Symmetry and degeneracy Let m= degeneracy (=number of basis functions) of irrep i: From ( irrep) 1 one can obtain all the m irrep j by acting with off-diagonal R and orthogonalization. For instance in the triangle Group the degenerate irrep E has basis functions that can be taken to transform like (x,y), The x function is taken by rotations and reflections to a linear combination of x and y. This implies that the energy eigenvalues whose eigenfunctions belong to an irrep must be degenerate m times: Can they be even more degenerate?
Accidental degeneracy It is possible that by accident two states unrelated by symmetry come so close in energy to appear degenerate in low resolution experiment; however a mathematically exact degeneracy with no symmetry reason is miraculous. Simply, one was unaware of using a Subgroup of the actual Group, because some symmetry had still to be discovered
Famous example: hydrogen atom H p m k r The angular momentum L is conserved since: 1 r k r F ( ) k central force. 3 3 r r r r dl d dr dp ( r p) p r r F 0. dt dt dt dt Why are the Hydrogen eigenvalues independent of L? One cannot produce p orbitals by rotating s! Extra degeneracy on L explained by conservation of Laplace- Runge-Lenz vector. In classical physics, the conserved vector is: R p L r k m r 3
Pierre-Simon Laplace, marchese di Laplace (Beaumont-en-Auge, 3 marzo 1749 Parigi, 5 marzo 187), Carl David Tolmé Runge (German: [ˈʀʊŋə]; 1856 197) Wilhelm Lenz (February 8, 1888 in Frankfurt am Main April 30, 1957 in Hamburg) perihelion L p R p L r k m r During the motion, p L is in the xy plane. At perihelion, p r p L is parallel to r. Since it is conserved, R is pinned at the aphelion-perihelion direction, and this is why the orbits are closed. Orbits are closed in the 3d harmonic oscillator, in which case, separate energy conservation along the 3 axes is the extra symmetry.
p L r R k m r r Force F k L (0, 0, L) 3 r i j k x y Ly Lx 0 i j 3 3 3 3 r r r r 0 0 L 1 r r r 3 constant of motion: proof d p L F L ( Fx, Fy,0) (0,0, L) dt m m m d p L ( Fy, Fx,0) L ( y, x,0) L. 3 dt m m mr d 1 1 The derivative of the second term: start by v. dt r r d 1 1 v. r d r v v. r r v- r (v. r ) v. r. 3 3 3 dt r r r dt r r r r This can be simplified: ( x y )vx-x( xvx +yv y) (yv 3 3 x-xv y) 3 d x y Ly dt r r r mr ( x y )vx-y(xv x+ yvy ) (-yv 3 3 x+xv y) 3 d y x Lx dt r r r mr d r L( y, x,0) 3 dt r mr dr dt 0 5
W. Pauli, On the hydrogen spectrum from the standpoint of the new quantum mechanics, Z. Physik 36, 336-363 (196). Wolfgang Pauli in 196 first solved the SE for the H atom using the SO(4) symmetry. SO(4) is a Lie Group (see below). Therefore we must say more about Lie Groups. Some are already familiar from the theory of angular momentum in Quantum Mechanics. 6
Short Outline of the Pauli treatment of the H spectrum of bound states p L r p L L p r The classical R k is replaced by M k. m r m r H such that M =( ) (L + ) +k, where H=Hamiltonian m H [ M i, M j ] ( i ) ijk Lk [L i, Lj ] i ijk Lk m [L i, M j ] i ijkm k H To build the generators since [ M, M ] ( i ) L m LM. 0 M.L=0 M H =( ) (L + ) +k m m for E<0 introduce M ' i M i. Then E [ M ', M ' ] i L i j ijk k [L, M ' ] i M ' i j ijk k L M ' L M ' The generators of SO(4) are: I,K. [I,I ]=i I i j ijk k [K,K ]=i K with [I, K ] 0. i j ijk k i j i j ijk k
Extension to SO4 rotations in 4d, relevant to H atom : 6 planes ij i 1,4, j 1,4. One finds: 4 6 generators A1 A A3 B1 B B3 in 3d L, L, L A, A i A extends those in 3 d : L L i L i j _ ijk k B, B i B i j _ ijk k Ai, B j 0 _ x y z The extra symmetry is the reason why there is an extra conserved quantity 8
Quantum Runge-Lenz vector : p L L p r R k and one finds after long algebra that [ R, H ] 0 m r Therefore, H E HR ER LM LM LM LM [ R, L ] 0 R does not belong to L LM E does not depend on L Dirac s equation relativistic Runge-Lenz vector e [, ]_ 0 r Biederharn-Johnson-Lippman peudoscalar operator H cp mc Z H B i e e. r 5 B K ( c p Z ) Z, K [. L ] mc r c r E does not depend on sign of K E( s ) E( p ) 1 1 9
My own simple example matrice delle adiacenze h 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 Autovalori e autofunzioni di h 1 1 1 1 (,,, ) 0 1 1 0,1 (,0,0, ) 1 1 0, (0,,,0) 1 1 1 1 (,,, ) 10
Now we deform the square, e.g. by doubling the hopping integrals connecting site 3 to 1 and 4. Now, the Hamiltonian reads h 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 10 (,,, ) 10 10 5 1 1 0,1 (,0,0, ) 0 1 0, (0,,,0) 5 5 1 1 1 10 (,,, ) 10 10 5 Accidental degeneracy! 11
s 0 0 0 1 0 1 0 0 s s s 0 0 1 0 1 0 0 0 0,1 0,1 0, 0, 1 1 1 10 (,,, ) 10 10 5 1 1 0,1 (,0,0, ) 0 1 0, (0,,,0) 5 5 1 1 1 10 (,,, ) 10 10 5 s is a symmetry. To explain this degeneracy we want a non-abelian Group, however. Generalized rotation with S T =S -1 0 0 1 0 0 1 3 0 1 0 0 0 1 3 0 0 1 T S S S hs h, rotation 0 0 0 1 1 0 0 3 10 0 1 0 0 0 3 1 0 Sh hs 1
S does not produce a mere permutation of sites; however: Generalized rotation with S T =S -1 1 1 1 10 (,,, ) 10 10 5 1 1 0,1 (,0,0, ) 0 1 0, (0,,,0) 5 5 1 1 1 10 (,,, ) 10 10 5 S 4 = 1 S mixes ψ 0,1 and ψ 0, does not commute with σ and explains the degeneracy. The deformed problem has a lower geometrical symmetry, but actually is still C 4v symmetric because of a hidden dynamical symmetry. 13
What makes different representations really distinct? D Matrices are not unique, e.g. one can rotate frame a c b By any unitary transformation (not necessarily a symmetry) we can change basis and matrices within the E irrep. Let U be any rotation: then for all R,D(R) UD(R)U 1 is consistent with the multiplication table and does not modify the Dirac characters, which are eigenvalues. Up to this unitary transformation, the Irreps with the same characters are to be identified. 14
Example: we are familiar with the representing matrices of E irrep E : c a Diagonal b 1 0 c s c s D( E) D( C ) D( C ) 3 3 0 1 s c s c 1 0 c s c s D( s ) D( s ) D( s ) a c b 0 1 s c s c 1 3 c cos( ), s sin( ) 3 3 UU Find a new representation D in which D (C 3) is diagonal. 1 i atomic orbitals: U= is unitary : 1 -i 1 i px px 1 px ipy U = 1 1 1 i py 1 -i py px ipy 1 0 L -i i 1 -i 0 1 z eigenstates 15 Det U = -i U is no rotation
The rotations are diagonalized: 1 i 1 1 0 e i 3 c s e 0 3 3 1 -i s c -i i i 3 D (C ) =UD (C ) U But then 0 1 D( sa) =UD ( sa) U is off diagonal 1 0 Unitary transformations U leave traces invariant since: Tr(U DU)=Tr(UU D)=Tr(D) 16
Joning bases of irreps large block-diagonal D matrices can be obtained D( R )= A block-diagonal matrix can be transformed to a non- block-diagonal form by a generic unitary U U U + = Reducible representation But one should try to do the reverse, until the representation is irreducible! 17
Reduction of a representation: one should use the smallest possible D matrices,without mixing different irreps. The set of irrep matrices D cannot have the same block structure D( R )= because the blocks would make up irreps. In particular the irrep matrices cannot be all diagonal
Matrices of Dirac Characters within an irrep irrep E : 1 0 c s c s D( E) D( C ) D( C ) 3 3 0 1 s c s c 1 0 c s c s D( s ) D( s ) D( s ) a c b 0 1 s c s c 1 3 c cos( ), s sin( ) 3 3 E C C 3 3 3 3 3 E C3 C C E C s s s a b c s s s b c a s s s c a b s s s a b c s s s c a b s s s b c a 3 3 3 3 3 C3 E E C C C E C C How can the Dirac characters commute with every D? c s c s 1 0 D( C ) D( C ) R 3 3 s c s c 0 1 1 0 c s c s 0 0 D( s ) D( s ) D( s ) s a c b 0 1 s c s c 0 0 19
1 0 0 0 R s 0 1 0 0 How do constant matrices arise? Diagonalizing H within each subspace one finds eigenstates; Those eigenstates carry the Irrep label. Consider a pair of eigenfunctions (ψ x, ψ y ) transforming like (x,y); they belong to the Irrep E of C3v. H x = x x, H y = y y. constant matrices! S G : S =a +ß, but since [S,H] = 0, HS =SH = S x x x y - x x x x S must belong to the same energy eigenvalue as. x 0 = = H x, a constant matrix. y 0 Why are the characters also constant? x 0 0
Indeed, quite the same reasoning can be repeated for the Ω matrices. S G : S =a +ß, but since [S, ] = 0, S =S = S S must belong to the same character eigenvalue as. x 0. 0 x x x y - x x x x To sum up, the matrices of H and those of Dirac characters in the subspace spanned by an irrep have the remarkable property of being constants, that is, identity matrices I m m multiplied by numbers: H = I m m Ω C = ω C I m m 1
Schur Lemma. Any m m matrix M commuting with all the representative matrices D(R i ), i = 1, N G of an m-dimensional Irrep must be proportional to the Identity matrix I m m. What characterizes the irrep is that the matrices cannot be taken simultaneously to the same block form, because otherwise the representation would be reduced. We first prove the lemma for a diagonal matrix, then generalize to any matrix of any size.
Proof for diagonal matrices x matrix d1 0 a11 a1 0 a1( d d1) M A [A, M ] 0 d a a a ( d d ) 0 1 1 1 Condition for [A, ] 0 a ( d d ) 0 0 1 1 M Airrep a1( d d1) 0 0 0 If d were different from d 1, this would require a diagonal A. Since one cannot diagonalize simultaneously all the D(Ri) i = 1, N G ), we conclude that d = d 1. 1 0 M d 0 1 3
3x3 matrix d1 0 0 a11 a1 a13 M 0 d 0 A a a a 1 3 0 0 d 3 a31 a3 a 33 0 a1 ( d d1) a13 ( d3 d1) [ A, M ] a1( d1 d) 0 a3( d3 d) a31( d1 d3) a3 ( d3 d) 0 If all the d i are different, this vanishes only for diagonal A. If d 1 is different from the other diagonal elements, a11 0 0 A 0 a a 0 a a 3 3 33 which cannot be true for all the representative matrices in an irrep, because they cannot have a common block form. This is generalized formally to matrices of any size with the conclusion that M = λi m m for some λ. 4
nxn diagonal matrix: similar proof Extension to Hermitean matrices Assuming M hermitean we know that by a unitary transformation U it can be diagonalized. U sends M to U M U = diagonal and the set A to the set U A U which still represent the same irrep and commute with U M U. Then U M U = diagonal= λi m m and actually this implies that M is such in any basis since U I m m U =I m m. 5
Extension to any matrix Assumption : A basis of irrep, [ M, A] 0 Thesis : M I Proof : First we show that if MA AM 0 then M also commute AM M A. mxm Clearly MA AM A M M A, 0. 1 1 Since A A, MA AM 0 A implies [ M, A ] 0, A. 1 1 But since A ( R) A( R ) basis of irrep, [ M,A] 0, A. Out of these we can make two further commuting matrices: H M M 1 ( ) H i M M These are Hermitean, hence they are constants, and s, 1 ( ) M H1 ih I mxm 6
Traces and Characters of an Irrep Consider a basis of a m-dimensional irrep i and the representative matrices For class C, define the character ( C) TrD( R) D ( R), R C; si nce S X is unitary ( ) depends SX C only on C ( C) n n C 1 is R independent ( C) Dirac's character R I also depends on C; Tr m,m= size of matrices. R C mxm m n C RC C C n ( C) m C mc C ( C) ( C) character mean R eigenvalue n n m C C C Identity: (E)=m We shall prove that irrep dim are obtained from irrep characters by: m C NG C n C 7
Character Tables of common point groups Below we shall prove: Character tables are square 8
9
It is important to realize that Dirac s characters =constant matrices by a mere algebraic fact. Therefore I shall present an algebraic proof, which is very enlightening. 30
Great Orthogonality Theorem (GOT) N G RG D R D R N ( i) * ( j) G ( ) ( ) ij mi Each element of each D(R) in each irrep is a symmetry type and this leads to orthogonality when integrating over the Group. We know we must diagonalize H on a symmetry-adapted basis, labelled by Dirac s characters. The practical method to do this results from this abstract theorem! 31
Remark number 1 Recall irrep E of C : 1 0 c s c s D( E) D( C ) D( C ) 3 3 0 1 s c s c 1 0 c s c s D( s ) D( s ) D( s ) a c b 0 1 s c s c 3v 1 3 c cos( ), s sin( ) 3 3 Note that C 3 is the inverse of C 3 and its matrix is D(C 3 ) T. More generally, the operators are unitary and D ( R) D ( R ) ( i) * ( i) 1 3
Remark number is following theorem: Let O total-symmetric operator : O, R 0, R G j and component of irrep j i j then O O( i) ij Orthogonality on irreps i,j : different good quantum numbers (Dirac characters). Orthogonality on components and : For i = j the matrix {O μν } represents an operator that commutes with everything, so it must commute with all the D matrices; so Schur s lemma applies and diagonal elements are equal. Note dependence on i: the diagonal element depends on the irrep. For example, a spherically symmetric potential V(r) has vanishing matrix elements between states of different L and within a given L it has vanishing matrix elements between states of different M L ; the diagonal matrix elements are independent of M L and depend on L. 33 Another example: matrix of the square of the angular momentum on the basis of spherical harmonics.
Remark number 3 on symmetrization Theorem: 1 arbitrary operator O = R R invariant, T G, OT TO : RG Proof: 1 1 1 OT TT OT TT R RT RG 1 1 T ( RT ) RT T ( RT ) RT RG RTG ( from rearrangement theorem) 1 T R R TO RG So indeed O commutes with every T 34
Remark number 4: example on the working of GOT Let us form 6-component vectors with the elements of the D matrices. Recall irrep E of C : 3v 1 3 1 3 1 0 D( E) D( C ) D( C ) 3 3 0 1 3 1 3 1 1 3 1 3 1 0 D( s ) D( s ) D( s ) a b c 0 1 3 1 3 1 35
Remark number 4: example of GOT Recall irrep E of C : 3v 1 3 1 3 1 0 D( E) D( C ) D( C ) 3 3 0 1 3 1 3 1 1 3 1 3 1 0 s s s D( ) D( ) D( ) a b c 0 1 3 1 3 1 E 1 1 1 1 E E 6 NG v (1,,, 1,, ), v. v 3 11 11 11 m 36
Remark number 4: example of GOT irrep E of C : 3v 1 3 1 3 1 0 D( E) 0 1 D( C ) D( C ) 3 3 1 3 3 1 1 3 1 3 1 0 D( s ) D( s ) D( s ) a b c 0 1 3 1 3 1 1 1 1 1 N v (1,,, 1,, ), v. v 3 m E E E G 11 11 11 E 3 3 3 3 E E E E v1 (0,,,0,, ), v1. v1 3, v1. v11 0 37
Remark number 4: example of GOT irrep E of C : 3v 1 3 1 3 1 0 D( E) D( C ) D( C ) 3 3 0 1 3 1 3 1 1 3 1 3 1 0 s s s D( ) D( ) D( ) a b c 0 1 3 1 3 1 E 1 1 1 1 E E NG v11 (1,,, 1,, ), v11. v11 3 m E 3 3 3 3 E E E E v1 (0,,,0,, ), v1. v1 3, v1. v11 0 E 3 3 3 3 E E E E E E v1 (0,,,0,, ), v1. v1 3, v1. v11 0, v1. v1 0 38
Remark number 4: example of GOT irrep E of C : 3v 1 3 1 3 1 0 D( E) D( C ) D( C ) 3 3 0 1 3 1 3 1 1 3 1 3 1 0 s s s D( ) D( ) D( ) a b c 0 1 3 1 3 1 1 1 1 1 N v (1,,, 1,, ), v. v 3 m E E E G 11 11 11 E 3 3 3 3 E E E E v1 (0,,,0,, ), v1. v1 3, v1. v11 0 E 3 3 3 3 E E E E E E v1 (0,,,0,, ), v1. v1 3, v1. v11 0, v1. v1 0 E 1 1 1 1 E E E E E E E E v (1,,,1,, ), v. v 3, v. v11 0, v. v1 0 v. v1 0 39
Example of GOT: including the other irreps element irrep list squares 1 1 1 1 11 E (1,,. 1,, ) 3 3 3 3 3 1 E (0,,,0,, ) 3 1 E (0, 3 3 3 3,,0,, ) 3 E 1 1 1 1 (1,,.1,, ) 3 11 A (1,1,1,1,1,1) 6 1 11 A (1,1,1, 1, 1, 1) 6 Each element of each D(R) in each irrep leads to orthogonality when integrating over the Group. 40
Proof of GOT : RG N i * j G ij mi D R D R 1 Let some operator, to be specified later; build the invariant O = R R. j Then let basis function of irrep j component : O O( i) with Oi ( ) independent of i j ij Work out j j ( j) inserting R D ( R) m j RG i R 1 R j m j RG and R D ( R) i i ( i) * s s s m j RG s, D ( R) D ( R) O( i) i j ( i) * ( j) s s ij ( i) * ( j) i j We want Ds ( R) D ( R), must get rid of s 41
m j RG s, D ( R) D ( R) O( i) i j ( i) * ( j) s s ij ( i) * ( j) i j We want Ds ( R) D ( R), must get rid of s We are still free to choose as we please, so pick the irrep&component switch operator i j i i j j s s s i j m j ( i) * ( j) We get sds ( R) D ( R) O( i) ij RG, s ( i) * ( j) that is, D ( R) D ( R) O( i) ij. RG Oi ( ) is independent of, and we got the orthogonality on,. We need to show yet: Oi ( ) is independent of, We got the orthogonality on, as well, we may insert. 4
RG Orthogonality on first indices, proved by recalling D ( R) D ( R ) ( i) * ( i) 1 i * 1 j 1 D R D R O i ij, but RG RG R G This exchanges first and second indices, Clearly, no dependence of O i on is allowed. (they are second indices, now) ( i) * ( j) D ( R) D ( R) O( i) ij and also 1 RG i * j D R D R O() i ij Still, we must find O(i). 43
RG i * j D R D R O() i ij m RG To determine O(i) we set i = j, μ = ν and α = β. We get i * i () i 1 i RG D R D R O i D R D R Summing over α from 1 to m i, mi i 1 i i 1 RG i i O( i) m O( i) D R D R D R R i 1 i 1 i 1 X identity 1 order of G RG D R R m m D R R D R R N i i G RG NG mio( i) 1 NG O( i) m RG i RG N i * j G ij mi D R D R CVD 44
Many Groups of fundamental importance in Physics are Lie Groups. They are continuous (elements can be labeled by parameters) and continuously connected ( for every pair of elements a continuous path in parameter space can be found that joins them). Moreover, the parameters of the products are C 1 functions of those of the factors. Marius Sophus Lie 184-1899 A compact Lie Group has all parameters that vary over a closed interval; the Lorentz Group and the Group of all translations are noncompact Lie Groups, while the rotations are a compact Lie Group. 45
Example: Transformation of functions: ( x, y) ( x', y ') under the a b x e Group of d linear coordinate transformations : ( x, y) ( x ', y ') c d y f x ' ax by c, y ' dx ey f a b In order to be invertible : det 0 c d Typical element of Group: g(a,b,c,d,e,f) Identity element of Group: g(a=1,b=0,c=0,d=1,e=0,f=0) effect of Group element g: ( x, y) ( g( x, y)) a b x e g : ( x, y) ( x ', y ') ( ax by e, cx dy f ) c d y f
Typical element of Group g: ( x, y) ( g( x, y)) with g( a, b, c, d, e, f) One cannot study element by element, but one can study the local structure of the Group around the identity (then we can go to any other spot with the application of a Group element). Introduce generators that do infinitesimal transformations: a b x e g : ( x, y) ( x ', y ') ( ax by e, cx dy f ) c d y f Expand around identity : a=d=1, b=c=e=f=0 for example: g(1,0,0,1,0,0) g(1,0,0,1,0,0) generator linked to a X a lim 0 1 0 x ( x, y) ((1 ) x, y) ( x, y) ( (1 ) x, y) 0 1 y ( x ', y ') ( ax by c, dx ey f ) ((1 ) x, y) ( x, y) x ( x, y) x ( x x, y) ( x, y) Hence, X a( x, y) lim x ( x, y) 0 x
g(1,,0,1,0,0) g(1,0,0,1,0,0) generator X b lim 0 1 x ( x, y) ( x y, y) 0 1 y ( x, y) ( x y, y) ( x, y) y x Xb y x Hence, X a x, X b y x x In similar way: X c X d y X f x y y It is well known that translations c and f yield momentum components.
General theorem for Lie Groups: k k [ Xi, X ] j cij X k c ij=structure constants k Recall: L L i L where L generates O(3) R e The SO(3) Group of rotations in 3d (SO=Special Orthogonal) is a continuous Group. An element may be represented as a vector φ directed along the axis and with length equal to the angle of (say, counterclockwise) rotation φ; this corresponds to a sphere of radius π where, il however, each point of the surface is equivalent to the opposite one. All the rotations with the same φ belong to the same class..
Recall: L L i L where L generates O(3) R e il L i ( yp zp ) and cyclic permutations (3 components). x z y How many operators generate O(4)? 4 6 generators are needed. The generators of SO(4) obey : [I,I ]=i I i j ijk k [K,K ]=i K with [I, K ] 0. i j ijk k i j This is the algebra of SO(4). Fore more details see L.Schiff Quantum Mechanics page 35
The angular momentum operator L is the generator of infinitesimal rotations. For integer L one finds L+1 spherical harmonics Y LM (θ, φ) that are simultaneous eigenvectors of L and L z. Only the harmonics of a given L mix under rotations; they are the basis of an irrep labelled by L. Double valued spin representation and Covering Group 1 For J, rotation matrix around z is More generally, R e i. S ; however, i( ) i( ) 1 0 0 1 e 0 is valued : (0,0, ) i( ) (0,0, ) SU () covering Group of O(3) : identity for 4 0,0, 0 i i s z e 0 e i 0 e R R e R R e j m' j fill a sphere of radius in SU(), in SO(3) j m' m j R jm jm' D, D jm ' R jm m' m 51
Great Orthogonality Theorem (GOT) N G RG D R D R N ( i) * ( j) G ( ) ( ) ij mi Each element of each D(R) in each irrep is a symmetry type and this leads to orthogonality when integrating over the Group. dd I presented the proof for discrete Groups, however the GOT extends to continuous ones. ( j) 8 ( ) D ( ) j 1 ( J) * ab a' b' jj' aa' bb' ( J) Euler angles D Wigner matrices ( J) D jm R( ) jm' mm' d d d d sin( ) 0 0 0 mm' The line of nodes (N) is defined as the intersection of the xy and the XYcoordinate planes. 5