ECEN 5005 Crystals, Nanocrystals and Device Applications Class 20 Group Theory For Crystals Laporte Selection Rule Polarization Dependence Spin Selection Rule 1
Laporte Selection Rule We first apply this result to the case in which the electronic center possesses inversion symmetry (e.g. O h symmetry), i.e. H(r) = H(-r). - Since the kinetic energy part of the Hamiltonian is always symmetric under inversion, this means V(r) = V(-r). Then, inversion operation commutes with Hamiltonian and thus all wavefunctions are simultaneous eigenstates of both H and i. There are only two possible eigenvalues for i because i 2 = E. - Suppose ψ is an eigenstate of i, i.e. i ψ = λ ψ. 2 2 - Then, i ψ = λ ψ = ψ. However, λ must be real as eigenvalues of a Hermitian operator must be real. Therefore, the only possible values for λ are 1 and 1. - The states with an eigenvalue 1 remain unchanged upon inversion. - The states with an eigenvalue -1 change sign upon inversion. The way a wavefunction transforms upon inversion operation is called parity. - When the wavefunction remains unchanged upon inversion, then it is said to have even parity. - When the wavefunction changes sign upon inversion, then it is said to have odd parity. The electric dipole moment operator, µ = er i, has odd parity. i 2
Laporte Selection Rule Recall that we denote the parity of irreducible representations by subscripts u and g for odd and even parities, respectively. Since Γ u Γ u = Γ g and Γ u Γ g = Γ, the electric dipole matrix element is zero when the two states have the same parity. In other words, the electric dipole transition is allowed only between the states with opposite parity. - This is known as the Laporte selection rule or the parity selection rule. e The magnetic dipole moment operator, µ = ( li + 2si ), on the 2m other hand, has even parity. Therefore, magnetic dipole transitions are allowed only when the initial and final states have the same parity. It is illustrative to perform actual integration of some matrix elements. Let us consider a hydrogen-like atom as an example. Ignoring spin-orbit interaction, the wavefunctions are R () r ( θ,φ) u i nl Y lm as defined in Class 13. The electric dipole operator can be expressed in terms of spherical harmonics, Y lm with l = 1. ex = er 4π Y1, 1 Y 3 2 1,1, ey = er 4π Y1, 1 + Y 3 2 1,1, ez = er 4π Y 3 1,0 3
Laporte Selection Rule Then, the matrix element for an electric dipole transition will involve an integral of the type R nlylm ry1 m Rn l Yl m = Rnl r Rn l Ylm Y1 m Yl m The radial integral, R nl r R n l, depends on n and l and it does not automatically become zero for particular values of n and l. The angular integral, Y lm Y m Yl m 1, contains three spherical harmonics functions and is non-zero only for particular values of l and m. As discussed in Class 16, Y lm Y m Yl m 1 is non-zero only when - m m = m - 1 + l + l = even - l l 1 l + l From the second and third conditions, we find l l = l = ±1 This is the Laporte selection rule for hydrogen-like atom. As discussed before, Laporte selection rule forbids any electric dipole transition between states with the same parity. The parity of spherical harmonics functions is ( 1) l. That is, Y lm with even l have even parity and odd l have odd parity. Therefore, the requirement l = ± 1 is equivalent to the requirement for the two states to possess opposite parity. 4
Laporte Selection Rule We can reformulate the above results in the language of group theory. The electric dipole moment belongs to the D (l=1) irreducible representation of the full rotation group. - which is why we could express the electric dipole operator in terms of Y lm with l = 1. - Y lm form basis functions for irreducible representations D (l). Also, recall that the quantum number l is the irreducible representation index and m is the row index. That is, the state R nl Y lm belongs to the mth row of the irreducible representation D (l). The transition matrix element is then where (l) Γ = D, (l ) Γ = D, and ( Γ, γ) Γ γ Γ, γ X, (1) Γ = D. Then, the general selection rule we derived earlier in this class states that the matrix element is non-zero only when the product representation Γ Γ contains Γ. Without proving it, we ll give ( 1) ( l ) ( l + 1) ( l ) ( l 1) D D = D + D + D if l' 0 D (1) D ( l ) = D (1) if l' = 0 The above reduction scheme for D ( l) ( l ) D is why the total angular momentum j has to satisfy l l j l + l when we add two angular momenta l and l'. 5
Laporte Selection Rule Therefore, it is now apparent that we need l l = l = 0, ± 1 for an allowed electric dipole transition. In the special case of l' = 0, l must be 1. Thus, the general selection rule for hydrogen-like atom is l l = l = 0, ± 1 with l = l' = 0 forbidden. Adding the requirement of parity eliminates the case of l = 0. Thus we arrive at the Laporte selection rule for hydrogen-like atom: l l = l = ±1 Note that we derived the same selection rule without evaluating any matrix elements. Our arguments are based solely on the symmetry properties of the system described in the language of group theory. Now let us consider Mn 2+ ion in SrS (an octahedral crystal field). Mn 2+ ion has 5 electrons in the 3d shell and their energy levels may be found in the Tanabe-Sugano diagram. The ground state is 6 A 1g and the first excited state is 4 T 1g. Let us apply the selection rule to the transition between 4 T 1g and 6 A 1g. Among the irreducible representations of O h, the electric dipole operator belongs to T 1u. Then, the selection rule requires that the direct-product T 1u A 1g must contain 4 T 1g. 6
Laporte Selection Rule We immediately notice that this transition is forbidden by the Laporte selection rule which prohibits electric dipole transition between same parity states. Let us then consider Mn 2+ ion in ZnS (a tetrahedral crystal field), which lacks inversion symmetry. The energy levels are the same as the octahedral field case, except that we don t have subscripts u and g. Thus, the ground state is 6 A 1 and the first excited state is 4 T 1. Since the system lacks inversion symmetry, the states do not have definite parity and therefore the Laporte selection rule does not apply. So, we need to invoke the general selection rule. To do that, we proceed to calculate the direct product: T 1 A 1 = T 1 The reduction of the direct product does not contain T 2. Thus, the transition is still forbidden. The 4 T 1-6 A 1 transition of Mn 2+ is forbidden by spin selection rule as well, as we will see later. However, in real crystals there exist many perturbations that relax the selection rule and make the transition weakly allowed. Examples of such perturbations are lattice vibration, defects, strain, and spin-orbit interaction. - Thus, the Mn 2+ ion produces green luminescence in SrS and orange luminescence in ZnS with long lifetimes (~ ms). 7
Polarization Dependent Emission & Absorption If the symmetry is lower than cubic, each component of the electric (or magnetic) dipole operator transforms according to a different irreducible representation. Consider, as an example, F 3 center (3 vacancies coupled together) which possesses C 3v symmetry. The character table for C 3v is (isomorphic to D 3 ): (Imbusch, pp. 318) C 3v E 2C 3 3σ v x 2 + y 2, z 2 Z A 1 1 1 1 R z A 2 1 1-1 2 2 ( x y, xy) ( x, y) ( R, ) ( xz, yz) x R y E 2-1 0 Recall that the electric dipole operator is ( xeˆ + yeˆ ze ) µ = e r = e + ˆ x y z (Tinkham, Appendix B.) 8
Polarization Dependent Emission & Absorption From the character table of C 3v, we can see that - µ x and µ y belong to the irreducible representation E, - and µ z belongs to A 1. Notice that the direction of electric dipole operator (or its components) determines the direction of E-field in the resulting electromagnetic field polarization. Consequently, we obtain different selection rules for polarizations in the xy-plane and along z-axis. (Imbusch, pp. 175) The linear polarization along the primary axis (z-axis in the present example) is called the π-polarization. - The transversality requires k-vector (light propagation direction) be in the plane perpendicular to the primary axis. 9
Polarization Dependent Emission & Absorption The linear polarization in the plane perpendicular to the primary axis is called the σ-polarization if the k-vector is in the same plane. The linear polarization in the plane perpendicular to the primary axis is called the α-polarization if the k-vector is along the primary axis. The plane polarized light (π- or α-polarization) may also form circular polarization in which the E-field vector rotates in the plane perpendicular to the primary axis. Now consider the energy levels of F 3 center which is a system of three electrons, one from each anion vacancy. The single electron crystal field theory predicts that the ground state is a 1 and the first excited state is e. - To indicate single-electron state, we used lowercase letters. The strong field configuration for the ground state is therefore (a 1 ) 2 e. Taking the direct-product to find the ground state gives A A E = E 1 1 - The ground state of F 3 center is E. The lowest excited state configuration is obtained by promoting one electron in the a 1 single-electron state to e, yielding a new strong field configuration, a 1 e 2. 10
Polarization Dependent Emission & Absorption To find the energy levels arising from this string field configuration, we once again take the direct-product: A E E = A + A + E 1 1 2 - There exist three distinct excited levels belonging to A 1, A 2 and E. We can now determine the allowed polarization for the absorption transitions to the various excited levels. For absorption to the A 1 level, the transition matrix elements are E E A 1 for σ-polarization and E A 1 A1 for π-polarization. We need to calculate the appropriate direct-products in order to invoke the Wigner-Ekart theorem. E A 1 = E and A 1 A1 = A1 Since E contains E, the matrix element E E A1 is non-zero. But A 1 A doe not contain E and so the matrix element E A 1 A1 is zero. 1 A 1 - The absorption (or emission) transition between the ground level E and the excited level A 1 is allowed only for σ-polarization. Similarly, we find the allowed transitions as follows. Excited levels A 1 A 2 E E (ground level) ED(σ,π) ED(σ) ED(σ) 11
Spin Selection Rule Another important selection rule concerns the spin quantum numbers of the initial and final states. As discussed previously, the electric dipole operator is a purely orbital operator that can be expressed with Y lm (l=1) and thus belongs to l=1 irreducible representation of the full rotation group. Consequently, the electric dipole operator does not affect the spin states. Due to the orthogonality between spin states, the electric dipole matrix element becomes zero unless the initial and final states have the same spin quantum numbers, i.e. S = 0. e The magnetic dipole operator, µ = ( li + 2si ), explicitly involves 2m spin. This operator mixes states with different M S but the matrix elements between states with different S are all zero. Thus, the same spin selection rule, S = 0, applies to the MD process. The same is true for electric quadrupole operator. Spin selection rule: Electronic transition is forbidden between two electronic states with different spin quantum number, S. In other words, transitions are allowed between only the states with the same spin, i.e. S = 0. i 12
Spin Selection Rule Example: V 3+ ion in KCl (octahedral crystal field), - - 1 T 2 3 T 1 : spin-forbidden transition 3 T 2 3 T 1 : spin-allowed transition It should be noted that spin-selection rule holds only when spin-orbit interaction is ignored. - Because spin-orbit interaction mix the states with different spin quantum numbers according to their total angular momentum quantum numbers J. In real crystals, spin-orbit interaction does induce transitions between two states with different spin quantum numbers. - This is a second order process and therefore the transition strength is usually one or two orders of magnitude smaller than allowed ED transitions. - However, it can still be much larger than the strengths of allowed MD or EQ transitions and thus become the dominant transition mechanism. - Example: 4 T 1 6 A 1 transition of Mn 2+ in SrS (octahedral field) 13