1.27: The Dirac Notation Applied to Variational Calculations

The particle-in-a-box problem is exactly soluble and the solution is calculated below for the first 20 eigenstates. All calculations will be carried out in atomic units.

\[\psi(n,x) = \sqrt{2} \sin(n \pi x) \nonumber \]

\[ E_n = \dfrac{n^{2} \pi^{2}}{2} \nonumber \]

with \(n = 1, 2, ..., 20\)

The First five EigenValues

The first five energy eigenvalues are:

\(E_{1} = 4.935\) \(E_{2} = 19.739\) \(E_{3} = 44.413\) \(E_{4} = 78.957\) \(E_{5} = 123.37\)

The first three eigenfunctions are displayed below:

%matplotlib inline

import matplotlib.pyplot as plt
import numpy as np

t = np.linspace(0,1,100)
t1 = t*np.pi
t2 = t*np.pi*2
t3 = t*np.pi*3
a = np.sin(t1)
b = np.sin(t2)
c = np.sin(t3)
plt.xlim(0,1)
plt.plot(t,a,color = "red", label= "\u03C8 (1,x)")
plt.plot(t,b,color = "blue",label = "\u03c8 (2,x)")
plt.plot(t,c,color = "limegreen",label = "\u03c8 (3,x)")
plt.plot(t,(t*0), color ="black")

plt.xticks([0,0.5,1])
plt.yticks([-.5,0,.5],[])
plt.xlabel("x")
leg = plt.legend(loc = "center", bbox_to_anchor=[-.11,.5],frameon=False)
plt.tick_params(top=True,right=True,direction="in")

plt.show()

The set of eigenfunctions forms a complete basis set and any other functions can be written as a linear combinations in this basis set. For examples, \(\Phi\), \(\chi\), and \(\Gamma\) are three trial functions that satisfy the boundary conditions for the particle in a 1 bohr box.

\[\Phi(x) = \sqrt{30}(x-x^{2}) \nonumber \]

\[\chi(x) = \sqrt{105}(x^{2}-x^{3}) \nonumber \]

\[\Gamma(x) = \sqrt{105}x(1-x)^{2} \nonumber \]

In Dirac bra-ket notation we can express and of these functions as a linear combination in the basis set as follows:

\[ \begin{align} \langle x | \Phi \rangle &= \sum_{n}^{\infty} \langle x | \psi_{n} \rangle \langle \psi_{n} | \Phi \rangle \\[4pt] &= \sum_{n}^{\infty} \langle x | \psi_{n} \rangle \int_{0}^{1} \langle \psi_{n} | x \rangle \langle x | \Phi \rangle dx \end{align} \nonumber \]

The various overlap integral for the three trial function are evaluated below.

\[a_{n} = \int_{0}^{1} \psi(n,x) \Phi(x)dx \nonumber \]

\[b_{n} = \int_{0}^{1} \psi(n,x) \chi(x)dx \nonumber \]

\[c_{n} = \int_{0}^{1} \psi(n,x) \Gamma(x)dx \nonumber \]

%matplotlib inline

import matplotlib.pyplot as plt
import numpy as np
import math

t = np.arange(0,1,.001)
plt.plot(t,math.sqrt(2)*np.sin(t*np.pi),color = "red", label = "\u03C8 (1,x)")
plt.plot(t,math.sqrt(30)*(t-t**2),color ="blue", linestyle = "--",label = "\u03A6(x)")
plt.plot(t,math.sqrt(105)*t*(1-t)**2,color = "lime", linestyle = "--", label = "\u0393 (x)")
plt.plot(t,math.sqrt(105)*(t**2 - t**3),color = "magenta", linestyle = "-.", label = "\u03A7 (x)")
plt.xticks([0.2,0.4,0.6,0.8])
plt.yticks([.5,1,1.5],[ ])
plt.tick_params(direction="in")
plt.xlabel("x")
leg = plt.legend(loc = "center",bbox_to_anchor=[-.11,.5], frameon=False)
plt.tick_params(top=True, right=True)
plt.xlim(0,1)
plt.ylim(0,2)
plt.show()

The figure shown below demonstrate that only \(\Phi\) is a reasonable representative for the ground state wavefunction.

First Five Particle in a BOx EigenFunctions

If \(\chi\) is written as a linear combination of the first 5 PIB eigenfunctions, one gets two functions that are essentially indistinguishable from one another.

The same, of course, is true for \(\chi\) and \(\Gamma\), as is demonstrated in the graphs shown below.

Traditionally we use energy as a criterion for the quality of a trial wavefunction by evaluating the variational integral in the following way.

\[
\int_{0}^{1} \Phi(x)-\frac{1}{2} \cdot \frac{d^{2}}{d x^{2}} \Phi(x) d x=5 \quad \int_{0}^{1} \chi(x)-\frac{1}{2} \cdot \frac{d^{2}}{d x^{2}} \chi(x) d x=7 \quad \int_{0}^{1} \Gamma(x) \cdot \frac{1}{2} \cdot \frac{d^{2}}{d x^{2}} \Gamma(x) d x=7
\nonumber \]

In Dirac notation we write:

\[
\langle E\rangle=\langle\Phi|\hat{H}| \Phi\rangle=\sum_{n}\langle\Phi|\hat{H}| \Psi_{n}\rangle\left\langle\Psi_{n} | \Phi\right\rangle=\sum\langle\Phi | \Psi_{n}\rangle E_{n}\left\langle\Psi_{n} | \Phi\right\rangle=\sum_{n} a_{n}^{2} E_{n}
\nonumber \]

Thus we easily show the same result.

\[
\sum_{\mathrm{n}}\left[\left(\mathrm{a}_{\mathrm{n}}\right)^{2} \cdot \mathrm{E}_{\mathrm{n}}\right]=5 \quad \sum_{\mathrm{n}}\left[\left(\mathrm{b}_{\mathrm{n}}\right)^{2} \cdot \mathrm{E}_{\mathrm{n}}\right]=6.999 \quad \sum_{\mathrm{n}}\left[\left(\mathrm{c}_{\mathrm{n}}\right)^{2} \cdot \mathrm{E}_{\mathrm{n}}\right]=6.999
\nonumber \]

We now show, belatedly, that the three trial functions are normalized by both methods.

\[
\int_{0}^{1} \Phi(x)^{2} d x=1 \quad \int_{0}^{1} \chi(x)^{2} d x=1 \quad \int_{0}^{1} \Gamma(x)^{2} d x=1
\nonumber \]

In Dirac notation this is formulated as:

\[
\langle\Phi | \Phi\rangle=\sum_{n}\langle\Phi | \Psi_{n}\rangle\left\langle\Psi_{n} | \Phi\right\rangle=\sum_{n} a_{n}^{2}
\nonumber \]

\[
\sum_{\mathrm{n}}\left(a_{n}\right)^{2}=1 \quad \sum_{n}\left(b_{n}\right)^{2}=1 \quad \sum_{n}\left(c_{n}\right)^{2}=1
\nonumber \]

We now calculate some over-lap integrals:

\[
\int_{0}^{1} \Phi(x) \cdot \chi(x) d x=0.935 \quad \quad \int_{0}^{1} \Phi(x) \cdot \Gamma(x) d x=0.935 \quad \int_{0}^{1} \chi(x) \cdot \Gamma(x) d x=0.75
\nonumber \]

In Dirac notation this is formulated as:

\[
\langle\Phi | \Gamma\rangle=\sum_{n}\langle\Phi | \Psi_{n}\rangle\left\langle\Psi_{n} | \Gamma\right\rangle=\sum_{n} a_{n} c_{n}
\nonumber \]

\[
\sum_{\mathrm{n}}\left(\mathrm{a}_{\mathrm{n}} \cdot \mathrm{b}_{\mathrm{n}}\right)=0.935 \quad \sum_{\mathrm{n}}\left(\mathrm{a}_{\mathrm{n}}-\mathrm{c}_{\mathrm{n}}\right)=0.935 \quad \sum_{\mathrm{n}}\left(\mathrm{b}_{\mathrm{n}}-\mathrm{c}_{\mathrm{n}}\right)=0.75
\nonumber \]