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


Expert's answer

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

−ℏ22m(∂2∂x2+∂2∂y2+∂2∂z2)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

−ℏ22mΔψ+(V−E)ψ=0-\frac{\hbar^2}{2m} \Delta\psi + (V-E)\psi=0 , where Δ\Delta is Laplace operator

Δψ+2mℏ2(E−V)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 :

1r2∂∂r(r2∂ψ∂r)+1r2sin⁡θ∂∂θ(sin⁡θ∂ψ∂θ)+1r2sin⁡2θ∂2ψ∂ϕ2+2mℏ2ψ=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=∂r∂x∂∂r+∂θ∂x∂∂θ+∂ϕ∂x∂∂ϕ\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=sin⁡θcos⁡ϕ∂∂r−cos⁡θcos⁡ϕr∂∂θ−sin⁡ϕrsin⁡θ∂∂ϕ\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=sin⁡θsin⁡ϕ∂∂r−cos⁡θsin⁡ϕr∂∂θ+cos⁡ϕrsin⁡θ∂∂ϕ\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=cos⁡θ∂∂r−sin⁡θr∂∂θ\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 ∂2∂x2=(∂∂x)2\frac{\partial^2}{\partial x^2} = (\frac{\partial}{\partial x})^2 and same for other coordinates) :

Δ=1r2∂∂r(r2∂∂r)+1r2sin⁡θ∂∂θ(sin⁡θ∂∂θ)+1r2sin⁡2θ∂2∂ϕ2\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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