Løsningsforslag Eksamen 18. desember 2003 TFY4250 Atom- og molekylfysikk og FY2045 Innføring i kvantemekanikk

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1 Eksamen TFY desember løsningsforslag 1 Oppgave 1 Løsningsforslag Eksamen 18. desember 003 TFY450 Atom- og molekylfysikk og FY045 Innføring i kvantemekanikk a. With Ĥ = ˆK + V = h + V (x), m x equation on the form we can write Schrödinger s time-independent h ψ(x) = [E V (x)]ψ(x) that is, m x d ψ/dx ψ = m [V (x) E]. h (i) In classically allowed regions (where E > V (x)), we see that the curvature d ψ/dx is negative when ψ is positive (and vice versa). This means that ψ must curve towards the x-axis. Examples: (ii) In classically forbidden regions (where E < V (x)), the curvature has the same sign as ψ. ψ then will curve away from the axis. Examples: For one-dimensional potentials V (x) the energy levels are non-degenerate, with only one eigenstate ψ n (x) for each energy level E n. (The degeneracy is g n = 1.) When the potential is symmetric (with respect to the origin x = 0), the parity operatotor will commute with the Hamiltonian, and it is possible to show that ψ n is also an eigenfunction of the parity operator, with parity +1 (ψ n symmetric) or 1 (ψ n antisymmetric). One also finds that the ground state is symmetric, the first excited state is antisymmetric, the second excited state is symmetric, and so on. b. For x > a, the time-independent Schrödinger equation, has the general solution ψ = m m [V (x) E]ψ = h h (V 0 E)ψ κ ψ, ψ(x) = C e κx + D e +κx. Since the last term diverges in the limit x, we have to choose D = 0 to get an acceptable solution. Thus, ψ(x) = C e κx for x > a, with κ 1 h m(v 0 E), q.e.d.

2 Eksamen TFY desember løsningsforslag The penetration depth may be defined as the depth at which ψ is reduced by a factor 1/e: e κl p.d. = e 1 = l p.d. = 1 κ. c. When the number N of bound states is large (>> 1), the energies E 1 and E of the ground state and the first excited state will be much smaller than V 0. Therefore, κ i = 1 h m(v 0 E i ) 1 h mv 0 for i = 1,. Since 8mV 0 a / h π N, we find that l p.d. a = 1 κ i a h 8mV 0 a 1 πn << 1, showing that the penetration depths for ψ 1 and ψ are almost equal and much smaller than a. Inside the well, the two solutions behave as ψ 1 = A 1 cos k 1 x and ψ = A sin k x. Since the penetration depths are small, we see from the figure that k 1 a π and k a π. Thus the energies are only a little bit lower than the corresponding energies for a box with width a: E 1 = h k 1 m π h 8ma and E = 4 h k m π h ma 4E 1, d. When b is small compared to l p.d., we have b κ i = κ b i 4 = b/4 << 1, i = 1,. l p.d. Then the solutions for the region 1 b < x < 1 b, ψ 1 = B 1 (e κ 1x + e κ 1x ) and ψ = B (e κ x e κ x ), q.e.d. will not curve very much over the interval 1 b < x < 1 b, even less than shown in the figure, which exaggerates the effect:

3 Eksamen TFY desember løsningsforslag 3 We then understand that the wave number k 1 and hence the energy E 1 will be slightly larger than for the case b = 0. We also see that k and E will be slightly smaller than for b = 0. e. When b is large compared to l p.d., on the other hand, the two wave functions are strongly suppressed in the barrier region in the middle, and ψ 1 and ψ in the well regions are very similar to the ground state for an isolated well of width a: Here, we see that the two wave numbers are almost equal, both being approximately equal to k for the case b = 0. Thus the two energy levels are almost degenerate, E 1 of course being slightly smaller than E : E > E1 π h ma. Oppgave a. From the formula for the current density we find for region III (x > L): [ ] j III = Re t ikx h d e im dx t eikx = hk m t. Similarly, with ψ i = exp(ikx) alone, or ψ r = r exp( ikx) alone, we would find j i = hk m 1 and j r = hk m r, respectively. With ψ I = exp(ikx) + r exp( ikx), we find [ (e j I = Re ikx + r e ikx) hk ( e ikx re ikx)] m = hk [ 1 r + Re ( ] r e ikx re ikx) m }{{} = j i + j r, q.e.d., since the underbraced quantity is purely imaginary. b. For a stationary state, the probability current density (and the probability density) are time-independent. Then there can be no accumulation of probability anywhere, and

4 Eksamen TFY desember løsningsforslag 4 since we are here dealing with a one-dimensional problem, the current density has to be constant, not only in time but also along the x-direction. Thus j I = j II = j III. This means that j i = j r + j III = j r + j III. Our interpretation is that the incoming probability current is divided into a reflected current and a transmitted current, and that the transmission and reflection probabilities are respectively. c. With T = j III j i = t and R = j r j i = r, k = me/ h, q = m(e V 0 )/ h and k q = m(e E + V 0 )/ h = mv 0 / h, we have T = t 4k q = 4k q cos ql + (k + q ) sin ql = 4k q 4k q + (k q ) sin ql 4E(E V 0 ) = 4E(E V 0 ) + V0 sin ql, q.e.d. In the limit E/ V 0, we have T = lim E/ V 0 1 V E(E V 0 ) sin ql = 1, in accordance with classical mechanics (which states that transmission takes place whenever E > V 0 ). For finite values of E/V 0 (> 1), we see that the transmission probability T is smaller than 1, contrary to the classical result. However, there are exceptions: For values of E and V 0 such that ql = L h m(e V 0 ) = nπ, n = 1,,, we get complete transmission also quantum mechanically. Since q = π/λ II, we see that T equals 1 whenever the width L of the barrier or well is an integer multiple of 1 λ II, where λ II is the wavelength in region II. (We are here supposing that E > V 0.) d. With a = πa 0 and k π/a = 1/a 0, we have an energy that is smaller than the height V 0 of the barrier, E = h k m e h 8m e a 0 < V 0 = 5 h k. 8m e a 0 In the formula for T we must then replace q by iκ, where κ = me V 0 h m ee h = 1 a 0.

5 Eksamen TFY desember løsningsforslag 5 With sin ql = [sin(iκl)] = sinh (κl), we then have a (tunneling) transmission probability 4E(V 0 E) T = 4E(V 0 E) + V0 sinh (κl). Since κl = 1 a 0 5a 0 = 5 is rather large, we have approximately sinh (κl) 1 4 (eκl e κl ) 1 4 eκl >> 1. This means that the second term in the denominator is much larger than the first one. Thus T 16E(V 0 E) e Lκ, V0 which is much smaller than 1. With E/V 0 = 1/5 we find T = 64 5 e 10 = To estimate the lifetime τ, we must find the semiclassical velocity and collision frequency of the particle. The velocity is of typical atomic size: v = This gives a collision frequency and a time E/m e = h m e a 0 = ν = v a = e 4πɛ 0 hc c = 1 αc. αc 8πa 0 = s 1, t 1 = 1 ν = s between each collision. The probability to find the particle still in jail at time t then is (1 T ) t/t 1. This means that the lifetime τ is given by Oppgave 3 (1 T ) τ/t 1 = 1/e = τ = t 1 T = s. a. The existence of a simultaneous set of eigenfunctions of a set of operators requires that the operators commute among themselves. In the present case we have for example: [Ĥ, ˆL ] = 0 = [Ĥ, ˆL z ] = [ˆL, ˆL z ]. The magnetic quantum number m l is restricted to the values 0, ±1, ±,..., ±l. This means that there are l + 1 spherical harmonics for a given value of the quantum number l. The magnetic quantum number m l does not enter the radial equation, which determines the energies. Therefore, the energy eigenvalues (E nl ) in this problem can be characterized by the quantum numbers n and l, and each of these levels will have a degeneracy l + 1, which is typical for a spherically symmetric potential.

6 Eksamen TFY desember løsningsforslag 6 Since the wave function ψ must be zero for r > a, where the potential is infinite, we must have u nl (a) = 0 to get a continuous wave function, just as for the one-dimensional box. Using the normalized spherical harmonics, we have from the normalization condition: a a 1 = ψ nlml d 3 r = Y lm dω [R nl (r)]r dr = 1 [u nl (r)] dr, q.e.d., when we work with real radial functions. 0 b. We see that the radial equation has one-dimensional form, and for l = 0 we have d u dr = me h u = k u, with E h k and u(0) = u(a) = 0, m that is, an ordinary box of width a. The general solution is u = A sin kr + B cos kr, where the condition u(0) = 0 gives B = 0, and the condition u(a) = 0 gives ka = nπ, or k n0 = nπ/a, with n = 1,, 3,. We get a normalized solution ( a 0 [u n0(r)] dr = 1) by choosing A = /a. The energies and the complete solutions for the s-waves then are E n0 = h kn0 m = h π ma n = n E 10 and ψ n00 = u n0 r Y 00 = 1 sin(nπr/a), n = 1,,. πa r c. The figure shows the effective potential, which in this case consists only of the centrifugal barrier h l(l + 1)/mr, for l = 1 and l =. 0 We note that the centrifugal barrier is proportional to l(l + 1) and makes the well more shallow and also more narrow for increasing l. Based on this we must expect that the energies for a given number n of nodes increase in the order of increasing l: E n0 < E n1 < E n <. We also expect the energy to increase when the number of nodes increases for a fixed l: E 11 < E 1 < E 31 <, as we have already verified for the s-waves. This is because an increasing number of nodes means increasing curvature and increasing kinetic energy. From this kind of reasoning, we expect the ground state to be an s-wave, with no zeros except those for r = 0 and r = a, that is, ψ 100.

7 Eksamen TFY desember løsningsforslag 7 d. With kr = x we have for small r: u a = sin kr kr cos kr = x 1 (x x 3 /3! + O(x 5 )) (1 x /! + O(x 4 )) = x /3 O(x 4 ), cos kr u b = kr sin kr = x 1 (1 x /!+O(x 4 )) (x x 3 /3!+O(x 5 )) = 1/x x/+o(x 3 ). Only u a behaves as (kr) l+1 r l+1 = r for small r, which is acceptable, while u b behaves unacceptably for small r and can not be normalized. Since u a is a solution of the radial equation and behaves as it should for small r, it only remains to require that u(a) = 0 : l = 1 : u(a) = sin ka ka cos ka = 0 = tan ka = ka, q.e.d. e. In c, we concluded that the ground state must correspond to nl = (1, 0), and u 10 sin(k 10 r), with k 10 = π a and E 10 = h k 10 m = h π ma. Based on the discussion in c, we must expect that the first excited level corresponds either to nl = 1, 1 or nl =, 0. In the latter case we have already found the energy: nl = 0 : k 0 = π a = k 10 = E 0 = 4E 10. To find the energy of the states ψ 11m = r 1 u 11 Y 1m, corresponding to nl = 1, 1, we must find the smallest value of k which gives u a a zero at x = a; sin kr sin ka cos kr = 0 = cos ka = 0, kr r=a ka corresponding to the condition tan ka = ka. To find this k-value it would be instructive to plot x 1 sin x cos x as a function of x (see the Comment below). However, it is fairly easy to locate the first zero using the calculator. We already know that this function is positive for small x, starting out as x /3. For x = π it is still positive (=1). For x = π it is equal to 1, so the first zero is somewhere between π and π. Using the calculator, it is fairly easy to find that the first zero occurs for x = ka = , corresponding to k 11 = a = π a π = k 10, and E 11 = (1.4303) E 10 =.046 E 10, which is lower than E 0. Thus the first excited level is E 11 (for n = 1 and l = 1), with the wave functions ( ) sin ψ 11m = Cr 1 k11 r cos k 11 r Y 1m, m = 0, ±1. k 11 r Comment: The dashed curve in the figure below shows ( ) sin k11 r u 11 (r) = C cos k 11 r k 11 r

8 Eksamen TFY desember løsningsforslag 8 (plotted with the E 11 -line as axis). Note that u 11 has a turning point where the E 11 - line crosses the centrifugal barrier for l = 1. Also shown is the E 1 -line (n = 1, l = ), which is in fact the second excited level (with energy E E 10 ), and the corresponding function u 1, which turns out to be u 1 = ( ) 3 (k 1 r) 1 sin(k 1 r) 3 k 1 r cos(k 1r). In addition we see that the s-waves u 10 and u 0 are ordinary box curves. We also observe that u 0 corresponds to the third excited level.