Question #154872

. Derive Schrodinger wave equation from Cartesian coordinates into Spherical polar coordinates. 1 𝑟 2 𝜕 𝜕𝑟 (𝑟 2 𝜕𝜓 𝜕𝑟 ) + 1 𝑟 2𝑠𝑖𝑛𝜃 𝜕 𝜕𝜃 (𝑠𝑖𝑛𝜃 𝜕𝜓 𝜕𝜃) + 1 𝑟 2𝑠𝑖𝑛2𝜃 𝜕 2𝜓 𝜕𝜙2 + 2𝑚 ℏ 2 (𝐸 − 𝑈)𝜓 = 0


1
Expert's answer
2021-01-14T10:40:36-0500

Let's first write (stationary) Schrodinger equation in cartesian coordinates :

22m(2x2+2y2+2z2)ψ+Vψ=Eψ-\frac{\hbar^2}{2m} (\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2})\psi +V\psi = E\psi

22mΔψ+(VE)ψ=0-\frac{\hbar^2}{2m} \Delta\psi + (V-E)\psi=0 , where Δ\Delta is Laplace operator

Δψ+2m2(EV)ψ=0\Delta\psi + \frac{2m}{\hbar^2} (E-V)\psi=0

Now it is just enough to use the expression of Δ\Delta in spherical coordinates :

1r2r(r2ψr)+1r2sinθθ(sinθψθ)+1r2sin2θ2ψϕ2+2m2ψ=0\frac{1}{r^2} \frac{\partial}{\partial r}(r^2\frac{\partial \psi}{\partial r}) + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial\theta} (\sin\theta \frac{\partial\psi}{\partial\theta}) + \frac{1}{r^2\sin^2\theta} \frac{\partial^2\psi}{\partial\phi^2}+\frac{2m}{\hbar^2}\psi =0

The expression of Δ\Delta in spherical coordinates can be found, for example, in Wikipedia : Laplace operator - Wikipedia . We can also, of course, calculate it directly :

x=rxr+θxθ+ϕxϕ\frac{\partial}{\partial x} = \frac{\partial r}{\partial x}\frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta} + \frac{\partial \phi}{\partial x}\frac{\partial}{\partial \phi} (chain rule)

x=sinθcosϕrcosθcosϕrθsinϕrsinθϕ\frac{\partial}{\partial x} = \sin\theta \cos\phi \frac{\partial}{\partial r} -\frac{\cos\theta \cos\phi}{r}\frac{\partial}{\partial\theta}-\frac{\sin\phi}{r\sin\theta} \frac{\partial}{\partial\phi} (using the expressions of spherical coordinates in cartesian coordinates)

The same calculation for y,zy, z gives :

y=sinθsinϕrcosθsinϕrθ+cosϕrsinθϕ\frac{\partial}{\partial y} = \sin\theta \sin\phi \frac{\partial}{\partial r} -\frac{\cos\theta \sin\phi}{r}\frac{\partial}{\partial\theta} + \frac{\cos\phi}{r\sin\theta} \frac{\partial}{\partial\phi}

z=cosθrsinθrθ\frac{\partial}{\partial z} = \cos\theta \frac{\partial}{\partial r} -\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}

And now we find by direct calculation (as 2x2=(x)2\frac{\partial^2}{\partial x^2} = (\frac{\partial}{\partial x})^2 and same for other coordinates) :

Δ=1r2r(r2r)+1r2sinθθ(sinθθ)+1r2sin2θ2ϕ2\Delta = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r}) + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial\theta} (\sin\theta \frac{\partial}{\partial\theta}) + \frac{1}{r^2\sin^2\theta} \frac{\partial^2}{\partial\phi^2}


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