## 3D Harmonic oscillator

Set $x = r/\alpha$The Schrodinger equation is

$\displaystyle \left(-\frac{\hbar^2}{2m} \nabla^2 + \frac{1}{2} m \omega^2 r^2 \right)\Psi = E \Psi$

in Cartesian coordinate, it is,

$\displaystyle -\frac{\hbar^2}{2m}\left( \frac{d^2}{dx^2}+\frac{d^2}{dy^2}+\frac{d^2}{dz^2} \right) \Psi + \frac{1}{2} m \omega^2 (x^2+y^2+z^2) \Psi = E \Psi$

We can set the wave function to be $\Psi(r, \Omega) = X(x) Y(y) Z(z)$

$\displaystyle \left( -\frac{\hbar^2}{2m}\frac{d^2X}{dx^2} + \frac{1}{2} m \omega^2 x^2 X \right) YZ + \\ \left( -\frac{\hbar^2}{2m}\frac{d^2Y}{dy^2} + \frac{1}{2} m \omega^2 y^2 Y \right) XZ + \\ \left( -\frac{\hbar^2}{2m}\frac{d^2Z}{dz^2} + \frac{1}{2} m \omega^2 z^2 Z \right) XY = E XYZ$

we can see, there are three repeated terms, we can set

$\displaystyle -\frac{\hbar^2}{2m}\frac{d^2X}{dx^2} + \frac{1}{2} m \omega^2 x^2 X = E_x X$

We decoupled the X, Y, Z. Each equation is a quadratic equation with energy

$\displaystyle E_x, E_y, E_z = \left(\frac{1}{2} + n \right) \hbar \omega$

and

$\displaystyle E_x + E_y + E_z = \left(\frac{3}{2} + n_x + n_y + n_z \right) \hbar \omega = \left(\frac{3}{2} + n \right) \hbar \omega = E$

The number of states for each energy level is

$\displaystyle C^{n_x+n_y+n_z+2}_2 = C^{n+2}_2 = \frac{(n+2)!}{n!2!}$

The first few numbers of states are 1, 3, 6, 10, 15, 21, 28, … The accumulated numbers of states are 1, 4, 10, 20, 35, 56, 84, … Due to the spin-state, the accumulated numbers of particles are 2, 8, 20, 40, 70, 112, 168, … The few magic numbers are reproduced.

The wave function is the product of the Hermite functions $H_n(x)$ and exponential function

$\Phi(x,y,z) = N H_{n_x} (x) H_{n_y}(y) H_{n_z}(z) \exp(-r^2/2)$

If we simply replace $(x,y,z) \rightarrow r( \cos(\phi) \sin(\theta), \sin(\phi) \sin(\theta) , cos(\theta) )$, we can see the ground state consists of s-orbit, the 1st excited state consists of p-orbit, and the 2nd excited state consists of d-orbit.

To see more clearly, we can project the function onto spherical harmonic, which is fixed angular momentum, i.e.

$\langle \Phi(x,y,z) | Y_{lm}(\theta,\phi) \rangle = C_{lm} R_{nl}(r)$.

where $C_{lm}$ is coefficients and $R_{nl}(r)$ is radial function.

To have a better understanding, the radial function has to be solved. The procedure is very similar to solving Coulomb potential.

$\displaystyle \left(-\frac{\hbar^2}{2m}\left(\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)\right) + \frac{\hbar^2}{2m} \frac{L^2}{r^2} + \frac{1}{2} m \omega^2 r^2 \right)\Psi = E \Psi$

separate the radial part and angular part.

$\displaystyle \left( \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right) - \frac{l(l+1)}{r^2} - \frac{m^2 \omega^2}{\hbar^2} r^2\right) R = - (2n+3) \frac{m \omega}{\hbar} R$

Set $\alpha^2 = \frac{\hbar}{m \omega}$

$\displaystyle \left( \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right) + \frac{2n+3}{\alpha^2} - \frac{l(l+1)}{r^2} - \frac{r^2}{\alpha^4}\right) R = 0$

Set $x = r/\alpha$

$\displaystyle \left( \frac{1}{x^2}\frac{d}{dx}\left(x^2\frac{d}{dx}\right) + (2n+3) - \frac{l(l+1)}{x^2} - x^2\right) R = 0$

Set $u(x) = x R(x)$

$\displaystyle \frac{d^2 u }{d x^2} + \left( (2n+3) - x^2 - \frac{l(l+1)}{x^2} \right) u = 0$

as usual, the short range behaviour is $r^{l+1}$, long range behaviour is $\exp(-x^2/2)$, as stated in the Cartesian coordinate. Thus, we set

$\displaystyle u(x) = f(x) \exp\left(-\frac{x^2}{2}\right) x^{l+1}$

$\displaystyle x\frac{d^2f}{dx^2} + (2(l+1) -2x^2) \frac{df}{dx} + 2x(n-l) f(x) = 0$

with change of variable $y = x^2$, the equation becomes

$\displaystyle y\frac{d^2f}{dy^2} + \left( l + \frac{1}{2} + 1 - y \right)\frac{df}{dy} + \frac{n-l}{2} f(y) = 0$

This is our friend, the Laguerre polynomial! In the Laguerre polynomial, $(n-l)/2 \geq 0$ must be non-negative integer. Now we set $k = (n-l)/2$, than the energy is

$\displaystyle E = \hbar \omega \left( 2k + l + \frac{3}{2} \right)$

In order to have $n, l, k$ are integer, when

$n = 0 \rightarrow l = 0 \\ n = 1 \rightarrow l = 1 \\ n= 2 \rightarrow l = 0, 2 \\ n = 3 \rightarrow l = 1, 3 \\ n = 4 \rightarrow l = 0,2,4$

The overall solution without a normalization factor is

$\displaystyle \Psi_{nlm}(r, \theta, \phi) = r^l \exp\left(-\frac{r^2}{2\alpha^2}\right) L_{k}^{l+\frac{1}{2}}\left( \frac{r^2}{\alpha^2} \right) Y_{lm}(\theta, \phi)$

The normalization constant can be calculate easily using the integral formula of Laguerre polynomial.

$\displaystyle \int R^2(r) r^2 dr = 1$

$\displaystyle N^2 \int r^{2l} \exp\left(-\frac{r^2}{2\alpha^2}\right) \left( L_{k}^{l+\frac{1}{2}}\left( \frac{r^2}{\alpha^2} \right)\right)^2 r^2 dr = 1$

change of variable $y = r^2/\alpha^2 \rightarrow r=\alpha \sqrt{y}$

$\displaystyle dr = \frac{\alpha}{2\sqrt{y}}dy$

$\displaystyle N^2 \frac{\alpha^{2l+3}}{2} \int y^{l+\frac{1}{2}} \exp(-y) \left( L_k^{l+\frac{1}{2}} \right)^2 dy = 1$

The integration is

$\displaystyle N^2 \frac{\alpha^{2l+3}}{2} \frac{(k+l+\frac{1}{2})!}{k!} = 1$

$\displaystyle N^2 = \frac{2}{\alpha^{2l+3}} \frac{k!}{(k+l+\frac{1}{2})!}$

we can use

$\displaystyle \left(n+\frac{1}{2}\right) = \frac{(2n+1)!}{2^{2n+1}n!}\sqrt{\pi}$

Thus,

$\displaystyle N^2 = \frac{1}{\sqrt{\pi}\alpha^{2l+3}} \frac{k! (k+l)! 2^{2k+2l+3}}{(2k+2l+1)!}$

replace $k = (n-l)/2$. The total wave function is

$\displaystyle \Psi_{nlm}(r, \theta, \phi) \\ =\sqrt{ \frac{1}{\sqrt{\pi}\alpha^{2l+3}} \frac{(\frac{n-l}{2})! (\frac{n+l}{2})! 2^{n+l+2}}{(n+l+1)!}} r^l \exp\left(-\frac{r^2}{2\alpha^2}\right) L_{k}^{l+\frac{1}{2}}\left( \frac{r^2}{\alpha^2} \right) Y_{lm}(\theta, \phi)$

Here are some drawing of the square of the wave functions. From the below is $n = 0, 1, 2, 3$, from left to right, are s-orbit, p-orbit, d-orbit, f-orbit.

With the LS coupling, the spatial function does not affected, unless the coupling has spatial dependence. With the LS coupling, the good quantum numbers are $n, l, j, m_j$

## 1-D Harmonic Oscillator

It is quite surprised that This is not here!

The Hamiltonian is

$\displaystyle H=\frac{P^2}{2m} + \frac{m \omega^2}{2} X^2$

Lets define creation operator $A^\dagger$ and destruction operator $A$, or the ladder operators, such that,

$\displaystyle A^\dagger =\sqrt{\frac{1}{2m}} P + i \sqrt{\frac{m \omega^2}{2}} X$

$\displaystyle A =\sqrt{\frac{1}{2m}} P - i \sqrt{\frac{m \omega^2}{2}} X$

Then we can see

$\displaystyle H = A^\dagger A + \frac{\hbar \omega}{2} = A A^\dagger - \frac{\hbar \omega}{2}$

and the commutative relations

$\displaystyle A^\dagger A - A A^\dagger = -\hbar \omega$

$\displaystyle HA^\dagger - A^\dagger H = \hbar \omega A^\dagger$

$\displaystyle HA - A H = -\hbar \omega A$

From the last two equations, we can see

$\displaystyle H |n\rangle = E_n |n\rangle \\ A H|n\rangle = (HA+\hbar \omega A) |n\rangle = E_n A|n\rangle \\ HA|n\rangle = (E_n-\hbar \omega) A|n\rangle$

This shows that, if $|n\rangle$ is a state with energy $E_n$, then $A|n\rangle$ is also a state with energy $E_n - \hbar \omega$. Similar for $A^\dagger |n\rangle$, the energy will be $E_n + \hbar \omega$.

Suppose the ground state is $|0\rangle$, which is is lowest energy state. If we apply the destruction operator, it will be gone, i.e.

$\displaystyle A|0 \rangle = 0 \\ A^\dagger A|0\rangle = 0 \\ \left(H- \frac{\hbar\omega}{2}\right)|0\rangle \\ H|0\rangle = \frac{\hbar\omega}{2}|0\rangle = E_0 |0\rangle$

Thus, the ground state energy is $\hbar \omega /2$!

We can donate

$\displaystyle A^\dagger |n\rangle = U_n |n+1\rangle \\ A |n\rangle = D_n |n-1\rangle$

Using $\displaystyle A^\dagger A - A A^\dagger = -\hbar \omega$, we have

$\displaystyle (A^\dagger A - A A^\dagger ) |n \rangle \\ = A^\dagger D_n |n-1 \rangle - A U_n|n+1 \rangle \\ = (U_{n-1}D_n- U_n D_{n+1} )|n\rangle = - \hbar \omega |n\rangle$

or

$\displaystyle U_n D_{n+1} = U_{n-1}D_n + \hbar \omega$

Consider above equation on $|0\rangle$, such that $A|0\rangle = 0$, thus,

$\displaystyle (A^\dagger A - A A^\dagger ) |0 \rangle = - U_0 D_{1} )|0\rangle = - \hbar \omega |0\rangle$

Thus, $U_0 D_1 = \hbar \omega$, then, in general, we have

$\displaystyle U_n D_{n+1} = (n+1) \hbar \omega$

Then

$\displaystyle A^\dagger A |n\rangle = U_{n-1} D_n |n\rangle \\ A A^\dagger |n\rangle = U_{n} D_{n+1} |n\rangle$

The simplest solution is

$U_n = \sqrt{\hbar \omega}\sqrt{n+1} \\ D_n =\sqrt{\hbar \omega}\sqrt{n}$

And, if we treat the state $|n\rangle$ contains $n$ oscillators, we have a number operator $N = A^\dagger A$, such that

$A^\dagger A |n \rangle = n \hbar \omega |n \rangle$

some people will normalized the ladder operators with $\sqrt{\hbar \omega}$. For me, I don’t want to make the starting point be so “artificial”. We can summarized the 1-D Harmonic oscillator as

Now, we are going to solve the wave function. One way is use the ladder operator,

$\displaystyle A|0\rangle = 0 \\ \left( \sqrt{\frac{1}{2m}} P - i \sqrt{\frac{m \omega^2}{2}} X \right) \phi_0 = 0 \\ \frac{d\phi_0}{dx} = - \frac{m \omega}{\hbar} x \phi_0 = 0$

The solution is

$\displaystyle \phi_0 = N \exp\left(-\frac{m \omega}{2\hbar}x^2\right)$

Another way is the Schrodinger equation is

$\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\phi}{dx^2} + \frac{1}{2}m\omega^2 x^2 \phi = E \phi = \frac{2n+1}{2}\hbar \omega \phi$

$\displaystyle \frac{d^2\phi}{dx^2} - \frac{m^2 \omega^2}{\hbar^2} x^2 \phi = (2n+1)\frac{m \omega}{\hbar} \phi$

Define $\alpha = \sqrt{\frac{\hbar}{m \omega}}$, the dimension of $\alpha$ is length. Set $u = x/ \alpha$, then

$\displaystyle \frac{d^2\phi}{du^2} + ((2n+1) - u^2) \phi = 0$

Since we know that the function must decay outside the well, set $\phi = f(y) \exp(-y^2/2)$, then

$\displaystyle \frac{d^2f}{du^2} -2 u \frac{df}{du} +2 nf = 0$

The solution is Hermite polynomial $H_n(u)$ of order $n$.

Assume

$\displaystyle H_n(x) = \sum_{i = 0}^{\infty} a_i x^i$

sub into the Hermite equation,

$\displaystyle a_{i+2} = \frac{2(i-n)}{(i+1)(i+2)} a_i$

First,  the even and odd series are separated. And either the odd-series or the even-series are converge, as the ratio

$\displaystyle \lim_{i->\infty} \frac{a_{i+2}}{a_i} = 2$

However, the series can be terminated when $i = n$. Thus, the Hermite polynomial has either even terms or odd terms, but not mixed.

When $n = 0$, $a_2 = 0$ and all later even terms are zero. The equation is $f''(x) - 2 x f'(x) = 0$, which only support even function. Thus, $H_0(x) = a_0$ is a constant. Thus, we can see, when $n$ is even (odd), Hermite polynomial is even (odd) with only even (odd) $n$.

Hermite polynomial are orthogonal with $\exp(-x^2)$, thus, the wave function $\phi_n(x) = H_n(x) \exp(-x^2/2)$ are orthogonal.

The normalized wave function is

$\displaystyle \phi_n(x) = \frac{1}{\sqrt{2^n n! \sqrt{\pi}}} H_n(x) \exp\left(-\frac{x^2}{2}\right)$

or

$\displaystyle \phi_n\left(\frac{x}{\alpha}\right) = \frac{1}{\sqrt{\alpha 2^n n! \sqrt{\pi}}} H_n\left(\frac{x}{\alpha}\right) \exp\left(-\frac{x^2}{2\alpha^2}\right)$