양자물리 4. 삼차원 양자역학

오주환·2023년 8월 14일

Schrödinger equation in 3 dim

According to Schrödinger eq,

i\hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}+V\psi \tag{1}

So in 3 dimension,

i\hbar \frac{\partial \psi}{\partial t}=-\nabla^2\frac{\partial^2 \psi}{\partial x^2}+V\psi \tag{2}

works

Separating variables to solve this

\Psi(r,\theta,\phi)=R(r)Y(\theta,\phi)

And we can make formula (2) as (formula of r)+(formula of $\theta,\phi$ )=0
And if we deem each part as $\pm l(l+1)$ formula (2) can be divided as following two formulas.

\frac{1}{R}\frac{d}{dr}(r^2\frac{dR(r)}{dr})-\frac{2mr^2}{\hbar^2}[V(r)-E]=l(l+1) \tag{2.1}

\frac{1}{Y}[\frac{1}{sin(\theta)}\frac{\partial}{\partial \theta}(sin \theta\frac{\partial Y}{\partial \theta})+\frac{1}{sin^2\theta}\frac{\partial^2 Y}{\partial \phi^2}]=-l(l+1)

And if we separate again the formula (2.1) as

Y(\theta, \phi)= \Theta(\theta)\Phi(\phi)

We can get following two formula,

\frac{1}{\Theta}[sin(\theta)\frac{d}{d\theta}(sin \theta \frac{d\Theta}{d\theta})]+l(l+1)sin^2(\theta)=m^2 \tag{2.1.1}

\frac{1}{\Phi}\frac{d^2\Phi}{d\Phi^2}=-m^2 \tag{2.1.2}

Solution of (2.1.2) is $\Phi(\phi)=e^{im\phi}$
and $m=0,\pm 1, \pm2.....$ becauese $\Phi$ has periodic function
For formula (2.1.1), its sol was turned out to be a legendre function expressed as

\Theta(\theta)=AP^m_l(cos(\theta))

P^m_l(cos(x))=(-1)^m((1-x^2)^{m/2}(\frac{d}{dx})^mP_l (x))

So in order to this function meaningful, |m|< l must be satisfied

By the way, for radial part of formula (2.1)
it can be transformed to

-\frac{\hbar^2}{2 m} \frac{d^2 u}{d r^2}+\left[V+\frac{\hbar^2}{2 m} \frac{\ell(\ell+1)}{r^2}\right] u=E u \tag{3}

If we define $u(r)=rR(r)$

for next part, i will solve this in various V conditino

finite potential well

Lets solve equation (3) for finite potential well
Its solution is known as bessel funciton

u(r)=Aj_l(kr), k=\frac{\sqrt{2mE}}{\hbar}

So if $\beta_{Nl}$ is N th position that $j_l$ becomes zero

E_{Nl}=\frac{\hbar^2}{2ma^2}\beta^2_{Nl}

coulomb potential (Hydrogen atom)

For Harmonic oscillator potential, the equation (3) becomes

-\frac{\hbar^2}{2 m_e} \frac{d^2 u}{d r^2}+\left[-\frac{e^2}{4 \pi \epsilon_0} \frac{1}{r}+\frac{\hbar^2}{2 m_e} \frac{\ell(\ell+1)}{r^2}\right] u=E u \tag{4}

Here if we define few variables as

\kappa=\frac{\sqrt{-2m_eE}}{\hbar}, \rho=\kappa r, \rho_0=\frac{m_ee^2}{2\pi\epsilon_0 \hbar^2 \kappa}

The equaiton (4) can be expressed as

\frac{d^2 u}{d \rho^2}=\left[1-\frac{\rho_0}{\rho}+\frac{\ell(\ell+1)}{\rho^2}\right] u

I will skip the process of solving this eqaution.
If we use same logic for 1-dim harmonic oscillator,the result is

u(\rho)=\rho^{l+1}e^{-\rho}v(\rho)

v(\rho)=\Sigma^\infty_{j=0}c_j \rho^j

c_{j+1}=\left\{\frac{2(j+\ell+1)-\rho_0}{(j+1)(j+2 \ell+2)}\right\} c_j

If the series has infinit terms, it diverges so there must be finite term in the series.
To make $c_N=0$
$2(N+l)=\rho_0$
$2n=\rho_0$

As a result,

E_n=-\left[\frac{m_e}{2 \hbar^2}\left(\frac{e^2}{4 \pi \epsilon_0}\right)^2\right] \frac{1}{n^2}=\frac{E_1}{n^2}

and

\kappa=\left(\frac{m_e e^2}{4 \pi \epsilon_0 \hbar^2}\right) \frac{1}{n}=\frac{1}{a n}

here $a$ is constant of
a=0.529 x 10^(-10) m
which is said to Bohr radious

To sum up,

$\psi_{n \ell m}(r, \theta, \phi)=R_{n \ell}(r) Y_{\ell}^m(\theta, \phi) \text {. }$

$R_{n \ell}(r)=\frac{1}{r} \rho^{\ell+1} e^{-\rho} v(\rho)$

$c_{j+1}=\frac{2(j+\ell+1-n)}{(j+1)(j+2 \ell+2)} c_j$

for $v(\rho)$ , It was already found by mathematicians to be Laguerre polinomial of
$v(\rho)=L_{n-\ell-1}^{2 \ell+1}(2 \rho)$

Angular momentum

As you know, angular momentum is described as $L= r\times p$
so angular momentum operator must be

L_x=tp_z-zp_y

p_x=-i\hbar\frac{\partial}{\partial x}

Now we can derive some features for these operators

[L^2,L_x]=0

L{\pm}=L_x{\pm} L_y

then

[L_z,L{\pm}]=\pm \hbar L_{\pm}

and

L_z(L_{\pm }f)=(\mu\pm\hbar)(L_{\pm}f)

So if ladder operator is operated on eig fun f, its eigvalue for $L_z$ increased by $\hbar$

using conditions of maximum and minimum angular momentum on z direction, eigvalue for angular and directional angular momentum can be decided as

L^2 f_{\ell}^m=\hbar^2 \ell(\ell+1) f_{\ell}^m ; \quad L_z f_{\ell}^m=\hbar m f_{\ell}^m

l=1 / 2,1,3 / 2, \ldots ; \quad m=-\ell,-\ell+1, \ldots

If we decribe L operator in Spherical coordinate system,

L_z=-i\hbar\frac{\partial}{\partial \phi}

L^2=-\hbar^2\left[\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial}{\partial \theta}\right)+\frac{1}{\sin ^2 \theta} \frac{\partial^2}{\partial \phi^2}\right]

and it is exactly same formula with angular equation in this chapter we saw when we solve schrödinger eq in harmonic potential situation. so the eig function $f_l^m$ is spherical harmonics function.
the different point is that in harmonics, half integer number was not allowed but in angular momentum eigfunction, it is allowed. Let me see for this situation, half integer angular momentum.

spin

lets see $l=1/2$ case.
In this case, particle's space can be expressed as

\chi=a\chi_{+}+b\chi_{-}

each $\chi$ is spin state of m=1,-1
for this basis states, we can dervie operators

$S^2=\frac{3}{4} \hbar^2\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
$S_z=\frac{\hbar}{2}\left(\begin{array}{ll}1 & 0 \\ 0 & -1\end{array}\right)$
$S_x=\frac{\hbar}{2}\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)$
$S_y=\frac{\hbar}{2}\left(\begin{array}{ll}0 & -i \\i & 0\end{array}\right)$

오주환

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