Quantum Mechanics II Lecture 11 ( David Ritchie

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1 Quantum Mechanics II Lecture ( David Ritchie Michaelmas.

2 So far we have found solutions to Section 4:Transitions Ĥ ψ Eψ Solutions stationary states time dependence with time independent. Apply a time-dependent perturbation to a stationary state, system no longer in an eigenstate so transitions liely. Investigate two problems: Atomic transitions time varying electric field (photon) can cause transitions, enhanced if photon energy matches transition energy. Scattering an electron, wavefunction encountering a scattering potential e.g. an atom, can be scattered into a different -state. = Ĥ exp( i r) ψ () t = ψ () e iet/! 4. Two state system 4. Time dependent perturbation theory 4. Radiative transitions in atoms and molecules 4.4 Selection rules 4.5 Scattering Michaelmas.

3 4.4 Selection rules From the previous Lecture, the Einstein coefficients are: Bj π d ω d = = ε A j j, A! πεc! Neither spontaneous ( ) or stimulated ( ) transitions will occur between and ψ unless a component of ˆ is non-zero ψ j Often possible to show that matrix elements are zero for certain pairs of states (in electric dipole approx.), summarise results in selection rules. Dipole operator changes sign under parity operation hence if ψ and have the same parity. Selection rule: Parity of wavefunction changes in electric dipole approx. Wavefunctions separate into spatial and spin components: Dipole operator acts only on matrix element becomes: Michaelmas. B ψ d ψ dˆ = qr r r ψ ˆdψ = j ψ j φ (r) ψ ˆ dψ j = χ χ j φ () r qrφj() r dτ j ψ = φ (r) χ

4 From the last slide: χ Spin term vanishes unless and are the same. Hence selection rule: Selection rules () ψ ˆ dψ j = χ χ j φ () r qrφj() r dτ χ χ χ j j s = ; m s = So spin state not altered in electric dipole transition. Consider a single electron atom with electric dipole moment operator: dˆ = qr = qr(sinθ cosφ, sinθsinφ, cos θ) Electron wavefunction will have the form: To evaluate the matrix element, tae z-component of the integral is: π im im φ jφ unless m = m e e dφ = j For the x-component of expanding in exponentials, part of integral: Same result for y. So - to get non-zero component Michaelmas.4 R() ry ( ) () () im m θ, φ = RrF m θ e φ ˆd φ ˆd cosφ φ π imφ iφ iφ imjφ e ( e + e ) e dφ = unless mj m =± m = or±

5 From the last slide we have selection rules: These different changes in are related to polarization of EM field For in z-direction, selection rule applies. For in x or y-direction, selection rule applies. Consider selection rule for orbital angular momentum quantum no. Parity operator changes Y Selection rules () m = or± m ε m = ε m ± r r so:, ( π θφ, + π) = ( ) Y ( θφ, ) and: hence parity eigenvalue m m Since the parity of the wavefunction must change in electric dipole transition and we must have: m = or± =± This is summarized by saying that the photon carries off (or brings in for absorption) one unit of angular momentum. It is also possible for the EM field to interact with electric quadrupole moment or magnetic dipole moment, however electric dipole transitions are much more liely. Michaelmas.5 θ π θ φ φ+ π P = ( )

6 Selection rules (4) When orbital and spin angular momenta are combined to mae total angular momentum, selection rules for imply rules for J = m = or± m s Since and then: s, For given and Thus does not imply a change in and we can have as well as For example: and can both mae But if initially then and the final state: So if we can only have For a single electron, while is an integer so we cannot have but in atoms with an even no. of electrons resultant angular momentum can be zero. To summarize:, sm,, ms = or± Michaelmas.6 m j j can tae any of the values: j = ( + s),( + s ),, s, j jm =±, s = j j = j =± j = =, s = =, s = j = j = j = = s j = s = ± s = s = j = j =,± ; not

7 From the last slide: Selection rules (5) j =,± ; not Could arrive at this rule by considering the photon as carrying one unit of angular momentum and using the rules for combining angular momentum. It applies even in cases where initial and final states are not separate eigenstates of or. ˆL Another example a charged particle in a D harmonic potential. Wavefunctions, quantum number, energies Oscillating electric field in x-direction induces transitions with matrix elements: since: Matrix element is: Ŝ ( n + )! ω n n n x n! ( ) Where ˆ ˆ n n xˆ = aˆ+ aˆ aa, are ladder operators mω! n x n = n+ n n+ + n n n mω n= n ± n =± ( ) Which vanishes unless so in electric dipole approx. Michaelmas.7

8 A different and important transition occurs in scattering processes Initial state a free particle approaching origin from large distance, momentum Particle interacts with potential Final state particle moving away with deflected momentum Use Fermi s Golden Rule for the transition rate : Energy of the particle elastic collision Density of final states Initial and final wavefunction: 4.5 Scattering V () r Assume the wavefunctions are normalised to one particle in a box of volume so: a Michaelmas.8 p =! i π Γ i f = ψ V r ψ g E! ge ( ) ψ = A = / a () ( ) = p / = p / E m m ir ψ () r = Ae, ψ () r = j Ae i r = θ p =!

9 Scattering () The matrix element in Fermi s Golden Rule: i i ψ V() ψ = r r r e V() r e dr a i = Kr V(r) e d r where K = a!k is the momentum transfer in the collision, and, K K = sin θ θ Where θ is the particle scattering angle. So the matrix element is the D Fourier transform of the potential. Analogous to Fraunhofer diffraction:- The integral adds up scattered waves from each point in the potential, taing into account relative phases. V () r The weights of components are proportional to. θ K = = Michaelmas.9

10 Scattering () Density of states calculated as follows: For a particle, mass in a cubical box of side. Wavevector has possible values: Each state occupies a volume of -space. Number of states in solid angle and between and Density of states in : To convert to density of states in use Hence m π n π n x y π nz =,,, nx, ny, nz =,,,... a a a ( π/a) dω + d ddω is: ( π/ a) dn a g ( ) = = d Ω d π E ( ) dn ( ) d a md Ω ge = = g = de de 8π! Michaelmas. E a! de! = = m d m

11 Scattering (4) In scattering measure no. of particles scattered per second per unit flux of incident particles. Incident flux measured in particles per unit area per second. Measured ratio has dimensions of area called scattering crossection ( ) Scattering crossection can be interpreted as effective area presented by scattering centre to incoming beam. To loo at no. of particles scattered as a function of angle we use the differential crossection:- The incident flux, taing is : Where dσ Number of particles scattered / unit time into dω dω Incident flux dω / iz ψ = a e v=! / m i! v ψ ψ ψ ψ m =! ma = a is the incident particle s velocity. Michaelmas. σ

12 So we have and since: we have: Scattering (5) dσ Number of particles scattered / unit time into dω Γi f = dω Incident flux dω / ma dω (! ) π ikr amd π Ω Γ i f = ψ V() r ψ g( E) = V(r) e dr!! a 8π! ikr amd V(r) e d r π Ω dσ! a 8π! m i = Kr V(r) e d r d ma d π! (! / ) Ω Ω This is the Born approximation. Based on first order perturbation theory - a good approximation when the scattering potential is a small correction to the Hamiltonian. The Born approximation becomes better as energy increases, there will be more about this in TP and Nuclear Physics next term. Michaelmas.

13 Born Approximation a screened potential Applying this to a screened Coulomb potential: A decent approximation for an atom, as we get pure a Coulomb potential. The Born Approximation: Now: V ( r) Ze = e 4πε r Michaelmas. λr λ dσ m i = V(r) e d r Kr dω π! λr ikr Ze e i V(r) e d r = Kr e r sinθdrdθdφ 4πε r K = θ π π λr ikrcosθ Ze λr ikru Ze = re dr sinθe dθ dφ = re dr e du 4πε ε ikru λr λr ikr ikr Ze e Ze = re dr e e e dr ε ikr = iε K u = cosθ

14 Born Approximation a screened potential () From the last slide: ikr Ze λr ikr ikr V(r) e d r = e e e dr iε K And since: V(r) e ( λ) ( λ) ik r ik + r Ze ( ik λ) r ( ik + λ) r Ze e e = e e dr = iε K iε K + ik λ ik + λ Ze Ze ik = iε K + = ik λ ik + λ iε K K + λ K = sin θ ikr d r Ze = ε K + λ Ze = ε 4 sin θ / + λ ( ) K = θ Michaelmas.4

15 Born Approximation a screened potential () From the last slide: Hence: V(r) e ikr d r Ze = ε 4 sin θ / + λ ( ) ikr V(r) e d r ( ) dσ m m Ze = = dω π! π! ε 4 sin θ / + λ Zme = 8πε! sin ( θ /) + λ dσ Zme Zme = = d 8πε! sin θ / 8πε p sin θ / ( ) ( ) Ω This is identical to the Rutherford scattering formula as derived from classical orbits. Here we have derived it from Quantum Mechanics using the Born Approximation. Michaelmas.5 And if λ - unscreened Coulomb where p =!

16 Lecture The End!! ( Michaelmas.6

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