Potential depth and potential barrier [ continue ]

Cont...

In region ( III) \(x>a), \psi(x)= \psi_{III}(x), V(x)=0\), Equation (2) becomes,

$$\frac{d^2\psi_{III}(x)}{dx^2}+ K_1^2 \psi_{III}(x)=0\dotsm(7)$$

Solution of equation (7) is, \(\psi_{III}(x)= Fe^{ik_1 x}+ Ge^{-ik_1 x}\)

Since, there is no barrier for x>a. So G must be zero

$$\therefore\; \psi_{III}(x)= Fe^{ik_1 x}\dotsm(8)$$

Using boundary condition at x=0

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

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

and

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

$$or,\;\; ik_1 (A-B)= ik_2 (C-D)$$

$$or,\;\; (A-B)= \frac{k_2}{k_1}(C-D)\dotsm(10)$$

Adding equation (9) and (10)

$$2A= C+D+\frac{k_2}{k_1}(C-D)$$

$$or,\;\; A= \frac C2\biggl[1+\frac{k_2}{k_1}\biggr]+ \frac D2 \biggl[1-\frac{k_2}{k_1}\biggr]\dotsm(2)$$

Substracting (10) from (9) we get,

$$or,\; B= \frac C2 \biggl[ 1- \frac{k_2}{k_1}\biggr]+ \frac D2 \biggl[1+\frac{k_2}{k_1}\biggr]\dotsm(12)$$

Using boundary conditions at x=a,

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

$$ce^{ik_2 a}+ De^{-ik_2 a}= Fe^{ik_1 a}\dotsm(13)$$

and,

$$\frac{d\psi_{II}(x)}{dx}\biggl|_{x=a}= \frac{d\psi_{III}(x)}{dx}\biggr|_{x=a}\dotsm(13)$$

$$ik_2 \biggl( ce^{ik_2 a}- De^{ik_2 a}\biggr)= ik_1 Fe^{ik_1 a}$$

$$or,\; ce^{ik_2 a}- De^{-ik_2 a}= \frac{K_1}{K_2} Fe^{ik_1 a}\dotsm(14)$$

Adding (13) and (14), we get,

$$C= \biggl(1+\frac{k_1}{k_2}\biggr) \frac{Fe^{ik_1 a}}{2}e^{-ik_2 a}\dotsm(15)$$

Similarly, subtracting (14) from (13)

$$D= \biggl(1-\frac{k_1}{k_2}\biggr) \frac{F}{2} e^{ik_1 a}\cdot e^{ik_2 a}\dotsm(16)$$

Substituting value of C and D from equation (15) and (16) in equation (11)

$$A=\frac{F}{4} e^{ik_1 a}\cdot e^{-ik_2 a} \biggl(1+ \frac{k_1}{k_2} \biggr) \biggl(1+ \frac{k_2}{k_1}\biggr)+ \frac{F}{4} e^{ik_1 a}\cdot e^{ik_2 a}\biggl(1- \frac{k_1}{k_2}\biggr)\biggl(1-\frac{k_2}{k_1}\biggr)$$

$$=\frac F4 e^{ik_1 a} \biggl[ e^{-ik_2 a}\biggl[ 1+1 + \biggl(\frac{k_1}{k_2}+ \frac{k_2}{k_1}\biggr)\biggr]+ e^{ik_2 a}\biggl(1+1- \biggl(\frac{k_4}{k_2}+ \frac{k_2}{k_1}\biggr)\biggr]$$

$$=Fe^{ik_1 a}\biggl[ \frac{2[ e^{ik_2 a}+ e^{-ik_2 a}}{4}\biggr]$$

$$=Fe^{ik_1 a}[ cosk_2 a- \frac{1}{2}\biggl(\frac{k_1}{k_2}+ \frac{k_2}{k_1}\biggr)sin k_2 a ]$$

$$\frac FA=\frac{e^{-ik_1 a}}{cos k_2 a- \frac{i}{2} \biggl(\frac{k_1}{k_2}+ \frac{k_2}{k_1}\biggr)sink_2 a}\dotsm(17)$$

$$\biggl(\frac{F}{A}\biggr)^*=\frac{e^{ik_1 a}}{cosk_2 a+ \frac{i}{2}\biggl(\frac{k_1}{k_2}+\frac{k_2}{k_1}\biggr)sink_2 a}\dotsm(18)$$

the probability of transmission or transmission coefficient is the ratio of transmitted probability current density to incident probability current density. So

$$T=\frac{\frac{\hbar k_1}{m}|F|^2}{\frac{hk_1}{m}|A|^2}$$

$$=\biggl|\frac{F}{A}\biggr|^2$$

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

Using (17) and (18) in (19), we get,

$$T=\frac{1}{cos^2k_2 a+ \frac14 \biggl(\frac{k_1}{k_2}+\frac{k_2}{k_1}\biggr)^2 sin^2 k_2 a}$$

$$or,\; T=\frac{1}{1-sin^2k_2 a+ \frac14\biggl(\frac{k_1}{k_2}+\frac{k_2}{k_1}\biggr)^2 sin^2k_2 a}$$

$$or,\;\; T= \frac{1}{1+\frac{sin^2k_2 a}{4}\biggl[\biggl(\frac{k_1}{k_2}\biggr)^2+ \biggl(\frac{k_2}{k_1}\biggr)^2+ 2\frac{k_1}{k_2}\cdot \frac{k_2}{k_1}-4\biggr]}$$

$$\therefore\; T=\frac{1}{1+\frac14 \biggl(\frac{k_1}{k_2}-\frac{k_2}{k_1}\biggr)^2\cdot sin^2k_2 a}\dotsm(21)$$

T in term of E and \(V_0\):

$$K_1= \sqrt{\frac{2mE}{\hbar^2}}\; and \; k_2= \sqrt{\frac{2m(E-V_0}{\hbar^2}}$$

$$T=\frac{1}{1+\frac14\biggl(\frac{k_1^2-k_2^2}{k_1k_2}\biggr)^2 sin^2 k_2 q}$$

$$=\frac{1}{1+\frac14 \frac{V_0^2}{E(E-V_0} sin^2 k_2 a}$$

$$=\frac{1}{1+\frac14\frac{V_0^2}{E(E-V_0)}sin^2k_2 a}$$

$$=\frac{4E(E-V_0)}{4E(E-V_0)+ V_0^2 sin^2 k_2 a}\dotsm(22)$$

Case (1) :

$$[sink_2a]_{max}=1$$

$$T_{min}= \frac{4E(E-V_0}{4E(E-V_0)+V_0^2}\dotsm(23)$$

Case (2):

$$[sin k_2 a]_{min}= 0$$

$$T_{max}= \frac{4E(E-V_0)}{4E(E-V_0)+0$$

$$=1\dotsm(24)$$

For fixed value of E and \(V_0\) and \( sink_2a=0\):

$$sink_2a = sin n\pi, n=0,1,2,3,..$$

$$\Rightarrow a= \frac{n\pi}{k_2}= \frac{n\pi}{2\pi}\lambda_2$$

Where, \(\lambda_2\) is deBroglie wave legth of particle inside potential barrier.

$$\therefore\; a= \frac n2\lambda_2= n\frac{\lambda_2}{2}$$

If the barrier width is integral multiple of half of de-Broglie wavelength of particle inside barrier then the value of transmission probability is 1 and reflection probability is zero. (Barrier becomes transparent to incident particle) this condition is known as resonance scattering.

fig

According to classical mechanics, for \(E>V_0\), all particles must be transmitted in region (III). But in Quantum mechanics for \(E>V_0\) value of T is less than 1 (i.e <1) in some cases as shown in above figure. The probability of reflection i.e. R= 1-T is non-zero in the case, which is not explained by classical mechanics.

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.