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


More precisely, the operator


$$-\frac{d^2}{dx^2}+x^2$$ 


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?



Harmonic oscillator in spherical coordinates
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](https://www.researchgate.net/publication/228813691_Creation_and_annihilation_operators_symmetry_and_supersymmetry_of_the_3D_isotropic_harmonic_oscillator), equation 16.


