Application of Boundary condition

Application of Boundary conditions:

Potential step:

It is defined as the form of potential which has finite discontinuity at a point and it continuous upto infinity.

Mathematically,

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

\(=V_0 \; For \; x\geq 0\dotsm(1)\)

Consider a flux of non interacting particles each having same mass 'm' and same K.E 'E' be incident on the potential step as shown in figure.

The moving particle behave as wave and the motion of particle is described by time independent one-dimensional schrodinger equation.

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

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

In region (I) \(\psi(x)=\psi_1(x), V(x)=0\) , equation (2) becomes,

$$or,\;\; \frac{d^2\psi_I(x)}{dx^2}+\frac{2mE}{\hbar^2}\psi_1(x)=0$$

$$or,\;\; \frac{d^2\psi_I(x)}{dx^2}+K_1^2\psi_1(x)=0\dotsm(3)$$

Where, \(K_1=\sqrt{\frac{2mE}{\hbar^2}}\dotsm(4)\), \(K_1\) real and +ve .

Solution of equation (3)is

$$\psi_I(x)= Ae^{ik_1x}+ Be^{-ik_1(x)}\dotsm(5)$$

Where, A and B are constant to be determined.

Case (I): For region (II) \(V(x)=V_0<E.\; \psi(x)= \psi_{II}(x)\)

Equation (2) becomes,

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

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

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

Solution of equation (6) is oscillatory and given by

$$\psi_{II}(x)= Ce^{ik_2x}+ De^{-ik_2 x}\Rightarrow ( reflected \; from \; x>0)$$

There is no any boundary in region II so value of D must be zero.

$$D=0$$

$$\therefore; \psi_{II}(x)= Ce^{ik_2 x}\dotsm(8)$$

Using boundary conditions at x=0, i.e.

$$\psi_I(x)|_{x=0}= \psi_{II}(x)|_{x=0}$$

$$A+B=C\dotsm(9)$$

And

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

$$\Rightarrow A-B=\frac{k_2}{k_1}\cdot C\dotsm(10)$$

Adding (9) and (10)

$$2A= c\biggl(1+\frac{k_2}{k_1}\biggr)$$

$$=\biggl(\frac{k_1+k_2}{k_1}\biggr)\cdot C$$

$$\Rightarrow \frac{C}{A}= \frac{2k_1}{k_1+k_2}\dotsm(11)$$

Dividing equation (9) by 'A' we get,

$$1+\frac{B}{A}= \frac{C}{A}$$

$$or,\; \frac{B}{A}=\frac{2k_1}{k_1+k_2}-1$$

$$=\frac{2k_1-k_1-k_2}{k_1+k_2}$$

$$\Rightarrow \frac{B}{A}=\frac{k_1-k_2}{k_1+k_2}\dotsm(12)$$

Probability current density in region \(I^{st}\; (x_0<0)\)

$$J_1(x)=\frac{\hbar}{2im}\biggl[\psi_I^*(x)\frac{d\psi_I(x)}{dx}-\psi_I(x)\frac{d\psi_I^*(x)}{dx}\biggr]$$

$$=\frac{\hbar}{2im}(A^*e^{-ik_1 x}+ B^* e^{ik_1 x})\frac{d}{dx}[Ae^{ik_1 x}+ Be^{-ik_1 x}]- ( Ae^{ik_1 x}+ Be^{-ik_1 x})\frac{d}{dx}[A^* e^{-ik_1 x}+B^*e^{ik_1 x}]$$

$$=\frac{\hbar}{2im}ik_1[(A^*e^{-ik_1x}+B^*e^{ik_1x})(Ae^{ik_1x}-Be^{-ik_1 x})+ (Ae^{ik_1 x}+ Be^{-ik_1 x})(A^* e^{-ik_1 x}-B^*e^{ik_1x})]$$

$$=\frac{\hbar ik_1}{2im}[|A|^2-A^*Be^{-2ik_1 x}+ B^*Ae^{2ik_1 x}-|B|^2 + |A|^2- AB^* e^{2ik_1 x}+ BA^*e^{-2ik_1x}\cdot |B|^2]$$

$$=\frac{\hbar k_1}{m}|A|^2+\biggl[\frac{-\hbar k_1}{m}|B|^2\biggr]$$

$$\vec J_1(x)= \vec J_{inc}(x)+ \vec J_{ref}(x)$$

Where incident probability current density is,

$$\vec J_{inc}(x)=\frac{\hbar K_1}{m}|A|^2$$

And reflected probability current density is,

$$\vec J_{ref}(x)=\frac{\hbar K_1}{m}|B|^2\dotsm(13)$$

Similarly, transmitted probability current density is,

$$\vec J_{tans}(x)=\frac{\hbar k_2}{m}|c|^2\dotsm(14)$$

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.