Quantum Mechanics 1 Lecture 11: More on the Hydrogen atom

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(1)Quantum Mechanics Lecture 11: More on the Hydrogen atom Massimiliano Grazzin. 1. i. h. University of Zuric.

(2) Introduction In Lecture 7 we have solved the problem of the Hydrogen atom by nding the energy levels and the corresponding eigenfunctions ψnlm(r, θ, ϕ, t) = e −iEnt/ℏ Rnl(r)Ylm(θ, ϕ) Angular dependence entirely expressed in terms of Spherical harmonics 1/2. χ(r) (n − l − 1)!(2κ) l −κr 2l+1 Rnl(r) = =− (2κr) e Ln+l (2κr) 3 r [ 2n((n + l)!) ] 3. 1 κ= ℏ. 2m(−E). Associated Laguerre polynomials. Energy levels depend only on the principal quantum number n ≡ nr + l + 1 = 0,1,2.... Radial quantum number 2 fi. 2 (Zα) En = − mc 2 2n 2.

(3) Introduction In Lecture 7 we have solved the problem of the Hydrogen atom by nding the energy level and the corresponding eigenfunctions This is only an approximation because, besides the Coulomb interaction, there is a number of additional terms in the Hamiltonian that need to be considered Relativistic corrections - relativistic kinetic energy. Can be obtained from a relativistic wave equation: Dirac equation. - spin-orbit coupling - Darwin term Further corrections - Lamb shift - Hyper ne structure. fi. fi. 3.

(4) Relativistic kinetic energy corrections p2 We recall that the expression we have used for the kinetic energy T = can 2m. only be considered an approximation for small velocities with respect to the velocity of light of the relativistic expression E=. p2c 2 + m 2c 4. Expanding in series this expression we obtain E = mc 2. 2 2 2 p2 p (p ) 2 1 + 2 2 ∼ mc + − + ... 3 2 mc 2m 8m c. Dropping the rest energy mc 2 we see that, besides the non-relativistic term we have (p2)2 H1 = − 8m 3c 2. This term is suppressed by a factor v 2 /c 2 with respect to p2 Ze 2 H0 = − 2m r. and can be treated as a perturbation 4.

(5) We note that since. 2. 1 Ze H1 = − H0 + 2 2mc ( r ) 2. commutes with L2 and L3 we can use non-degenerate perturbation theory The shift of the energy levels is 2. 1 Ze | n, l, m⟩ ΔE1 = − ⟨n, l, m | H0 + 2 2mc r ) ( 2. 2 2 4 1 Ze Z e (0) 2 (0) | n, l, m⟩ + ⟨n, l, m | 2 | n, l, m⟩ =− (En ) + 2En ⟨n, l, m | 2 2mc ( r r ). So, we need to compute the expectation values of 1/r and 1/r 2 These expectation values could be computed by evaluating the corresponding integrals but we will use another method: we will nd Z 2e 4 n (0) 2 ⟨ 2 ⟩ = 4(En ) r l + 1/2. Ze 2 ⟨ ⟩ = − 2En(0) r. (En(0))2 4n ΔE1 = − −3 + 2 2mc ( l + 1/2 ) 5 fi. and thus we have.

(6) 2 (Zα) By using En(0) = − mc 2 2 we nd 2n. mc 2(Zα)2 (Zα)2 n 3 ΔE1 = − − 2 2 2n n ( l + 1/2 4 ). Comments: The correction is always negative (for any n and l) consistently with the fact that H1 is negative de nite The correction is of order (Zα)2 with respect to En(0) where e2 1 α= ∼ ℏc 137. is the ne structure constant. This could be expected by noting that the size of the atom is of order aB /Z where aB = ℏ2 /me 2 is the Bohr radius and typical velocities are Δp ℏ ℏZ e 2Z v∼ ∼ ∼ = = (Zα)c m mΔr maB ℏ. fi. fi. fi. 6.

(7) Intermezzo: the Feynman-Hellman and Virial Theorems An ef cient way to compute the expectation values we need is to use the Feynman-Hellmann theorem The theorem states that if the Hamiltonian depends on a continuous parameter λ and a stationary state | ψλ⟩ corresponds to the energy eigenvalue Eλ we have the relation ∂Hλ ∂Eλ Feynman-Hellmann | ψλ⟩ = ⟨ψλ | ∂λ ∂λ theorem The proof is straightforward and it exploits the fact that Hλ | ψλ⟩ = Eλ | ψλ⟩. ⟨ψλ | ψλ⟩ = 1. We have ∂Eλ ∂ψλ ∂ψλ ∂Hλ ∂ | Hλ | ψλ⟩ + ⟨ψλ | Hλ | | ψλ⟩ = ⟨ψλ | Hλ | ψλ⟩ = ⟨ ⟩ + ⟨ψλ | ∂λ ∂λ ∂λ ∂λ ∂λ ∂ψλ ∂ψλ ∂Hλ | ψ ⟩ + ⟨ψλ | | ψλ⟩ = Eλ ⟨ ⟩ + ⟨ψλ | ( ∂λ λ ) ∂λ ∂λ n. fi. 7.

(8) ∂Hλ ∂Hλ ∂ | ψλ⟩ = ⟨ψλ | | ψλ⟩ = Eλ ⟨ψλ | ψλ⟩ + ⟨ψλ | ∂λ ∂λ ∂λ e2 We now apply the theorem to compute ⟨ ⟩ r. ✅. We use the ne structure constant α as parameter, which enters only in the potential term Ze 2 Zℏcα En(0). We have. =−. 2. 2 (Zα) mc 2n 2. and multiplying by α. V(r) = −. and. r. =−. r. ∂Hα ∂En(0) Zℏc Z 2mc 2α ⟨ ⟩ = ⟨− ⟩= =− ∂α r ∂α n2 Ze 2 mc 2(αZ )2 (0) ⟨ ⟩= = − 2E n r n2. Note that this result could also be obtained by using the virial theorem We start from the general form of the Hamiltonian p2 H= + V(x) = T + V 2m fi. 8.

(9) 1 2 [H, x ⋅ p] = [p , x ⋅ p] + [V(x), x ⋅ p] 2m p2 = − iℏ − x ⋅ ∇V (m ). We have. If we evaluate the expectation value of [H, x ⋅ p] over an energy eigenstate we of course nd ⟨[H, x ⋅ p]⟩ = 0. and therefore. p2 = ⟨x ⋅ ∇V⟩ ⟨m⟩. virial theorem. 1 For the Coulomb potential ⟨x ⋅ ∇V⟩ = Ze ⟨ ⟩ = − ⟨V⟩ r 1 2. use ∇ = − r. x r3. and thus 2⟨T⟩ = − ⟨V⟩ which together with ⟨T⟩ + ⟨V⟩ = En(0) leads to Ze 2 ⟨ ⟩ = − 2En(0) r. and we con rm the result obtained with the Feynman-Hellmann theorem. fi. fi. 9.

(10) Z 2e 4 We now apply the Feynman Hellmann theorem to compute ⟨ 2 ⟩ r. To do so we use as parameter the quantum number l, which is present only in the centrifugal term ℏ2l(l + 1) Vl(r) =. 2mr 2. We have ∂Hl ∂El ℏ2 2l + 1 ∂ 1 2 1 1 2 (0) 2 2 2 ⟨ ⟩= = = − mc (Zα) = mc (Zα) = − 2E /n n 2 2 3 ⟨ ⟩ ( ) ∂l 2m r ∂l ∂l 2 (nr + l + 1) 2 n. therefore. Z 2e 4 n (0) 2 ⟨ 2 ⟩ = 4(En ) r l + 1/2. 1 For later use we also compute 3 ⟨r ⟩. This expectation value can be obtained from the previous one by using the Ehrenfest theorem (recall Lecture 3) d i ∂V ⟨pj⟩ = ⟨[H, pj]⟩ = − dt ℏ ⟨ ∂xj ⟩ 10.

(11) The time derivative of the expectation value of the momentum operator is equal to the expectation value of the force But on a stationary state the left-hand side of the above equation must vanish ∂V =0 ⟨ ∂xj ⟩ dV In particular we must have ⟨ ⟩ = 0 dr. but dV d Ze 2 ℏ2l(l + 1) Ze 2 ℏ2l(l + 1) = − + = 2 − 2 dr dr ( r 2mr ) r mr 3. Imposing that the expectation value of this quantity vanishes we obtain 1 m Ze 2 m 3c 3α 3Z 3 = 2 = 3 3 3 2 ⟨ r ⟩ ℏ l(l + 1) ⟨ r ⟩ ℏ n l(l + 1/2)(l + 1) 11. (1).

(12) Spin-orbit interaction Another correction to the energy levels originates from the magnetic interaction of the electron spin with the magnetic-moment generated by its orbital motion In the rest frame of the electron the nucleus moves with velocity −v and charge −Ze and therefore generates a magnetic eld Ze v × r Ze r × p Ze L B=− = = 3 3 c r mc r mc r 3. e Such eld interacts with the magnetic moment of the electron μe = − S and the mc. interaction hamiltonian is. Ze 2 H2 = − μe ⋅ B = S⋅L 2 2 3 2m c r. In the last equation we have added a factor of two at the denominator: this factor takes into account the fact that the electron rest frame is not inertial Such factor is called Thomas factor and emerges naturally if one starts from the beginning from a relativistic wave equation, the Dirac equation We now observe that. 1 1 2 2 2 2 S ⋅ L = ((S + L) − L − S ) = (J − L2 − S2) 2 2 fi. fi. 12.

(13) Therefore the states | j = l ± 1/2,mj, l⟩ diagonalise the operator S ⋅ L and we have ℏ2 S ⋅ L | j, mj, l⟩ = (j( j + 1) − l(l + 1) − 3/4) | j, mj, l⟩ 2. this is equal to l (for j = l + 1/2) or to (−l − 1) for j = l − 1/2. The correction to the eigenstates is thus. Ze 2 ℏ2 1 l ΔE2 = ⟨H2⟩n,j=l±1/2,l,mj = 2m 2c 2 2 ⟨ r 3 ⟩ (−l − 1). and using the result in Eq.(1) for the expectation value of 1/r 3. we nd. 1 m 3c 3α 3Z 3 = ⟨ r 3 ⟩ ℏ3n 3l(l + 1/2)(l + 1) mc 2(Zα)4 l ΔE2 = 3 4n l(l + 1/2)(l + 1) (−l − 1). l≠0. We note that ⟨1/r 3⟩ is divergent when l = 0 so the above result is correct only for l ≠ 0 When l = 0 the matrix element of S ⋅ L vanishes so we can set ΔE2 = 0. l=0. This is in agreement with what one would obtain from the Dirac equation fi. 13.

(14) Darwin term By considering the non-relativistic limit of the Dirac equation a further term arises 2 2 πℏ Ze (3) H3 = δ (x) 2 2 2m c. (2). can affect only s orbitals !. The origin of this term can be understood from the trembling motion of the electron (Zitterbewegung) According to the relativistic theory the position of the electron uctuates with λe ℏ δr = = mc 2π. where λe is the Compton wavelength of the electron (recall Lecture 1) Therefore the electron feels on average the potential 1 ∂ 2V ⟨V(x + δx)⟩ = V(x) + ⟨δx ⋅ ∇V⟩ + ⟨δxiδxj ⟩ + .... 2 ∂xi∂xj vanishes by isotropy. 1 6. 14. fl. can be rewritten as (δr)2 ∇2 V.

(15) By recalling that we nd. 21. ∇. r. = − 4πδ (3)(x). 2πℏ2Ze 2 (3) ⟨V(x + δx)⟩ = V(x) + δ (x) 2 2 3m c. which is in qualitative agreement in form, sign and magnitude, with the Darwin term in Eq.(2) The correction to the energy levels in rst order perturbation theory is πℏ2Ze 2 2 | | ΔE3 = ψ (0) 2m 2c 2 nl. More precisely, writing ψnl(x) = Rnl(r)Ylm(θ, ϕ) 1 δ(r) and using δ (x) = 2 4πr (3). we have. 1 1 2 2 2 2 | | | | | | d x | ψnl(x) | δ (x) = R (r) r δ(r)dr Y (θ, ϕ) dΩ = R (0) ∫ ∫ 4πr 2 nl ∫ lm 4π nl 3. 2 (3). fi. fi. 15.

(16) The radial wave function Rnl(r) vanishes in r = 0 for l > 0 By using the explicit expressions for Rnl (Lecture 7) we nd Z Rn0(0) = 2 ( naB ). 3/2. e2 and therefore by using as usual α = we obtain ℏc (αZ )4 2 ΔE3 = mc δl0 3 2n. The results for the three relativistic corrections can be combined in a single formula valid at the rst perturbative order mc 2(Zα)2 (Zα)2 3 n ΔErel = − 2 2 ( 2n n 4 j + 1/2 ). fi. fi. 16.

(17) To denote the energy levels we use the spectroscopic notation 2s+1. Lj. L = S, P . . . . s = 1/2 for the hydrogen atom. The relativistic corrections remove the degeneracy for the levels l ≥ 1 in j = l ± 1/2 The levels 2S1/2 and 2P1/2, having the same j are still degenerate: this is true at all orders in α in the Dirac equation Such degeneracy is lifted by the additional effects we are now going to discuss. 17.

(18) Further corrections Lamb shift Due to the fact that the electromagnetic eld is quantised, a shift in the electron position occurs, which leads to a correction to the energy level which is similar to the one of the Darwin term but further suppressed by a power of α ln(Zα) ΔELamb ∼. En(0)(Zα)2α ln (. 1 δl0 ) Zα. An heuristic derivation of this formula was provided by Welton. fi. The effect is a positive shift of the 2S1/2 level with respect to 2P1/2 by about 1057 MHz.

(19) Hyper ne structure The nucleus spin I leads to a nuclear magnetic moment egN μ= I 2MN c. where MN and gN are the nucleus mass and gyromagnetic ratio The generated vector potential is. and by using the formula. μ×r 1 A= 3 =−μ×∇ (r) r. ∇ × (a × b) = a( ∇ ⋅ b) − b( ∇ ⋅ a) + (b ⋅ ∇)a − (a ⋅ ∇)b. We nd for the corresponding magnetic eld. fi. fi. fi. 1 B=∇×A=−∇× μ×∇ ( r )) ( 1 21 = − μ ∇ − (μ ⋅ ∇) ∇ ( r r).

(20) We only consider s-wave states: in this case the hyper ne interaction is given by the interaction of such magnetic eld with the electron spin Ze 2gN e 1 21 Hhyp = − S⋅B= S ⋅ −I ∇ + (I ⋅ ∇) ∇ 2 ( mc 2mMN c r r). Since s-wave states are spherically symmetric we obtain 1 2 8π 2 21 (3) −I ∇ + (I ⋅ ∇) ∇ = − I⟨ ∇ ⟩ = I ⟨δ (x)⟩ = I | Rn0(0) |2 ⟨ r r⟩ 3 r 3 3 21. Therefore using again we nd. Z Rn0(0) = 2 ( naB ). 3/2. 3 4 Ze 2gN Z 4 4 2 m 1 S⋅I ⟨Hhyp⟩ = S ⋅ I = gN (αZ ) mc 2 ( ) 3 mMN c naB 3 MN n 3 ℏ2. fi. fi. fi. We see that compared to the relativistic corrections this effect is suppressed by a factor m /MN.

(21) We now need to evaluate the expectation value of the spin product S ⋅ I We introduce the total spin F = S + I S⋅I 1 1 3 2 2 2 = 2 (F − S − I ) = F(F + 1) − − I(I + 1) 2 ( ) ℏ 2ℏ 2 4 =. 1 I 2 1 (−I 2. for F = I + − 1) for F = I −. 1 2 1 2. For the hydrogen atom I = 1/2 and, therefore, the s-wave states are either in a singlet state F = 0 (ground state) or in a triplet state F = 1 (excited state). 1 1 2I + 1 The splitting in the n-level for s-wave states is therefore I − (−I − 1) = (2 2 2 ) 2 4 m mc 2I + 1 4 ΔEn,1/2 = gN (αZ ) 3 MN n 3 2. For the lowest levels we have (gN = gp = 5.59) 1S1/2 : ν = 1420 MHz. λ = c/ν = 21 cm. 1 2S1/2 : ν = × 1420 MHz = 177 MHz 8.

(22) The λ = 21 cm line is very important in astronomy It is located in the microwave spectrum Electromagnetic energy in this range can easily pass through the Earth's atmosphere and be observed from the Earth with little interference. Assuming that the hydrogen atoms are uniformly distributed throughout the galaxy, each line of sight through the galaxy will reveal a hydrogen line The only difference between each of these lines is the Doppler shift that each of these lines has One can calculate the relative speed of each arm of our galaxy. .. It is then possible to use the plot of the rotation curve and the velocity to determine the distance to a certain point within the galaxy..

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