Tunneling effect [ continue ]

Cont..

Using boundary condition at x=a

$$\psi_{II}(x)|_{x=a}= \psi_{III}(x)|_{x=a}$$

$$Ce^{k_2a}+ De^{-k_2a}= Fe^{ik_1 a}\dotsm(15)$$

and,

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

$$Ce^{k_2a}- De^{-k_2 a}= \frac{ik_1}{k_2}Fe^{ik_1 a}\dotsm(16)$$

Adding (15) and (16)

$$\therefore\; C= \biggl(1+\frac{ik_1}{k_2}\biggr)\frac F2 e^{ik_1 a}\cdot e^{-k_2a}\dotsm(17)$$

Subtracting (16) from (15)

$$\therefore\; D= \biggl(1- \frac{-ik_1}{k_2}\biggr)\frac F2 e^{ik_1 a}\cdot e^{k_2 a}\dotsm(18)$$

Substituting from (17) and (18) in (13), we get,

$$A=\frac F4 e^{ik_1 a } e^{-k_2 a}\biggl(1+\frac{k_2}{ik_1}\biggr)\biggl(1+\frac{ik_1}{k_2}\biggr)+ \frac F4 e^{ik_1 a} e^{k_2a}\biggl(1-\frac{k_2}{|k_1}\biggr)\biggl(1-\frac{ik_1}{|k_1}\biggr)$$

$$=\frac F4 e^{ik_1 a}\biggl[\biggl(2+ (\frac{k_2}{ik_1}+ \frac{ik_1}{k_2})\biggr)e^{-k_2 a}+e^{k_2 a}\biggl(2-\biggl(\frac{k_2}{ik_1}+ \frac{ik_2}{k_2}\biggr)\biggr]$$

$$=\frac F4 e^{ik_1 a}\biggl[2(e^{k_2a}+ e^{-k_2a})- \biggl(\frac{k_2}{ik_1}+ \frac{ik_1}{k_2}\biggr)(e^{k_2a- e^{-k_2 a}})\biggr]$$

$$=Fe^{k_1a}\biggl[\frac{e^{k_2a}+ e^{-k_2 a}}{2}- \frac12 \biggl(\frac{k_2}{ik_1}+ \frac{ik_1}{k_2}\biggr)\biggl(e^{k_2a- e^{-k_2a}}{2}\biggr)\biggr]$$

$$= Fe^{ik_1 a}[ coshk_2a- \frac{(k_2^2- k_1^2)}{2ik_1k_2} sinhk_2 a]$$

$$\Rightarrow \frac FA= \frac{e^{-ik_1 a}}{coshk_2a- \frac{(k_2^2- k_1^2)}{2ik_1 k_2}sinhk_2a}\dotsm(19)$$

$$\biggl(\frac FA\biggr)^*= \frac{e^{ik_1 a}}{coshk_2a+ \frac{(k_2^2- k_1^2)}{2ik_1k_2}sinhk_2a}\dotsm(20)$$

Transmission coefficiet (T)= \(\frac{ J_{trans}(x)}{J_{inc}(x)}=\biggl|\frac{F}{A}\biggr|^2\)

$$\Rightarrow \biggl(\frac FA\biggr)\biggl(\frac FA\biggr)^*= \frac{1}{cosh^2k_2a+ \frac{(k_2^2-k_1^2)^2}{4k_1^2 k_2^2}sinh^2k_2a}$$

$$\therefore\; T= \frac{1}{1+ \frac{(k_2^2+ k_1^2)^2}{4k_1^2 k_2^2}sinh^2k_2a}\dotsm(21)$$

Similarly, reflection coefficient is,

$$R= \frac{ J_{ref}(x)}{J_{inc}(x)}$$

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

$$= \biggl|\frac BF\biggr|^2\cdot \biggl|\frac FA\biggr|^2$$

Solving above equation, we get,

$$R= \frac{\frac{(K_2^2+ k_1^2)^2}{4k_1^2 k_2^2}\cdot sinh^2 k_2 a}{1+\frac{(k_2^2+ k_1^2)^2}{4k_1^2k_2^2}sinh^2k_2a}$$

$$\therefore\;\; R= \frac{(k_1^2+ k_2^2)^2sinh^2k_2a}{4k_1^2k_2^2+ (k_1^2+k_2^2)^2\cdot sinh^2_2 a}\dotsm(22)$$

In terms of \(V_0\) and E, substituting \(k_1^2=\frac{2mE}{\hbar^2}, k_2^2= \frac{2m(V_0- E)}{\hbar^2}\)

$$R= \frac{V_0^2 sinh^2k_2a}{4E(V_0-E)+ V_0^2sinh^2 k_2a}\dotsm(23)$$

$$and $$

$$ T= \frac{ 4E(V_0-E}{4E(V_0-E)+ V_0^2 sinh^2 k_2 a}\dotsm(24)$$

Special case:-

If \(k_2a\) is large:

$$sinhk_2a= \frac{e^{k_2a}- e^{-k_2a}}{2}$$

$$e^{-k_2a}\Rightarrow Too\; samll$$

$$\therefore\; sinhk_2a\approx\frac{e^{k_2a}}{2}$$

$$\therefore\; sinh^2k_2a\approx \biggl[\frac{e^{k_2a}}{2}\biggr]^2= \frac{e^{2k_2a}}{4}\dotsm(25)$$

Here,

$$T= \frac{1}{1+\frac{V_0^2sinh^2k_2a}{4E(V_0-E)}}\dotsm(24)$$

Neglecting 1 wirth respect to \(\frac{e^{2k_2a}}{4}\) in the denominator of equation (24) we get,

$$T\approx \frac{1}{\frac{V_0^2}{4E(V_0- E)}\frac{e^{2k_2a}}{4}}$$

$$\therefore\;= 16\frac{E}{V_0}\biggl(1- \frac{E}{V_0}\biggr)e^{-2k_2a}\dotsm(26)$$

$$OR$$

$$T= 16\frac {E}{V_0}\biggl(1- \frac{E}{V_0}\biggr)e^{-2a\sqrt{\frac{2m(V_0-E}{\hbar^2}}}$$

Physical significance:

1. In case of classical mechanics for \(E<V_0\), the transmission problem across potential barrier is zero. So x>0 in our case is classically forbidden region. But Quantum Mecahnically the value of transmission coefficient is not zero, it is finite as given by equation (24) and (26)

2. The value of transmission problem as due to wave nature of incident particle. This phenomenon of crossing high energy potential barrier by particle of low K.E is called Tunneling effect.

3. As the width of potential barrier ( a) increases, then probability of transmission decreases.

4. As the height of potential barrier increases, then the probability of transmission also decreases.

5. For particle of larger mass, the probability of transmission is samll.

In the limit of \(V_0\Rightarrow \infty\)

\(\therefore\; T\Rightarrow 0\)

If \(a\Rightarrow \infty\)

\(\therefore\; T\Rightarrow 0\)

\(m\Rightarrow large\; T\Rightarrow small\)

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.