Ch 5: Central Potential

For this example, we’ll explore 1 particle systems under the influence of an external force field with Potential

(1)   \begin{equation*} V(r)=\frac{q1*q2}{4\pi \epsilon _0*r} \end{equation*}

Where q1 is the charge of an electron, and q2 is the charge of a simulated atomic nucleus. Since we’re working in Atomic Units, the Potential reduces to

(2)   \begin{equation*} V(r)=-\frac{N}{r} \end{equation*}

Where N is the number of protons in the nucleus (for Hydrogen N=1). The equation for the electron in a Hydrogen atom becomes:

(3)   \begin{equation*} i\frac{\partial \Psi }{\partial t}+\frac{1}{2}\nabla ^2\Psi +\frac{1}{r}\Psi =0 \end{equation*}

The way to solve this is to express it in Spherical Coordinates, apply Separation of Variables (as usual) and lots of Algebra. Separable solutions wind up having the form:

    \[ \Psi _{n,l,m}(r,\theta ,\phi ) = \sqrt{(\frac{2}{n})^3\frac{(n-l-1)!}{2n ((n+l)!)}}*(\frac{2r}{n})^le^{-r/(n)}* \]

(4)   \begin{equation*} LaguerreL(n-l-1,2l+1,2r/n)*SphericalHarmonicY(l,m,\theta ,\phi ) \end{equation*}

There are many interesting avenues that we could explore here. For now, I’m just going to show a few of the possible states.

To start with, let’s see the ground state:

(5)   \begin{equation*} \Psi _{1,0,0}(r,\theta ,\phi ) = \frac{1}{\sqrt{\pi }}e^{-r+\frac{i t}{2}} \end{equation*}

Figure 5.1: Ground State

Figure 5.1: Ground State

Next, let’s check out \Psi_{2,1,0}:

(6)   \begin{equation*} \Psi _{2,1,0}(r,\theta ,\phi ) = \frac{e^{-\frac{r}{2}+\frac{i t}{8}} r \text{Cos}[\theta ]}{4 \sqrt{2 \pi }} \end{equation*}

Figure 5.2: Psi[2,1,0]

Figure 5.2: Psi[2,1,0]

And, \Psi_{2,1,1}:

(7)   \begin{equation*} \Psi _{2,1,1}(r,\theta ,\phi ) = -\frac{e^{i \text{ph}-\frac{r}{2}+\frac{i t}{8}} r \text{Sin}[\text{th}]}{8 \sqrt{\pi }} \end{equation*}

Figure 5.3: Psi[2,1,1]

Figure 5.3: Psi[2,1,1]

Notice that this state does have nonzero fluid momentum, despite having a constant density (and being an energy eigenstate, where nothing is supposed to happen).

(8)   \begin{equation*} Momentum=\vec{p}=\nabla S= Im[\frac{\nabla \Psi }{\Psi }]= 0 \overset{\wedge }{r}+0 \overset{\wedge }{\theta }+\frac{\text{Csc}[\text{th}]}{r} \overset{\wedge }{\phi } \end{equation*}

The actual swirling motion of the electron fluid explains why this state posses a magnetic moment. Here’s a plot of the momentum field.

Figure 5.4: Momentum field of Psi[2,1,1]

Figure 5.4: Momentum field of Psi[2,1,1]

And here’s an animation showing the motion (all animations in this chapter take place over 500 Atomic Time units, or about 1 80 Trillionth of a second):

By combining Eigenstates, we can get some pretty impressive motion. Here are a few:

Below is \Psi_{1,0,0}+\Psi_{2,0,0}. Spherically symmetric states bounce back and forth:

Below is \Psi_{2,1,0}+\Psi_{3,0,0}. States without a magnetic moment do not spin, but tend to bounce up and down.:

Below is \Psi_{2,1,1}+\Psi_{3,1,0}:

Below is \Psi_{2,1,1}+\Psi_{3,1,1}. Note the similarity to the Classical system from Chapter 2. Not a coincidence.:

Below is \Psi_{3,1,1}+\Psi_{4,3,3}. States with 2 different magnetic moments have sections of the fluid that spin at different rates.:

Below is \Psi_{3,2,-2}+\Psi_{4,3,3}.

 

For more animations, check out the Gallery.

 

Chapter 6

Sources

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