Suppose we have a hydrogen atom in a specific state $\psi_{nlm}$. Of course, the state is given by $$\psi_{nlm} = Ce^{-r/na}r^l L_{n-l-1}^{2l+1}(2r/na)Y_l^m(\theta, \phi) $$

I am being asked to compute expectations of some quantities (for example, $E$ and $L^2$). Ordinarily I'd just compute $$\int \psi_{nlm}^* L^2 \psi_{nlm} r^2\sin{\theta} \,\;dr\;d\theta \;d\phi $$

but this seems like an exceedingly ugly integral, and moreover I'm not sure what the operator forms of $E, L^2$ are. What are the operator forms, and is there an alternative to evaluating the integral directly?

  • 2
    $\begingroup$ Hint: the eigenstates of the hydrogen atom are eigenstates of the Hamiltonian and $L^2$. $\endgroup$
    – Praan
    Commented Dec 17, 2015 at 14:25

1 Answer 1


The operator form of $\hat H$ (total energy) is of course $\hat p^2/2m + \hat u$ where $\hat p = -i\hbar \nabla$ as usual and $\hat u = k e/r$ is your usual Coulomb potential. You should find that this integral ultimately depends only on $n$, as that is the "total energy quantum number." You may prefer to use mathematical properties of the Laguerre polynomials (in particular, notice the differential equation that we invented them to solve!) and note please that the spherical harmonics totally disappear due to spherical symmetry, getting integrated away from the very start.

The operator form of $L^2$ is much more complicated. It comes from the operator form of $\vec L = \vec r \times \vec p.$ Written out in spherical coordinates $\vec p = -i \hbar \nabla$ becomes $$\nabla = \hat r ~\partial_r + \hat\theta ~r^{-1} ~\partial_\theta + \hat\varphi ~(r \sin\theta)^{-1} ~\partial_\varphi.$$ Crossing with $\vec r = r~\hat r$ we get $$\vec L = -i\hbar \big(\hat\varphi ~\partial_\theta - \hat\theta ~(\sin\theta)^{-1} ~\partial_\varphi\big).$$

Now here is a subtle point about squaring this operator: the unit vectors change with the coordinates. At a polar position $(r, \theta, \varphi)$ we have Cartesian position $[x, y, z] = [r\sin\theta\cos\varphi,~ r\sin\theta\sin\varphi,~ r\cos\theta]$ with unit vectors $$\begin{array}{ccccc} \hat r = &[&\sin\theta\cos\varphi,&~ \sin\theta\sin\varphi,&~\cos\theta&],\\ \hat\theta = &[&\cos\theta\cos\varphi,&~\cos\theta\sin\varphi,&~-\sin\theta&],\\ \hat\varphi =& [&-\sin\varphi,&~\cos\varphi,&~ 0&]. \end{array}$$ So therefore $\partial_\theta \hat\theta = -\hat r,\;\partial_\theta \hat\varphi = 0,\;\partial_\varphi\hat\theta = \hat\varphi~\cos\theta,\;\partial_\varphi\hat\varphi =-\hat r\sin\theta - \hat\theta\cos\theta.$ Comparing these with $\vec L$ above all that we're really concerned with for calculating $L^2$ is $\hat \theta\cdot\partial_\varphi\hat\varphi = -\cos\theta,$ everything else fails to contribute. With this we can use the product rule as usual to say: $$\begin{align}\vec L \cdot (\vec L f)/(-\hbar^2) ~&= \Big((\hat\varphi\cdot) ~\partial_\theta - (\hat\theta\cdot) ~(\sin\theta)^{-1} ~\partial_\varphi\big)\big(\hat\varphi ~f_\theta - \hat\theta ~(\sin\theta)^{-1} ~f_\varphi\big)\\ ~&=f_{\theta\theta} + (\sin\theta)^{-1}~\cos\theta~f_\theta + (\sin\theta)^{-2} ~f_{\varphi\varphi}.\end{align}$$We can then reverse-apply the chain rule to this to get the form:$$L^2 = -\hbar^2\left({1\over\sin\theta}{\partial\over\partial\theta}\left(\sin\theta {\partial\over\partial\theta}\right) + {1\over\sin^2\theta}{\partial^2\over\partial\varphi^2}\right).$$ Good luck! You should get that this combination forces $r$ and hence $n$ to immediately drop out of the integral as a constant, giving you something which turns out to only depend on $\ell$ and not on $m$ either.

The two expressions are actually classics of the field, the Rydberg formula $1/n_0^2 - 1/n_1^2$ and the characteristic $\sqrt{\ell(\ell+1)} > m$ expression which tells you, "hey, not all of this angular momentum can be directed about the $z$-axis at any one time! The $L_z$ eigenstate has some nonzero $L_x, L_y$ components to it!"


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