)$ instead of $J$.

I'm making use of the analogy of the particle in a box energy where


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    $\begingroup$ You can't compare the eigenvalues (or eigenvectors, for that matter) for the particle-in-a-box Schroendinger equation with that for a rigid rotor, for starters. $\endgroup$ – Todd Minehardt Jan 22 '18 at 1:55
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    $\begingroup$ Because it is $J(J+1)$. Those are the eigenvalues of the squared angular momentum operator in quantum units, given three dimensions of space, $J^2$ just isn't except for $J=0$. And that's the truth. $\endgroup$ – Oscar Lanzi Jan 22 '18 at 2:01

This arises from the quantum mechanical properties of the angular momentum operator (vide infra). Since the operators for angular momentum and translational energy are different, there is no reason to expect a similarity between the form of the eigenvalues.

Spectra of the angular momentum operators

First we note the classical relationship between the angular momentum $L$ and the rotational kinetic energy

$$E_\mathrm{rot} = \frac{1}{2}I\omega^2 = \frac{L^2}{2I} \tag{1}$$

since $L = I\omega$, with $I$ being the moment of inertia and $\omega$ being the angular velocity. By analogy, the quantum mechanical Hamiltonian for rotational motion is

$$\hat{H} = \frac{\hat{L}^2}{2I} \tag{2}$$

where $\hat{L}^2$ is the operator for (the square modulus of) orbital angular momentum. Because $L$ is a vector, we have

$$\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 \tag{3}$$

where $L_x$ is the operator for the projection of orbital angular momentum onto the $x$-axis, and so on. The classical angular momentum is given by $\vec{L} = \vec{r} \times \vec{p}$; by expanding the cross product we can obtain the equations $L_x = yp_z - zp_y$, $L_y = zp_x - xp_z$, and $L_z = xp_y - yp_x$. Using these relations as well as the classical commutation relationship $[x,p_x] = [y,p_y] = [z,p_z] = \mathrm{i}\hbar$, we can derive the commutators of all four operators in equation $(3)$:

$$\begin{align} [L_x,L_y] &= \mathrm{i}\hbar L_z; & [L_y,L_z] &= \mathrm{i}\hbar L_x; & [L_z,L_x] &= \mathrm{i}\hbar L_y; \tag{4} \\ [L^2,L_x] &= 0; & [L^2,L_y] &= 0; & [L^2,L_z] &= 0 \tag{5} \end{align}$$

Conventionally, angular momentum eigenstates are chosen to be simultaneous eigenstates of $L^2$ and $L_z$: since $[L^2,L_z] = 0$, it is guaranteed that there is a complete set of states which are eigenstates of both operators. We denote these states by $|\lambda,\mu\rangle$ where $\lambda$ and $\mu$ are related to the eigenvalues of these states:

$$\begin{align} L^2|\lambda,\mu\rangle &= \lambda\hbar^2 \tag{6} \\ L_z|\lambda,\mu\rangle &= \mu\hbar \tag{7} \end{align}$$

In this paradigm, we then define the raising and lowering operators $L_+$ and $L_-$ (the meaning will become clear in due time)

$$L_\pm \equiv L_x \pm \mathrm{i}L_y \tag{8} $$

Now consider the effect of $L_\pm$ on any arbitrary state $|l,m\rangle$ which is simultaneously an eigenstate of $L^2$ and $L_z$. Since $L^2$ commutes with both $L_x$ and $L_y$, we have

$$\begin{align} L^2L_\pm|\lambda,\mu\rangle &= L_\pm L^2|\lambda,\mu\rangle \\ &= L_\pm \lambda\hbar^2|\lambda,\mu\rangle \\ &= \lambda\hbar^2 L_\pm|\lambda,\mu\rangle \tag{9} \end{align}$$

or in other words, $L_\pm|\lambda,\mu\rangle$ is also an eigenstate of $L^2$ with the same eigenvalue $\lambda\hbar^2$ as $|\lambda,\mu\rangle$ itself. Also, using the commutators in equation $(4)$,

$$\begin{align} L_z L_\pm |\lambda,\mu\rangle &= (L_zL_x \pm \mathrm{i}L_zL_y)|\lambda,\mu\rangle \\ &= [(\mathrm{i}\hbar L_y + L_xL_z) \pm \mathrm{i}(L_yL_z - \mathrm{i}\hbar L_x)]|\lambda,\mu\rangle \\ &= [\hbar(\mathrm{i}L_y \pm L_x) + (L_x \pm \mathrm{i}L_y)L_z]|\lambda,\mu\rangle \\ &= [\pm\hbar(L_x \pm \mathrm{i}L_y) + (L_x \pm \mathrm{i}L_y)L_z]|\lambda,\mu\rangle \\ &= L_\pm(L_z \pm \hbar)|\lambda,\mu\rangle \\ &= L_\pm(\mu \pm 1)\hbar|\lambda,\mu\rangle \\ &= (\mu \pm 1)\hbar L_\pm |\lambda,\mu\rangle \tag{11} \end{align}$$

which tells us that $L_\pm|\lambda,\mu\rangle$ is also an eigenstate of $L_z$, but with a different eigenvalue: the new eigenvalue is $(\mu \pm 1)\hbar$. The raising operator (for example) therefore has the effect of conserving the total angular momentum $L^2$, but increasing the $z$-projection of this angular momentum $L_z$.

However, for this to make physical sense, there has to be a limit to how far this can go: or else we will have no maximum value of $L_z$, even for a given value of $L^2$. Note that since $L_x^2$ and $L_y^2$ are both real and non-negative (which follows from the Hermiticity of $L_x$ and $L_y$), equation $(3)$ sets a maximum limit on $L_z$:

$$L_z^2 \leq L^2 \tag{12}$$

Mathematically, the only way this can be fulfilled is if there is a "highest" state $|\lambda,\mu_\mathrm{max}\rangle$ which the raising operator turns into zero:

$$L_+|\lambda,\mu_\mathrm{max}\rangle = 0 \tag{13}$$

Now consider

$$\begin{align} L_-L_+|\lambda,\mu_\mathrm{max}\rangle &= (L_x - \mathrm{i}L_y)(L_x + \mathrm{i}L_y)|\lambda,\mu_\mathrm{max}\rangle \\ &= [L_x^2 + L_y^2 + \mathrm{i}(L_xL_y - L_yL_x)]|\lambda,\mu_\mathrm{max}\rangle \\ &= [L^2 - L_z^2 + \mathrm{i}(\mathrm{i}\hbar L_z)]|\lambda,\mu_\mathrm{max}\rangle \\ &= (L^2 - L_z^2 - \hbar L_z)|\lambda,\mu_\mathrm{max}\rangle \\ &= (\lambda\hbar^2 - \mu_\mathrm{max}^2\hbar^2 - \mu_\mathrm{max}\hbar^2)|\lambda,\mu_\mathrm{max}\rangle \tag{14} \end{align}$$

However, we already know that $L_+|\lambda,\mu_\mathrm{max}\rangle = 0$. When $L_-$ acts on zero, it simply returns zero, and so the "eigenvalue" in equation $(14)$ must be equal to zero:

$$\begin{align} \lambda\hbar^2 - \mu_\mathrm{max}^2\hbar^2 - \mu_\mathrm{max}\hbar^2 &= 0 \\ \lambda &= \mu_\mathrm{max}^2 + \mu_\mathrm{max} \\ &= \mu_\mathrm{max}(\mu_\mathrm{max}+1) \tag{15} \end{align}$$

Fundamentally, that is where the term comes from. There are several steps now to close some loopholes. I will not go through them in detail because it is far too long, but I will outline what is needed.

  1. You need to show that for orbital angular momentum the value of $\mu_\mathrm{max}$ is restricted to non-negative integer values; we define this quantity to be $l$ and we can replace the quantum number $\lambda$ with $l$ because $\lambda$ is related to $\mu_\mathrm{max}$, which is in turn equivalent to $l$. The eigenvalues of the total angular momentum operator $L^2$ are $\lambda \hbar^2 = \mu_\mathrm{max}(\mu_\mathrm{max}+1)\hbar^2 \equiv l(l+1)\hbar^2$.

    Note that for orbital angular momentum only integer values of $l$ are allowed; half-integer values are not allowed. This condition can only be derived by solving the differential equations corresponding to the eigenvalue equations of the angular momentum operators (see further reading). For example, you need to solve this:

    $$\begin{align} L_z f(x,y,z) &= (xp_y + yp_x)f(x,y,z) \\ &= \left(-x\mathrm{i}\hbar\frac{\partial}{\partial y} - y\mathrm{i}\hbar\frac{\partial}{\partial x}\right)f(x,y,z) = \mu\hbar f(x,y,z) \tag{16} \end{align}$$

    Half-integer values for spin are allowed because the exact form of the spin angular momentum operators are not known. However, they obey the same commutation relations as the orbital angular momentum operators, so the rules governing them are the same except for the constraint on integer values.

  2. You need to show that the allowed values of $\mu$ are integers between $-l$ and $l$ respectively. We have already shown that $\mu$ increases in integer steps, and that $\mu \leq l$, so what remains is to show that $\mu \geq -l$: the proof is exactly analogous to that in equations $(13)$ through $(15)$, except that you need to consider $L_+L_-|\lambda, \mu_\mathrm{min}\rangle$.

    Conventionally the quantum number $\mu$ is renamed to $m_l$ and the eigenstates are labelled with $|l,m_l\rangle$. So now we have shown that $l$ is the maximal value of $m_l$, and that $m_l$ ranges from $-l$ to $l$ in integer steps, as we expected

  3. The total angular momentum $J$ is the sum of the orbital and spin angular momenta, i.e. $J = L + S$. For the rigid rotor system where you ignore spin, $J = L$ and hence we can simply replace $L$, $l$, and $m_l$ with $J$, $j$, and $m_j$: we therefore have $J^2|j,m_j\rangle = j(j+1)\hbar^2|j,m_j\rangle$.

  4. Finally, the energies are then given by

    $$\begin{align} E &= \langle j, m_j|H|j,m_j \rangle \\ &= \left< j,m_j \middle| \frac{J^2}{2I} \middle| j,m_j \right> \\ &= j(j+1)\frac{\hbar^2}{2I} \tag{17} \end{align}$$

    where $j$ is a non-negative integer (i.e. $j = 0,1,2, \ldots$).

Further reading

Chapter 4 of Molecular Quantum Mechanics, 5th ed., Atkins.

Chapter 7 of The Physics of Quantum Mechanics, Binney & Skinner for an alternative derivation.

  • $\begingroup$ Why define $L_x = yp_z - zp_y$? Is it just one of those "we do it because it works" sort of things? Or is there a deeper rationale? (Should I ask this as a new question?) $\endgroup$ – hBy2Py Jan 23 '18 at 18:38
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    $\begingroup$ @hBy2Py because $\vec{l} = \vec{r} \times \vec{p}$, evaluating the cross product gets us to the three equations. Good point, though: in hindsight "define" was a really poor word choice so I've edited it out. $\endgroup$ – orthocresol Jan 23 '18 at 19:08
  • $\begingroup$ This is exactly the sort of thing that drives me bonkers about a lot of quantum chem derivations: things get asserted that do actually have a rational basis, but that basis isn't even mentioned in passing. Thanks! $\endgroup$ – hBy2Py Jan 23 '18 at 19:21
  • $\begingroup$ why can't I upvote this twice? $\endgroup$ – user37142 Jan 24 '18 at 9:44

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