Harmonic Oscillator

Many physical systems, including molecules with configurations near their equilibrium positions, can be described (at least approximately) by the Hamiltonian of the harmonic oscillator:R4(483) R1(62), click here
$\displaystyle \hat{H} = \frac{\hat{P}^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2.$
In order to find the eigenfunctions of $ \hat{H}$ we introduce two operators called creation$ \hat{a}^+$ and annihilation$ \hat{a}$, which are defined as follows:

$ \hat{a}^+ \equiv \frac{1}{\sqrt{2}}(\tilde{x} - i \tilde{p})$, and $ \hat{a} \equiv \frac{1}{\sqrt{2}}(\tilde{x} + i \tilde{p})$, where $ \tilde{x}=\hat{x}\sqrt{\frac{m\omega}{\hbar}}$, and $ \tilde{p} = \frac{\hat{p}}{\sqrt{m\omega \hbar}}$.

Using these definitions of $ \hat{a}^+$ and $ \hat{a}$, we can write $ \hat{H}$ as follows,

$ \hat{H}=(\hat{a}^+\hat{a} + \frac{1}{2})\hbar \omega.$

Introducing the number operator $ \hat{N}$, defined in terms of $ \hat{a}^+$ and $ \hat{a}$ as follows,

$ \hat{N}\equiv \hat{a}^+\hat{a}$,

we obtain that the Hamiltonian of the Harmonic Oscillator can be written as follows,

$ \hat{H} = ( \hat{N} +1/2) \hbar \omega.$

Exercise 15: Show that if $ \Phi_{\nu}$ is an eigenfunction of $ \hat{H}$ with eigenvalue $ E_{\nu}$, then $ \Phi_{\nu}$ is an eigenfunction of $ \hat{N}$ with eigenvalue $ \nu=\frac{E_\nu}{\hbar\omega} -\frac{1}{2}$. Mathematically, if $ \hat{H}\vert\Phi_{\nu}>=E_{\nu}\vert\Phi_{\nu}>$, then $ \hat{N}\vert\Phi_{\nu}> = \nu \vert\Phi_{\nu}>$, with $ \nu=\frac{E_\nu}{\hbar\omega} -\frac{1}{2}$.

Theorem I:

The eigenvalues of $ \hat{N}$ are greater or equal to zero, i.e., $ \nu \geq 0$.

Proof:

$ \int dx \vert<x\vert\hat{a}\vert\Phi_{\nu}>\vert^2 \geq 0,$

$ <\Phi_{\nu}\vert\hat{a}^+ \hat{a}\vert\Phi_{\nu}> \geq 0,$

$ \nu <\Phi_{\nu}\vert\Phi_{\nu}> \geq 0.$

As a consequence: $ \hat{a}\vert\Phi_0> =0,$

$ \frac{1}{\sqrt{2}}[\hat{x}\sqrt{\frac{m\omega}{\hbar}} + i \frac{\hat{p}}{\sqrt{m\omega \hbar}}]\vert \Phi_0> = 0,$

$ \hat{p} = - i \hbar \frac{\partial }{\partial x},$

$ x\Phi_0 (x) + \frac{\hbar}{m\omega} \frac{\partial \Phi_0(x)}{\partial x} =0,$

$ \partial$   ln$ \Phi_0 (x) = - \frac{m\omega}{\hbar}x\partial x,$

$\displaystyle \boxed{\Phi_0(x) = A \ \text{exp}(-\frac{m\omega}{\hbar 2}x^2),}$
where $ A=\sqrt[4]{\frac{m\omega}{\pi \hbar}}$. The wave function $ \Phi_0(x)$ is the eigenfunction of $ \hat{N}$ with $ \nu =0$ (i.e., the ground state wave function because $ \nu \geq 0$).

Theorem II:

If $ \nu > 0$, state $ \hat{a}\vert\Phi_{\nu}>$ is an eigenstate of $ \hat{N}$ with eigenvalue equal to ($ \nu$ -1).

Proof:

In order to prove this theorem we need to show that,

$\displaystyle \hat{N} \hat{a} \vert\Phi_{\nu} > = (\nu-1) \hat{a} \vert \Phi_{\nu}>.$ (25)

We first observe that,

$\displaystyle [\hat{N}, \hat{a}] = -\hat{a}.$
Therefore,

$ [\hat{N}, \hat{a}] = \hat{a}^+\hat{a}\hat{a}-\hat{a}\hat{a}^+\hat{a},$

$ [\hat{N}, \hat{a}] = [\hat{a}^+, \hat{a}] \hat{a},$

$ [\hat{N}, \hat{a}] = -1 \hat{a}$, because $ [\hat{a}^+, \hat{a}] = -1,$

$ [\hat{a}^+, \hat{a}] = \frac{1}{2\hbar}(\hat{x}\hat{x}+i\hat{x}\hat{p}-i\hat{p......\hat{p} - (\hat{x}\hat{x} - i\hat{x}\hat{p} +i\hat{p}\hat{x} +\hat{p}\hat{p})),$

$ [\hat{a}^+, \hat{a}] = \frac{i}{2 \hbar}2 [\hat{x}, \hat{p}] = -1,$   since$ \ [\hat{x}, \hat{p}] = i\hbar.$

Applying the operator $ -\hat{a}$ to state $ \vert\Phi_{\nu}>$ we obtain,

$ (\hat{N}\hat{a} - \hat{a}\hat{N}) \vert \Phi_{\nu}>= -\hat{a} \vert\Phi_{\nu}>,$

and, therefore,

$ \hat{N}\hat{a}\vert \Phi_{\nu}> - \hat{a} \nu\vert \Phi_{\nu}> = -\hat{a} \vert\Phi_{\nu}>$, which proves the theorem.

A natural consequence of theorems I and II is that $ \nu$ is an integer number greater or equal to zero. The spectrum of $ \hat{N}$ is therefore discrete and consists of integer numbers that are $ \geq 0$. In order to demonstrate such consequence we first prove that,

$\displaystyle \hat{N}\hat{a}^p\vert \Phi_{\nu}> = (\nu - p)\hat{a}^p\vert \Phi_{\nu}>.$ (26)

In order to prove Eq. (26) we apply $ \hat{a}$ to both sides of Eq. (25):

$\displaystyle \hat{a}\hat{N}\hat{a}\vert \Phi_{\nu}> = (\nu -1) \hat{a}^2\vert \Phi_{\nu}>,$
and since $ [\hat{N},\hat{a}]= -\hat{a}$ we obtain,
$\displaystyle (\hat{a}+\hat{N}\hat{a})\hat{a}\vert\Phi_{\nu}>=(\nu-1)\hat{a}^2\vert\Phi_{\nu}>,$
and
$\displaystyle \hat{N}\hat{a}^2\vert\Phi_{\nu}>=(\nu-2)\hat{a}^2\vert\Phi_{\nu}>.$ (27)

Applying $ \hat{a}$ to Eq. (27) we obtain,

$\displaystyle \hat{a}\hat{N}\hat{a}^2\vert\Phi_{\nu}>=(\nu-2)\hat{a}^3\vert\Phi_{\nu}>,$
and substituting $ \hat{a}\hat{N}$ by $ \hat{a}+\hat{N}\hat{a}$ we obtain,
$\displaystyle \hat{N}\hat{a}^3\vert\Phi_{\nu}>=(\nu-3)\hat{a}^3\vert\Phi_{\nu}>.$
Repeating this procedure $ p$ times we obtain Eq. (26). Having proved Eq. (26) we now realize that if $ \nu=n$, with $ n$ an integer number, 
$\displaystyle \hat{a}^p\vert\Phi_n > =0,$
when $ p > n$. This is because state $ \hat{a}^n\vert\Phi_n >$ is the eigenstate of $ \hat{N}$ with eigenvalue equal to zero, i.e., $ \hat{a}^n\vert\Phi_n > = \vert\Phi_0 >$. Therefore $ \hat{a}\vert\Phi_0 > = \hat{a}^p\vert\Phi_n > =0 $, when $ p > n$. Note that (Eq. 26) would contradict Theorem I if $ \nu$ was not an integer, because starting with a nonzero function $ \vert\Phi_{\nu}>$ it would be possible to obtain a function $ \hat{a}^p\vert\Phi_{\nu} >$ different from zero with a negative eigenvalue.

Eigenfunctions of $ \hat{N}$

In order to obtain eigenfunctions of $ \hat{N}$ consider that,

$\displaystyle \hat{N}\vert\Phi_{\nu} > = \nu \vert\Phi_{\nu}>,$
and 
$\displaystyle \hat{N}\hat{a}\vert\Phi_{\nu +1} > = \nu \hat{a}\vert\Phi_{\nu +1} >.$
Therefore, $ \hat{a}\vert\Phi_{\nu+1}>$ is proportional to $ \vert\Phi_{\nu}>$,
$\displaystyle \hat{a}\vert\Phi_{\nu+1} > = C_{\nu+1} \vert \Phi_{\nu} >$ (28)

Applying $ \hat{a}^+$ to Eq. (28) we obtain,

$\displaystyle \hat{N}\vert\Phi_{\nu +1} > = C_{\nu+1} \hat{a}^+\vert\Phi_{\nu}>,$
$\displaystyle \vert\Phi_{\nu +1} > = \frac{C_{\nu+1}}{(\nu +1)} \hat{a}^+\vert\Phi_{\nu}>,$
$\displaystyle <\Phi_{\nu +1}\vert\Phi_{\nu +1} > = 1= \frac{C_{\nu+1}^2}{(\nu +1)^2} <\Phi_{\nu}\vert\hat{N} +1 \vert\Phi_{\nu}>,$
$\displaystyle C_{\nu +1} = \sqrt{\nu +1}.$
Therefore,
$\displaystyle \boxed{ \vert\Phi_{\nu +1} > = \frac{1}{\sqrt{\nu+1}} \hat{a}^+\vert\Phi_{\nu} >= \frac{(\hat{a}^{+})^{\nu +1}}{\sqrt{(\nu +1)!}} \vert \Phi_0> }$
The eigenfunctions of $ \hat{N}$ can be generated from $ \vert\Phi_0>$ as follows,
$\displaystyle \vert\Phi_{\nu}> = \frac{1}{\sqrt{\nu !}}\left(\hat{x} \sqrt{\fra......{\hbar}} - i \frac{\hat{p}}{\sqrt{\hbar \omega m}} \right)^{\nu} \vert \Phi_0>,$
$\displaystyle \Phi_{\nu}(x) = \frac{1}{\sqrt{\nu !}} \left(x \sqrt{\frac{m\omeg......r }{\sqrt{\hbar \omega m}} \frac{\partial }{\partial x}\right)^{\nu} \Phi_0(x).$
For example,
$\displaystyle \Phi_1(x) = \left(x\sqrt{\frac{m\omega}{\hbar}} + \sqrt{\frac{\hb......m\omega}} \frac{m \omega }{\hbar} x \right) A e^{-\frac{m \omega}{2 \hbar}x^2},$
$\displaystyle \Phi_1(x) = \underbrace{ 2x\sqrt{\frac{m\omega}{\hbar}} \sqrt[4]{\frac{m\omega}{\pi \hbar}} } e^{-\frac{m \omega}{2 \hbar}x^2}.$
The pre-exponential factor is the Hermite polynomial for $ \nu =1$.

Time Evolution of Expectation Values

In order to compute a time-dependent expectation value, 

$\displaystyle \bar{A}_t=<\psi_t\vert\hat{A}\vert\psi_t>,$
it is necessary to compute $ \vert\psi_t>$ by solving the time dependent Schrödinger equation, $ i\hbar\partial\vert\psi_t>/\partial t=\hat{H}\vert\psi_t>$. This can be accomplished by first finding all eigenstates of $ \hat{H}$$ \Phi_n$, with eigenvalues $ E_n$, and then computing $ \vert\psi_t>$ as follows,
$\displaystyle <x\vert\psi_t>=\sum_n C_n e^{-\frac{i}{\hbar}E_nt} <x\vert\Phi_n>,$
where the expansion coefficients $ C_n$ are determined by the initial state $ <x\vert\psi_0>$. The time dependent expectation value $ <\psi_t\vert\hat{A}\vert\psi_t>$ is, therefore,
$\displaystyle \bar{A}_t = \sum_{nm} C_m^* C_n e ^{-\frac{i}{\hbar} \hbar \omega (n-m)t} <\Phi_m\vert\hat{A}\vert\Phi_n>.$
Note that this approach might give you the wrong impression that the computational task necessary to solve the time dependent Schrödinger equation can always be reduced to finding the eigenstates and eigenvalues of $ \hat{H}$ by solving the time independent Schrödinger equation. While this is possible in principle, it can only be implemented in practice for very simple problems (e.g., systems with very few degrees of freedom). Most of the problems of interest in Chemical Dynamics, however, require solving the time dependent Schrödinger equation explicitly by implementing other numerical techniques. For animations see for example the following references.