# Expression of the potential in the hydrogen atom while solving radial part of the wave function

I was reading about the derivation of wavefunction of Hydrogen atom from Atkins book. After the separation of variables and writing the wave function, $$\psi_{(r,\theta,\phi)}=R_{(r)}Y_{(\theta,\phi)}$$, and further solving the equation, they reach the expression of potential as

$$V_{eff}=-\frac{Ze^2}{4\pi\epsilon_0r}+\frac{l(l+1)\hbar^2}{2\mu r^2}$$

The first term $$(-\frac{Ze^2}{4\pi\epsilon_0r})$$ represent the electrostatic potential between the electron and the proton. I am not able to understand the interpretation of the second term $$(\frac{l(l+1)\hbar^2}{2\mu r^2})$$. In book it is written that it arises from centrifugal effect.

Also why the second term is considered in the potential as potential is defined for a conservative force field? I know that second term in potential comes in picture as a result of the derivation. But the second term doesn't make sense to me as in Bohr's model, in the expression of the potential we only consider the electrostatic potential. Please explain about the second term in potential.

• Stop thinking of it as a potential for a moment. We separated the variables, and we got this; which step of it do you disagree with? Jun 21, 2020 at 11:41
• Basically the problem is that while doing the derivation, we got this. So the expression of the potential is mathematically consistent. But I wonder that second term in potential basically is $L^2/2I$, and while solving the hamiltonian of angular part of the wave function (taken under the different heading as "particle on sphere" in some books) $L^2/2I$ is included in the energy of it.
– Manu
Jun 21, 2020 at 12:04
• So why we again consider $L^2/2I$ in radial part of energy. Intuitively only radial kinetic energy and electrostatic potential energy should be considered in the total energy. I know that my intution is wrong as it doesn't met with the mathematical analysis of the radial Hamiltonian equation. But I am not able to find the reason for the same.
– Manu
Jun 21, 2020 at 12:04
• That's why I started like this: stop thinking of it as a potential. Once you've separated the variables, there are no energy, potential, or other words with direct physical meaning. There is just the mathematical derivation, and it is what it is. Follow the steps, check that the result is right. So it goes. Why people call it the effective potential is another story. Jun 21, 2020 at 12:20
• There is only one energy and no innate way to split it into the radial and angular parts. You may do so, but that's a matter of convenience, not the law of nature. Jun 21, 2020 at 12:40

The 'centrifugal' term appears when the coordinates are changed from $$(x,y,z)$$ to spherical $$(r, \theta, \varphi)$$. This is not to say that it is artificial, or a mathematical trick, but is a consequence of the fact that the electron has orbital angular momentum, and this is made clear by changing coordinate systems.

To summarize: the equation is $$H\varphi(\pmb{r})=E\varphi(\pmb{r})$$ where $$\displaystyle H=-\frac{\hbar^2}{2\mu}\Delta +V(r)$$. Changing to spherical coordinates produces,

$$\Delta =-\frac{1}{r}\frac{\partial ^2}{\partial r^2}+\frac{1}{r^2}\left(\frac{\partial ^2}{\partial \theta^2}+\frac{1}{\tan(\theta)}\frac{\partial}{\partial \theta}+\frac{1}{\sin^2(\theta)}\frac{\partial^2}{\partial \varphi^2} \right)$$

which means that eigenfunctions of $$\varphi(\pmb{r})$$ are variables in $$r, \theta, \varphi$$

The second term is almost the same as that for $$\pmb{L^2}$$ and substituting for this makes the Hamiltonian

$$\displaystyle H=-\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{\partial ^2}{\partial r^2}r+ \frac{1}{2\mu r^2}\pmb{L^2}+V(r)$$

As $$\pmb{L^2}\varphi(\pmb{r})=l(l+1)\hbar^2\varphi(\pmb{r})$$ then

$$\displaystyle H=-\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{\partial ^2}{\partial r^2}r + \frac{l(l+1)\hbar^2}{2\mu r^2}+V(r)$$

The potential is unchanged and still $$V(r) = -e^2/r$$ and the Hamiltonian is still a function of $$r$$ but now depends on variable $$l$$ thus the eigenvalues must depend on $$l$$.

The centrifugal term is so called because the corresponding force, the negative slope, always repels the electron from the centre and labelled a centrifugal potential or barrier. The term is always positive or zero and can change the shape of the potential the electron experiences and thus forms an effective potential written as $$\displaystyle V_{eff}=\frac{l(l+1)\hbar^2}{2\mu r^2}+V(r)$$. When $$l>0$$ the centrifugal term dominates at small $$r$$ causing $$V_{eff}$$ to be repulsive. As larger $$r$$ the potential becomes attractive.