Schrodinger Equation of a Hydrogen Atom in a Rydberg State

Question

I am not a mathematician, so I really appreciate it if someone could explain it in a simple way. In Rydberg Atoms book by Thomas F. Gallagher, the Schroedinger equation for the H atom in atomic units is written as:

$$(-\frac{\nabla^2}{2} - \frac{1}{r})\psi=W\psi$$

If we assume that this equation is separable and write $\psi$ as a product of radial and angular functions, we get in spherical coordinates:

$$\frac{r^2}{R}[\frac{\partial^2R}{\partial r^2}+\frac{2}{r}\frac{\partial R}{\partial r}+2(W+\frac{1}{r})R]+\frac{1}{Y}[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta\frac{\partial Y}{\partial\theta})+\frac{1}{\sin^2\theta}\frac{\partial^2 Y}{\partial^2\phi}]=0$$

The two terms, which depend only on $R$ and $Y$, respectively, are independent of each other and therefore must be separately equal to $±\lambda$, where $\lambda$ is a constant. So we can write that:

$$\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta\frac{\partial \Theta}{\partial\theta})\Phi+\frac{\Theta}{\sin^2\theta}\frac{\partial^2 \Phi}{\partial^2\phi}=- \lambda\Phi\Theta $$

The solution to this equation is normalized spherical harmonics.

I have two questions:

1. I didn't understand the idea of 'since $R$ and $Y$ are independent of each other, they must be separately equal to $±\lambda$.' How can we separate the two terms and how it is related to 'independent of each other?'

2. What do we mean here by normalized?

Any help is much appreciated.

Answer 1

1. I didn't understand the idea of 'since R and Y are independent of each other, they must be separately equal to ±λ.' How can we separate the two terms and how it is related to 'independent of each other?'

2. What do we mean here by normalized?

For point 1., let's take a much simpler separable PDE as an example; Pascal's Equation:

$$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$$

Or in shorthand:

$$u_{xx}+u_{yy}=0$$

Assume a solution of the shape:

$$u(x,y)=X(x)Y(y)$$

Inserted into the original PDE and with minimum reworking we get separation of variables:

$$\frac{X''(x)}{X(x)}=-\frac{Y''(y)}{Y(y)}$$

Now, since as the LHS is a function of $x$ alone and the RHS a function of $y$ alone, the only way that statement can be true is that both expressions equal a Real Number, often denoted as $-\lambda^2$ and called the separation constant, so: $$\frac{X''(x)}{X(x)}=-\frac{Y''(y)}{Y(y)}=-\lambda^2$$

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As for your second point, I'm not sure you understand what normalisation really means.

Suppose we have a wave function $\psi(x)$, bound to the $x$ domain $[0,L]$. This means that the particle will ALWAYS be found somewhere on $[0,L]$ and NEVER outside of it.

With the Born rule this means that the probability of finding the particle on a $[x,x+\delta x]$ interval is:

$$P(\delta x)=|\psi(x)|^2\delta x$$

And the probability over the whole domain $[0,L]$:

$$\int_0^L|\psi(x)|^2\text{d}x=1$$

because the probability of finding the particle on $[0,L]$ is $1$. If this is true then $\psi(x)$ is said to be normalised.

Hope this helps.

Answer 2

Given the equation you wrote we can rearrange the two terms as follows$$\frac{r^2}{R(r)}\left[\frac{\partial^2R(r)}{\partial r^2}+\frac{2}{r}\frac{\partial R(r)}{\partial r}+2\left(W+\frac{1}{r}\right)R(r)\right]= -\frac{1}{Y(\theta,\phi)}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial Y(\theta,\phi)}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2 Y(\theta,\phi)}{\partial^2\phi}\right]$$ Now you can clearly see that the LHS is a function solely of the radial component $r$ while the RHS is only a function of the angular components $\theta,\phi$ $$f(r) = g(\theta,\phi)$$ For the two expressions to be equal they can only be equal to the same constant, which is going to be independent both on $r$ and on $\theta,\phi$. This explains why you set both equal to a given constant $\pm\lambda$.

Normalised means exactly what you would expect. If $Y(\theta,\phi)$ is a normalised solution to the equation $$\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial Y(\theta,\phi)}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2 Y(\theta,\phi)}{\partial^2\phi}=-\lambda Y(\theta,\phi)$$ then $$\int\mathrm{d}\Omega\, |Y(\theta,\phi)|^2 = 1$$