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Quantum Mechanics-II Problem Set - 6 Quantum Mechanics-II Problem Set - 6 1. Use WKB approximation to determine energy levels of the potential 8 V if 0 < x < a=2 <> 0 V (x) = 0 if a=2 < x < a :>1 otherwise 0 Express your answer in terms of V0 and the energy En of a particle in 0 a box. Assume E1 > V0. Solution: 1 1 Figure 1: Figure for Q.1 This is a problem with two walls and the corresponding quantization R a condition is 0 pdx = nπ~. The momentum of the particle is given by p = p2m[E − V (x)] where V (x) is as given above. Now Z a p Z a=2 p p Z a p pdx = 2m E − V0dx + 2m Edx 0 0 a=2 p p a p a = 2m[ E − V + E ] = 0 2 2 Thus p a p p 2m [ E − V + E] = nπ 2 0 ~ 1 Square both sides ma2 [E − V + E + 2pE(E − V )] = n2π2 2 2 0 0 ~ which simplifies to 2 2E − V − n2π2 2 = −2pE(E − V ) 0 ma2 ~ 0 n2π2 2 Define E = ~ , which is the energy of the n− th state of particle n 2ma2 in a box. The above equation becomes p 2E − V0 − 4En = −2 E(E − V0) square it again 2 2 2 4E + V0 − 4EV0 + 16En − 8En(2E − V0) = 4E(E − V0) which gives E2E E4 E2 n = V 2 + n − n 4 0 256 8 Simplifying, we get, 2 V0 V0 E = En + + 2 16En V The first order perturbation theory gives E = E + 0 . The extra n 2 2 V0 term goes to zero for small V0 (which gives the correct limit for 16En particle in a box) or in the classical limit for which n is very large. 2. Consider a particle in one dimension in the presence of a potential V = αjxj; α > 0. Using WKB approximation find its energy. Solution: V (x) E −c +c x 2 This is a problem with no wall but with two turning points at ±c where E c = . The momentum is given by α r E p(x) = p2m(E − V (x)) = 2mα( − j x j) α Thus we have, using Z +c 1 p(x)dx = (n + )~ −c 2 s s Z +c p Z 0 E p Z +c E pdx = 2mα + x dx + 2mα − x dx −c −c α 0 α " #0 " #+c p 2 E 3=2 p 2 E 3=2 = 2mα + x + 2mα − − x 3 α 3 α −c 0 " # p 2 E 3=2 E 3=2 E 3=2 E 3=2 = 2mα − − c + + − c 3 α α α α p 2 E 3=2 = 2 2mα 3 α Thus we have p 2 E 3=2 1 2 2mα = (n + ) 3 α 2 ~ which simplifies to 1 3 2=3 1 E = ~ α(n + )2=3 (2mα)1=3 8 2 3. Tunneling: Consider a potential barrier of width a and arbitrary shape. A free particle with energy E comes from the left and strikes it at x = 0. Find the probability of its tunnelling to region III. 3 E Region I Region Region III II x = 0 x = a Figure 2: Figure for Q.3 Solution: 2k2 A free particle with energy E = ~ is incident on the barrier from 2m the left and strikes the barrier at x = 0. The wavefunctions in three regions are as follows. Region I : ikx −ikx I (x) = e + rI e where rI is the reflection coefficient at the barrier at x = 0. If tI I is the transmission fraction at x = 0, the wave function in region II : −α(x) II (x) = tII (x)e where we have assumed that the barrier is thick enough so there is no reflection from the other end at x = a and Z x r2m α(x) = 2 (V (x) − E)dx 0 ~ 1 1 tII (x) = tII (0) = tII (0) j p(x) j r2m (V (x) − E) ~2 In the third region, which is to the right of x = a, the wavefunction is once gain a free particle with no reflected component, ikx III (x) = Ae 4 The transmission coefficient is the ratio of the square of the amplitude transmitted into the region III to that incident from region I (which we have taken to be 1). j A j2 T = =j A j2 1 Continuity of the wavefunction at x = 0 gives α(0) I (0) = tII (0)e Thus 1 Z xr2m (x) = (0) exp( (V (x) − E)dx) Ii 2m I 2 [ (V (x) − E)]1=4 0 ~ ~2 Continuity at x = a gives II (a) = III (a) which gives 1 Z ar2m (a) = (0) exp( (V (x) − E)dx) III 2m I 2 [ (V (a) − E)]1=4 0 ~ ~2 The tunnelling probability is given by 2 j III (a) j −2γ(a) T = 2 = Ae j I (0) j where Z x r2m γ(x) = 2 (V (x) − E)dx 0 ~ 4. Assume that the potential function in a metal can be approximated by the following potential V (x) = Θ(x)[V1 − eEx] where E is constant electric field. The metal occupies the space x < 0 with its surface at x = 0. Since applied field does not penetrate inside the metal, the energy of electrons below the surface are unaffected 5 by the field. Find the probability that an electron with an energy 0 < E < V1 can be ejected from the surface of the metal. V = Φ + EF − eEx EF Filled electrons x = 0 x = Φ=eE Figure 3: figure for Q.4 Solution: This problem is a simple application of the tunnelling of the previous problem, known as \cold emission". We know that the electrons which lie below the surface of a metal do not feel the effect of the electric field as the field does not penetrate inside the metal. The electrons near the surface, however, may be regarded as being in a potential well of depth φ (with , where φ is the work function of the metal, which we have come across in photoelectric effect. If the potential just at the surface is V , the electrons would get ejected with an energy E = V − φ. In the presence of an external field, the new potential outside the surface of the metal is (measured from the bottom of the conduction band) V (x) = φ + EF − eEx ≡ V1 − eEx where V1 = φ+EF , Thus the electrons at the Fermi level see a triangular V − E potential barrier extending from x = 0 to x = 1 F . We are 1 2 eE looking for the tunnelling probability of electrons at the Fermi surface 6 which we take to have energy E (in place of EF mentioned above). The tunnelling probability was shown to be given by T = exp(−γ) where r Z x2 2mp γ = 2 2 φ − eEx dx x1 ~ The integral is straight forward and we get r 3 2 2mV1 E 3=2 γ = 2 2 2 [1 − ] 3 ~ e E V1 5. Find the probability of transmission through a parabolic barrier 8 x2 <V 1 − jxj < a V (x) = 0 a2 :0 otherwise V (x) E -a -b 0 b a Figure 4: Figure Q. 5 Solution: Turning points are given by x2 E = V [1 − ] 0 a2 7 which gives p V − E x = ±a 0 = ±b V0 Thus (see Problem 3) Z x r2m γ = 2 (V (x) − E)dx 0 ~ r2m Z b x2 1=2 = 2 V0 1 − 2 − E dx ~ −b a r 2 π 2mV0a E = 2 1 − 2 ~ V0 The tunnelling probability is given by e−2γ. 6. A ball of mass m is acted on by uniform gravity g bouncing up and down an elastic floor. Take the floor to be the zero of the potential energy. Using WKB approximation obtain the quantized energy level of the system. Solution Suppose the ball is dropped from a height z = a above the floor (z = 0). The potential energy is V (z) = mgz. We have one here, one, the floor and the other at z = a. The other turning point depends on the energy of the ball (assume that the coefficient of restitution is unity). If the energy is E, the turning point is E=mg. The quantization condition is Z E=mg 3 pdz = n + π~ 0 4 p2 where, since E = + mgz, we have 2m p r mgz p(z) = 2mE 1 − E Thus Z E=mg p Z E=mg r mgz pdz = 2mE 1 − dz 0 0 E 8 mg E Integral is done by substituting y = 1 − z, so that dz = − dy. E mg The quantization condition then becomes p E Z 1 p p E 2 8E3 1=2 3 2mE ydy = 2mE × = 2 = n + π~ mg 0 mg 3 9mg 4 Thus for the energy levels are given by 9π2 31=3 E = mg2π2 2 n + 8 ~ 4 1 7. A truncated harmonic oscillator has potential m!2(x2 −a2) for jxj < a 2 and is zero outside it. Find the WKB estimate for the bound state. Under what condition (connecting m; ! and a, there is only one bound state? Solution: This is a problem with two turning points. The turning points are symmetric and given by x = ±c 1 E = m!2(c2 − a2) 2 r 2E so that c = + a2. We have m!2 Z +c 1 p(x)dx = n + π~ −c 2 r 2E p Since p(x) = p2m(E − V ) = m! + a2 − x2 = m! c2 − x2 is m!2 even, the integral may be written as Z c p π 1 2 c2 − x2dx = ~ n + 0 m! 2 The integral on the left is πc2=4.
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