Quantum number

   A detail derivation of schrodinger equation under spherical coordinate, which can obtain quantum number.

According to Schrodinger equation \[\left(-\frac{{\hslash }^2}{2m}{\nabla }^2\right)\psi \left(r,t\right)=E\psi \left(r,t\right)\] If we consider ${\mathrm{\nabla }}^{\mathrm{2}}$ under spherical coordinate, we have \[\left(-\frac{{\hslash }^2}{2m}\left(\frac{1}{r^2}\frac{\partial }{\partial r\ }\left(r^2\frac{\partial }{\partial r}\right)+\frac{1}{r^2{{sin}^2 \theta \ }}\left(sin\theta \frac{\partial \left(sin\theta \frac{\partial }{\partial \theta }\right)}{\partial \theta }+\frac{{\partial }^2}{\partial {\phi }^2}\right)\right)+V\left(r,t\right)\ \right)\psi \left(r,t\right)=E\psi \left(r,t\right)\] Divided by $\mathrm{-}\frac{\mathrm{2}{\mathrm{mr}}^{\mathrm{2}}}{{\hslash }^2}$ \[\left(\left(\frac{\partial \left(r^2\frac{\partial }{\partial r}\right)}{\partial r\ }-\frac{1}{{{\mathrm{sin}}^2 \theta \ }}\left(sin\theta \frac{\partial \left(sin\theta \frac{\partial }{\partial \theta }\right)}{\partial \theta }+\frac{{\partial }^2}{\partial {\phi }^2}\right)\right)-\frac{2mr^2}{{\hslash }^2}V\left(r,t\right)\ \right)\psi \left(r,t\right)=-\frac{2mr^2}{{\hslash }^2}E\psi \left(r,t\right)\] Using separation to solve the PDE} \[\psi \left(r,t\right)=R\left(r\right)\mathit{\Theta}\left(\theta \right)\mathit{\Phi}\left(\phi \right)=R\left(r\right)Y\ \ \ ,\ Y\left(\theta ,\phi \right)=\mathit{\Theta}\left(\theta \right)\mathit{\Phi}\left(\phi \right)\] Divided by }$R\left(r\right)Y$ \[\frac{1}{R}\frac{\partial }{\partial r\ }\left(r^2\frac{\partial }{\partial r}\right)-\frac{2mr^2}{{\hslash }^2}\left(V-E\right)-\frac{1}{Ysin^2\theta }\left(sin\theta \frac{\partial }{\partial \theta }\left(sin\theta \frac{\partial }{\partial \theta }\right)+\frac{{\partial }^2}{\partial {\phi }^2}\right)=0\] \[\left\{ \begin{array}{c} \frac{1}{R}\frac{\partial }{\partial r\ }\left(r^2\frac{\partial }{\partial r}\right)-\frac{2mr^2}{{\hslash }^2}\left(V-E\right)=l\left(l+1\right) \\ \frac{1}{Ysin^2\theta }\left(sin\theta \frac{\partial }{\partial \theta }\left(sin\theta \frac{\partial }{\partial \theta }\right)+\frac{{\partial }^2}{\partial {\phi }^2}\right)=-l\left(l+1\right) \end{array} \right.\] according to Schrodinger equation and separable variable method, we can derive following partial differential equation:} \[sin\theta \frac{\partial }{\partial \theta }\ \left(sin\theta \frac{\partial Y}{\partial \theta }\right)+\frac{{\partial }^2Y}{\partial {\phi }^2}=-l\left(l+1\right){{sin}^2 \theta \ }Y\] with variable $Y\ $are function of $\mathrm{\Theta }$ and $\mathit{\Phi}$ \[Y\left(\theta ,\phi \right)=\mathit{\Theta}\left(\theta \right)\mathit{\Phi}\left(\phi \right)\] Plugging this, and divide by $\mathrm{\Theta }$ and $\mathit{\Phi}$ \[sin\theta \frac{\partial }{\partial \theta }\ \left(sin\theta \frac{\partial }{\partial \theta }(\mathit{\Theta}\left(\theta \right)\mathit{\Phi}\left(\phi \right))\right)+\frac{{\partial }^2}{\partial {\phi }^2}\mathit{\Theta}\left(\theta \right)\mathit{\Phi}\left(\phi \right)=-l\left(l+1\right){{sin}^2 \theta \ }\mathit{\Theta}\left(\theta \right)\mathit{\Phi}\left(\phi \right)\] \[\mathit{\Phi}\left(\phi \right)sin\theta \frac{\partial }{\partial \theta }\ \left(sin\theta \frac{\partial \ \mathit{\Theta}\left(\theta \right)}{\partial \theta }\right)+\mathit{\Theta}\left(\theta \right)\frac{{\partial }^2\mathit{\Phi}\left(\phi \right)}{\partial {\phi }^2}=-l\left(l+1\right){{sin}^2 \theta \ }\mathit{\Theta}\left(\theta \right)\mathit{\Phi}\left(\phi \right)\] \[\frac{1}{\mathit{\Theta}\left(\theta \right)}sin\theta \frac{\partial }{\partial \theta }\ \left(sin\theta \frac{\partial \ \mathit{\Theta}\left(\theta \right)}{\partial \theta }\right)+\frac{1}{\mathit{\Phi}\left(\phi \right)}\frac{{\partial }^2\mathit{\Phi}\left(\phi \right)}{\partial {\phi }^2}=-l\left(l+1\right){{sin}^2 \theta \ }\] Consider the parts of $\mathit{\Phi}$ We define a quantum number $m_l$ as following \[{-m}^2_l=\frac{1}{\mathit{\Phi}\left(\phi \right)}\frac{{\partial }^2\mathit{\Phi}\left(\phi \right)}{\partial {\phi }^2}\] And the solution of differentiate equation related to }$\mathrm{\phi }$ are \[\mathit{\Phi}\left(\phi \right)=A\times e^{-m_l\phi }\] Plugging in and we get \[\frac{1}{\mathit{\Theta}\left(\theta \right)}sin\theta \frac{\partial }{\partial \theta }\ \left(sin\theta \frac{\partial \mathit{\Theta}\left(\theta \right)}{\partial \theta }\right){-m}^2_l=-l\left(l+1\right){{sin}^2 \theta \ }\] After some algebra \[sin\theta \frac{\partial }{\partial \theta }\ \left(sin\theta \frac{\partial \mathit{\Theta}\left(\theta \right)}{\partial \theta }\right)+\mathit{\Theta}\left(\theta \right)\left({l\left(l+1\right){{sin}^2 \theta \ }-m}^2_l\right)=0\] Using situational variable \[cos\theta =x\] we get following usable relation \[sin\theta =\frac{dx}{d\theta }=\sqrt{1-x^2}\] plugging those relation, we get \[\sqrt{1-x^2}\frac{\partial }{\partial \theta }\ \left(\sqrt{1-x^2}\frac{\partial \mathit{\Theta}\left(\theta \right)}{\partial \theta }\right)+\mathit{\Theta}\left(\theta \right)\left({l\left(l+1\right)(1-x^2)-m}^2_l\right)=0\] We can change the differentiate variable theta to x by chain rule \[\frac{\partial }{\partial \theta }\ \left(\sqrt{1-x^2}\frac{\partial \ \mathit{\Theta}\left(\theta \right)}{\partial \theta }\right)=\frac{\partial }{\partial \theta }\ \left(L\right)=\frac{dL}{dx}\frac{dx}{d\theta }\] And \[\frac{\partial \ \mathit{\Theta}\left(\theta \right)}{\partial \theta }=\frac{\partial \ \mathit{\Theta}\left(\theta \right)}{\partial x}\frac{dx}{d\theta }\] Therefore \[\sqrt{1-x^2}\frac{\partial \left(\sqrt{1-x^2}\frac{\partial \ \mathit{\Theta}\left(\theta \right)}{\partial x}\frac{dx}{d\theta }\right)}{\partial x}\ \frac{dx}{d\theta }+\mathit{\Theta}\left(\theta \right)\left({l\left(l+1\right)\left(1-x^2\right)-m}^2_l\right)=0\] Substitute$\frac{dx}{d\theta }$ with$\sqrt{1-x^2}$ \[\sqrt{1-x^2}\frac{\partial \left(\sqrt{1-x^2}\frac{\partial \ \mathit{\Theta}\left(\theta \right)}{\partial x}\sqrt{1-x^2}\right)}{\partial x}\ \sqrt{1-x^2}+\mathit{\Theta}\left(\theta \right)\left({l\left(l+1\right)(1-x^2)-m}^2_l\right)=0\] Merge the $\sqrt{1-x^2}$ \[\left(1-x^2\right)\frac{\partial \left(\left(1-x^2\right)\frac{\partial \ \mathit{\Theta}\left(\theta \right)}{\partial x}\right)}{\partial x}\ +\mathit{\Theta}\left(\theta \right)\left({l\left(l+1\right)(1-x^2)-m}^2_l\right)=0\] Using the chain rule to expansion the differential item} \[{\left(1-x^2\right)}^2\frac{{\partial }^2\mathit{\Theta}\left(\theta \right)}{\partial x^2}+\left(1-x^2\right)\frac{\partial \left(1-x^2\right)}{\partial x}\frac{\partial \ \mathit{\Theta}\left(\theta \right)}{\partial x}\ +\mathit{\Theta}\left(\theta \right)\left({l\left(l+1\right)(1-x^2)-m}^2_l\right)=0\] After some algebra \[{\left(1-x^2\right)}^2\frac{{\partial }^2\mathit{\Theta}\left(\theta \right)}{\partial x^2}+\left(1-x^2\right)(-2x)\frac{\partial \ \mathit{\Theta}\left(\theta \right)}{\partial x}\ +\mathit{\Theta}\left(\theta \right)\left({l\left(l+1\right)(1-x^2)-m}^2_l\right)=0\] Dividing $\left(1-x^2\right)$ \[\left(1-x^2\right)\frac{{\partial }^2\mathit{\Theta}\left(\theta \right)}{\partial x^2}-2x\frac{\partial \ \mathit{\Theta}\left(\theta \right)}{\partial x}\ +\mathit{\Theta}\left(\theta \right)\left(l\left(l+1\right)-\frac{m^2_l}{1-x^2}\right)=0\] Which is known as Legendre equation and the solution for ml=0 are \[P_l\left(x\right)=\frac{1}{2^ll!}{\left(\frac{d}{dx}\right)}^l{\left(x^2-1\right)}^l\] And for more general cases, while $m_l\neq 0,$ we get associated legendre function \[P^{m_l}_l\left(x\right)={{\left(-1\right)}^{m_l}\left(1-x^2\right)}^{\frac{\left|m_l\right|}{2}}\ {\left(\frac{d}{dx}\right)}^{\left|m_l\right|}P_l\left(x\right)\] With associated legendre function condition of \[0\le m\le l\] An example of : \[P^0_2=\frac{1}{2}(3{{cos}^2 \theta \ }-1)\]