Reflection coefficient

Question:

• Calculate the value of R and T in the case of symmetric potential barrier of the form,

$$V(x)= V_0\; for \; -a\leq x\leq a$$

$$= 0\; otherwise$$

Solution:

Calculation of R

$$B= \frac c2 \biggl[1-\frac{k_2}{k_1}\biggr]+\frac D2 \biggl[1+\frac{k_2}{k_1}\biggr]\dotsm(12)$$

$$C= \biggl(1+\frac{k_1}{k_2}\biggr)\frac F2 e^{ik_1a}\cdot e^{-ik_2a}\dotsm(15)$$

$$D= biggl(1-\frac{k_1}{k_2}\biggr)\frac{F}{2}e^{ik_1a}e^{ik_2 a}$$

Using value of C and D in equation (12) we get,

$$B= \frac{F}{4}e^{ik_1a}\biggl[\biggl(1-\frac{k_2}{k_1}\biggr)\biggl(1+\frac{k_2}{k_2}\biggr)e^{-k_2a}+\biggl(1+\frac{k_2}{k_1}\biggr)\biggl(1-\frac{k_1}{k_2}\biggr)e^{ik_2 a}\biggr]$$

$$=\frac F4 e^{ik_1 a}\biggl[\biggl(1+\frac{k_1}{k_2}- \frac{k_2}{k_1}-1\biggr)e^{-ik_2a}+ \biggl(1-\frac{k_1}{k_2}+\frac{k_2}{k_1}-1\biggr)e^{ik_2a}\biggr]$$

$$=\frac F4 e^{ik_1 a}\biggl[\biggl(\frac{k_1}{k_2}-\frac{k_2}{k_1}\biggr)\biggl(e^{-ik_2a}-e^{ik_2 a}\biggr)\biggr]$$

$$=\frac F4 e^{ik_1 a}\biggl(\frac{k_1}{k_2}-\frac{k_2}{k_1}\biggr)(2i)\frac{[e^{ik_2a}- e^{-ik_2a}]}{2i}$$

$$\therefore\; B= \frac{-iF}{2}e^{ik_1a}\biggl(\frac{k_1}{k_2}- \frac{k_2}{k_1}\biggr)sink_2 a\dotsm(25)$$

From equation (17),

$$F=\frac{A}{e^{ik_1a}[cosk_2a- \frac i2\biggl(\frac{k_1}{k_2}+\frac{k_2}{k_1}\biggr)sink_2a]}$$

$$or,\; B=\frac{\frac{-i}{2}A\biggl(\frac{k_1}{k_2}-\frac{k_2}{k_1}\biggr)sink_2a}{cosk_2a-\frac i2 \biggl(\frac{k_1}{k_2}+\frac{k_2}{k_1}\biggr)sink_2a}$$

$$or,\;\; \frac BA= \frac{(\frac{-i}{2})\biggl(\frac{k_1}{k_2}-\frac{k_2}{k_1}\biggr)sink_2a}{cosk_2a- \frac{i}{2}\biggl(\frac{k_1}{k_2}+\frac{k_2}{k_1}\biggr)sink_2a}\dotsm(26)$$

$$and$$

$$\biggl(\frac BA\biggr)^*= \frac{\frac i2 \biggl(\frac{k_1}{k_2}-\frac{k_2}{k_1}\biggr)sink_2a}{cosk_2a+ \frac i2 \biggl(\frac{k_1}{k_2}+\frac{k_2}{k_1}\biggr)sink_2a}\dotsm(27)$$

$$now,$$

$$R= \biggl|\frac BA\biggr|^2$$

$$=\biggl(\frac BA\biggr)^*\biggl(\frac BA\biggr)$$

$$R= \frac{\frac14\biggl(\frac{k_1}{k_2}-\frac{k_2}{k_1}\biggr)^2\cdot sin^2k_2a}{1+}\dotsm(28)$$

R in terms of E and \(V_0\),

$$R= \frac{V_0^2 sin^2 k_2 a}{4E(E-V_0)+V_0^2 sin^2 k_2 a}\dotsm(29)$$

From (22) and (29)

$$R+T=\dotsm(30)$$

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.