# 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.

• In short, that's because the problem has spherical symmetry, and you want to use the coordinates which retain as much as possible of that symmetry. Not a chemical question, really. – Ivan Neretin Oct 26 '15 at 10:08
• I'm voting to close this question as off-topic because it's purely about the mathematical methods used to solve equations and not about conceptual chemistry. – bon Oct 26 '15 at 10:14
• I am not sure if this would be eligible for migration though. – Martin - マーチン Oct 26 '15 at 11:28
• I’m going to vote to leave open even though it is a maths/physics question technically. Reasoning being that as far as I know all the basic quantum stuff is usually handled in chemistry courses. – Jan Oct 26 '15 at 12:39
• The question is really just wrong. There is no "need" to use polar coordinates, it is just simpler to do the math in polar coordinates. – MaxW Oct 28 '15 at 6:53

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.

• But there is still an $r$ in the expression (in $\tfrac{e^2}{r}$), and that's the same $\sqrt{x^2+y^2+z^2}$ you mentioned as a problem for the Cartesian version, right? – Yoda Oct 27 '15 at 18:35
• @AndersMB Yes, there is still the same $r$ (and it still equals $\sqrt{x^2+y^2+z^2}$) in the expression,. However the beauty of polar co-ordinates is that the variables depend on $r, ~\theta, ~\phi$ not $x, ~y,~ z$. Therefore we just have to group all the $r$'s together then solve the equation independently. There is no need to split $r$ into terms of $x,~y,~z$ as they aren't variables in the polar equation. – Nanoputian Oct 28 '15 at 3:34

For an extensive answer, check out Nanoputian’s one.

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.