c n ψ n (r)e ient/ h (2) where E n = 1 mc 2 α 2 Z 2 ψ(r) = c n ψ n (r) = c n = ψn(r)ψ(r)d 3 x e 2r/a0 1 πa e 3r/a0 r 2 dr c 1 2 = 2 9 /3 6 = 0.

Transcription

1 Poblem {a} Fo t : Ψ(, t ψ(e iet/ h ( whee E mc α (α /7 ψ( e /a πa Hee we have used the gound state wavefunction fo Z. Fo t, Ψ(, t can be witten as a supeposition of Z hydogenic wavefunctions ψ n (: Ψ(, t n c n ψ n (e ient/ h ( whee E n mc α Z n In the sudden appoximation, Ψ is continuous at t, so eqns. ( and ( give ψ( c n ψ n ( n c n ψn(ψ(d x In paticula, c ψ (ψ(d x π ( a / e /a πa e /a 4π d / 4 a 9/ e /a d So the pobability the He ion is in the gound state is {b} Using the hint, the mean enegy adiated is c 9 / 6.7 Ē ad c (E E + c (E E + c (E E + Each tem is the pobability of the He winding up in an excited state times the enegy it would adiate in etuning to the gound state. (The fist tem is zeo. Rewite this as Ē ad c n E n E c n n Ψ H Ψ t E Hee we have used the fact that second sum is unity. Now H p m e whee the is in the potential tem since Z fo t +. Since Ψ(t + Ψ(t ψ(, we get ( p Ē ad ψ ( m e ψ(d x E ψ ( (H t e ψ(d x E n

2 Now ψ (H t ψ(d x ψ (E (Z ψ(d x and So ψ ( E (Z ( e ψ(d x e 4e a e a πa e /a 4π d e /a d mc α ( E (Z Ē ad mc α mc α + mc α mc α (.6 ev Poblem The initial state is Ψ(x, t Ψ(x, t, the gound state of the hamonic oscillato. Afte t, the state is a fee paticle, with geneal solution a supeposition of plane waves (eqn.. in Giffiths: Ψ(x, t π A(ke i(kx Ωt dk ( whee Ω hk /m. (We need to distinguish Ω fom the symbol ω used to descibe the oscillato. So Inveting the Fouie tansfom: A(k π Ψ(x, π Ψ(x, e ikx dx A(ke ikx dk ( mω /4 ( exp mω π hπ h x e ikx dx ( mω ( /4 / π h e hk /mω, π hπ mω whee we have used the integal in the hint. Substitute this in eqn. (: Ψ(x, t π π π ( mω ( /4 / π h hπ mω ( mω ( /4 / π h hπ mω ( mω ( /4 / π h hπ mω ( mω /4 exp hπ + iωt exp exp ( hk mω + ikx i hk t m dk [ h ] mω ( + iωtk + ikx dk ] [ mωx h( + iωt π h( + iωt/mω exp ] [ mωx h( + iωt whee we have used the integal in the hint again. Note that Ψ(x, t is a Gaussian centeed on the oigin whose width inceases with time: the pobability distibution speads.

3 Poblem {a} By popety (i, Ψ(t c ψ + c ψ Since an enegy measuement yields E and E with equal pobability, c c /. The oveall phase of a state is abitay, so let s choose c eal: c, c e iφ Ψ(t ( ψ + e iφ ψ Now let s use popety (ii: x ( ψ + e iφ ψ ˆx ( ψ + e iφ ψ Substitute eqn. (?? fo ˆx. Now the matix elements of ˆx between the same states give zeo because ψ â ψ etc. So x h ( e iφ ψ â + â ψ + e iφ ψ â + â ψ mω Note that ψ â + â ψ ψ â + â ψ, so the second tem is the complex conjugate of the fist. x h [ e iφ ( ψ â ψ + ψ â ψ + c.c. ] mω Since â ψ ψ and â ψ ψ, only the fist tem is nonzeo. x h [ e iφ + c.c. ] mω h mω cosφ The lagest positive value is fo φ, so Ψ(t ( ψ + ψ {b} Now put in the time dependence: Ψ(t (e iet/ h ψ + e iet/ h ψ (e iωt/ ψ + e iωt/ ψ {c} mω h p Ψ(t i (â â Ψ(t i mω h ( ( e iωt/ ψ + e iωt/ ψ (â â e iωt/ ψ + e iωt/ ψ The only tems that contibute ae ψ â ψ ψ ψ and ψ â ψ ψ ψ, so p i mω h ( e iωt/ e iωt/ + e iωt/ e iωt/ mω h sinωt

4 The lagest positive value occus when sinωt. The smallest t fo which this is tue is ωt π T π ω Poblem 4 In classical mechanics, the motion of a paticle with enegy E in a potential V ( is constained by the equation E T + V. The maximum distance attainable fom the cente of an attactive potential is a tuning point of the motion, found by setting T, so that all the enegy is potential enegy. Any distance geate than this is the classically fobidden egion. E V max mc α e max Hee we want to eplace the expession fo the gound-state enegy of hydogen with the equivalent expession e /a, so Poblem 5 {a} max a Pob( > max 4 a a R ( d 4 a e /a d e y y dy (y /a e y ( y y 4 /e 4.8 S (s + s s + s + s s so H c(s s s Now the states ˆp x sm ae eally states s s s m with s and s fixed (in this case, s s. So Fo ˆp x, we get Fo ˆp x m, we get {b} In the ˆp x m m basis, we have H ˆp x sm c h [ s(s + 4 4] ˆpx sm E c h [ ] 4 c h E c h [ ] 4 c h (tiply degeneate s s s x s x + s y s y + s z s z 4 (s + + s (s + + s 4 (s + s (s + s + s z s z s +s + s s + + s z s z Recall that So s +, s+ h, s h, s H ˆp x 4 c hˆp x H ˆp x c hˆp x 4 c hˆp x H ˆp x c hˆp x 4 c hˆp x H ˆp x 4 c hˆp x

5 So fo example H ˆpx 4 c h and the matix epesentation of H in this basis is H 4 c h The eigenvalues of this ae the enegy levels (solutions of Hψ Eψ: E det E E E gives E twice and E E, ( + E 4, E o Putting back the facto of 4 c h, we have the same answe as pat (a: E 4 c h o 4 c h ( times {c} Wite the geneal solution as a supeposition of stationay states: Ψ(t n c n ψ n e ient/ h c ˆp x e iet/ h + c ˆp x e iet/ h + c ˆp x e iet/ h + c 4ˆp x e iet/ h {d} The easiest way to do this is to ewite the solution (c in tems of the states ˆp x etc. using ˆp x ˆp x ˆp x (ˆpx + ˆp x ˆp x ˆp x ˆp x (ˆpx ˆp x (Remembe that we ae being lazy hee: ˆp x eally means, but since the fist two quantum numbes s s stay fixed, we e omitting them. Similaly, ˆp x eally means the poduct state ˆp x ˆp x whee we dop the fist quantum numbe in each ket. So solution (c becomes Ψ(t A (ˆp x ˆp x e ie t/ h + [ Bˆp x + C (ˆp x + ˆp ] x + Dˆpx e ie t/ h whee we have edefined some constants to absob the facto of /. Now at t, Ψ( ˆp x B D, A + C, A + C Thus A C and Ψ(t (ˆpx ˆp x e ie t/ h + (ˆpx + ˆp x e ie t/ h {e} P(, Ψ(t P(, Ψ(t (e iet/ h + e iet/ h e ict h/4 + e ict h/4 4 e ict h/4 e ict h/4 + e ict h/4 4 ( ct h cos

6 Poblem 6 The fist-ode coection to the gound state enegy is E H (4 whee is the gound state ket. To find H, we need to fist find the potential inside and outside a unifom spheical chage distibution of adius R. The petubation H is the diffeence between this potential and the Coulomb potential used when the poton is teated as a point paticle. (Moe pecisely, the diffeence in the potential enegies. The quickest way to find the potential is to use Gauss s Law to get the E field and then integate the field to get the potential. Gauss s Law is E ds 4π ρ d x S In spheical symmety the E field is adial and we choose the Gaussian suface to be a sphee of adius. This gives 4π E( 4π V ρ(d x E( q( whee q( is the chage enclosed by a sphee of adius. Since e, > R q( ( e, < R R we get e E(, > R e R, < R (Note that e is positive. The chage on an electon is e and on a poton is +e. The potential enegy of the electon is ( V ( ( e E( d Fo < R we get R V ( e d + e R R d ( e R + R R while fo > R we have V ( e /. So H V ( ( e, > R ( e + R, < R R Eqn. (4 gives R ( ( E 4π d πa e /a e + R R Since R a, a thoughout the egion of integation. So we can set the exponential tem to unity. This gives E 4e a R 5 er a d ( + 4 R R

7 Since e /a.6 ev and R/a cm/.59 8 cm, we get E.9 9 ev Note that since this is much smalle than the distance between the gound state enegy and the fist excited state, petubation theoy is a good appoximation. (In fact, E is much smalle even than the hypefine splitting of the gound state, as we will see. Poblem 7 The unpetubed Hamiltonian of a -d otato is H L I whee I is the moment of inetia. Adding the petubation H E d Ed cosθ gives the full Hamiltonian H H + H. The unpetubed enegies satisfy The enegy levels and eigenfunctions ae E l L I ψ Eψ l(l + h, ψ lm Y lm (θ, φ I (Note that thee is no adial dependence. Although each state is (l + -fold degeneate, since the petubation doesn t depend on φ, it couples only the m states (as we ll see explicitly below. So we can use non-degeneate petubation theoy. To second ode, the gound state enegy is E E + ψ H ψ + l,m Note that we can wite H in tems of spheical hamonics as 4π H Ed Y So the fist-ode coection is 4π ψ H ψ Ed ψ lm H ψ E E l YY Y dω Y Y dω Hee dω sinθ dθ dφ and we have used the othogonality of the Y lm s. (Note that Y / 4π is a constant. The leading ode coection to the gound state enegy is theefoe second-ode. The matix elements ae 4π ψ lm H ψ Ed Y lmy Y dω Ed δ lδ m whee we have used othogonality again. So E l,m ψ lm H ψ ψ H ψ E d / E E l E E h /I E d I h Second-ode petubation theoy is a good appoximation povided E E E E d I h h I whee we have dopped the facto of /. Poblem 8 E h di

8 {a} The inteaction enegy of a chage e in an extenal potential φ is H eφ. Hee the electic field is (Adding a constant doesn t change the physics. So {b} The fist-ode coection to the gound state enegy is Thee ae seveal ways to see this is zeo: E Ee z φ φ Ez H eez ee cosθ E ee cosθ ( It is E times the expectation value of a dipole moment in a state of definite paity, which we showed in class is zeo using a paity agument. ( It involves the integal of Y Y Y. Teat the last facto, Y, as a constant. Then the fist two tems give zeo by othogonality. ( Do the θ integal explicitly. (Needless to say, this is not the ecommended method. {c} The second-ode coection to the gound-state enegy is The matix element is E n l m H (5 E E n n l,m n l m H n l m ee cosθ 4π ee n l m Y ee Ylm Y dω R nl R d ee δ l δ m R nl R d Using the Konecke deltas to do the l and m sums in (5 gives E e E n z n E E n with z n R nl R d and α E E e n z n E E n {d} In a matix element like z n, the states ae nomalized and so thei dimensions dop out of the integal. The dimensions ae just the dimensions of z, and the scale is set by the size of the atom: z n a. The enegies scale as E E n E e /a. (This is a bette expession to use hee than the equivalent mc α. Thus Poblem 9 The petubation Hamiltonian is α e a e /a a H eez ee cosθ We need to find the eigenvalues of the matix of H in the basis of n degeneate eigenstates. Label the states as,,, 4

9 whee the quantum numbes ae n, l, m. We want to find H ij i H j. We use symmety aguments to detemine which of these matix elements must be zeo. Conside the paity opeato P. This is not a symmety opeato fo H, but H has a definite tansfomation popety: P n l m ( l n l m (fom the Y lm PH P H (fom the z So n l m H n l m n l m (PH P n l m ( n l m P H ( P n l m ( l+l + n l m H n l m Thus n l m H n l m if (l + l is even. Anothe way of seeing this is fom the explicit integal fo the matix element. Since cosθ Y, the angula pat of the integal involves Y lmy Y l m dω Unde an invesion, each Y lm gives a facto of ( l, giving a total of ( l+l + as befoe. So the possible nonzeo matix elements ae {H, H, H 4 } and thei Hemitian conjugates. Thee is also a selection ule fom the φ-integal, which involves π e i( m+m φ dφ unless m m Anothe way to see this is fom the symmety of the petubation unde otations about the z-axis. The geneato of these otations is J z, and the symmety is expessed by [H, J z ]. (This is just the statement that [z, J z ], one of the equations saying that (x, y, z fom the components of a vecto opeato. Now n l m H J z n l m m h n l m H n l m Using the commutato elation, this is also equal to n l m Jz H n l m m h n l m H n l m So the matix element is nonzeo only if m m. This educes the set of nonzeo matix elements to just H H. So we have only one integal to compute explicitly: H ee ee 4π 4π Y Y Y dω 4π Y Y dω R R d [ 48a a ( ] e /a d Hee we have used the fact that Y is eal so we can easily see that the angula integal is unity. Put x /a in the adial integal to get H eea (x 4 x5 e x dx ( eea 4! 5! eea So finally Diagonalizing the matix yields the eigenvalues eea H eea E,, ±eea Thus in the pesence of the extenal electic field, the states and eceive no fist-ode coection to thei enegies. The two linea combinations of and that ae the good eigenstates ae split, one with a positive coection and one with a negative coection. a

10 state n : 4 states E states, l, m ± state E The way to undestand these esults without all the fomalism is as follows: The s state has even paity (l, the p states have odd paity (l. The petubation couples states of diffeent paity, since it s a pola vecto. So only matix elements of s with p ae nonzeo. In addition, otational symmety about the z-axis means angula momentum about that axis is conseved. So the m value of a state isn t changed by the petubation only states with the same m ae coupled.