It is probably the most well-known result in quantum mechanics that the harmonic oscillator can be solved by supersymmetry.

More precisely, the operator


can be decomposed as

$$-\frac{d^2}{dx^2}+x^2 = \left(-\frac{d}{dx}-x\right)\left(\frac{d}{dx}-x\right)=:a^*a.$$

When I tried to see whether the same holds true in spherical coordinates I noticed a complete suprise

$$\left(-\frac{d}{dr}-\frac{n-1}{r}-r\right)\left(\frac{d}{dr}-r\right) = -\frac{d^2}{dr^2} -\frac{n-1}{r} \frac{d}{dr} +r^2+(n-1).$$

The first three terms comprise the harmonic oscillator in spherical coordinates that is

$$-\Delta_r + r^2.$$

For convenience, I discarded the angular part in this calculation.

But for some reasons I seem to get this superficial term $(n-1)$. Why is that? And can I do anything smarter to avoid this term?


Indeed, the supersymmetric operators do not factorize the Hamiltonian of the three-dimensional harmonic oscillator, there is an additional term. See Creation and annihilation operators, symmetry and supersymmetry of the 3D isotropic harmonic oscillator, equation 16.


Supplementing @CarloBeenakker's answer: a different sort of "factorization" of the Laplacian can be done in terms of the Dirac operator, which in some regards better preserves the nature of the situation, but/and has scalar coefficients in an appropriate Clifford algebra, which in general is not commutative. E.g., on $\mathbb R$, it is $\mathbb C$, which is convenient. But already in two dimensions the relevant Clifford algebra is the Hamiltonian quaternions...

Depending what you want to do, this might be a useful direction to consider, in addition to the "scalarization" of immediate reduction to the spherically symmetric case.


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