Why is there a need of polar coordinates to solve the Schrödinger wave equation for the hydrogen atom?

Why is there a need of polar coordinate to solve the Schrödinger wave equation for the hydrogen atom?  

I went through some standard text books but I am feeling rather confused about the explanation.




**Tl;DR**


When solving the Schrödinger equation for the hydrogen atom, its potential energy is dependent on $r$. Using Cartesian co-ordinates $r$ is expressed as $\sqrt{x^2+y^2+z^2}$ which greatly complicates the process of solving the equation. Therefore we use polar co-ordinates as the variables are$~r,\theta,\psi$ which means that there is no nasty square root sign as we don't have to split $r$ into terms of $x,y,z$. 


**Long Version**


This is because to solve the Schrödinger wave equation for the hydrogen atom you require to use the technique known as, separation of variables. Separation of variables is a method for solving partial differential equations. For a partial differential equation in a function $\Phi~(x, y~...)$ and variables $x, y,~...$ we assume that it can be factored into functions which is function of only one variable: 
$$\Phi~(x, y,\dots) \equiv X(x)~Y(y)~\dots$$ 


By breaking the resulting equation into a set of independent ordinary differential equations, we are able to easily solve for $X~(x), Y~(y), ~...$ and then plug them back into the equation to get our original function, $\Phi~(x, y~...)$. For example, lets consider this differential equation of 
$$\frac{\partial^2 U(x,y)}{\partial x^2} - \frac{\partial U(x,y)}{\partial y} = 0$$


To solve this partial differential equation we use the separation of variables method. We assume that $U(x,y)$ can be written as $X(x)~Y(y)$. Therefore the equation becomes: 
$$\frac{\partial^2 (XY)}{\partial x^2} = \frac{\partial (XY)}{\partial y} $$


Now, the beauty of this is that $X$ and $Y$ are independent of each other meaning that $X$ is not a function of $Y$ and $Y$ is not a function of $X$. Therefore we can rearrange that equation to give: 
$$\frac{Y~\partial^2 X}{\partial x^2} = \frac{X~\partial Y}{\partial y} $$ 


The equation can be rearranged to give: 
$$\frac{X''}{X} = \frac{Y'}{Y}$$ 


Since each side of the equation is a constant with respect to the other since one is a function of x and the other is a function of y, we can write: 
\begin{align}\frac{X''}{X} &= C, &\frac{Y'}{Y} &= C\end{align}


Now each differential equation can be solved independently of other to obtain $X(x)$ and $Y(y)$. From that, we can easily find $U(x,y)$.


So as you can see, for the method of separation of variables, **it is crucial that you are able to split the variables apart** so that you can solve each differential equation independently of other variables. This same technique can be applied to solve the Schrödinger equation for the particle in the $3$D box. However when we try to do this for the hydrogen atom we get a problem. The **Schrödinger equation for the hydrogen atom using Cartesian co-ordinates** looks like this: 
$$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right)\psi + \frac{2m}{\hbar^2}\left(\frac{e^2}{r}+E\right)\psi = 0 $$ 


This is a partial differential equation in three variables and we suspect that we need use separation of variables technique to solve it. So lets assume that $\psi(x,y,z) \equiv X(x)~Y(y)~Z(z)$. Now all we have to do is substitute $\psi$ with that and then separate the $x$, $y$ and $z$ variables so that they are together then solve them independently. 


But there is one problem. In the equation, $r$ which is the radius is equal to $\sqrt{x^2+y^2+z^2}$ That is a problem because due to the square root sign there is no way we can separate the $x$, $y$ and $z$ variables. 


Therefore to solve this equation more easily, we use polar co-ordinates. By using polar co-ordinates, **we take advantage of its spherical symmetry which is convenient as the entire system also has a spherical symmetry.** So we now transform the Cartesian Schrödinger equation to polar co-ordinates, so that it is dependent on the variables, $r, \theta, \phi$. The Schrödinger equation using polar co-ordinates is:
$$
 \frac{1}{r^2}\frac{\partial}{\partial r}
 \left(r^2 \frac{\partial}{\partial r}\psi\right) 
 + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta}
 \left(\sin\theta \frac{\partial}{\partial \theta}\psi\right) 
 + \frac{1}{r^2\sin^2\theta} \frac{\partial^2}{\partial \phi^2}\psi 
 + \frac{2m}{\hbar^2}\left(\frac{e^2}{r} + E\right)\psi = 0
$$


This may seem really complicated at first, however with some manipulation, we will be able to separate out all three variables, $r, \theta, \phi$.


We assume that $\psi ~(r, \theta, \phi) \equiv R(r)~\Theta(\theta)~\Phi(\phi)$. For simplicity we will write them as $R, \Theta, \Phi$ without showing the functionality. The Schrödinger equation becomes:


$$
 \frac{1}{r^2}\frac{\partial}{\partial r}
 \left(r^2 \frac{\partial R\ \Theta\ \Phi}{\partial r}\right) 
 + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta}
 \left(\sin\theta \frac{\partial R\ \Theta\ \Phi}{\partial \theta}\right) 
 + \frac{1}{r^2\sin^2\theta} \frac{\partial^2 R\ \Theta\ \Phi}{\partial \phi^2} 
 + \frac{2m}{\hbar^2}\left(\frac{e^2}{r} + E\right)R\ \Theta\ \Phi = 0
$$
This simplifies to:
$$
 \frac{\Theta\ \Phi}{r^2}\frac{\partial}{\partial r}
 \left(r^2 \frac{\partial R\ }{\partial r}\right) 
 + \frac{R\ \Phi}{r^2 \sin\theta} \frac{\partial}{\partial \theta}
 \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right) 
 + \frac{R\ \Theta}{r^2\sin^2\theta} \frac{\partial^2 \Phi}{\partial \phi^2} 
 + \frac{2m}{\hbar^2}\left(\frac{e^2}{r} + E\right)R\ \Theta\ \Phi = 0
$$
Divide both sides by $R\ \Theta\ \Phi$ and multiply by $r^2\sin^2\theta$:
$$
 \frac{\sin^2\theta}{R}\frac{\partial}{\partial r}
 \left(r^2 \frac{\partial R\ }{\partial r}\right) 
 + \frac{\sin\theta}{\Theta} \frac{\partial}{\partial \theta}
 \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right) 
 + \frac{1}{\Phi} \frac{\partial^2 \Phi}{\partial \phi^2} 
 + \frac{2mr^2\sin^2\theta}{\hbar^2}\left(\frac{e^2}{r} + E\right) = 0
$$
Inspection shows that no other terms except for the $3$ terms has a functional dependence on $\phi$. Therefore with respect to the other variables, the third term can be treated as a constant, which we will call $-n^2$ for convenience. So we obtain: $$\frac{1}{\Phi} \frac{\partial^2 \Phi}{\partial \phi^2} = -n^2$$ The general equation to this can be easily solved. So we have separated one of the three variables. Now the equation can be written in the form:
$$
 \frac{\sin^2\theta}{R}\frac{\partial}{\partial r}
 \left(r^2 \frac{\partial R\ }{\partial r}\right) 
 + \frac{\sin\theta}{\Theta} \frac{\partial}{\partial \theta}
 \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right) 
 + -n^2
 + \frac{2mr^2\sin^2\theta}{\hbar^2}\left(\frac{e^2}{r} + E\right) = 0
$$
Dividing by $\sin^2\theta$ and rearranging: 
$$
 \frac{1}{R}\frac{\partial}{\partial r}
 \left(r^2 \frac{\partial R\ }{\partial r}\right) 
 + \frac{2mr^2}{\hbar^2}\left(\frac{e^2}{r} + E\right)
 + \frac{1}{\Theta\sin\theta} \frac{\partial}{\partial \theta}
 \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right) 
 + \frac{-n^2}{\sin^2\theta}
 = 0
$$
Now by inspection, the first two terms depend on $r$ while the last two terms depend on $\theta$. As before, we will set the two terms equal to a constant, $\beta$, so that:
$$
 \frac{1}{R}\frac{\partial}{\partial r}
 \left(r^2 \frac{\partial R\ }{\partial r}\right) 
 + \frac{2mr^2}{\hbar^2}\left(\frac{e^2}{r} + E\right) = -\beta
$$
$$
 \frac{1}{\Theta\sin\theta} \frac{\partial}{\partial \theta}
 \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right) 
 + \frac{-n^2}{\sin^2\theta} = -\beta
$$
Now we have successfully separated all 3 variables into 3 separate **single variable** differential equations that can be solved relatively easily. 


 10

For an extensive answer, check out [Nanoputian’s one](https://chemistry.stackexchange.com/a/39688/7475).


In a nutshell it boils down to the following:


* Mathematically, it does not matter which coordinate system you use.
* Since your entire system is of spherical symmetry (the effective charge decreases proportionally to $\frac{1}{r}$) you benefit from choosing a coordinate system which has the same inherent symmetry.
* Cartesian coordinates *do not* have this symmetry why polar coordinates *do.*
* Thus, polar coordinates allow for easier calculations.


 7