Boundary condition [ Continue ]

Boundary Condition:

2. At the point, surface or boundary having infinite value of P.E, the wave function must be zero or it must vanish in that region.

Consider a particle of mass m and energy E is incident on the potential step of the form

$$V(x)=0 \; For x<0$$

$$= \infty For x\geq 0\dotsm(1)$$

To derive the boundary condition in this case we assume P.E is finite in region II ( i.e. x>0 ).

So,

$$V(x)=0\; For \; x<0\dotsm(2)$$

$$= V_0 (assumed for x\geq 0\dotsm(2)$$

Finally, we must substitute, $V_0=\infty$

The time independent on-dimensional Schrodinger equation for particle,

$$or,\frac{-\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)= E\psi(x)$$

$$or,\;\frac{d^2\psi(x)}{dx^2}= \frac{-2m}{\hbar^2}[E-V(x)]\psi(x)$$

$$or\; \frac{d^2\psi(x)}{dx^2}+\frac{2m}{\hbar^2}[E-V(x)]\psi(x)=0\dotsm(3)$$

In region (I): (x<0) , V(x)=0, $\psi(x)= \psi_1(x)$

Equation (3) become

$$\frac{d^2\psi_I(x)}{dx^2}+\frac{2m}{\hbar^2} E\psi_I(x)=0$$

$$or,\; \frac{d^2\psi_1(x)}{dx^2}+K_1^2 \psi_1(x)=0\dotsm(4)$$

Where, $K_1=\sqrt{\frac{2mE}{\hbar^2}}$

Solution of equation (4) is,

$$\psi_1(x)= Ae^{ik_1 x}+ Be^{-ik_1 x}\dotsm(5)$$

Where, A and B are constant..

For region (II): (x>0) $V(x)=V_0>E), \psi(x)= \psi_{II}(x)$

Equation (3) becomes,

$$\frac{d^2\psi_{II}(x)}{dx^2}+ \frac{2m(E-V_0)}{\hbar^2}\psi_{II}(x)=0$$

$$or,\;\; \frac{d^2\psi_{II}(x)}{dx^2}- \frac{2m(V_0- E)}{\hbar^2} \psi_{II}(x)=0$$

$$or,\;\; \frac{d^2\psi_{II}(x)}{dx^2}- K_2^2 \psi_II(x)=0\dotsm(6)$$

Where, $K_2= \sqrt{\frac{2m}{\hbar^2}(V_0-E)}\dotsm(7)$

$K_2$ is real and +ve.

Solution of equation (6) is,

$$\psi_{II}(x)= Ce^{-k_2 x}+ De^{k_2 x}[Non\; oscillating \; solution]$$

At infinity i.e. $x=+\infty$, the wave function must be zero. Here, $e^{k_2 x}$ gives divergent result at $x=+\infty$ So D must be zero ( For x>0).

$$\therefore\; \psi_{II}(x)= ce^{-k_2 x}\dotsm(8)$$

Using boundary condition for $v(x)= v_0(finite)$. $\psi(x)$ is continuous across x=0.

$$i.e.\; \psi_1(x)\biggl|_{x=0}=\psi_{II}(x)\biggr|_{x=0}$$

$$Ae^0+ Be^0=C\dotsm(9)$$

Again, Gradient of $\psi(x)$ is continuous across the boundary,

$$\frac{d\psi_I (x)}{dx}\biggl|_{x=0}= \frac{d\psi_{II}(x)}{dx}\biggr|_{x=0}$$

$$ik_1(A-B)e^0 =-k_2 ce^0$$

$$or,\; C=\frac{-k_2(A-B)}{K_2}\dotsm(10)$$

$$or,\; A-B= \frac{-k_2 C}{ik_1}\dotsm(11)$$

Adding (9) and (11)

$$2A= C\biggl[1-\frac{k_2}{ik_1}\biggr]$$

$$or,\; \frac{c}{A}= \frac{2}{\biggl(1-\frac{k_2}{ik_1}\biggr)}$$

$$or,\; C= \frac{2}{\biggl(\frac{1+ik_2}{k_1}\biggr)}A$$

$$or, C^*C= \frac{2}{\biggl(1-\frac{ik_2}{k_1}\biggr)}\frac{2}{\biggl(1+\frac{ik_2}{k_1}\biggr)}AA^*$$

$$or,\;\; |C|^2= \frac{4}{1+\frac{k_2^2}{k_1^2}}|A|^2$$

When, $V_0\to\infty$ ( infinite potential )

$$|C|^2= \frac{4}{1+\frac{2m(V_0- E) |\hbar^2}{2mE|\hbar^2}}|A|^2$$

$$=\frac{4}{1-\biggl(\frac{V_0-E}{E}\biggr)}|A|^2$$

$$=\frac{4E|A|^2}{V_0}$$

$$=\frac{4E|A|^2}{\infty}=0$$

$$\therefore\; |C|=0$$

Hence the 2^nd wave function for $v_0=\infty$ is always zero. The wave function vanish at the surface of infinite potential.

$\psi_1(x)$at x=0 is given by

$$\psi_1(x)\biggl|_{x=0}= Ae^{ik_1\cdot 0}+ Be^{-ik_1 0}$$

$$= A+B$$

$$=C$$

$$=0$$

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.