3D Harmonic oscillator

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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.

3-D Harmonic Oscillator.png

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

 

 

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1-D Harmonic Oscillator

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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

HarmonicOsciilator.png


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)