Finding irreducible representation of symmetry operators linearly
aa r X i v : . [ m a t h - ph ] A ug Finding irreducible representation of symmetry operators linearly
Young-Chung Hsue
Department of Physics, National Cheng Kung University, Tainan 70101, Taiwan andNational Center for Theoretical Sciences, National Tsing Hua University, Hsinchu 30043, Taiwan
The main purpose of this paper is providing a simple method to generate the matrices of irreduciblerepresentations because it is useful to reduce the computational time of solving the eigenvalueproblems. The only information we need to provide for this method is the group-multiplicationtable, and the proof of validity of this method is also shown in this paper.
I. INTRODUCTION
When we study the quantum mechanics [1, 2], we usu-ally solve schr¨odinger equation or Kohn-Sham equationin many body system and we solved them through theequation ˆ Hψ = εψ where ˆ H , ε and ψ are Hamiltonian,eigenvalue and eigenfunction, respectively. In this equa-tion, ψ is expanded by an independent bases usually andˆ H will be a N × N matrix. For large N case, computingthe eigenvalues and eigenstates of ˆ H will be time con-suming. One of the solution to save time is the use ofsymmetry. The symmetry can not only be applied onthe charge density and potential but also on the eigen-functions. However, if we want to apply the symmetryon the eigenfunctions, we need to have irreducible rep-resentation first [3]. Although people already prove thatone character table is related to one kind of irreduciblerepresentation, we still need to find a way to find theirreducible representation from the character table. Be-sides, the equations people used to find out the charactertable is the multiplication relation of ˆ χ ( ˆ R i ) which is thetrace of representation of ˆ R i [4]. Unfortunately, it is nota set of linear equations and will be not so easy to findthe solution numerically. Hence, I tried to find a methodwhich is linear and simple, and show it in the follow-ing. Although you need to compute the eigenvalues andeigenstates of a N × N matrix where N is the same asthe number of symmetry operators, the irreducible repre-sentations are derived when you get the eigenstates. Themethod to derive the irreducible representation is shownin Sec. II. I also show the proof of the validity of thismethod in Sec. III. Finally, I will show a brief conclusionin Sec. IV. II. ONE SIMPLE WAY TO DERIVE THEIRREDUCIBLE REPRESENTATIONS
The irreducible representation is useful when we tryto find a system’s eigenfunctions. We can use it to pre-dict the degeneracy and to reduce the calculation time.Following is a simple way to derive the irreducible repre-sentation and character table.
1. group-multiplication table
First of all, we need the symmetry operators. Oncewe get the symmetry operators, we can get the group-multiplication table.For example, the square lattice case, the symmetryoperators for vectors areˆ R = (cid:18) (cid:19) , ˆ R = (cid:18) − (cid:19) , ˆ R = (cid:18) −
11 0 (cid:19) , ˆ R = (cid:18) − − (cid:19) , ˆ R = (cid:18) − (cid:19) , ˆ R = (cid:18) − − (cid:19) , ˆ R = (cid:18) (cid:19) , ˆ R = (cid:18) − (cid:19) . Then we can get the group-multiplication tableRight l e f t C v ˆ R ˆ R ˆ R ˆ R ˆ R ˆ R ˆ R ˆ R ˆ R R R R R R R R → R → ˆ R and they equal ˆ R left ˆ R right , respectively. ˆ H After we get the group-multiplication table, we canuse it to generate a Hamiltonian ˆ H which obeys thissymmetry for finding irreducible representation. HereI choose the plane wave as basis and define H G , G ′ = R e i ( G ′ − G ) · r H ( r ). Because of symmetry, ˆ H ( r ) = ˆ H ( ˆ R r ),we will get H ˆ R G , ˆ R G ′ = H G , G ′ , (1)so that we can classify H G , G ′ and just need to provideone value for each class. The value I provided for eachclass is H G , G ′ = 1 /t (depends on you), where t means t th class. The value I chosen is for breaking the unnecessarydegeneracy and then each kind of eigenvalue related toone kind of irreducible representation.What is the G ? In fact, for deriving the irreduciblerepresentation we don’t need to know what it is, wejust need the group-multiplication table. We can defineˆ R i G = G i , so that H G i , G j = H ˆ R ˆ R i G , ˆ R ˆ R j G for allˆ R . Hence, the H G , G ′ for square lattice generated by thisway is H G , G ′ = / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / , where I force ˆ H to be a hermitian operator, whichmeans H G , G ′ = H ∗ G ′ , G and ˆ H ( r ) ∈ R . For example,you can find H , = H , in this case; the reason isˆ R ˆ R = ˆ R which is shown in group-multiplication tableand H G , G ′ = H ˆ R G , ˆ R G ′ , we get H , = H , ; besides, be-cause of the requirement of hermitian, H , = H , , wefinally get H , = H , . Remember, ˆ H is unnecessaryto be a hermitian operator. In some cases, such as C symmetry, its traces of irreducible representations can becomplex numbers and this kind of result is impossible tobe generated through a real and hermitian operator andusually the complex conjugate terms will mix together tobe a reducible representation. The reason why I choosea real and hermitian operator for C v symmetry calcu-lation is that I want to get the representations with realnumbers.
3. get irreducible representation from eigenfunctions of ˆ H This part is simple, you just need to find out the eigen-functions with the same eigenvalues which are related tothe same irreducible representation ˆΓ α and it obeysˆΓ α ( ˆ R i )ˆΓ α ( ˆ R j ) = ˆΓ α ( ˆ R i ˆ R j ) . (2)If we set X m ′ α Γ αm α m ′ α ( ˆ R ) ψ m ′ α ( r ) = ψ m α ( ˆ R r ) ,ψ m α ( r ) = X G C m α , G e i G · r , where m α denotes the index of eigenfunction in α th irre-ducible representation , we get C m α , ˆ R G = X m α m ′ α Γ αm α m ′ α ( ˆ R ) C m ′ α , G , (3) because of the independence of plane waves. Since C m α , G is the eigenfunction, C m α , ˆ R G is just the rear-rangement of C m α , G based on group-multiplication ta-ble. For example, if we want to get ˆΓ α ( ˆ R ) of squarelattice case, we need to provide Eq. 3 { C m α , G | m α , G } and { C m α , ˆ R G | m α , G } , where ˆ R { G } = ˆ R ˆ R ··· G =[ G , G , G , G , G , G , G , G ] which is derived fromthe third row of group-multiplication table. Since weget { C m α , G | m α , G } as the eigenfunctions with the sameeigenvalues and { C m α , ˆ R G | m α , G } is just the rearrange-ment of { C m α , G | m α , G } , it is no doubt that we can getˆΓ α ( ˆ R ). Hence, once we get the eigenfunctions, we canget the irreducible representation from Eq. 3.
4. character table
Since we can get the irreducible representation, whatwe need to do is taking the trace of irreducible represen-tation is enough.
5. one interesting behavior
When we derive an irreducible representation by thisway, we get each irreducible representation l α timeswhere each ˆΓ α ( ˆ R ) is a l α × l α matrix; besides, we need l α independent eigenstates with the same eigenvalue toderive ˆΓ α . Hence, there are l α eigenstates are related toan irreducible representation ˆΓ α . When we consider allthe irreducible representations, we get n R = X α l α , (4)where n R is the number of symmetry operators and ˆ H is a n R × n R matrix. Hence, we get the relationship predictedby group theory. The reason is shown as follows: Whenwe substitute Eqs. 1 and 3 into ˆ Hψ = εψ where ˆ H isthe one I used to generate irreducible representations, wewill get X m ′ α X ˆ R H G , ˆ R G Γ α ( ˆ R ) m α ,m ′ α C m ′ α , G = C m α , G , (5)where P ˆ R H G , ˆ R G ˆΓ α ( ˆ R ) will be a l α × l α matrix andits eigenvalues will be l α different ones and C G will be a l α × l α orthogonal eigenvectors for each eigenvalue throughEq. 3 since we know C G . This is the reason that we willget l α states related to one irreducible representation. Infact, we can use this way to reduce the computationaltime based on this concept. For example, the originalmatrix size of the ˆ H of C v is 8 ×
8, now you can reduceit into four 1 × × III. THE REASON WHY WE CAN GET ALLTHE IRREDUCIBLE REPRESENTATIONS BYTHIS WAY
If we want to prove this thing, we need to prove fourthings first.1. The functions belong to different irreducible repre-sentation will orthogonal to each other.2. If the functions of { φ i ( ˆ R r ) | ˆ R } are independent toeach other, all the irreducible representations canbe derived from this set. Hence, we don’t need todeal with infinite number of functions to find theirreducible representations and the number of setis the same as of symmetry operators.3. If a Hamiltonian ˆ H obeys ˆ H ( r ) = ˆ H ( ˆ R r ) , theeigenfunctions with the same eigenvalue belong tothe same representation of { ˆ R } . Hence, we canget the representations through solving eigenvalueproblem. If the unnecessary degeneracy are broken,we even can get the irreducible representations di-rectly.4. One kind of trace value of irreducible representa-tion is just related to one kind of irreducible rep-resentation. Hence, once we find the irreduciblerepresentations through one Hamiltonian, they arewhat we need. A. Proof of the first part
Assume a function set obeys ψ n α ( ˆ R r ) = X n ′ α Γ αn α ,n ′ α ( ˆ R ) ψ n ′ α ( r ) , (6) Z d r ψ ∗ n α ( r ) ψ n α ( r ) = 1 , and we want to find the value of R d r ψ ∗ n α ( r ) ψ m β ( r ),where α and β denote the index of irreducible represen-tations and m and n are the index of eigenfunctions ineach irreducible representation, respectively. Because theintegration is over the whole space, we get Z d r ψ ∗ n α ( r ) ψ m β ( r ) = Z d r ψ ∗ n α ( ˆ R r ) ψ m β ( ˆ R r ) . Now, let’s expand ψ n α ( ˆ R r ) by Eq. 6 and consider X ˆ R Γ α ∗ n α ,n ′ α ( ˆ R )Γ βm β ,m ′ β ( ˆ R ) = n R l α δ α,β δ n α ,m β δ n ′ α ,m ′ β , (7) which is shown in the textbook, we will get Z d r ψ ∗ n α ( r ) ψ m β ( r ) = 1 n R X ˆ R Z d r ψ ∗ n α ( ˆ R r ) ψ m β ( ˆ R r )= X ˆ R Γ α ∗ n α ,n ′ α ( ˆ R )Γ βm β ,m ′ β ( ˆ R ) X n ′ α ,m ′ β R d r ψ ∗ n ′ α ( r ) ψ m ′ β ( r ) n R = δ α,β δ n α ,m β , (8)where n R and l α are the number of symmetry operatorsand of n α which means n α = 1 → l α , respectively. Hence,they are orthogonal to each other. B. Proof of the second part
Assume all the functions can be expanded by { φ m ( r ) | m } , where { φ m ( r ) | m } are independent to eachother and for arbitrary n and ˆ R , φ n ( ˆ R r ) ⊂ { φ m ( r ) | m } .For example, plane waves obey this requirement for ro-tational symmetry in periodic structure. Hence, we candefine that φ ∆ i ( r ) obeys { φ ∆ i ( r ) | i } ⊂ { φ m ( r ) | m }{ φ ∆ i ( ˆ R r ) | i, ˆ R } = { φ m ( r ) | m } , and { φ ∆ i ( r ) | i } is a set with minimal number of members.Therefore, an arbitrary function ψ n α ( r ) can be writtenas ψ n α ( r ) = X i, ˆ R C in α ( ˆ R ) φ ∆ i ( ˆ R − r ) , (9)where n α means the n th states of α th irreducible repre-sentation. If we let ψ n α ( ˆ R r ) = X n ′ α Γ αn α ,n ′ α ( ˆ R ) ψ n ′ α ( r ) , expand ψ by Eq. 9 and use the concept of ˆ R { ˆ R } = { ˆ R } ,we have X i, ˆ R C in α ( ˆ R ˆ R ) φ ∆ i ( ˆ R − r ) = X n ′ α ,i, ˆ R Γ αn α ,n ′ α ( ˆ R ) C in ′ α ( ˆ R ) φ ∆ i ( ˆ R − r ) . Because of the independence of { φ ∆ i ( r ) | i } , we can get X ˆ R C in α ( ˆ R ˆ R ) φ ∆ i ( ˆ R − r ) = X n ′ α , ˆ R Γ αn ′ α ,n α ( ˆ R ) C in ′ α ( ˆ R ) φ ∆ i ( ˆ R − r ) , (10)which means the representations can be just related to i th subset { φ ∆ i ( ˆ R r ) | ˆ R } . For finding all the irreduciblerepresentations, we need to require that the elements of { φ ∆ i ( ˆ R r ) | ˆ R } are independent to each other. If not, it ispossible C in α ( ˆ R ) = 0 for all ˆ R . Besides, when we requirethat { φ ∆ i ( ˆ R r ) | ˆ R } are independent to each other, Eq. 10can be simplified to be C in α ( ˆ R ˆ R ) = X n ′ α Γ αn α ,n ′ α ( ˆ R ) C in ′ α ( ˆ R ) , (11)that’s why we get Eq. 3. Here we can find that the for-mula of bases is not important.Besides, in the following, when we provide C in α forEq. 11, we can prove that we just can find one kind of Γ α when they are independent to each other and l α ≤ n R ,where each ˆΓ α ( ˆ R ) is a l α × l α matrix. Please note that C in α ( ˆ R ) are independent to each other for different n α because of the first proof. If they are not independentto each other, we can prove that the related ˆΓ α derivedfrom Eq. 3 is reducible. The proof is as follows:If we define (cid:12)(cid:12) C in α (cid:11) and (cid:12)(cid:12)(cid:12) C in α ( ˆ R ) E as C in α ( ˆ R ) and C in α ( ˆ R ˆ R ) and they are n R × (cid:0)(cid:12)(cid:12) C i (cid:11) (cid:12)(cid:12) C i (cid:11) · · · (cid:1) ˆΓ T ( ˆ R ) = (cid:16)(cid:12)(cid:12)(cid:12) C i ( ˆ R ) E (cid:12)(cid:12)(cid:12) C i ( ˆ R ) E · · · (cid:17) , and (cid:0)(cid:12)(cid:12) C i (cid:11) (cid:12)(cid:12) C i (cid:11) · · · (cid:1) ˆ χ ˆ χ − ˆΓ T ( ˆ R ) ˆ χ = (cid:16)(cid:12)(cid:12)(cid:12) C i ( ˆ R ) E (cid:12)(cid:12)(cid:12) C i ( ˆ R ) E · · · (cid:17) ˆ χ, (12)where Γ Ti,j = Γ j,i and ˆ χ is an arbitrary l α × l α matrixwhose determine is not zero. If { (cid:12)(cid:12) C in α (cid:11) | n } is a dependentset, we always can find a ˆ χ whose determine is not zeroand (cid:0)(cid:12)(cid:12) C i (cid:11) (cid:12)(cid:12) C i (cid:11) · · · (cid:1) ˆ χ = (cid:0)(cid:12)(cid:12) C ′ i (cid:11) · · · Ø (cid:1) . Besides, if (cid:12)(cid:12) C ′ il α (cid:11) = Ø, because of rearrangement, | C ′ il α ( ˆ R ) > = Ø. Equation 12 will become (cid:0)(cid:12)(cid:12) C ′ i (cid:11) · · · Ø (cid:1) ˆΓ ′ T ( ˆ R ) = (cid:16)(cid:12)(cid:12)(cid:12) C ′ i ( ˆ R ) E · · · Ø (cid:17) , where ˆΓ ′ T ( ˆ R ) = ˆ χ − ˆΓ T ( ˆ R ) ˆ χ . Because of this equation,when we choose ˆΓ( ˆ R − ) = ˆΓ( ˆ R ) † and ˆ χ − = ˆ χ † , we getΓ ′ m,l α = Γ ′ l α ,m = 0 where m is 1 to l α − l α is larger than the numberof symmetry operators, { (cid:12)(cid:12) C in α (cid:11) | n } must be a dependentset.If the elements of { (cid:12)(cid:12) C in α (cid:11) | n } are independent to eachother, this set will just be related to a unique represen-tation. The simple proof is as follows: If ˆΓ and ˆΓ ′ obeyEq. 11, we get Ø = (cid:0)(cid:12)(cid:12) C i (cid:11) · · · (cid:1) (ˆΓ T − ˆΓ ′ T ) . Because the elements of { (cid:12)(cid:12) C in α (cid:11) | n } are independent toeach other, ˆΓ − ˆΓ ′ = Ø. Therefore, we just can get onekind of ˆΓ when (cid:12)(cid:12) C in α (cid:11) are provided. Hence, we can find the irreducible representations from { φ ∆ i ( ˆ R r ) | ˆ R } whose number equals of symmetry opera-tors.Here, let me summarize this proof. Assume afunction set used to find the representation are de-scribed by { C in α ( ˆ R ) | i, n, ˆ R } . Based on the first proof, { C in α ( ˆ R ) | i, n, ˆ R } ≡ {| C n α i | n } must be independent toeach other where | C n α i is a n R n i × n R and n i denote the number of symmetry operators and of φ ∆ i , re-spectively. If the elements of { C n α ( ˆ R ) | n, ˆ R } are alreadyindependent to each other, we can get an unique ˆΓ fromthis set and ignore all other contributions from i = 0terms. That’s why we just need { φ ∆ i ( ˆ R r ) | ˆ R } to find theirreducible representations. C. Proof of third part
The problem we want to study isˆ H ( r ) ψ n α ( r ) = ε α ψ n α ( r ) , (13)where ˆ H ( r ) is provided, and ε α and ψ n α are the eigen-value and eigenstate of ˆ H ( r ). Because ˆ H ( r ) = ˆ H ( ˆ R r ),we have ˆ H ( r ) ψ n α ( ˆ R r ) = ε α ψ n α ( ˆ R r ) , and that means { ψ n α ( ˆ R r ) | ˆ R } are degenerate states.Hence, ψ n α ( ˆ R r ) = P n ′ α Γ αn α ,n ′ α ( ˆ R ) ψ n ′ α ( r ) , and ˆΓ α willbe one kind of representation because it obeys Eq. 2.Therefore, for the states with the same eigenvalue ε α ,we should get the representation from them. If the un-necessary degenerate are broken, they are related to theirreducible representation. D. Proof of the fourth part
It is easy to prove that the same irreducible representa-tions will relate to the same trace. The same means thatthey are { ˆ χ − ˆΓ ˆ χ | det ( ˆ χ ) = 0 } , and their trace will bethe same because of Tr (cid:16) ˆ χ − ˆΓ ˆ χ (cid:17) = Tr (cid:16) ˆΓ (cid:17) . If we wantto prove that the same trace will be correspond to thesame irreducible representations, we can prove it throughthe orthogonality of trace , i.e. X R ˆ χ α ∗ ( ˆ R ) ˆ χ β ( ˆ R ) = n R δ α,β , (14)where ˆ χ β ( ˆ R ) = P j Γ βj,j ( ˆ R ) is the trace of β th irreduciblerepresentation. It is easy to prove this equation becausethe left hand side of Eq. 14 can be expanded as X R ˆ χ α ∗ ( ˆ R ) ˆ χ β ( ˆ R ) = X i,j, ˆ R Γ α ∗ i,i ( ˆ R )Γ βj,j ( ˆ R ) , and it is the same as P i,j δ i,j δ α,β n R l α = n R δ α,β , and fi-nally we get Eq. 14.Therefore, the first and second proof tell us that wecan get irreducible representations through a set of in-dependent functions; the third proof tell us that we canget this independent functions as the eigenfunctions of ahamiltionian which obeys the symmetry and the fourthpart help us to classify the irreducible representations. IV. CONCLUSION
Although the proof is based on the real space, the in-formation we need is just the group-multiplication table as shown in Sec. II. Hence, once we know the group-multiplication table, the related irreducible representa-tions can be obtained and it is independent of what kindof bases we choose. Since that, it is not restricted in Liegroup. The code of this paper is submitted to matlaband welcome you to test it. In that case, I use Buck-minsterfullenerene (C60) whose symmetry is icosahedral-inversion symmetry [5] as an example and get its char-acter table as1 E C ′ C C C C ′ C C ′ C ′ i − − − − − − . − . − − .
618 0 0 1 .
618 35 − − − − − .
618 1 . − .
618 0 0 − .
618 33 1 1 . − . − .
618 0 0 − . −
35 1 0 0 1 0 − − − − − −
44 0 − − − − − .
618 1 . − − .
618 0 0 0 . −
31 1 1 1 1 1 1 1 1 1where the first row denotes the number of elements ofeach class with the class type and the following is thetrace of each irreducible representation. Besides, thevalues shown as − .
618 and 0 .
618 are (1 − √ / √ / − , respectively. Acknowledgments
I would like to acknowledge NCTS and the financialsupport from NSC of Taiwan under Grant No. NSC 95- 2745-M-006-004-. [1] J. Zeng,
Quantum mechanics Roll II (Chinese Language) (Beijing: Scientific Publications, Beijing, 2000).[2] P. Atkins and R. Friedman,
Molecular quantum mechanics (Oxford University Press Inc., New York, 2005).[3] W. Rudge, Phys. Rev. , 1024 (1969). [4] M. Tinkham,
Group Theory and Quantum Mechanics (Dover Publications, INC, Mineola, New York, 2003).[5] B. Judd, Proc. R. Soc.A241