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The kinetic energy of a system $p^2/2m$ in real space representation takes the form $-\hbar^2\nabla^2/2m$. I want to express this in cylindrical coordinates via the representation of the laplace operator Wikipedia. This is $$\nabla^2 = \frac{1}{r}\partial_r(r\partial_r)+\frac{1}{r^2}\partial^2_\varphi + \partial^2_z.$$ Expanding the radial part yields $\frac{1}{r}\partial_r+\partial^2_r$. While I know that the second derivative is a hermitian operator, this is not true for the first derivative. The factor if $i$ that comes with $$p = -i\hbar\nabla$$ takes care that the operator $p$ is really hermitian, however here it is missing. How can I see that the radial component is still hermitian in this representation?

Bonus: Eventually I want to implement this numerically using finite differences method. The stencil for $\partial^2_r$ is $[1,-2,1]$ which results in a hermitian matrix. For $\partial_r$ however, the stencil is $[-1/2,0,1/2]$ which takes a minus sign if I do hermitian conjugate.

Qmechanic
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Suppenkasper
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  • Related: https://physics.stackexchange.com/q/9349/2451 and links therein. – Qmechanic Apr 15 '19 at 08:10
  • The Laplacian is hermitian and so is the third term of the RHS of your equation. The second term in the RHS is also hermitian, since $1/r$ and $\partial_\phi$ commute. Hence the remaining term, the radial part, is hermitian. – lcv Apr 15 '19 at 09:55
  • The relevant Hilbert space is $L^2((0,+\infty), r^2 dr)$... that $r^2$ makes $(1/r)\partial_r(r \partial_r)$ Hermitian. – Valter Moretti Apr 15 '19 at 13:42

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