Probability of Transitions in a Hydrogen Atom under the Influence of an Electric Field, Assignments of Physics

Download Probability of Transitions in a Hydrogen Atom under the Influence of an Electric Field and more Assignments Physics in PDF only on Docsity! Physics 512 Winter 2003 Homework Set #10 – Solutions 1. This is based on Sakurai, Chapter 5, Problem 28. A hydrogen atom is initially in its ground state (1s). At time t = 0 we turn on a spatially uniform electric field as follows:
E(t) = { 0 t < 0 E0e−t/τ ẑ t ≥ 0 a) Using first-order time dependent perturbation theory, compute the probability for the atom to be found in each of the three 2p states at time t  τ . You need not evaluate the radial integrals, but perform all other integrations. This electric field corresponds to a potential V = −eE0ze−t/τ Thus we calculate Ak←s = − i h̄ ∫ t 0 eiωkst ′〈k(−e)E0ze−t′/τ |s〉 dt′ = ieE0 h̄ ∫ t 0 e(iωks−1/τ)t ′〈k|z|s〉 dt′ = ieE0 h̄ e(iωks−1/τ)t − 1 iωks − 1/τ 〈k|z|s〉 The transition probability is given by Pk←s = |Ak←s|2 = e 2E20 h̄2 1 − 2e−t/τ cosωkst+ e−2t/τ ω2ks + 1/τ2 |〈k|z|s〉|2 For t τ , the decaying exponentials may be dropped, and we find Pk←s(t τ) = e 2E20 h̄2 |〈k|z|s〉|2 ω2ks + 1/τ2 (1) which is independent of time t. We now note that this matrix element corresponds to a dipole transition, with corresponding selection rule ∆ = ±1. Furthermore, in terms of spherical tensors, z = r cos θ = √ 4π 3 rY 01 (θ, φ), which is an m = 0 component of a spherical vector. As a result, we furthermore have a selection rule δm = 0. So the only possible transition from the 1s state is to the m = 0 2p state, ie |100〉 → |210〉. In this case, the relevant hydrogen wavefunctions are ψ100 = ( 1 πa30 )1/2 e−r/a0 ψ210 = ( 1 2πa30 )1/2 r 4a0 e−r/2a0 cos θ 1 Thus 〈210|z|100〉 = 1√ 2πa30 ∫ ( r 4a0 e−r/2a0 cos θ ) r cos θ ( e−r/a0 ) r2 sin θ dr dθ dφ = 2π 4 √ 2πa40 ∫ ∞ 0 r4e−3r/2a0 dr ∫ π 0 cos2 θ sin θ dθ = 1 3 √ 2a40 ∫ ∞ 0 r4e−3r/2a0 dr Although the problem does not require it, this radial integral is not too difficult to evaluate. We find 〈210|z|100〉 = 128 √ 2 243 a0 On dimensional grounds, this is a reasonable result, since 〈z〉 has to give a length, and the Bohr radius a0 is what governs the size of the hydrogen atom. This may now be substituted into (1) to obtain the transition probability P|210〉←|100〉(t τ) = ( 8 9 )5 e2E20a20 h̄2 1 ω2ks + 1/τ2 where h̄ωks = E210 −E100 = −E0/22 +E0 = 34E0 (with E0 = 13.6 eV). b) What would happen if instead the atom was in the 2s state to begin with? Since we have the selection rules δ = ±1 and δm = 0, transitions to the 1s state would be forbidden. However it would still be possible to end up in the 2p |210〉 state. Note, however, that in this case, ωks ≈ 0, so the probability would behave as P (t τ) ≈ e 2E20 τ2 h̄2 |〈210|z|200〉|2 where a calculation indicates 〈210|z|200〉 = −3a0. Note, however, that this ex- pression may be large for large τ , and thus must be handled with care. This mixing between the 2s and 2p state is similar to that which happens for the linear Stark effect, where degenerate time independent perturbation theory was required. 2. Consider a two-photon transition from an initial state of angular momentum 1 = 0 to a lower energy intermediate state of angular momentum 2 = 1 to the ground state of the system with angular momentum 3 = 0. Both photons are emitted via electric dipole transitions. a) Using second order time-dependent perturbation theory, show that the amplitude for this transition is proportional to ̂1 · ̂2 where ̂1 and ̂2 are the polarization vectors of the two photons (ignore identical particle effects). The second order transition amplitude is given by A = ( − i h̄ )2 ∑ n ∫ t 0 ∫ t′ 0 eiωknt ′ eiωnst ′′〈k|V |n〉〈n|V |s〉 dt′′ dt′ 2 Calculate the rate of ionization of the atom as a function of ω. We use Fermi’s Golden Rule for harmonic perturbations w = 2π h̄ |〈k|V †|s〉|2ρ(Ek) ∣∣∣ Ek≈Es+h̄ω where the perturbation is given by V (t) = V eiωt + V †e−iωt The given electric field in the ẑ direction corresponds to a linear electric potential φ(t) = −E0z cos(ωt). Multiplying by the (negative) charge of the electron gives the potential energy V (t) = eE0z cos(ωt) = 12eE0z(eiωt + e−iωt) Therefore we need to calculate the matrix element of V where V = V † = 12eE0z The initial state of the atom is given by the 1s wavefunction ψ100(r) = ( 1 πa30 )1/2 e−r/a0 while for the final (ionized) electron state, we may take a plane wave ψk(
r) = 1 L3/2 ei k·r Note that, for normalization, we have placed the ionized electron into a box with peri- odic boundary conditions. Thus the matrix element is 〈k|V †|s〉 = 12eE0〈k|z|s〉 = eE0 2 √ πL3a30 ∫ e−ik·rze−r/a0 d3
r This integral is easier to perform when the plane wave state is decomposed in terms of spherical waves. Recall (eg from scattering theory) that ei k·r = ∑ l (2l + 1)iljl(kr)Pl(k̂ · r̂) = 4π ∑ l,m iljl(kr)Y ml (k̂) ∗Y ml (r̂) Hence 〈k|V †|s〉 = 2eE0π 1/2 (La0)3/2 ∑ l,m (−i)lY ml (k̂) ∫ jl(kr)Y ml (r̂) ∗(r cos θ)e−r/a0 r2drdΩ 5 Since z = r cos θ is the m = 0 component of a vector operator, and since we start from the 1s state, the Wigner-Eckart selection rules indicates that the final state can only have l = 1, m = 0. Thus 〈k|V †|s〉 = −2ieE0π 1/2 (La0)3/2 Y 01 (k̂) ∫ j1(kr)e−r/a0 r3dr ∫ Y 01 (r̂) ∗ cos θ dΩ = − 3ieE0 cos θk 2π1/2(La0)3/2 ∫ j1(kr)e−r/a0 r3dr ∫ cos2 θ dΩ = −2ieE0π 1/2 cos θk (La0)3/2 ∫ ∞ 0 j1(kr)e−r/a0 r3dr The integral may be performed using the explicit form of the spherical Bessel function ∫ ∞ 0 j1(kr)e−r/a0 r3dr = k−4 ∫ ∞ 0 j1(z)e−z/(ka0)z3dz = k−4 ∫ ∞ 0 ( sin z z2 − cos z z ) e−z/(ka0)z3dz = −k−4Re ∫ ∞ 0 (iz + z2)eize−z/(ka0) dz = −k−4Re ∫ ∞ 0 (iz + z2)e(i−1/(ka0))z dz = −k−4Re ( i (i− 1/(ka0))2 − 2 (i− 1/(ka0))3 ) = k−4Re 3 + i/(ka0) (i− 1/(ka0))3 = k −4 8(ka0) 5 ((ka0)2 + 1)3 As a result, we have 〈k|V †|s〉 = −16ieE0π 1/2 cos θk k4(La0)3/2 (ka0)5 ((ka0)2 + 1)3 Finally, this matrix element must be squared and combined with the density of states ρ(Ek) = m3/2E 1/2 k L 3 √ 2π2h̄3 dΩ 4π (for plane wave electrons in three dimensions). The result is w = 2π h̄ 256e2E20π cos2 θk k8L3a30 (ka0)10 ((ka0)2 + 1)6 m3/2E 1/2 k L 3 √ 2π2h̄3 dΩ 4π = 256me2E20 h̄3k4 (ka0)7 ((ka0)2 + 1)6 cos2 θk dΩ 4π 6 More properly, this is the differential rate dw for emission of an electron into a solid angle dΩ (at an angle θk from the ẑ axis). This cos2 θk behavior is standard for a dipole. Integrating over all angles then gives a total ionization rate of w = 256E20 3h̄k3 (ka0)6 ((ka0)2 + 1)6 where Ek = h̄2k2 2m = h̄ω − E0 (E0 = 13.6 eV) and we have also made use of the relation a0 = h̄2/me2. Note that typically ka0 1, in which case the ionization rate is proportional to k3 ∼ E3/2k w ≈ 256E 2 0a 6 0 3h̄ k3 4. Given two distinguishable spin-1 particles with vanishing orbital angular momenta, one can form states of total angular momentum j = 0, 1 and 2. Now suppose the two particles are identical. What restrictions do we get? What about two spin-2 particles? What is the general rule for allowed values of j? [See Sakurai, Chapter 6, Problem 2]. For spin-1, it is easy to see that the Clebsch-Gordan combinations are symmetric for j = 0 and 2, and antisymmetric for j = 1. Since spin-1 particles are bosons, and since the spatial wavefunction is symmetric (due to vanishing orbital angular momentum), only the symmetric spin combinations are allowed. Hence only states of j = 0 or 2 are allowed for the combination of two identical spin-1 particles. Similarly, only states of total j = 0, 2, or 4 are allowed for two identical spin-2 particles. In general, we note the following symmetry property of Clebsch-Gordan coefficients under interchange 〈j1j2m1m2|j1j2JM〉 = (−)J−j1−j2〈j2j1m2m1|j2j1JM〉 For identical particles, with j1 = j2 = s, this reads 〈ssm1m2|ssJM〉 = (−)J−2s〈ssm2m1|ssJM〉 Thus interchanging both particles (m1 ↔ m2) is symmetric for J − 2s even and antisymmetric for J − 2s odd. Hence for identical bosons (s integer), only even J combinations are allowed, while for identical fermions (s half-integer) antisymmetry also demands that only even J combinations are allowed. This results in a simple rule that combinations of any two identical particles with vanishing orbital angular momenta can only result in even J . Note that for integer spin this result is closely related to the symmetry properties of the spherical harmonics. 7