Quantum Mechanics I. 21 April, x=0. , α = A + B = C. ik 1 A ik 1 B = αc.

Transcription

1 Quatum Mechaics I 1 April, 14 Assigmet 5: Solutio 1 For a particle icidet o a potetial step with E < V, show that the magitudes of the amplitudes of the icidet ad reflected waves fuctios are the same Fid the phase shift that the wave fuctio acquires o reflectio Aswer For the give case of E < V, the time-idepedet Schrodiger equatio solved i each regio yields the wave fuctios: Vo E Vo ψ Ae ik 1 + Be ik 1 < where ψ Ce α > k 1 me mv E, α Note α > Applyig the boudary coditios cotiuity of the wave fuctio ad its first derivative at yields A + B C ik 1 A ik 1 B αc A is the amplitude of the icidet wave, B is the amplitude of the reflected wave ad C is the amplitude of the trasmitted wave Solvig these equatios, obtai ik1 + α B A ad C ik 1 ik 1 α ik 1 α A This implies that B ik 1 + α ik 1 α A ik 1 + α ik 1 α A Due Date: April 8, 14, 1: am 1

2 Quatum Mechaics I 1 April, 14 α + k1 A α + k1 A Hece the magitudes of amplitudes of the icidet ad reflected waves fuctios are the same For calculatig the phase shift B A where we have used, eta 1k1/α e ta 1 k 1 /α eta 1 k 1 /α ta 1 k 1 /α e ta 1 k 1 /α, ta 1 y 1 ta y 1 ta 1 y ta y ta y ta y 1 Now sice k 1 α me E mv E V E, the phase shift acquired upo reflectio is ta 1 E/V E Use the Heiseberg idetermiacy relatio to estimate the groud state eergy of a particle i a ifiite square well potetial Aswer The ucertaity priciple is p / If the particle is cofied i the well leadig to p / The mometum p has to be at least of the order of p, hece p > / which meas that the eergy is at least p /m /8m The groud state eergy is give by π /8m, showig a qualitative agreemet with the miimum eergy predicted by Heiseberg s ucertaity At t a particle i a ifiite square well potetial {, < <, V elsewhere, is i a state described by the wave fuctio { Ψ, si π 4 +i si π, < <, elsewhere, 1 Due Date: April 8, 14, 1: am

3 Quatum Mechaics I 1 April, 14 Determie, a the probability P E, t that a measuremet of the eergy will yield the value E b E t, c t, d p t Aswer For a particle i a ifiite square well potetial described by ormalized wave fuctio [Eq ], the iitial quatum state i the Hilbert space ca be writte as a superpositio of two eergy eigestates At time t, the state becomes ψ c ψ +c ψ 1 ψ +i ψ ψt e iĥt/ ψ e ie t/ ψ +i e ie t/ ψ, Here E is the eergy of the state ψ which is a eigestate of the Hamiltoia These eigestates are foud by solvig the time-idepedet Schrodiger equatio iside the well d m d ψ E ψ For a ifiite well the wavefuctioig correspodig to the eigestates of Hamiltoia ψ ψ π si a The probability of measurig E at time t is P E, t ψ ψt Now ψ ψt e ie t/ ψ ψ +i e ie t/ ψ ψ e ie t/ δ + i e ie t/ δ Due Date: April 8, 14, 1: am

4 Quatum Mechaics I 1 April, 14 Here δ jk is a Kroecker delta equal to 1 whe j k ad otherwise Therefore, P E, t ψ ψt e ie t/ e δ i eie t/ ie t/ δ δ + i e ie t/ δ 1 δ + δ Sice δ δ δ δ Hece the probability of measurig the eergy E at time t is 1/ ad probability of measurig the eergy E at time t is / Oe ever measures the eergy E k with k, b To fid the epectatio value of E, first we eed to determie Ψ, t; Ψ, t c ψ e ie t/ 1 ψ e iet/ + i ψ e iet/ I hope you would uderstad how Eq is obtaied Give here is first a derivatio of Eq The Schrodiger equatio is solved through ψt e iĥt/ ψ I the positio represetatio, Ψ, t ψt e iĥt/ ψ 4 The iitial state ψ ca be writte as a superpositio of the eigestates of the Hamiltoia which form a complete basis, ψ c ψ, whereupo isertig the above ito Eq 4 results i Ψ, t e iĥt/ c ψ c e iet/ ψ Due Date: April 8, 14, 1: am 4

5 Quatum Mechaics I 1 April, 14 which is the desired form Eq Now, we preset three methods for calculatig E First, we have E Ψ, t Ĥ Ψ, t where dψ, tĥψ, t 5 1 d ψ e iet/ i ψ e ie t/ 1 Ĥ ψ e ie t/ +i ψ e ie t/ 1 d ψ e iet/ i ψ e ie t/ E ψ e iet/ + i E ψ e ie t/ 1 [E ψ d + E ψ ] d +, ψ ψ d So the epectatio value of E is π si si π E 1 [E 1 + E 1] E + E d Now the secod method Sice the eergy is quatized i discrete steps, oe ca fid its average value through the discrete sum give below ad use probabilities determied from the previous part; E P E i E i i P E E + P E E 1 E + E I yet a third approach, oe ca solve this part by directly performig the calculatio i the Hilbert space as see below: ψt c e iĥt/ ψ E ψt Ĥ ψt to be foud Ĥ ψt c Ĥe ie t/ ψ c E e ie t/ ψ Due Date: April 8, 14, 1: am 5

6 Quatum Mechaics I 1 April, 14 ψt Ĥ ψt c m ψ m e iĥt/ c E e iet/ ψ m c m c E e ie m E t/ ψ m ψ m c m c E e iem Et/ δ m m c E E + E, sice c whe, Appedical discussio Here we digress motivatig the studet towards Eq5 From the discussio o page 4 ad 5 of Beck s book, you will recogize that, ˆp ψ i ψ i ψ i ψ Furthermore, it is straightforward to realize that, ˆ ψ ψ Therefore, the epectatio value of mometum ca be foud i the wave mechaical formulatio as, Similarly, we have ˆp ψ ˆp ψ ˆp The epectatio value of positio is ψ ˆ ψ d ψ ˆp ψ dψ i ψ dψ i ψ ad so o d ψ ˆ ψ dψ ψ Now the eergy E p + V, m p E m + V i / dψ + V ψ m dψ Ĥψ, Due Date: April 8, 14, 1: am 6

7 Quatum Mechaics I 1 April, 14 which is Eq 5 This is how you perform epectatio value measuremets i the wavefuctio represetatio c Ψ, t ˆ Ψ, t Now from Eq, dψ, tψ, t d Ψ, t 6 Ψ, t 1 ψ e iet/ + i ψ e ie t/ Ψ, t 1 ψ e iet/ i ψ e ie t/ 1 ψ e iet/ + i ψ e ie t/ 1 ψ + ψ +i ψ ψ e ie E t/ i ψ ψ e ie E t/ Sice ψ ψ si π ad ψ ψ si π ψ ψ ψ ψ π π si si 1 [ ] π 5π cos cos Ψ, t 1 ψ + ψ +i 1 [ cos π cos 5π ] i si Isertig this epressio ito Eq 6, we ca evaluate as give below The first itegral is ψ d ad the secod itegral is also, π si ψ d, ikewise the third itegral becomes [ π 5π cos cos ]d π 4, + 5π 48 5π Due Date: April 8, 14, 1: am 7 E E t,

8 Quatum Mechaics I 1 April, 14 Hece the epectatio value is [ π si + 5π Note that depeds o time ] E E t E E t si d p Ψ, t ˆp Ψ, t i i dψ, t i [ Ψ, t d ψ e iet/ i ψ e ie t/ d ψ ψ ψ ψ eie E t/ ] + ψ ψ +i Here both terms i the first itegral vaishes, dψ ψ 4π π si cos ad Therefore, dψ ψ dψ ψ 6π 4 5 p i i 16 5 π si cos 4 5 E E t cos π ψ e ie t/ + i ψ d π d, e ie t/ ψ ψ eie E t/ d 6π 4 4 5π 5, [ ] e ie E t/ + e ie E t/ 4 At t a particle i a ifiite square well potetial [Eq 1] is i a state described by the wave fuctio Ψ, { 5, < <, elsewhere, Due Date: April 8, 14, 1: am 8

9 Quatum Mechaics I 1 April, 14 which is depicted i the figure overleaf Determie a Ψ, t, b the probability P E, t that a measuremet of the eergy at time t will yield the value E, c E t, d t, e p t Aswer ets first verify if Ψ, is ormalized Hece Ψ, Ψ, 5 ϵ [, ] d Ψ, 5 [ d ] 5 is properly ormailzed a I Q, we have already see that Ψ, t c e ie t/ ψ, Due Date: April 8, 14, 1: am 9

10 Quatum Mechaics I 1 April, 14 where ψ si π are the eigestates of the Hamiltoia ad c ψ Ψ, d ψ Ψ, dψ Ψ, π d si 6 π d si 6 5 You ca leave this itegral as it is, call it some I, or solve to obtai π, for eve I d si 4 for odd π 7 Hece Ψ, t 6 π I 6 e iet/ si 1 1 π 1 I 7 e iet/ si 7 1,,5 4 π e ie t/ si π Note that the iitial state Ψ, is a superpositio of all the eigestates, ad is ot a statioary state of the Hamiltoia b The probability is ψ 1 Ψ, t Now ψ 1 Ψ, t ψ 1 e iĥt/ Ψ, ψ 1 e iĥt/ c ψ 1 c ψ 1 e iĥt/ ψ 1 c ψ 1 ψ e iĥt/ 1 c 1 e ie t/ ψ 1 Ψ, t c π Due Date: April 8, 14, 1: am 1

11 Quatum Mechaics I 1 April, 14 c E t Ψ, t Ĥ Ψ, t Ψ, t e iĥt/ Ψ, t Ψ, Ĥ Ψ, Sice Ψ, c ψ E t c m ψ m Ĥ c ψ m c mc E ψm ψ m c E 1,,5 E t 1,,5 E π π 4 m 1,,5 The epectatio value of eergy is idepedet of time π 166 m 6 π 6 d The easiest way to fid the epectatio value of is through wavefuctio, Sice Ψ, t oly eists for odd Ψ, t ˆ Ψ, t dψ, tψ, t 1 7 I e iet/ si π, where I is give i Eq 7 ad 1 mπ π d I 7 m e iemt/ si I e iet/ si m 1 mπ π I 7 m I e iem Et/ d si si Now + m mπ π d si si cosm π + m π sim π m [ π 1 m + 1 m + cosm + π + m + π sim + π m + Due Date: April 8, 14, 1: am 11 ]

12 Quatum Mechaics I 1 April, 14 [ ] 1 π m + 1 cosm π cosm + π + m + m m + [ ] 4m 4m cosmπ cosπ + π m m + m m + m [ ] cosmπ cosπ 1 π m m + 1 e iem Et/ m [ ] I 7 m I cosmπ cosπ 1 π m m + m 4 I π 5 m I e [ ie m E t/ cosmπ cosπ 1 ] I m If m, are both eve, cosmπ cosπ 1, I If m, are both odd, eve the cosmπ cosπ 1, I If either m is eve ad is odd or viceversa, either I m or I the I Hece e Similarly, for the epectatio value of the mometum, p Ψ, t ˆp Ψ, t Ψ, t i Ψ, t i1 mπ π d I 7 m e iemt/ si I e iet/ si m i1π mπ π I 8 m I e iem Et/ d si cos m Now mπ π d si cos 1 d si m + π + si m π 1 [ ] cosm + π/ cosm π/ m + π/ + m π/ [ ] cosm + π 1 cosm π 1 + π m + m [ ] m cosmπ cosπ m π m m π [ cosmπ cosπ 1 m ] p i1 [ ] I 7 m I e ie m E cosmπ cosπ 1 t/ m m m For m, p I all other cases p Due Date: April 8, 14, 1: am 1

13 Quatum Mechaics I 1 April, 14 5 A electro is iside a aowire, cofied to oe dimesio The legth of the wire is The quatum state of the electro is described by a particle i a ifiite well A measuremet of the eergy of the electro yields π /m, idicatig that the electro is i the groud state, ψ 1 Suddely, a mechaical force is applied to the aowire doublig its legth The ew potetial well at the istat of stretchig t is show i Fig b I our otatio ψ refers to the th eigestate i a well of legth, ad ψ refers to the th eigestate i a well of legth The groud state has 1 Similar otatio is used for eergies V oo V oo V oo V oo ψ 1, a b a Is ψ 1 is a statioary state of the ew cofiguratio? b Write ψ 1 i the positio basis, ie fid ψ 1 ψ 1 c What is the groud state eergy of the particle i the well of legth? Call it E 1 What is the groud state wavefuctio ψ? d Now the eergy is measured at t +, right after the well epads What is the probability of measurig the eergy to be E 1? e What is the probability of measurig the eergy to be E 4E 1 ad E 9E 1? f Sketch the wavefuctio E 1 ad E 1, E, E iside the well of legth Predict if you would have some probability of measurig the eergy E is eve, or odd? Justify your aswer whe g Calculate the time depedet wavefuctio Ψ, t ψt i the well of legth Due Date: April 8, 14, 1: am 1

14 Quatum Mechaics I 1 April, 14 Aswer The iitial state ψ1, is ot a eigestate of the epaded well So oe eeds to write this as a superpositio of the eigestates of the ew well, ψ et ψ1, c ψ, Therefore, ad c where ψ si 1 c ψ ψ 1, π 1 si d ψ ψ1, 1 π d si 1 π d si Ψ 1, 1 si π π Note that we have chaged the limits of itegratio from [, ] to [, ] because Ψ 1, i ϵ [/, ] Proceedig further, c 1 π π d si si 1 [ ] π π d cos 1 cos [ si 1 π 1 π si + 1 π ] + 1 π 1 [ si 1 π π 1 si + 1 π ] + 1 Now si ± 1 π π si ± π π π si cos π ± cos π 1 c 1 π si 1 π π si π si π si Due Date: April 8, 14, 1: am 14

15 Quatum Mechaics I 1 April, π π si 4 1 π π 4 si 4 4 π4 si π 4 π4 4 π4, 1, 5, 9,,, 7, 11, eve With this backgroud we ca attempt the solutio a No it is ot The statioary state of the ew cofiguratio are ψ π 1 π si si b ψ 1 si π c E1 π m π 8m ψ 1 ψ 1 1 si π see part a d From the previous questio, we ve already leart that the probability of measurig the state s eergy E Hece right after the well epads is ψ, ψ 1, c Prob measure E1 c1 1 8 π e Prob E, Prob E 4 π π Due Date: April 8, 14, 1: am 15

16 Quatum Mechaics I 1 April, 14 f The wavefuctios are show below V oo V oo ψ ψ 1 X, ψ It ca be see that Ψ, is a superpositio of oly ψ 1, ψ, ψ 5,, i the eact proportio, adjusted such that Ψ, is zero i the regio [, ] This shows that there is zero probability of detectig the eergies E, E 4, E 6 etc g Ψ, t c e iet/ ψ 4 [ 1 π 4 e iet/ ψ 4 π 4 π 1,5,9,1,,1, e ie +1t/,7,11 ] 1 4 e iet/ ψ ψ t/8m e i+1 + 1π si Due Date: April 8, 14, 1: am 16