1. Hydrogen Atom: 3p State

Transcription

1 7633A QUANTUM MECHANICS I - solutio set - autum. Hydroge Atom: 3p State Let us assume that a hydroge atom is i a 3p state. Show that the radial part of its wave fuctio is r u 3(r) = e r 3 r(6 r). The radial part of the hydroge eergy eigefuctios is solved i the lecture otes. We recall that the radial part is r u l(r), where ( ) r u l (r) = N l r l+ e r/ L l+ l. (.) The ormalizatio factor N l = By employig Eq. (.), we obtai ( l )! Γ(l + + ) ( ) l+3. (.) N 3 = 7 6. (.3) Let us determie the eeded Laguerre polyomial by employig the Rodrigues formula (cf. lecture otes). We obtai Thus L 3 (x) = e x x 3 d ( e x x 4) dx = x + 4. (.4) L 3 Employig Eqs. (.), (.3), ad (.5), we obtai Thus the radial part ( ) r = (6 r). (.5) 3 3 u 3 (r) = e r 3 r (6 r). (.6) r u 3(r) = e r 3 r(6 r). (.7)

2 . Normalizatio Factor for the Hydroge Atom Calculate the ormalizatio factor N or the Hydroge atom. I = N l Use the recurrece relatio r l+ exp r [ L l+ l ( )] r dr tl α k = (k + α + )L α k (k + )L α k+ (k + a)l α k. This problem is solved o the lecture otes pp

3 3. Hydroge Atom: Certai Expectatio Values Calculate the expectatio value of the potetial eergy i the hydroge atom eergy eigestates. Calculate also v usig V + T = E. Fially, calculate the speed v = v i SI-uits. The hydroge atom is discussed i the lecture otes chapter 6. foud out that the potetia atural uits is There it is V (r) = r. (3.) Moreover, the eergy eigefuctios are obtaied at p. 48. Employig them, we obtai Clearly, ψ lm V (r) ψ lm = [ ( )] r N l r l+ e r/ L l+ l dr π π Y l m (θ, ϕ) Y l m (θ, ϕ) dϕ dθ } {{ } = [ ( )] r = N l r l+ e r/ L l+ l dr r = t ( ) l+ ] = N l t l+ e t L l+ l dt cf. p. 38 ( ) l+ Γ(l + + ) = N l ( l )! cf. p. 49 O the other had, =. (3.) ψ lm H ψ lm = ψ lm T ψ lm + ψ lm V ψ lm = E. (3.3) v = p m p = m m = T. (3.4) m Thus ψ lm v ψ lm = m ψ lm T ψ lm = m (E ψ lm V ψ lm ) E = = m. (3.5) 3

4 By expressig the eergy eigevalue ad the potetial eergy expectatio value i SI uits, we obtai the result i SI uits. Thus, i SI uits The speed ψ lm v ψ lm = v = v = e 4 (4πɛ ). (3.6) 4πɛ. (3.7) e 4

5 4. Two-Dimesioal Hydroge Atom Determie the eergy levels of a two-dimesioal hydroge atom. That is, determie the eergy levels of a two-dimesioal system comprisig a proto ad a electro. Note that the potetiauctio of the system is V (r) = 4πɛ e r, where r is the distace betwee the proto ad the electro. Let us cosider the lecture otes discussio o the hydroge atom (pp ). We ote that it ca be applied to our case up to Eq. (3). Thus, by usig a aalogous otatio with the lecture otes, we have ( ψ(r) + E + ) ψ(r) =. (4.) r Sice the expressio i the parethesis depeds oly o r, it is useful to write the Laplacia i polar coordiates. By doig so we obtai ( r ψ ) + ( ψ r r r r ϕ + E + ) ψ =. (4.) r Let us seek a solutio of the form ψ(r, θ) = R(r)Y (ϕ). As we substitute it ad divide both sides by it, we obtai d R r dr (rr ) + Y r Y + E + =. (4.3) r As we carry out the differetiatio, multiply both sides by r ad rearrage, we obtai rr R + r R R + r + Er = Y Y. (4.4) Now the left-had side depeds oly o r, whereas the right-had side depeds oly o ϕ. Thus both sides must be equal to some costat, say m. Settig the right-had side equal to m yields Y (ϕ) = m Y (ϕ). (4.5) We obtai periodic solutios oly if m > (i.e. m is real). Now, by takig ito accout the requiremet that Y (ϕ) must be π-periodic, we obtai Y (ϕ) = Ae imϕ + Be imϕ, m Z. (4.6) We obtai the polar represetatio of the Laplacia by omittig the term / z i the cylidrical Laplacia. 5

6 Settig the left-had side of Eq. (4.4) equal to m yields rr + r R + rr + Er R = m R. (4.7) Substitutig R = u/ r yields ( u + E + ( ) ( m r m + ) ) u =. (4.8) That is, we obtai the lecture otes Eq. (33) with the differece that l has bee replaced by m /. Aalogously with the lecture otes, we demad that R() must be fiite ad that the itegral rr(r) dr = r must coverge. That is, the fuctio u must satisfy u(r) dr (4.9) u() =, (4.) lim u(r) =. (4.) r Moreover, for large r Eq. (4.8) ca be approximated by The geeral solutio to this is u + Eu =. (4.) u(r) = Ae Er + Be Er. (4.3) Sice u(r) must vaish at ifiity, we decide to seek a solutio of the form u(r) = w(r)e Er. (4.4) Substitutig this ito Eq. (4.8) yields a differetial equatio for the ukow fuctio w, amely w ( Ew + r (m )(m + ) ) r w =. (4.5) For this equatio, we decide to seek a solutio of the form w(r) = a k r k. (4.6) Substitutig ad rearragig, we obtai (cf. problem 3 i this exercise) [t(t ) (m )(m + ) ] a tr t k=t k=t + [(k(k + ) (t )t) a k+ + ( k ] E)a k r k =. (4.7) 6

7 Sice this must hold for all r [, ), the coefficiets must vaish separately. Settig the coefficiet of r t equal to zero yieds t = m + or t = m +. (4.8) That is, the summatio i Eq. (4.6) must begi from either of these values. I order to meet the boudary coditio (4.), we choose t = m / +. By settig the remaiig coefficiets equal to zero, we obtai k E a k+ = k(k + ) (t )t a k k : k {t, t +, }. (4.9) If E is such that oe of the coefficiets a k is zero, the a k+ lim = E. (4.) k a k k That is, the the fuctio w(r) behaves like e Er for large r. Thus the fuctio u(r) does ot satisfy the boudary coditio (4.) if the series defiig w(r) does ot termiate. I order for the series to termiate, the eergy must be of the form [cf. Eq. (4.9)] E = k, k {t, t +, }. (4.) We recall that t is a positive half-iteger ad i particular t = / whe m = [cf. Eqs. (4.8) ad (4.6)]. Thus the eergy levels of a two-dimesioal hydroge atom are E = ( +, {,,, }. (4.) ) 7

8 5. Selectio rules for electroic trasitios i the hydroge atom Derive the dipole selectio rules l = ± ad m = ± by calculatig the matrix elemets for the x ad y compoets of the dipole operator D = er. You may eed the recurrece relatio (l + ) t P m l (t) = Pl+ m (t) P l m (t). I the lecture otes sectio 6.3 we studied the electroic trasitio rules i the hydroge atom. We bega by statig the probability of spotaeous emissio per uit time, amely W sp. em. = 4e ω 3 if 3 c 3 f r i. (5.) The we stated that the matrix elemet f r i is ozero oly if the azimuthal quatum umber chages by oe, i.e. l = = ±. (5.) However, we showed oly that the matrix elemet f z i vaishes if the selectio rules l = ± ad m = are ot satisfied. Thus, let us ext examie the matrix elemets f x i ad f y i. Let us cosider the matrix elemet of the operator x = r si θ cos ϕ betwee some hydroge atom eergy eigestates f ad i. Usig the lecture otes otatio, we obtai f x i = r 3 R i R f dr π π ( si θ Y m f ) Y m i cos ϕ dθdϕ. (5.3) } {{ } =:I θ,ϕ As we write the cosie i the expoetiaorm ad substitute the spherical harmoic fuctio, we obtai I θ,ϕ = N lim i N lf m f ( ) mi+m f π si θ P mi π (cos θ)p m f } {{ } =:I θ e i(mi m f +)ϕ + e i(mi m f )ϕ dϕ (cos θ)dθ. (5.4) We easily see that the itegral over the azimuth agle is ozero if ad oly if m f = m i ± (i.e. m = ±). O the other had, by makig the the chage of variables t = cos θ, we obtai I θ = t P mi (t)p m f (t) dt. (5.5) 8

9 As we apply a recurrece relatio to the factors t P mi (t) ad t P m f (t), we obtai I θ = [ + [ = + P mi+ + (t)p m f (t) dt P mi+ (t)p m f ] (t) dt ] P mi (t)p m f + + (t) dt P mi (t)p m f + (t) dt, (5.6) respectively. Whe m f = m i ±, it follows from the orthogoality of the associated Legedre polyomials that I θ is ozero if ad oly if = ±. To summarize, we have show that the matrix elemet f x i vaishes if the selectio rules l = ± ad m = ± are ot satisfied. Similarly we ca show that the matrix elemet f y i obeys the same selectio rules. I the lecture otes we showed that the matrix elemet f z i vaishes if the selectio rules l = ± ad m = are ot met. Thus it follows from Eq. (5.) that oly such trasitios are possible i which l = ± ad m = ±,. 9