Central Potential

Consider a two-particle system represented by the following diagram,R1(123) R3(168)

where $x$, $y$ and $z$ represent distances between the two particles along the three Cartesian axes, where $\vec{r} = (x,y,z) = \vec{r}_2 -\vec{r}_1$, with $\vec{r}_1$ and $\vec{r}_2$ the position vectors of particles 1 and 2, respectively.

The central potential $V(x,y,z)$ is a function of $\vert\bar{r}\vert=\sqrt{x^2+y^2 +z^2}$, rather than a function of the individual Cartesian components. Assuming that such function defines the interaction between the two particles, the Hamiltonian of the system has the form,

$$H=\frac{P_1^2}{2m_1} +\frac{P_2^2}{2m_2} + V(\vert\vec{r}_2 - \vec{r}_1\vert) = T +V(\vert\vec{r}_2 - \vec{r}_1\vert),$$

where, $T=\frac{m_1}{2}\vert\dot{\vec{r}}_1\vert^2 + \frac{m_2}{2}\vert\dot{\vec{r}}_2\vert^2$, with $\vert\dot{\vec{r}}_1\vert^2=\dot{\vec{r}}_1 \cdot \dot{\vec{r}}_1$.

Changing variables $\vec{r}_1$, and $\vec{r}_2$, by the center-of-mass coordinates $\vec{R}$, and the relative coordinates, $\vec{r} = \vec{r}_2 - \vec{r}_1$, where,

$$\vec{R}\equiv \frac{m_1 \vec{r}_1 +m_2 \vec{r}_2}{m_1+m_2}; \qquad \vec{r} = \vec{r}_2-\vec{r}_1,$$

we obtain,

$$\vec{r}_1 = \vec{R} - \frac{m_2}{m_1+m_2} \vec{r}, \qquad \vec{r}_2 = \vec{R} + \frac{m_1}{m_1+m_2} \vec{r}.$$

Therefore,

$$T=\frac{m_1}{2}\left(\dot{\vec{R}}- \frac{m_2}{m_1+m_2} \dot{\vec... ...t{\vec{r}}\right)\left(\dot{\vec{R}}+ \frac{m_1}{m_1+m_2} \dot{\vec{r}}\right),$$

or,

$$T=\frac{m_1+m_2}{2}\vert\dot{\vec{R}}\vert^2 + \frac{1}{2} \frac{... ...frac{1}{2}M\vert\dot{\vec{R}}\vert^2 + \frac{1}{2}\mu\vert\dot{\vec{r}}\vert^2,$$

where $M=m_1 +m_2$ is the total mass of the system, and $\mu \equiv \frac{m_1m_2}{m_1 + m_2}$ is the reduced mass of the two-particle system. Therefore, the total Hamiltonian of the system can be written as follows,

$$H=\frac{1}{2}M\vert\dot{\vec{R}}\vert^2 +\frac{1}{2} \mu\vert\dot... ...=\frac{\vec{P}_M^2}{2M} + \frac{\vec{P}_{\mu}^2}{2 \mu} + V(\vert\vec{r}\vert),$$

where the first term corresponds to the kinetic energy of a particle of mass $M$, and the second and third terms constitute the Hamiltonian of a single particle with coordinates $r$. Therefore, the time-independent Schrödinger equation for the system is,

$$\left [ \frac{\vec{P}_M^2}{2M} + \frac{\vec{P}_{\mu}^2}{2 \mu} + V(\vert\vec{r}\vert) \right ] \psi(\vec{R}, \vec{r}) =E \psi(\vec{R}, \vec{r}).$$

Trying a factorizable solution, by separation of variables,

$$\psi(\vec{r}, \vec{R})=\psi_{\mu}(\vec{r}) \psi_M(\vec{R}),$$

we obtain,

$$\underbrace{-\frac{\hbar^2 \psi_{\mu} {\nabla_R}^2 \psi_M}{\psi_{... ...i_{M}} V(\vert\vec{r}\vert) } = E \frac{\psi_{\mu}\psi_M}{\psi_{\mu} \psi_{M}}.$$

                         depends on R                 depends on r

Therefore, each one of the parts of the Hamiltonian have to be equal to a constant,

$$-\frac{\hbar^2 }{2M}\frac{1}{ \psi_M} {\nabla_R}^2 \psi_M =E_M,$$ (38)

$$-\frac{\hbar^2 }{2\mu}\frac{1}{ \psi_{\mu}} {\nabla_r}^2 \psi_{\mu} + V(\vert\vec{r}\vert)=E_{\mu}, \hspace{.3cm} \ E_M +E_{\mu} =E.$$ (39)

Eq. (38) is the Schrödinger equation for a free particle with mass $M$. The solution of such equation is,

$$\psi_M(R) =(2\pi \hbar)^{-3/2} e^{i\bar{k}\bar{R}}, \hspace{.3cm} \$$   where$$\ \frac{\vert\bar{k}\vert^2 \hbar^2}{2M}=E_M.$$

According to Eq. (39), the energy $E_{\mu}$ is found by solving the equation,

$$\boxed{-\frac{\hbar^2}{2 \mu}{\nabla_r}^2 \psi_{\mu}+V(\vert\bar{r}\vert)\psi_{\mu}=E_{\mu}\psi_{\mu} }.$$ (40)

Eqs. (38) and (39) have separated the problem of two particles interacting according to a central potential $V(\vert\bar{r}_2-\bar{r}_1\vert)$ into two separate one-particle problems that include:

(1) The translational motion of the entire system of mass M.

(2) The relative (e.g., internal) motion.

These results apply to any problem described by a central potential (e.g., the hydrogen atom, the two-particle rigid rotor, and the isotropic multidimensional harmonic-oscillator).

Consider Eq. (40), with $\nabla^2 \equiv \frac{\partial^2}{\partial x^2} +\frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2}$, and $V(\vert\bar{r}\vert)$ a spherically-symmetric potential, i.e., a function of the distance $r=\vert\bar{r}\vert$. It is natural to work in spherical coordinates. Exercise 29: Prove that the Laplacian $\nabla^2$ can written in spherical coordinates as follows,

$$\nabla^2 = \frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial }{\partial r}- \frac{1}{r^2 \hbar^2} \hat{L}^2,$$   where$$\hspace{.20cm} \hat{L}^2=-\hbar^2 \left( \frac{\partial^2}{\parti... ...\theta} + \frac{1}{\text{sin} \theta}\frac{\partial^2}{\partial \phi^2}\right).$$

It is important to note that the commutator

$$[\nabla^2, L^2]= \left[\frac{\partial^2 }{\partial r^2}+ \frac{2}... ...\hat{L}^2 \right] -\left[ \frac{1}{r^2 \hbar^2}\hat{L}^2, \hat{L}^2 \right] =0,$$

because $\hat{L}^2$ does not involve $r$, but only $\theta$ and $\phi$. Also, since $\hat{L}^2$ does not involve $r$, and $V$ is a function of $r$,

$$[V, L^2]=0.$$

Consequently,

$$[H, L^2]=0,$$

whenever the potential energy of the system is defined by a central potential. Furthermore, $[H, L_z] =0$, because $\hat{L}_z= -i \hbar \frac{\partial}{\partial \phi}$.

Conclusion: A system described by a central-potential has eigenfunctions that are common to the operators $H$, $L^2$ and $L_Z$:

$\hat{H}\psi_{\mu}= E_{\mu} \psi_{\mu},$

$\hat{L}^2 \psi_{\mu} = \hbar^2 l (l +1) \psi_{\mu}, \qquad l=0, 1, 2, ...$

$\hat{L}_z \psi_{\mu}= \hbar m \psi_{\mu}, \qquad m=-l, -l+1, ..., l.$

Substituting these results into Eq. (40) we obtain,

$$-\frac{\hbar^2}{2 \mu} \left( \frac{\partial^2}{\partial r^2} + \... ...hbar} l (l+1) \psi_{\mu} + V(\vert\bar{r}\vert)\psi_{\mu} = E_{\mu} \psi_{\mu}.$$

Since the eigenfunctions of $\hat{L}^2$ are spherical harmonics $Y_l^m(\theta, \phi)$, we consider the solution,

$$\psi_{\mu} = R(r) Y_l^m(\theta, \phi),$$

and we find that $R(r)$ must satisfy the equation,

$$-\frac{\hbar^2}{2 \mu} \left( \frac{\partial^2 R}{\partial r^2} +... ...ght) + \frac{\hbar^2}{2 \mu r^2} l (l+1) R + V(\vert\bar{r}\vert)R = E_{\mu} R.$$ (41)