there! I am studying tensor analysis and now try to apply it to solving a quantum physics problem. Here I am trying to calculate angular momentum squared written in terms of the spherical coordinates directly. Here is how it looks:

$\hat L^2=-\hbar^2(1/sin\theta\ \partial_\theta(sin\theta\ \partial_\theta)+1/sin^2\theta\ \partial_{\phi\phi})=-\hbar^2r^2(\nabla^2-1/r^2\ \partial_r(r^2\partial_r))$

And I can easily begin here:

$\begin{align} \hat L^2&=(-i\hbar\vec r\times\nabla)^2\\ &=-\hbar^2r^2(\hat e_r\times\nabla)\bullet(\hat e_r\times\nabla)\\ &=-\hbar^2r^2(\hat e_r\bullet(\nabla\times(\hat e_r\times\nabla)))\\ &=-\hbar^2r^2(\hat e_r\bullet(\hat e_r \nabla^2-\nabla(\hat e_r\bullet\nabla)))\\ &=-\hbar^2r^2(\nabla^2-\hat e_r\bullet\nabla(\hat e_r\bullet\nabla)) \end{align}$

So there I confused, for time independent schrodinger function $\Phi(r,\theta,\phi)$ is a scalar function, and sperical coordinates is orthongonal everywhere, the form of $\nabla$ should be as simple as $\vec g^i\partial_{v^i}$ and then through a dot product it remains a scalar function $\partial_r$ again, and thus it finally turns into $\partial_{rr}$. So I cannot derive the correct expression.

Where am I wrong in deduction? And how can I derive the expression swiftly using tools of tensor analysis?

Thanks in advance!

  • $\begingroup$ The issue is that $\hat e_r$ is not constant. Its direction changes with $\theta$ and $\phi$. $\endgroup$
    – Andrei
    Oct 24, 2019 at 17:14
  • $\begingroup$ @Andrei Yeah, I noticed that. But it is always orthogonal to the other two local axis and has a unit length. So I wonder if the dot product eradicates this change, and if not, how? ah, very confusing. $\endgroup$ Oct 24, 2019 at 17:19
  • $\begingroup$ See for example thphys.nuim.ie/Notes/MP469/Laplace.pdf $\endgroup$
    – Andrei
    Oct 24, 2019 at 17:29

1 Answer 1


Having learning a few more chapters in tensor analysis, now I can solve this question on my own! Here I post my answer.

The point is that covariant derivatives and covariant/contravariant as functions of coordiantes opponents are not commutable, so we should compute it totally in tensor language, or explicitly write down every commutation:

\begin{align} \hat L^2&=(-i\hbar\vec r\times\nabla)^2\\ &=-\hbar^2\vec \epsilon:\vec r\nabla\cdot\vec \epsilon:\vec r\nabla\\ &=-\hbar^2\epsilon^{ijk}r_i\nabla_j\epsilon_{lmk}r^l\nabla^m\\ &=-\hbar^2(\delta^i_l\delta^j_m-\delta^i_m\delta^j_l)r_i\nabla_jr^l\nabla^m\\ &=-\hbar^2(r_i\nabla_jr^i\nabla^j-r_i\nabla_jr^j\nabla^i)\\ \end{align}

since $r^i=0, i\ne1$ in spherical coordinates, and $\nabla_jr^1=0, j\ne1$, so apply multiplication rule to it: \begin{align} r_i\nabla_jr^i\nabla^j-r_i\nabla_jr^j\nabla^i&=r_1\nabla_jr^1\nabla^j-r_1\nabla_1r^1\nabla^1\\ &=r^2\nabla_j\nabla^j-r^2\nabla_1\nabla^1\\ \end{align}

From this we give

$\hat L^2=-\hbar^2r^2(\nabla^2-1/r^2\ \partial_r(r^2\partial_r))$


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