HW10.nb 1. It is basically quadratic around zero and is very steep. One may guess that the harmonic oscillator is a pretty good approximation.


 Cecilia York
 2 years ago
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1 HW0.nb HW #0. Varatonal Method Plot of the potental The potental n the problem s V = 50 He x  L. V = 50 HE x  L 50 H + x L 8x, , <D Ü Graphcs Ü It s bascally quadratc around zero and s very steep. One may guess that the harmonc oscllator s a pretty good approxmaton. Intal guess 8x, 0, 3<D 50 x  50 x If we try to dentfy the frst term wth the harmonc oscllator potental ÅÅÅÅ m w x, we fnd w = 0 because m =. Then one may hope that t would gve the groundstate energy ÅÅÅÅ w = 5. However, ths hope s not qute fulflled. Usng the groundstate wave functon of the harmonc oscllator, y = J ÅÅÅÅÅÅÅÅÅ m w ê4 p N E m w x êh L  m x w ÅÅÅÅÅÅÅÅÅÅÅÅÅ H ÅÅÅÅÅÅ m w ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Lê4 p ê4 ÅÅÅÅÅ
2 HW0.nb y = y ê. 8m Ø, w Ø 0, Ø < 5 x J ÅÅÅÅÅÅÅ 0 Nê4 p K = ÅÅÅÅÅÅÅÅÅ  8x, <D, 8x, , <D ê. 8 Ø, m Ø < m 5 ÅÅÅÅ P = V, 8x, , <D Unque::usym : 5ê s not a symbol or a vald symbol name. The error s 50 H  ê40 + ê0 L Ebar = K + P 5 ÅÅÅÅ + 50 H  ê40 + ê0 L %  39 ê 8 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅ 39 ê and s bgger than 5%, but s pretty good, accurate wthn 7.%. Therefore, we are motvated to try a Gaussan as a tral functon. Gaussan tral functon y = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ p ê4 D ê Ex êh D L  ÅÅÅÅÅÅÅÅÅ x D ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ pê4 è!!! D K = ÅÅÅÅÅÅÅÅÅ  8x, <D, 8x, , <, Assumptons Ø D > 0D ê. 8 Ø, m Ø < m Unque::usym : 5ê s not a symbol or a vald symbol name. ÅÅÅÅÅÅÅÅÅ 4 D P = V, 8x, , <, Assumptons Ø D > 0D Unque::usym : êh4*d^l s not a symbol or a vald symbol name. 50 J  D ÅÅÅÅÅÅ 4 + D N
3 HW0.nb 3 Ebar = K + P ÅÅÅÅÅÅ 50 J  D 4 + D N + ÅÅÅÅÅÅÅÅÅ 4 D 8D, 0 ê <D , 8D Ø <<  39 ê 8 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 39 ê It has mproved, but not yet wthn 5%. Lnear tmes Gaussan One way to mprove t further s the followng. Note that the potental s not party symmetrc, and hence the groundstate wavefuncton s not expected to be an even functon. Because the potental s lower on the rght, we expect the wave functon s skewed towards the rght. Therefore, we can try y = H + k xl E x êh D L  ÅÅÅÅÅÅÅÅÅÅ x D H + k xl It s not normalzed at ths pont, and we calculate ts norm norm = 8x, , <, Assumptons Ø D > 0D Unque::usym : êh4*d^l s not a symbol or a vald symbol name. è!!! ÅÅÅÅ p D H + k D L K = ÅÅÅÅÅÅÅÅÅ  8x, <D, 8x, , <, Assumptons Ø D > 0D ê. 8m Ø, Ø < m Unque::usym : êh4*d^l s not a symbol or a vald symbol name. è!!! p H + 3 k D L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅ 8 D P = V, 8x, , <, Assumptons Ø D > 0D Unque::usym : á30à s not a symbol or a vald symbol name. 5 è!!! p D ÅÅÅÅÅÅÅ J  4 D 4 + D ÅÅÅÅÅÅÅ + 4 D 4 k D  4 D k D + k D ÅÅÅÅÅÅ  D 4 k D + D k D ÅÅÅÅÅÅ  D 4 k D 4 + D k D 4 N
4 HW0.nb 4 Ebar = K + P ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ norm j è!!! p H + 3 k j D L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅ + 5 è!!! ÅÅÅÅÅÅÅ p D J  4 D 4 + k k 8 D D ÅÅÅÅÅÅÅ + 4 D 4 k D  4 D k D + k D ÅÅÅÅÅÅÅ  D 4 k D + D k D ÅÅÅÅÅÅ  D 4 k D 4 + D k D 4 N y z y z ì H è!!! p D H + k D LL soluton = 8D, 0 ê <, 8k, 0<D , 8D Ø , k Ø <<  39 ê 8 ÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅ 39 ê Ths s correct at the % level! The wave functon s therefore y ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ è!!!!!!!!!!!!! ÅÅÅÅÅ norm ê. x H xl y PlotA ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅ è!!!!!!!!!!!! norm ê. 8x, , <E Ü Graphcs Ü As expected, t s more or less a Gaussan, but skewed to the rght. Obvously, there are many ways to mprove Gaussan. I hope you found one successfully. The analytc soluton Ths problem s actually a specal case of the Morse potental V = D HE a u  L D H + a u L
5 HW0.nb 5 Ths potental s a form of the nteratomc potental n datomc molecules proposed by P.M. Morse, Phys. Rev. 34, 57 (99). The varable u s the dstance between two atoms mnus ts equlbrum dstance u = r  r 0. The Schrödnger equaton can be solved analytcally. Expandng t around the mnmum, 8u, 0, 4<D a D u  a 3 D u ÅÅÅÅÅÅÅ a4 D u Harmonc oscllator approxmaton gves w = "########### a D ÅÅÅÅÅÅÅÅÅÅÅÅÅ. The correct energy egenvalues are known to be m E n = whn + ÅÅÅÅ L  a ÅÅÅÅÅÅÅÅÅÅÅÅ m Hn + ÅÅÅÅ L, where the second term s called the anharmoncanharmonc correcton. For large n, the second term wll domnate and the energy appears to become negatve. Clearly, the bound state spectrum does not go forever, and ends at a certan value of n, qute dfferent from the harmonc oscllator. But ths s expcted because the potental energy asymptotes to D for u Ø + and hence states for E > D must be unbound and hence have a contnuous spectrum. The groundstate wave functon s known to have the form d Ea u b a uê y = E E d a u  ÅÅÅÅÅÅÅÅÅÅÅ a b u where b = d , d = è!!!!!!!!!!!! D m ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ. Let us see that t satsfes the Schrödnger equaton. a I ÅÅÅÅÅÅ m SmplfyA 8u, <D + D HEa u  L ym ÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ y è!!! a H4 è!!!!!!! D m + a L  ÅÅÅÅÅÅÅ ÅÅÅÅÅÅ 8 m ê. 8b Ø d  < ê. 9d > è!!!!!!!!!! D m ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ =E a The frst term s ÅÅÅÅ w = a "######## ÅÅÅÅÅÅÅ D m and the zeropont energy wth the harmonc oscllator approxmaton. The second term  a ÅÅÅÅÅÅÅÅÅÅÅÅ s the anharmonc correcton. 8 m The normalzaton s norm = 8u, , <, Assumptons Ø a > 0 && d > 0D Unque::usym : êh4*d^l s not a symbol or a vald symbol name. b d b ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ a Actually, ths wave functon could have been guessed f one pays a careful attenton to the asymptotc behavors. For u Ø, the potental asymptotes to a constant D, and hence the wave functon must damp exponentally e k u wth k = è!!!!!!!!!!!!!!! m» E» ë. For u Ø , the potental rses extremely steeply as D e  a u. It suggests that the energy egenvalue becomes quckly rrelevant, and the behavor of the wave functon must be gven purely by the rsng behavor of the potental. By droppng the energy egenvalue and lookng at the Schrödnger equaton,  ÅÅÅÅÅÅÅÅ d y ÅÅÅÅÅÅÅÅÅÅ + D e  a u y = 0, m d u and change the varable to y = e a u, we fnd  ÅÅÅÅÅÅÅÅ m I d y ÅÅÅÅÅÅÅÅÅÅ + ÅÅÅÅ ÅÅÅÅÅÅÅ d y d y y d y M + D y = 0. The second term n the parentheses s neglble for y Ø. Therefore the wave functon has the behavor y e è!!!!!!!!!!! m D yê a. Combnng the behavor on both ends, the wave functon has precsely the exact form gven above. Ths s the lesson: the onedmensonal potental problem s so smple that there are many ways to study the behavor of the wave functon. On the other hand, the realworld problem nvolves many more degrees of freedom. In many cases, the Hamltonan tself must be guessed.
6 HW0.nb 6 Back to the Morse potental. The case of the homework problem corresponds to D = 50, a =, m =, =. Therefore the groundstate energy s E 0 = ÅÅÅÅ a "######## ÅÅÅÅÅÅÅ D m  a ÅÅÅÅÅÅÅÅÅÅÅÅ 8 m = 5  ÅÅÅÅ 8 = ÅÅÅÅÅÅ SmplfyA y ê. 8b Ø d  < ê. 9d Ø è!!!!!!!!!!!! norm u  ÅÅÅÅÅÅÅÅÅ 9 u ÅÅÅÅÅÅÅÅÅ 567 è!!!!!!!!!! 43 9 u ÅÅÅÅÅÅÅÅÅ è!!!!!!!!!! D m ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ a PlotA u  ÅÅÅÅÅÅÅÅÅ 567 è!!!!!!!!!!, 8u, , <, PlotPonts Ø 50E 43 = ê. 8D Ø 50, a Ø, m Ø, Ø <E Ü Graphcs Ü The varatonal lnear tmes Gaussan wave functon was ` x PlotA ` H ` xl, 8x, , <, PlotPonts Ø 50E Ü Graphcs Ü
7 HW0.nb 7 %%D Ü Graphcs Ü Qute close.. Neutrno Oscllaton (a) The untarty matrx s 88, 0, 0<, 80, 3 D, 3 D<, 80, 3 D, 3 D<< êê MatrxForm 0 0 y 0 3 D 3 D j z k 0 3 D 3 D 3 D, 0, 3 D E I d <, 80,, 0<, 3 D E I d, 0, 3 D<< êê MatrxForm 3 D 0 Â d 3 D y 0 0 j z k  Â d 3 D 0 3 D D, D, 0<, D, D, 0<, 80, 0, << êê MatrxForm D D 0 y D D 0 j z k 0 0 U = 88, 0, 0<, 80, 3 D, 3 D<, 80, 3 D, 3 D<<. 3 D, 0, 3 D E I d <, 80,, 0<, 3 D E I d, 0, 3 D<<. D, D, 0<, D, D, 0<, 80, 0, << D 3 D, 3 D D, Â d 3 D<, 3 D D  Â d D 3 D 3 D, D 3 D  Â d D 3 D 3 D, 3 D 3 D<, 8 Â d D 3 D 3 D + D 3 D,  Â d 3 D D 3 D  D 3 D, 3 D 3 D<<
8 HW0.nb 8 Udagger = ê. 8d Ø d<d D 3 D, 3 D D  Â d D 3 D 3 D,  Â d D 3 D 3 D + D 3 D<, 3 D D, D 3 D  Â d D 3 D 3 D,  Â d 3 D D 3 D  D 3 D<, 8 Â d 3 D, 3 D 3 D, 3 D 3 D<< 88, 0, 0<, 80,, 0<, 80, 0, << For the twobytwo case, the untarty matrx s U3 = U ê. 8q Ø 0, q 3 Ø 0< 88, 0, 0<, 80, 3 D, 3 D<, 80, 3 D, 3 D<< % êê MatrxForm 0 0 y 0 3 D 3 D j z k 0 3 D 3 D The man pont n the calculaton s that the Hamltonan can be wrtten as H = UIc p + m c ÅÅÅÅÅÅÅÅÅÅ 3 Å p M U, and hence the tme evoluton operator s e  H tê = U e Hc p+m c 3 ê pl tê U e  m c 3 tê p 0 0 = e  c p tê U 0 e  m c 3 tê p 0 U j k 0 0 e  m 3 c 3 tê p z The ampltude of our nterest s y A = E I c p tê I80,, 0<.U3.DagonalMatrxA9E I m c 3 têh p L, E I m c 3 têh p L, E I m 3 c 3 têh p L =E.  Â c p t ÅÅÅÅÅÅÅÅÅÅÅÅÅ 8<, j  Â c 3 ÅÅÅÅ p 3 D +  Â c3 3 p 3 D y z k Astar = A ê. 8t Ø t< ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Â c p t j Â c3 k p 3 D + Â c3 3 p 3 D y z Psurv = AstarDDD 3 D 4 + CosA c3 t Hm  m 3 L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ E p 3 D 3 D + 3 D 4
9 HW0.nb 9 Ths can be rewrtten as P surv = cos 4 q 3 + cos q 3 sn q 3 coshm  m 3 L c 3 t ÅÅÅÅÅÅÅÅÅÅ p + sn4 q 3 = Hcos q 3 + sn q 3 L  cos q 3 sn q 3 + cos q 3 sn q 3 coshm  m 3 L c 3 =  cos q 3 sn q 3 I  coshm  m 3 L c 3 t ÅÅÅÅÅÅÅÅÅÅÅ p M =  4 cos q 3 sn q 3 sn Hm  m 3 L c 3 t ÅÅÅÅÅÅÅÅÅÅÅ 4 p =  sn q 3 sn Hm m 3 L c 3 t ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 4 p. Ths s nothng but Eq. (5) n the problem. (b) t ÅÅÅÅÅÅÅÅÅÅÅ p For p = GeV, m 3  m =.5ä03 ev ê c, and usng = 6.58 ä0  MeV sec = 6.58 ä06 ev sec and c = 3.00 ä0 0 cm sec , the argument of sn.5ä0 s 3 ev êc c 3 t ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅÅÅÅÅ 950 t = ÅÅÅÅÅÅÅÅÅÅÅÅÅ 950 c t = 37 L L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = 3.7 c ev ev sec sec c sec 0 0 ä03 ÅÅÅÅÅÅÅ cm km. Takng q 3 = ÅÅÅÅÅÅÅÅ è!!!! and hence sn q 3 =, we fnd P surv =  sn H3.7 ä03 ÅÅÅÅÅÅÅ L km L = cos H3.7 ä03 ÅÅÅÅÅÅÅ L km L. 03 LD, 8L, 0, 0000<E Ü Graphcs Ü The oscllaton occurs on the dstances of thousands of klometers. Ths s what had been observed by the SuperKamokande experment, Phys. Rev. 93, 080 (004), even though the data shows nearly washedout oscllaton due to the soso resoluton n energy and dstance. A very macroscopc quantum phenomenon. (c) For the threestate case, we keep all angles, and we calculate the oscllaton probablty from n m to n e,
10 HW0.nb 0 Amue = E I c p tê I8, 0, 0<.U.DagonalMatrxA9E I m c 3 têh p L, E I m c 3 têh p L, E I m 3 c 3 têh p L =E.Udagger.  Â c p t ÅÅÅÅÅÅÅÅÅÅÅÅÅ 880<, 8<, j Â d Â c3 3 ÅÅÅ p 3 D 3 D 3 D + k  Â c3 p D 3 D 3 D D  Â d D 3 D 3 DL +  Â c3 p 3 D D D 3 D  Â d D 3 D 3 DL y z Amuestar = % ê. 8t Ø t, d Ø d< ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Â c p t j Â d+ Â c3 3 k Â c3 p 3 D 3 D 3 D + p D 3 D 3 D D  Â d D 3 D 3 DL + Â c3 p 3 D D D 3 D  Â d D 3 D 3 DL y z and from n e to n m, Pmue = AmuestarDD D 3 D 3 D D  CosA c3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ p E D 3 D 3 D D  CosAd  c3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ p + c3 p E D 3 D 3 D D 3 D 3 D + CosAd  c3 p + c3 p E D 3 D 3 D D 3 D 3 D + D 3 3 D 3 D D 3 D 3 D  CosAd + c3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ p E D 3 3 D 3 D D 3 D 3 D  D 3 D 3 D D 3 3 D 3 D + CosAd  c3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ p + c3 p E D 3 D 3 D D 3 3 D 3 D + 3 D 3 D 3 D  CosA c3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ p E D 3 D 3 D 3 D + D 4 3 D 3 D 3 D  CosA c3 p E 3 D D 3 D 3 D + CosA c3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ p E D 3 D D 3 D 3 D + 3 D D 4 3 D 3 D
11 HW0.nb Aemu = E I c p tê I80,, 0<.U.DagonalMatrxA9E I m c 3 têh p L, E I m c 3 têh p L, E I m 3 c 3 têh p L =E.Udagger.  Â c p t ÅÅÅÅÅÅÅÅÅÅÅÅÅ 88<, 80<, j Â d Â c3 3 ÅÅÅ p 3 D 3 D 3 D + k  Â c3 p D 3 D 3 D D  Â d D 3 D 3 DL +  Â c3 p 3 D D D 3 D  Â d D 3 D 3 DL y z Aemustar = % ê. 8t Ø t, d Ø d< ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Â c p t j Â d+ Â c3 3 k Â c3 p 3 D 3 D 3 D + p D 3 D 3 D D  Â d D 3 D 3 DL + Â c3 p 3 D D D 3 D  Â d D 3 D 3 DL y z Pemu = AemustarDD D 3 D 3 D D  CosA c3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ p E D 3 D 3 D D  CosAd + c3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ p E D 3 D 3 D D 3 D 3 D + CosAd + c3 p E D 3 D 3 D D 3 D 3 D + D 3 3 D 3 D D 3 D 3 D  CosAd  c3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ p + c3 p E D 3 3 D 3 D D 3 D 3 D  D 3 D 3 D D 3 3 D 3 D + CosAd + c3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ p E D 3 D 3 D D 3 3 D 3 D + 3 D 3 D 3 D  CosA c3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ p E D 3 D 3 D 3 D + D 4 3 D 3 D 3 D  CosA c3 p E 3 D D 3 D 3 D + CosA c3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ p E D 3 D D 3 D 3 D + 3 D D 4 3 D 3 D  PemuD 4 3 D SnA c3 t Hm  m L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ E 4 p SnA c3 t Hm  m 3 L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ E SnA c3 t Hm  m 3 L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ E q 4 p 4 p D 3 D q 3 D Indeed, when d 0, two probabltes are dfferent.
12 HW0.nb (d) As we dd n the class, the tmereserval nvarance states that Xa» e  H tê» b\ = Xb è» e  H tê» a è \, and hecepha Ø bl = PHb è Ø a è L. Here, a è s the tmereversed state of a, and b è s the tmereversed state of b. In our case, the tmereversed state has the opposte momentum. (If you worry about the helcty, both the momentum and the spn flp, and remans the same, namely lefthanded.) However, the Hamltonan depends only on the magntude of the momentum and hence the oscllaton probabltes also depend only on the magntude of the momentum. Therefore PHb è Ø a è L = PHb Ø al. Choosng n e and n m of the same momenta to be the ntal and fnal states, we fnd that the tme reversal nvarance would predct PHn m Ø n e L = PHn e Ø n m L. The fact that the tmereversal volaton requres d 0makes sense because the tmereversal opeator nvolves the complex conjugaton, and d s the only complex parameter n the Hamltonan.
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