Need help understanding this figure on energy levels

[hmparticle9]

Homework Statement

I am having trouble interpreting this figure on energy levels

Relevant Equations

$$E_{Nl} = \frac{h^2}{2ma}\beta_{Nl}^2$$

$\beta_{Nl}$ is the $N^{th}$ zero of the $l^{th}$ spherical Bessel function.

This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure.

After the equation (4.50) it says

"It is customary to introduce the principal quantum number, $n$, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)"

I still don't understand the figure :(

Here is another figure from page 149, but I found an equation to relate $N,n,l$ in this case: $n= N +l$

But it is not the case that $n= N +l$

 

[kuruman]

The figure allows you to associate an energy level of an infinite spherical potential with its appropriate spherical Bessel function.

You have the spherical Bessel function that Griffiths writes as $j_l(x).$
For a given $l$, the function has zeros at different values of $x$ that are computed numerically.
Griffiths defines $\beta_{Nl}$ as the $N$th zero of the $l$th spherical Bessel function.

Look at the energy ladder that starts at the label $N=1$. This is the ladder of first zeroes (same value of $N$) for the Bessel functions $j_l(x)$ where the value of $l$ is shown on the horizontal axis. The first zero of $l=1$ has energy proportional to $\beta_{11}^2$ and is represented by the first tick mark on the vertical scale. The first zero of $l=2$ has energy proportional to $\beta_{12}^2$ and is represented by the second tick mark on the vertical scale. And so on for the rest of the $N=1$ ladder.

Similarly for the $N=2$ ladder and so on. The left vertical axis orders the energy levels in ascending label $n$ which, as Griffiths says, "simply orders the allowed energy levels". You can see for example that the $n=6$ energy level corresponds to the second zero ($N=2$) of $j_1(x).$

P.S. I own the Second Edition which does not have the figure you posted.

 

[kuruman]

Here is another figure from page 149, but I found an equation to relate $N,n,l$ in this case: $n= N +l$

View attachment 363568
But it is not the case that $n= N +l$

There can be no comparison here. The infinite spherical potential is not the same as Coulomb potential. The radial wavefunctions are completely different.

 

[hmparticle9]

@kuruman

Let us first look at the first image and your post #2. Your explanation makes perfect sense :) Thank you again for your help.

The equation 4.50 says $E_{Nl} = \frac{h^2}{2ma^2}\beta_{Nl}^2$. It all makes sense.

In the case of the second image, what is $N$ and what is $l$? Is it still the case that $n$ is the principle quantum number, which simply orders the allowed energies?

 

[kuruman]

In the case of the second image, what is $N$ and what is $l$? Is it still the case that $n$ is the principle quantum number, which simply orders the allowed energies?

I don't have that figure in my edition. I don't know what $N$ represents; read Griffiths and see how he defined it. You can see from the figure that here $N=n$ in other words the ordering of the energy levels is independent of the value of $l$ and $N$ is the same as the principal quantum number. You knew that already (I hope) because the energy levels for the hydrogen atom are given by $E_n=-\dfrac{13.6~\text{eV}}{n^2}.$

 

[hmparticle9]

I think I get it.

For instance if $n=2$, then from $n = N + l$ we have either $N = 2, l = 0$ or $N = 1, l = 1$. These match the bound states on the horizontal line marked $n= 2$.

If $n=3$ then we have $N = 1, l = 2$, $N = 2, l = 1$ or $N = 3, l = 0$ Again, these match the bound states on the horizontal line marked $n=3$

I suffer with a learning difficulty, sometimes I just need to talk about things for it to click.

 

[kuruman]

I suffer with a learning difficulty, sometimes I just need to talk about things for it to click.

We are here to help.