Particle in a Spherical-Box

Particle in a Spherical-Box:

The spherical box potential is given by

$$V(r)=0; r<a$$

$$V(r)=\infty; r\geq a$$

Inside the box particle is free and hence,

$$\psi_{klm}= J_l(k,r)Y_l^m (\theta,\phi); r<a\dotsm(2)$$

Where, k corresponds to teh energy term given by,

$$E_k= \frac{\hbar^2k^2}{2\mu}\dotsm(3)$$

Also \(V(r=a)=\infty\), NOw, using boundary condition, we get

$$J_l(k,a)=0$$

and $$J_{l+\frac12}(k,a)=0$$

i.e. $$J_0(\rho)=0 \; if \; sin\rho=0\Rightarrow \rho=n\pi$$

$$J_1(\rho)=0\; if \; tan\rho=0$$

$$J_2(\rho)=0\; if \; tan \rho= 3\rho(3-\rho)^{-1}$$

If \(X_{ne} ( n= 1,2,3,...)\) represents the non-zero roots of \(J_l(\rho)=0\)

$$\therefore\; \; x_{nl}=ka$$

Now the equation (3) becomes,

$$E_{nl}= \frac{1}{2\mu a^2}\hbar^2 X_{n1}^2\dotsm(4)$$

Since the energy value \(E_{nl}\) is discrete for each value of 'l' and principal quanatum number 'n' counting to zeros. Unlike columb potential, the energy value \(E_{nl}\) depends on l ( since there is no degeneracy with respect to l).

So the solution of l=0 exactly corresponds to the odd state of one dimensional problem.

So considering 3D spherical box we have,

$$H= H_x+ H_y+ H_z$$

Where,

$$H_x= \frac{P_x^2}{2\mu}+v(x)$$

$$H_y= \frac{P_y^2}{2\mu}+v(y)$$

$$H_z= \frac{p_z^2}{2\mu}+v(z)$$

Since,$$ V(x)= \infty ; x<0\\=0; 0<x<L\\ =\infty ; x>L\dotsm(5)$$

Now the Hamiltonian of the system is given by

If \(U_{En}(x,y,z)= E_n U_{En}(x,y,z)\)

$$U_{En}= sin\biggl(\frac{n_x \pi x}{L}\biggr)sin\biggl(\frac{n_y\pi y}{L}\biggr) sin\biggl(\frac{n-\pi z}{L}\biggr)\dotsm(6)$$

Where, $$E_n= \frac{n^2\hbar^2 \pi^2}{2\mu L^2}\dotsm(7)$$

and , \(n^2= h_x^2+ h_y^2+ h_z^2\)

Since each triplet of integers (\(n_x,n_y,n_z)\) represents the lattice point in three dimension for a large number of such points. We assume the continuous distribution of such points in the space having the sphere of radius 'R' defined by

$$n^2= n_x^2+ n_y^2 + n_z^2=\frac{2\mu^2E_n}{\hbar^2\pi}= R^2$$

Since each state occupies the unit volume, so these number \(n_i>0\) gives the volume of octant having energy \(E_f\) and giving by

$$=\frac18\biggl(\frac{4\pi}{3}R^3\biggr)=\frac16 \biggl(\frac{2\mu E_f}{\hbar^2 \pi^2}\biggr)^\frac32 L^3$$

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.