TDSE for Hydrogen Atom

Write down the 7 dimensional TDSE (time dependent Schroedinger Equation) for the hydrogen atom in the position space. Explain how the separation of the center of mass and relative position variables converts it into two 4 dimensional TDSE’s. Explain why solving the center of mass TDSE leads to the 3d Guassian Packet behavior of the center of mass. Explain why the radial variable r in the relative position TDSE can be separated from the angular variables θ and φ. Explain how this separation of r from θ  and φ leads to the effective potential for the hydrogen atom.

Let $(\overrightarrow{r_1})$ and $(\overrightarrow{r_2})$ be the coordinates of the nucleus ($m_1$) and of the electron ($m_2$) in the hydrogen atom. Then the total Hamiltonian for this system is

$H ̂=-ℏ^2/(2m_1 )*∇_{r_1}^2-ℏ^2/(2m_2 )*∇_{r_2}^2+V((r_2 )-(r_1 ))$

The time dependent Sch. equation using 7 dimensions $((r_1 ), (r_2 ),t)$ is $[Ψ=ψ((r_1 ),(r_2 ),t)]$

$iℏ* ∂Ψ/∂t=H ̂Ψ$

If one goes to the center of mass and relative position variables:

$R=(m_1 r_1+m_2 r_2)/(m_1+m_2)$  ,$r=r_2-r_1$

with $M=m_1+m_2$  and $μ=(m_1 m_2)/(m_1+m_2 )$

By computing the first and second derivatives above $(∂^2 Ψ)/(∂x_1^2)$,….,$(∂^2 Ψ)/(∂x_2^2)$  ,… :

$\frac{∂}{(∂r_1)}=-\frac{∂}{∂r}+\frac{m_1}{M}  \frac{∂}{∂R}$  

and    $\frac{∂^2}{(∂r_1^2)}=\frac{∂^2}{(∂r^2)}-\frac{(2m_1)}{M}*\frac{∂^2}{∂r∂R}+(\frac{m_1}{M})^2 \frac{∂^2}{(∂R^2)}$
$\frac{∂}{(∂r_2)}=-\frac{∂}{∂r}+\frac{m_2}{M}  \frac{∂}{∂R}$   and    $\frac{∂^2}{(∂r_2^2)}=\frac{∂^2}{(∂r^2)}-\frac{(2m_2)}{M}*\frac{∂^2}{∂r∂R}+(\frac{m_2}{M})^2 \frac{∂^2}{(∂R^2)}$One gets a new expression for the Hamiltonian:
$H ̂=-\frac{ℏ^2}{2M} ∇_R^2-\frac{ℏ^2}{2μ} ∇_r+V(r)$

The 7 dimensional time dependent Sch. equation $iℏ* ∂Ψ/∂t=H ̂Ψ$ can be “decoupled” into two 4 dimensional TDSE for the particular wavefunction $Ψ(R ⃗,r ⃗,t)=ψ_M (R)*ψ_μ (r)*f(t)$

$-(ℏ^2/2M)*∇_R^2 [ψ_M (R)f(t)]=iℏ (\frac{∂[ψ_M (R)f(t)])}{∂t})$     and
$-(ℏ^2/2μ)*∇_r^2 [ψ_μ (r)f(t)]+V(r)=iℏ (\frac{∂[ψ_μ (r)f(t)])}{∂t})$

By taking $f(t)=e^{-iωt}$ from above it remains just the time-independent part of each eq.

$(-ℏ^2/2M)*∇_M^2 ψ_M (R ⃗ )=E_1 ψ_M (R ⃗ )$      (1) 
and $(-ℏ^2/2μ) ∇_μ^2 ψ_μ (r ⃗ )+V(r ⃗ )=E_2 ψ_μ (r ⃗)$    (2)

The first equation (1) is satisfied by all plane waves
$ψ_M (R)=A*e^{ikR}$ 

and since $R$ is the coordinate of the center of mass of the hydrogen atom (which is strongly localized) one can consider the coefficient A equal to a Gaussian ($1/√a$ is the width of the packet)

$A=e^{-aR^2}$  so that $ψ(R)=e^{-aR^2+ikR}$

For the second time independent Schroedinger equation (2) one can again “decouple” it if one writes

$ψ_μ (r ⃗ )=ψ_μ (r,θ,φ)=R(r)Y(θ,φ)$

and uses the spherical coordinates to write the Laplacian $∇_r^2$

$-\frac{ℏ^2}{2μ}*[\frac{1}{r^2}  \frac {∂}{∂r} (r^2  \frac{∂}{∂r})+\frac{1}{(r^2  \sin⁡θ)}  \frac{∂}{∂θ}(\sin⁡ θ \frac{∂}{∂θ})+\frac{1}{(r^2  \sin^2⁡θ)}  \frac{∂^2}{(∂φ^2)} ]RY+V(r)RY=E*RY$


$-Y\frac{1}{r^2}   \frac{∂}{∂r} (r^2  \frac{∂R}{∂r})-R[\frac{1}{(r^2  \sin⁡θ)}  \frac{∂}{∂θ}(\sin⁡ θ \frac{∂Y}{∂θ})+\frac{1}{(r^2  \sin^2⁡θ)}  \frac{(∂^2 Y)}{(∂φ^2)}]+\frac{2μ}{ℏ^2} (V(r)-E)RY=0$

Divide by RY and multiply with r^2

$[-\frac{1}{R}*\frac{∂}{∂r} (r^2  \frac{∂R}{∂r})+\frac{(2μr^2)}{ℏ^2} (V(r)-E)]-\frac{1}{Y} [\frac{1}{\sin ⁡θ}   \frac{∂}{∂θ}(\sin⁡ θ \frac{∂Y}{∂θ})+\frac{1}{\sin^2⁡ θ}   \frac{(∂^2 Y)}{(∂φ^2)} ]=0$

This equation can be satisfied only if each term in square brackets is equal to the same constant $(C-C)=0$). We take the constant equal to $l(l+1)$ where $l$ is positive integer. Therefore

$-\frac{1}{R}*\frac{∂}{∂r} (r^2  \frac{∂R}{∂r})+\frac{(2μr^2)}{ℏ^2}  (V(r)-E)=-l(l+1)$   and

$\frac{1}{Y} [\frac{1}{\sin⁡θ}   \frac{∂}{∂θ} (\sin⁡ θ \frac{∂Y}{∂θ})+\frac{1}{\sin^2 ⁡θ}   \frac{(∂^2 Y)}{(∂φ^2)}]=-l(l+1)$

First equation above contains only the radial variable and is called the Radial equation. Second equation is the Angular equation. Let’s rewrite again the Radial equation like below

$-\frac{ℏ^2}{(2μr^2)}*\frac{∂}{∂r} (r^2  \frac{∂R}{∂r})+(V(r)+(ℏ^2 \frac{l(l+1))}{(2μr^2 )})R= E*R$

From here one can define the effective potential for the Radial equation as

$V_{eff} (r)=V(r)+\frac{ℏ^2 l(l+1))}{(2μr^2)}=-\frac{(ke^2)}{r}+\frac{ℏ^2 l(l+1))}{(2μr^2)}$