Schrodinger Equation for two interacting particle in Spherical co-ordinate

Schrodinger Equation for two interacting particle in Spherical co-ordinate:

The value of \(\hat L^2\) in the spherical polar co-ordinate is given by

$$\hat L= -\hbar^2 \nabla_{\theta\phi}^2\dotsm(1)$$

Consider two masses \(m_1\) and \(m_2\) is located at the \((x_1,y_1,z_1\)) & \(x_2, y_2 , z_2\) as shown in figure. the two body problem is 6D problem but it can be reduced to 3D problem by introducing the concept of centre of mass and reduced mass.

Now, the Hamiltonian of the system is given by

$$H= -\frac{\hbar^2}{2m_1}\nabla_1^2 - \frac{\hbar^2}{2m_2} \nabla_2^2+ V(r)$$

$$or,\;\; H= -\frac{\hbar^2}{2m_1}\biggl(\frac{\partial^2}{\partial x_1^2}+\frac{\partial^2}{\partial y_1^2}+ \frac{\partial^2}{\partial z_1^2}\biggr)- \frac{\hbar^2}{2m_2}\biggl(\frac{\partial^2}{\partial x_2^2}+\frac{\partial^2}{\partial y_2^2}+\frac{\partial^2}{\partial z_2^2}\biggr)+ V(r)\dotsm(2)$$

Now, we convert the \(x_1\Rightarrow \) x and x as,

$$\frac{\partial}{\partial x_1}= \frac{partial}{\partial x}\times \frac{\partial x}{\partial x_1}+\frac{\partial}{\partial x}\times \frac{\partial x}{\partial x_1}$$

Now,\(\vec r= \vec r_2- \vec r_1\)

$$R= \frac{m_1 r_1+ m_2 r_2}{m_1+ m_2}$$

Where, \(M= m_1+ m_2=\) Total mass of the system.

$$x= x_2-x_1$$

$$ X= \frac{m_1 x_1 + m_2x_2}{m_1+m_2}$$

$$y=y_2-y_1$$

$$Y= \frac{m_1y_1+ m_2 y_2}{m_1+m_2}$$

$$z=z_2-z_1$$

$$Z= \frac{m_1 z_1 + m_2 z_2}{m_1+m_2}$$

Now,

$$\frac{\partial}{\partial x}=\frac{\partial}{\partial x}\biggl(\frac mM\biggr)-\frac{\partial}{\partial x}$$

$$\therefore\; \frac{\partial^2}{\partial x_1^2}= \frac{\partial }{\partial x_1}\frac{\partial}{\partial x_1}= \biggl[\biggl(\frac{m_1}{M}\biggr)\frac{\partial}{\partial x}- \frac{\partial}{\partial x}\biggr] \biggl[\biggl(\frac{m_1}{M}\biggr)\frac{\partial}{\partial x}- \frac{\partial }{\partial x}\biggr]$$

$$or\; \frac{\partial^2}{dx_1^2}= \biggl(\frac{m_1}{M}\biggr)^2 \frac{\partial^2}{\partial x^2}- \frac{2m_1}{M}\frac{\partial }{\partial x}-\frac{\partial }{\partial x}+ \frac{\partial^2}{\partial x^2}\dotsm(4)$$

Similarly,

$$\frac{\partial^2}{\partial x_2^2}= \biggl(\frac{m_2}{M}\biggr)^2\frac{\partial^2}{\partial x^2}+ \frac{2m_2}{M}\frac{\partial }{\partial x}\frac{\partial }{\partial x} + \frac{\partial^2}{\partial x^2}\dotsm(5)$$

Multiplying equation (4) by \(\frac{1}{m_1}\) and equation (5) by \(\frac{1}{m_2}\) and adding we get,

$$\frac{1}{m_1}\frac{\partial^2}{\partial x_1^2}+\frac{1}{m_2}\frac{\partial^2}{\partial x_2}= \biggl(\frac{m_1+m_2}{M^2}\biggr)\frac{\partial^2}{\partial x^2}+ \biggl(\frac{1}{m_1}+\frac{1}{m_2}\biggr)\frac{\partial^2}{\partial x^2}$$

$$=\frac{1}{M} \frac{\partial^2}{\partial x^2} + \frac{1}{\mu}\frac{\partial^2}{\partial x^2}$$

$$i.e.\; \frac{1}{m_1}\frac{\partial^2}{\partial x_1^2}+\frac{1}{m_2}\frac{\partial^2}{x_2^2}= \frac{1}{M}\frac{\partial^2}{\partial x^2}+\frac{1}{\mu}\frac{\partial^2}{\partial x^2}\dotsm(6)$$

Similarly,

$$\frac{1}{m_1}\frac{\partial^2}{\partial y_1^2}+ \frac{1}{m_2}\frac{\partial^2}{y_2^2}= \frac{1}{M} \frac{\partial^2}{\partial y^2} + \frac{1}{\mu}\frac{\partial^2}{\partial y^2}\dotsm(7)$$

$$\frac{1}{m_1}\frac{\partial^2 }{\partial z_1^2}+ \frac{1}{m_1}\frac{\partial^2}{\partial z_2^2}= \frac{1}{M}\frac{\partial ^2}{\partial Z^2}+ \frac{1}{\mu}\frac{\partial^2}{\partial z^2}\dotsm(8)$$

$$H= \frac{-\hbar^2}{2m}\nabla_m^2 - \frac{\hbar^2}{2\nu}\nabla_\nu^2 + V(r)\dotsm(9)$$

Now the Schrodinger equation is given by

$$H|\psi(r,R)>= E|\psi(r,R)>\dotsm(10)$$

Now, equation (9) becomes,

$$\biggl[\frac{-\hbar^2}{2M}\nabla_m^2- \frac{\hbar^2}{2\mu}\nabla_\mu^2+ v(r)\biggr]|\psi(r,R)>= E|\psi(r,R)>\dotsm(11)$$

Now, we take,

$$|\psi(r,R)>= |psi_{M}(R)>|\psi_{\mu}(r)>$$

This is possible only if r and R are independent to each other.

Now the Hamiltonian for the system can be written as

$$H= H_M+ H_\mu$$

Where,

$$H_M= -\frac{\hbar^2}{2M}\nabla_m^2$$

$$H_\mu= \frac{-\hbar}{2\mu}\nabla_\mu^2+ V(r)\dotsm(13)$$

For the free particle, the Hamiltonian of the systme can be written as,

$$H_M|\psi_M(R)>= E_M|\psi_m(k)>\dotsm(13)$$

So the motion of centre of mass is same to that of motion of free particle of mass M.

For the reduced mass,

$$H_\mu |\psi_\mu(r)>= E_\mu |\psi_\mu(r)>$$

$$or,\; \biggl[-\frac{\hbar^2}{2\mu}\nabla_\mu^2+V(r)\biggr]|\psi_\mu(r)>= E_\mu |\psi_\mu(r)>\dotsm(14)$$

So that motion of reduced mass is motion including the interacting potential \(V(\vec r\))

Since the Hamiltonian of reduced mass carries information of V(r). So the \(\nabla_\mu^2\) carries the information for (\(r, \theta,\phi)\) components of the spherically symmetric potential.

Therefore equation (14) becomes,

$$\biggl[-\frac{\hbar^2}{2\mu}[ \nabla_r^2+ \frac{1}{r^2}\nabla_{\theta\phi}^2]V(r)\biggr]|\psi_{\mu}(r)>= E_\mu|\psi_\mu(r)>$$

$$or,\; \biggl[-\frac{\hbar^2}{2\mu}\nabla_r^2- \frac{\hbar^2}{2\mu r^2}\biggl(\frac{L^2}{\hbar^2}\biggr)+ v(r)\biggr]|\psi_\mu(r)>= E_\mu |\psi_\mu()>$$

$$or,\;\; (\nabla_r^2- \frac{L^2}{2\mu r^2}\biggr)|\psi_\mu(r)>= \frac{2\mu}{\hbar^2}[V(r)- E_\mu]|\psi_\mu(r)>=0$$

This is the required expression for schrodinger equation for the two interacting particles in terms of \(L^2\) in spherical polar co-ordinate system.

Reference:

1. Mathews, P.M and K Venkatesan.A Text Book of Quantum Mechanics.New Delhi: Tata McGraw Hill Publishing Co. Ltd, 1997.
2. Merzbacher, E.Quantum Mechanics .New York: John Wiley, 1969.
3. Prakash, S and S Salauja.Quantum Mechanics.Kedar Nath Ram Nath Publishing Co, 2002.
4. Singh, S.P, M.K and K Singh.Quantum Mechanics.Chand & Company Ltd., 2002.