QM I Exercise Sheet 2

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Gertrude Cox
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1 QM I Exercise Sheet 2 D. Müller, Y. Ulrich HS 7 Prof. A. Signer Issued: Due: 0./ Exercise : Finite Square Well (5 Pts.) Consider a particle in a finite well potential (V 0 > 0) V 0 a x a V (x) = 0 x > a. One can parametrise the even solution of this problem as F e x x > a Ψ(x) = D cos(lx) a x a F e +x x < a, () with 2mE 2m(E + V0 ) = and l =. (2) The allowed energies for the bound states (E < 0) are given by the relation = l tan(la). (3) a) Let us assume that the particle in the well is an electron and that the potential parameters are a = 0 9 m and V 0 = 0 8 J. Determine the lowest four energy states. This cannot be done analytically. You can use a computer program to solve this problem. What are the wavelengths of the photons emitted from this system? b) Calculate the probability to find a particle in one of the lowest two energy states in the interval [ a, a]. Do this both analytically and numerically with the values above. c) Find the odd solutions (Ψ( x) = Ψ(x)) of the bound states (E < 0). SOLUTION: a) In order to write down equation (3) in a compact way, we define the dimensionless parameters z la, z 0 a 2mV0. (4)

2 Using the fact that 2 + l 2 = 2mV 0 /, we can write tan(z) = When we use the numerical values, we obtain E = J E 2 = J E 3 = J z 2 0 z 2. (5) E 4 = J ( E5 = J ). The wavelength of photon emitted from this system can be calculated via E is the difference between two energy levels λ = hc E. (6) E 2 E : λ = 992 nm E 3 E : λ = 644 nm E 4 E : λ = 326 nm E 3 E 2 : λ = 970 nm E 4 E 2 : λ = 393 nm E 4 E 3 : λ = 66 nm. b) First we need to determine the coefficients D and F. Ψ(x) has to be continuous, i.e. Another condition is the normalisation of the wave function Ψ(x) 2 dx = 2 Combining both conditions we obtain 0 F e a = D cos(la). (7) D 2 cos 2 (lx)dx + 2 a F 2 e 2x dx =. (8) F = De a cos(la) (9) D =. (0) a + sin(2la) 2l + cos2 (la) Now we can compute the probability to find the particle in the intervall [ a, a] P [ a, a] = = a + 2l Ψ(x) 2 dx = 2 D 2 cos 2 (lx)dx () 0 cos 2 (la) 2la+sin(2la). (2)

3 If we take the numerical results from the previous exercise, we obtain E : P [ a, a] = E 2 : P [ a, a] = 0.99 E 3 : P [ a, a] = E 4 : P [ a, a] = E 5 : P [ a, a] = c) Since we are looking for an odd solution of the bound state (Ψ( x) = Ψ(x)), we take the ansatz F e +x x < a Ψ(x) = D sin(lx) a x a. (3) F e x x > a We then use the condition for the continuity to find the relation and in combination with the normalisation condition we obtain F = De ax sin(al) (4) D =. (5) a + sin2 (al) sin(2al) 2l We also have to consider the continuity of the derivative of Ψ(x) Inserting the condition for the continuity we obtain F e a = Dl cos(la). (6) = l cos(la) sin(la) = l cot(la). (7) We take the same numbers as in the previous exercise and we find the following solutions of the odd states E = J E 2 = J E 3 = J E 4 = J. Note that there are only four solutions contrary to the even wave function where we found five bound states.

4 Exercise 2: Potential Barrier and Tunneling effect (5 Pts.) This time we consider a potential of the form (V 0 > 0) V 0 0 x a V (x) = 0 otherwise. We have to distinguish the cases where the energy of the particle (E > 0) is E > V 0 or E < V 0. a) What is the behaviour of a particle in this system when E > V 0 and E < V 0, considering classical mechanics? How does this differ from quantum mechanical point of view? In the case where E < V 0 a solution reads Ae ikx + Be ikx x < 0 Ψ(x) = Ce x + De x 0 x a F e ikx a < x. (8) b) What is the physical interpretation of this situation? c) Compute the transition and reflection coefficients T = F 2 A 2 and R = B 2 A 2. (9) To do so, use the conditions for the continuity of the wave function and its derivative. d) Let us now consider this in a simplified version of the α-decay by modelling the atomic nucleus with the potential V (x) as above. For Polonium-22 V 0 20 MeV and a r22 Po 0 fm. Calculate T. SOLUTION: a) If E > V 0, the particle can pass the potential barrier. This is valid in QM and in classical mechanics. If E < V 0 classical mecanics forbids a particle to pass the barier. However, in QM this is possible. This is the so-called tunneling effect. b) For x < 0 the wave function is a superposition of in- and outgoing waves. In the region of the potential the amplitude is exponentially suppressed and for x > a the wave function is outgoing. We can interpret this as particles with flux A hitting the potential barrier. It is partially reflected (with amplitude B) and partially transmitted by tunneling and exiting the potential at x = a with amplitude F. c) The wave function and its derivative have to be continuous. This has to be especially the case at x = 0 and x = a. Therefore we are able to determine the coefficients x = 0 : A + B = C + D ika ikb = C + D x = a : Ce a + De a = F e ika Ce a + De a = ikf e ika.

5 So the transmition coefficient reads F 2 T = A 2 = V E(E V 0 ) sinh2 (a) R = B 2 A 2 = V E(E V 0 ) sinh2 (a) (20). (2) Note that R + T =. d) Polonium-22 decays to lead and α-particle with m α = kg and energy E = J. So we obtain for the transition probability T = 0.9. (22)

Quantum Mechanical Tunneling

The square barrier: Quantum Mechanical Tunneling Behaviour of a classical ball rolling towards a hill (potential barrier): If the ball has energy E less than the potential energy barrier (U=mgy), then