Chapter 13 - Hydrogen functions¶

Start with some imports fro Symbolic Python library:

In [1]:

from sympy.physics.hydrogen import R_nl
from sympy.functions.special.spherical_harmonics import Ynm
from sympy import *

Define some variables, radial, polar, azimuthal, time, and two frequencies:

In [2]:

var("r theta phi t w1 w2")

Out[2]:

(r, theta, phi, t, w1, w2)

Look at a few of the radial equations and the spherical harmonics¶

Notice that instead of Ylm the name is Ynm... the arguments to the function are still the quantum numbers l and m

In [3]:

R_nl(1, 0, r, 1)  # the n = 1, l = 0 radial function

Out[3]:

$\displaystyle 2 e^{- r}$

In [4]:

Ynm(0,0,theta,phi).expand(func=True)  # the l = 0, m = 0 spherical harmonic

Out[4]:

$\displaystyle \frac{1}{2 \sqrt{\pi}}$

Write the equation for the $|nlm\rangle = |100\rangle$ state. Use the sympy method .expand(func=True) to convert to the actual expression. To create this state, we combine the Radial function and the Ylm function. Make sure to set n, l, and m to the correct values. The fourth argument to R_nl is Z which we set to 1 since we are talking about a 1-proton nucleus.

The combination of R_nl and Ynm should look like the following (replace N, L, and M with the appropriate values):

R_nl(N, L, r, 1)*Ynm(L, M, theta, phi).expand(func=True)

In [5]:

# this is the |100> state:
psi100 = R_nl(1, 0, r, 1)*Ynm(0,0,theta,phi).expand(func=True)

In [6]:

psi100  # check to see how it looks as an expression

Out[6]:

$\displaystyle \frac{e^{- r}}{\sqrt{\pi}}$

Integrating over all space¶

Remember spherical coordinate integrals of function $f(r,\theta,\phi)$ over all space look like: $$\int_0^\infty\int_0^\pi\int_0^{2\pi}r^2\sin(\theta)drd\theta d\phi \,\,f(r,\theta,\phi)$$ so you alwasy need to add a factor of r**2*sin(theta) and then integrate r from 0 to infinity, theta from $0-\pi$ and phi from $0-2\pi$. As a check, you should integrate the square of the psi100 wavefunction over all space to see that it equals 1 (i.e. it is normalized)

In [7]:

integrate(r**2*sin(theta) * (psi100)**2 ,(r,0,oo),(theta,0,pi),(phi,0,2*pi))

Out[7]:

$\displaystyle 1$

Now do the $|210\rangle$ state:¶

In [8]:

psi210 = R_nl(2, 1, r, 1)*Ynm(1,0,theta,phi).expand(func=True)

In [9]:

psi210  # check how it looks

Out[9]:

$\displaystyle \frac{\sqrt{2} r e^{- \frac{r}{2}} \cos{\left(\theta \right)}}{8 \sqrt{\pi}}$

Note, if you compare these to listed solutions (for example at http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html#c3) you see that there are not any factors of $a_0$. This is because the R_nl function is defined in units of $a_0$. $a_0$ is the Bohr Radius: http://en.wikipedia.org/wiki/Bohr_radius

In [10]:

psi211 = R_nl(2, 1, r, 1)*Ynm(1,1,theta,phi).expand(func=True)
psi211

Out[10]:

$\displaystyle - \frac{r e^{- \frac{r}{2}} e^{i \phi} \sin{\left(\theta \right)}}{8 \sqrt{\pi}}$

Now calculate $\langle z \rangle$:¶

To calculate $\langle z \rangle$ we need to convert to spherical coordinates: $z = r\cos\theta$. The terms in the following integral are the $r^2\sin\theta$ then $z$ (in spherical coords) then the wave function squared.

In [11]:

expect = integrate(r**2*sin(theta)* (r*cos(theta)) * (psi100*psi100),(r,0,oo),(theta,0,pi),(phi,0,2*pi))

In [12]:

expect

Out[12]:

$\displaystyle 0$

No surprise, the average z position of the electron in the hydrogen atom is 0.

Now for problem 13.21¶

find $\langle z \rangle(t)$. Use the same integral, but add a time-dependent piece to each term in the wavefunction, add them together and multiply by the complex conjugate.

In [13]:

psi = 1/sqrt(2)*(psi100*exp(1j*w1*t) + psi210*exp(1j*w2*t))
psi_conj = 1/sqrt(2)*(psi100*exp(-1j*w1*t) + psi210*exp(-1j*w2*t))

In [17]:

outer = (psi*psi_conj).simplify()

In [18]:

outer

Out[18]:

$\displaystyle \frac{\left(\sqrt{2} r e^{r + 1.0 i t w_{1}} \cos{\left(\theta \right)} + 8 e^{\frac{r}{2} + i t w_{2}}\right) \left(\sqrt{2} r e^{r + 1.0 i t w_{2}} \cos{\left(\theta \right)} + 8 e^{\frac{r}{2} + i t w_{1}}\right) e^{- 3 r - 1.0 i t w_{1} - 1.0 i t w_{2}}}{128 \pi}$

In [19]:

expect2 = integrate(r**2 * sin(theta) * (r*cos(theta)) * outer,(r,0,oo),(theta,0,pi),(phi,0,2*pi))

In [21]:

expect2

Out[21]:

$\displaystyle 2 \pi \left(- \frac{\left(- 27390260577042432 \sqrt{2} \pi^{11} e^{3.0 i t w_{1}} e^{5.0 i t w_{2}} - 141371459682435072 \pi^{11} e^{4.0 i t w_{1}} e^{4.0 i t w_{2}} - 27390260577042432 \sqrt{2} \pi^{11} e^{5.0 i t w_{1}} e^{3.0 i t w_{2}}\right) e^{- 4.0 i t w_{1}} e^{- 4.0 i t w_{2}}}{415989582513831936 \pi^{12}} + \frac{\left(27390260577042432 \sqrt{2} \pi^{11} e^{3.0 i t w_{1}} e^{5.0 i t w_{2}} - 141371459682435072 \pi^{11} e^{4.0 i t w_{1}} e^{4.0 i t w_{2}} + 27390260577042432 \sqrt{2} \pi^{11} e^{5.0 i t w_{1}} e^{3.0 i t w_{2}}\right) e^{- 4.0 i t w_{1}} e^{- 4.0 i t w_{2}}}{415989582513831936 \pi^{12}}\right)$

In [22]:

expect2.simplify()

Out[22]:

$\displaystyle \frac{64 \sqrt{2} \left(e^{2.0 i t w_{1}} + e^{2.0 i t w_{2}}\right) e^{- 1.0 i t \left(w_{1} + w_{2}\right)}}{243}$

We need to interpret this result. First you should show that this expression is simply a constant amplitude factor times $\cos((w2-w1)t)$, in other words $\langle z \rangle$ oscillates at frequency w2-w1.

Your assignment:¶

Explore other combinations of states and draw conclusions about the z behavior from the results. You may not be able to get these expressions to simplify, but the important thing is to look for the time dependence and simplify that part.

Does $\langle z \rangle$ oscillate for any combination of two Hydrogen states $|nlm\rangle$?
Are there restrictions on what n values give oscillating $\langle z \rangle$ expressions? (hint, to keep it simple, always let one state be the n=1 state)
How does $\langle z \rangle$ change with different l and m values are used in the state?

Hints for interpreting your results:

What are the relavant frequencies in your expression for $\langle z \rangle$ and why?
Simplify one of your $\langle z \rangle$ expressions and write the time dependence in terms of the frequencies w2 and w1.

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