# What is the eigenvalue of the Hamiltonian?

##### 1 Answer
May 2, 2015

The eigenvalue is an energy result you get back such that it is the same every time you examine the same energy level.

The rest below is a bunch of Physical Chemistry.

First of all the Hamiltonian has many different forms. If, for example, we are using the basic particle-in-a-box model in one dimension (left or right), the Hamiltonian operator for quantized quantum number $n$ can be written like so:

$\hat{H} = - {h}^{2} / \left(8 {\pi}^{2} m\right) \left[\frac{{\partial}^{2}}{\partial {x}^{2}} + V \left(x\right)\right] {\psi}_{n} \left(x\right)$

$\frac{{\partial}^{2}}{\partial {x}^{2}}$ is called the second derivative. It operates on (affects) ${\psi}_{n} \left(x\right)$, where ${\psi}_{n} \left(x\right)$ is the wave function involving the quantum number $n$. ${\psi}_{n} \left(x\right)$ describes the state of a quantum mechanical system at energy level $n$. A quantum mechanical system is small enough that classical mechanics cannot properly describe it, such as that of an electron.

In a particle-in-a-box model, $V \left(x\right)$ is the potential energy of the electron in the box (orbital), and it is 0 because it is assumed that the electron is elastically colliding with its boundaries, x = 0 and x = a, where a is the length of the box.

The electron is assumed to be confined within the box and to only have kinetic energy. It cannot get out of the box because the box is said to be infinitely tall such that the 1D assumption is valid.

(Note that the electron can still get out 100% of the time if it has twice the energy it needs to get out. SCIENCE!)

Kinetic energy is represented by:

$K \left(x\right) = - {h}^{2} / \left(8 {\pi}^{2} m\right) \left[\frac{{\partial}^{2}}{\partial {x}^{2}}\right]$

The eigenvalue is a result you get back such that it is the same every time you examine the same energy level. This is important because you don't know how fast the electron is moving if you know where it is, and vice versa. For example, the condensed, time-independent Schroedinger equation is:

$\hat{H} {\psi}_{n} \left(x\right) = {E}_{n} {\psi}_{n} \left(x\right)$

where ${E}_{n}$ is the energy value at the energy level of quantum number $n$.

The full time-independent Schroedinger equation for the Hydrogen atom is (now we're getting serious!):

$- {h}^{2} / \left(8 {\pi}^{2} m\right) \left[\frac{{\partial}^{2}}{\partial {x}^{2}}\right] {\psi}_{n} \left(x\right) = {E}_{n} {\psi}_{n} \left(x\right)$

Now, the operator always operates to whatever is on its right. The wave function ${\psi}_{n} \left(x\right)$ is the following for the 1D particle-in-a-box model:

${\psi}_{n} \left(x\right) = \sqrt{\frac{2}{a}} \sin \left(\frac{n \pi x}{a}\right)$

Its second derivative is therefore, after two Chain Rule applications:

$\frac{{\partial}^{2}}{\partial {x}^{2}} \left\{\sqrt{\frac{2}{a}} \sin \left(\frac{n \pi x}{a}\right)\right\} = - \sqrt{\frac{2}{a}} \sin \left(\frac{n \pi x}{a}\right) \cdot \frac{{n}^{2} {\pi}^{2}}{a} ^ 2$
$= \frac{- {n}^{2} {\pi}^{2}}{a} ^ 2 \cdot \sqrt{\frac{2}{a}} \sin \left(\frac{n \pi x}{a}\right)$

When you evaluate the Schroedinger equation, you have to use the following integral:

${\int}_{0}^{a} {\psi}_{n}^{\text{*}} \left(x\right) \hat{H} {\psi}_{n} \left(x\right) \mathrm{dx}$

where ${\psi}_{n}^{\text{*}} \left(x\right)$ is the complex conjugate of ${\psi}_{n} \left(x\right)$. In this case, it doesn't matter because there is no $i$ in the function.

The integral is evaluated like follows.

${\int}_{0}^{a} {\psi}_{n}^{\text{*}} \left(x\right) \hat{H} {\psi}_{n} \left(x\right) \mathrm{dx}$

$= {\int}_{0}^{a} \sqrt{\frac{2}{a}} \sin \left(\frac{n \pi x}{a}\right) \left[\frac{- {h}^{2}}{8 {\pi}^{2} m} \frac{{\partial}^{2}}{\partial {x}^{2}}\right] \sqrt{\frac{2}{a}} \sin \left(\frac{n \pi x}{a}\right) \mathrm{dx}$

Operate on the rightmost function:

$= {\int}_{0}^{a} \sqrt{\frac{2}{a}} \sin \left(\frac{n \pi x}{a}\right) \frac{- {h}^{2}}{8 {\pi}^{2} m} \frac{{n}^{2} {\pi}^{2}}{a} ^ 2 \cdot \sqrt{\frac{2}{a}} \cdot - \sin \left(\frac{n \pi x}{a}\right) \mathrm{dx}$

Pull all the constants out of the integral:

$= \left(\frac{- {h}^{2}}{8 {\pi}^{2} m} \cdot \frac{2}{a}\right) \left(\frac{- {n}^{2} {\pi}^{2}}{a} ^ 2\right) {\int}_{0}^{a} \sin \left(\frac{n \pi x}{a}\right) \cdot \sin \left(\frac{n \pi x}{a}\right) \mathrm{dx}$

Simplify a little:

$= \frac{{h}^{2} {n}^{2} {\pi}^{2}}{4 {\pi}^{2} m {a}^{3}} {\int}_{0}^{a} {\sin}^{2} \left(\frac{n \pi x}{a}\right) \mathrm{dx}$

Use the identity for ${\sin}^{2} \left(x\right) = \frac{1 - \cos \left(2 x\right)}{2}$ and pull the 2 out of the integral:

$= \frac{{h}^{2} {n}^{2} {\pi}^{2}}{8 {\pi}^{2} m {a}^{3}} {\int}_{0}^{a} 1 - \cos \left(\frac{2 n \pi x}{a}\right) \mathrm{dx}$

Solve and cancel out wherever you have $\sin \left(k n \pi\right) = 0$ (when integer $k = 0$ or $2 n$), plugging in $a$ and doing some subtraction:

$= \frac{{h}^{2} {n}^{2} {\pi}^{2}}{8 {\pi}^{2} m {a}^{3}} \left[\left(a - \cancel{\sin \left(\frac{2 n \pi a}{a}\right) \cdot \frac{a}{2 n \pi}}\right) - \left(\cancel{0 - \sin \left(\frac{2 n \pi 0}{a}\right) \cdot \frac{0}{2 n \pi}}\right)\right]$

Thus, the eigenvalue we've been after the whole time in the particle-in-a-box model, for the Hydrogen atom, is:

$\textcolor{b l u e}{{E}_{n}} = \frac{{h}^{2} {n}^{2} \cancel{{\pi}^{2}}}{8 \cancel{{\pi}^{2}} m {a}^{2}}$

$= \textcolor{b l u e}{\frac{{h}^{2} {n}^{2}}{8 m {a}^{2}}}$

where $h$ is Planck's constant, $n$ is the quantized quantum number, $m$ is the mass of the electron, and $a$ is the width of the boundaries on the "box".

This is the exact energy for the ground-state Hydrogen atom.