Quantum Physics III (8.06) Spring 2005 Solution Set 5
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1 Quntum Physics III (8.06 Spring 005 Solution Set 5 Mrch 8, 004. The frctionl quntum Hll effect (5 points As we increse the flux going through the solenoid, we increse the mgnetic field, nd thus the vector potentil is time-dependent. This in turn induces n electric field, the mgnitude of which is E φ = A φ c t = Φ Q t πrc. This electric field in turn induces current, j = t l = σe. Integrting this over φ, we find the chrge tht flows through the circulr pth: Q = σ Φ dφ r πrc = σφ 0 c. We lerned on the previous problem set tht R H = σ H, so using σ H = e /3h, we find Q = e 3. The frctionl chrge of the excittion leds to lot of very interesting physics; one notble effect is tht the qusiprticles behve neither s fermions or s bosons, but rther s something in between ( nyons.. A delt-function bump in the center of n infinite squre well (0 points ( ( point The eigenfunctions for the unperturbed Hmiltonin re ( πnx ψn(x 0 = cos n odd ( πnx = sin n even (b ( points The first order correction to the energies due to the perturbtion is given by / E n ( = n H n = dx ψn(x 0 αδ(x / = α ψ 0 n(0. Thus E n ( = α if n is odd, nd E( n = 0 if n is even. The energies re unchnged for even n becuse the wve functions ψk 0 vnish t x = 0, nd therefore do not feel the delt function perturbtion, which hs support only t tht point. (c (3 points The correction to the ground stte wve function is, to first order, ψ ( H m m ψ E m. 0 The mtrix element of the perturbtion is 0 E0 m { / α H n = dx ψ(xψ 0 n(xαδ(x 0 = n odd 0 n even, / = which gives us ψ ( = n odd,n> 4mα ( πnx π h n cos. (
2 P P0 0.5 m_lph hbr pi Figure : First order computtion of the probbility of finding the prticle t x = 0. (d (4 points To first order in α, the probbility of the prticles in the ground stte being found t x = 0 is given by ψ (0 = ψ (0 (0 + Re[ψ ( (0ψ(0 (0]. From prt (c, this is ψ (0 = ( + 4mα π h n. We need to evlute the infinite sum. To do this, it is useful to write n = so tht our sum is Thus, to first order, n odd,n> n = n odd,n> ψ (0 = ( n = n +. ( mα π h. ( n n+ We plot in figure the rtio ψ (0 / ψ (0 (0. In the region where mα π h, our pproximtion mkes good physicl sense: For α > 0, the potentil αδ(x is repulsive, nd so the probbility of finding the prticle t x = 0 decreses. For α < 0, the potentil is ttrctive, nd so the probbility of finding the prticle t x = 0 increses. However, t the point mα = π h /, vlue of α/ is equl to the ground stte energy, hence αδ(x cnnot be considered s perturbtion ny more. Notice tht beyond these vlues our first order pproximtion begins producing negtive probbilities, which is not trustworthy. (e (3 points The second order correction to the energies is gin 0 for even energy levels. For odd energy levels, we hve ( α E n ( m = π h n k. k odd,k n,
3 The second order correction to the ground stte energy in prticulr is ( E ( α m = π h k = m (mα π h. k odd,k> The second order correction to the ground stte energy is equl to the first order correction when (mα m π h = m (mα, tht is, when π h = mα. This estimte of the brekdown of perturbtion theory grees, up to fctors of O(, with our estimte of prt (d. (f (6 points We now wnt to solve the problem exctly. The boundry conditions on our wve function re: ( ψ(± = 0, (b ψ is continuous t x = 0, (c ψ x (ɛ ψ x mα ( ɛ = ψ(0. h We use the tril wve function { ψ(x = A e ikx + B e ikx / x 0 A e ikx + B e ikx 0 x /, where k = me/ h. Applying the first of our three conditions tells us, A + B e ik = A e ik + B = 0. ( Applying the second condition gives, A + B = A + B, using which with ( gives (B A ( e ik = 0. Therefore either exp(ik = or B = A. Now using the third boundry condition, we obtin: (A B (A B = mα ik h (A + B. (3 If exp(ik = then using ( we obtin A = B nd A = B, which substituted into (3 gives A = A. Hence, in this cse we deduce our wvefunctions re sin(k n x, where n is even nd k n = nπ/. This cse is sme s the unperturbed solution with even n. Hence there re no chnges either to the energy or to the wvefunctions with n even due to the delt function perturbtion. If B = A A, then ( gives A = Ae ik nd B = Ae ik. Substituting for A, B, A nd B in (3 we get trncendentl eqution: ( k cot = mα k h. (4 3
4 This implicitly determines the llowed vlues of k n, nd therefore the energies E n. The exct wvefunction is given by (not required for credit: { ψ cos(kn x mα n(x = k n h sin(k n x / x 0 cos(k n x + mα. k n h sin(k n x 0 x / We would now like to expnd (4 for smll α. This mens we need to know the expnsion of cot (x round x 0, cot (x / x + x 3 /3 + O(x 5, where n is n odd integer. This expnsion is not unique, becuse the cotngent is multivlued; we hve to specify the integer n ppering in the first term of the expnsion. Therefore we cn mrk these eigensttes by n odd n. Using this expnsion, k n + mα h k n We need to solve this order by order. To zeroth order in α, k n (0. To first order in α, k n ( + mα h k n (0 Therefore, up to first order in α the energy is + mα nπ h. E n = h k n m = h π n m + α + O(α, which is precisely the expnsion in α tht we found using the first order perturbtion theory. Optionl prts: Expnding to the second order in α k n ( + mα h k ( n Therefore energy up to the second order in α is (5 + mα nπ h m α h 4 n 3 π 3. (6 E n = h kn m = h π n m + α mα n π h + O(α3, which is exctly the energy upto the second order in α. To find the first order corrections to the wvefunction, let us evlute ψ n (0 ψ, which re co-efficients of expnsion c n. For even n you cn check tht they re going to generte second order corrections only. For odd n > we get, c n = ψ (0 n ψ = πn sin( mα sin( nπ π h (7 + 4 mα (n h π. (8 ( mα π h 4nmα ( n π h sin ( nπ Note: When we write the series for ψ (, using these coefficients c n, wht we get is similr to but NOT the sme s wht we obtined using perturbtion theory, given in Eq. (. Tht mens tht, somewhere, I hve mde slip. Perhps one of you cn figure out where. 4
5 3. A delt-function interction between two bosons in n infinite squre well (5 points ( ( points The prticles re independent, so to find the two prticle wve functions we simply multiply the free wve functions nd symmetrize: ψ ground = ψ (x ψ (x = ( cos πx cos ψ first = [ψ (x ψ (x + ψ (x ψ (x ] = [ ( πx cos sin ( πx + sin ( πx ( πx ( πx ] cos. The energies of these wve functions re the sums of the free prticle energies, E ground = h π m E first = 5 h π m. (b (3 points The first order correction to the ground stte energy is E ( ground = V 0 / = V 0 ( = 3V 0. / dx dx ψ (x ψ (x δ(x x / / ( dx cos 4 πx The first order correction to the first excited stte energy is E ( first = V / 0 / ( / = V 0 = V 0. dx dx ψ (x ψ (x + ψ (x ψ (x δ(x x / ( dx cos πx ( πx sin 4. Anhrmonic oscilltor (5 points ( (3 points The ground stte wve function is even, while the perturbtion λx 3 is odd. Therefore the integrl dx ψ 0 λx 3 vnishes, nd the first order contribution to the ground stte energy is zero. To clculte the second-order shift, we need the expecttion vlue n ( We clculte ( = , so the only contributions to the shift in the ground stte energy will come from the sttes nd 3. We find, therefore, ( 3 E ( h 0 = λ k ( ( 3 [ h 9 = λ mω E 0 E k mω hω + ], hω k>0 5
6 V x m omeg 3 lmbd x Figure : The potentil λx 3. which gives E ( 0 = λ h 8m 3 ω 4. (b (3 points The first order correction to the ground stte wve function is ψ ( 0 = m m H 0 E 0 E m m. As in prt (, the only contributions re from m = 3 nd m = : ψ ( 0 = λ ( ( 3/ h 3 + hω mω 3 3. ( The ground stte wve function is then, to first order, ψ 0 = 0 λ h 3/ ( hω mω (c (3 points A sketch of the potentil is in figure. This potentil is unbounded from below, nd so there is no ground stte ny stte loclized ner x = 0 is unstble, s it will eventully tunnel through the brrier. Perturbtion theory cnnot see tunneling effects. It is good for exmining reltively smll, loclized chnges in the potentil, but cnnot del with cses like this, where we hve drsticlly chnged the symptotic behvior of the potentil. The bove qulittive rgument is enough for full credit. However, it is nice to see quntittively wht is going on. You lerned in 8.04 how to compute the probbility of tunneling through brrier. If you pply these methods to our problem, you will find tht the probbility for the prticle to tunnel from x = 0 to x = mω λ goes like exp( const/λ. This is nonzero, but we cnnot expnd it s Tylor series in λ. Thus, to ny finite order in perturbtion theory, we compute the probbility of tunneling to be zero. (d (6 points We now tke the perturbtion to be H = λx 4. To clculte the first order shift in the ground stte energy, we need ( = We find the 6
7 V x V x x x Figure 3: The potentil mω x + λx 4, with (left λ > 0, nd (right λ < 0. first order shift in the ground stte energy is ( E ( h 0 = 3λ. mω We sketch the behvior of the potentil for positive nd negtive λ in figure 3. If λ > 0, perturbtion theory is good pproximtion to the ground stte, which sees region of the potentil where λx 4 mω x. (Perturbtion theory will not work so well for higher sttes. The contribution to the ground stte energy t first order is positive, which mkes physicl sense we re confining the prticles with shrper potentil, which suggests tht their energy should increse. If, of the other hnd, λ < 0, perturbtion theory clerly fils. The chnge to the ground stte energy is negtive, so perturbtion theory does see tht the overll energy will be lowered. However, s in prt (c, perturbtion theory doesn t know bout the chnge in the symptotic behvior of the potentil, nd cnnot inform us bout the tunneling tht will result. In cses like this, where flipping the sign of the perturbtion prmeter lters the symptotic behvior of the perturbtion, strictly speking the rdius of convergence of perturbtion theory s series in λ is zero. Therefore, even when λ > 0, perturbtion theory is n symptotic expnsion: it cnnot cpture ll the physics, lthough it cn still give us good informtion bout the first few terms in the series. 5. Polrizbility of prticle on ring (5 points ( (3 points The unperturbed Hmiltonin H 0 = h hs (normlized eigenfunctions ψ n (φ = π e inφ, for ll integers n. The energy eigenvlues re E n = h n m. Since the energy m φ depends only on n, the energies E n nd E n re degenerte, so ll energy levels re doubly degenerte except n = 0, which is non-degenerte. (b (5 points We cn use the usul non-degenerte perturbtion theory to clculte the first order correction to the ground stte wve function. We need the mtrix elements n cos φ 0, where I hve introduced the nottion n = Now cos φ = (eiφ + e iφ, nd π e inφ. 7
8 e ±iφ n = n ±, so n cos φ 0 = (δ n, + δ n,. Thus, ψ ( 0 = m3 qɛ ( +. h The induced electric dipole moment is ψ 0 q cos φ ψ 0 = q ψ (0 0 some lgebr we find to be d induced = m4 q h ɛ. The polrizbility of the system is, therefore, P = m4 q h. cos φ ψ( 0, which fter (c (7 points Becuse there re three hydrogen toms, if we rotte one CH 3 group by π/3, then the overll system is left invrint. Therefore our perturbtion should hve 3-fold rottion symmetry. Since tht is the simplest wy to mke sure H (φ + π/3 = H (φ. The simplest Hermitin opertor we could thus write down would be H cos 3φ (the choice H sin 3φ only differs by choice of origin. The chnge in the ground stte energy due to the perturbtion H = b cos 3φ is to first order 0. To second order, we hve E ( 0 = k = m b 9 h. m h k b 4 (δ k,3 + δ k, 3 The chnge in the ground stte wve function is to first order ψ ( 0 = m b h k δ k,3 + δ k,3 k k = m b ( h In position spce, the ground stte wve function is ψ 0 = minimized where cos 3φ =, nd mximized where cos 3φ =. π ( m b 9 h cos 3φ, so ψ 0 is Thus there is higher probbility of finding the orienttion of the CH 3 group t φ = π/3, π, 5π/3, nd lower probbility of finding it t φ = 0, π/3, 4π/3. This is esy to understnd physiclly, if we look t the ethne molecule end-on (see figure 4. The molecule cn lower its energy by rotting in such wy s to minimize the electrosttic energy rising from the interction between the hydrogen toms in the two groups. 8
9 Figure 4: A crtoon of n ethne molecule in its most fvorble orienttion, seen end on. 9
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