1 / 15

We’ve found an exponential expression for operators

We’ve found an exponential expression for operators. n number of dimensions of the continuous parameter . Generator G. The order (dimensions) of G is the same as H. We classify types of transformations (matrix operator groups ) as. Orthogonal O (2) SO (2) O (3) SO (3)

lynn
Download Presentation

We’ve found an exponential expression for operators

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. We’ve found an exponential expression for operators n number of dimensions of the continuous parameter  Generator G The order (dimensions) of G is the same as H

  2. We classify types of transformations (matrix operator groups) as OrthogonalO(2) SO(2) O(3) SO(3) UnitaryU(2) SU(2) U(3) SU(3) groups in the algebraic sense: closed within a defined mathematical operation that observes the associative property with every element of the group having an inverse

  3. O(n) set of all orthogonal UT=U-1 (therefore real) matrices of dimension n×n SO(n) “special” subset of the above: unimodular, i.e., det(U)=1 group of all rotations in a space of n-dimensions Rotations in 3-dim space SO(3) ALL known “external” space-time symmetries in physics 4-dim space-time Lorentz transformations SO(4) Orbital angular momentum rotations SO(ℓ) (mixing of quantum mechanical states) cos  sin  0 Rz() = -sin  cos  0 0 0 1 U(n) set of all n×n UNITARY matrices U†=U-1 i.e. U†U=I new “internal” symmetries (beyond space-time) SU(n) “special” unimodular subset of the above det(U)=1

  4. SO(3) cos3sin3 0 cos20 -sin2 1 0 0 do not commute R(1,2,3)= -sin3cos3 0 0 cos1sin1 0 1 0 sin2 0 cos2 0 -sin1cos1 0 0 1 cos3cos2+sin3sin2cos1sin3-sin1sin2sin3sin1sin3-cos1sin2cos3 = -cos2sin3 cos1cos3-sin1sin2sin3sin1cos3-cos1sin2sin3 sin2-sin1cos2cos1cos2 do commute Contains SO(2) subsets like: acting on vectors like NOTICE: all real and orthogonal cossin 0 vx vy vz v = Rz() = -sincos 0 in the i, j, k basis 0 0 1 ^ ^ ^

  5. Obviously “reduces” to a 2-dim representation cossin vx vy Rv = -sincos Call this SO(2) What if we TRIED to diagonalize it further? ^ U†x seek a similarity transformation on the basis set: Uv  which transforms all vectors: URU† and all operators:

  6. An Eigenvalue Problem cos-l sin 0 -sin cos-l0 = 0 0 0 1-l = (1-l)[cos2-2lcos+l2+sin2]=0 (1-l)[1 - 2lcos + l2]=0 l=1 Eigenvalues:l=1, cos  + isin , cos -isin 

  7. To find the eigenvectors cos sin 0 aa -sin cos 0 b=lb 0 0 1 cc forl=1 acos + b sin = a -asin + b cos = b c = c a(1-cos) = bsin b(1-cos) = -asin a/b = -b/a ?? a=b=0 acos  + b sin  = a(cos+isin) -asin  + b cos  = b(cos+isin) c = c(cos+isin) forl=cos+isin b =i a, c = 0 since a*a + b*b = 1 a=b= forl=cos-isin b =-i a, c = 0 since a*a + b*b = 1 a=b=

  8. With < v | R | v > cos sin 0 0 -sin cos 0 0 0 0 1 0 1 0 URU† eigenvectors cos+isin 0 0 = 0 1 0 0 0 sin-icos

  9. cos+isin 0 0 = 0 1 0 0 0 sin-icos < v | R | v > and under a transformation to this basis (where the rotation operator is diagonalized) vectors change to: v1(v1+iv2)/ Uv = Uv2 = v3 v3(v1-iv2)/

  10. SO(3) cos3cos2+sin3sin2cos1sin3-sin1sin2sin3sin1sin3-cos1sin2cos3 R(1,2,3) = -cos2sin3 cos1cos3-sin1sin2sin3sin1cos3-cos1sin2sin3 sin2-sin1cos2cos1cos2 Contains SO(2) subsets like: acting on vectors like cossin 0 vx vy vz v = Rz() = -sincos 0 in the i, j, k basis 0 0 1 ^ ^ ^ which we just saw can beDIAGONALIZED: e+i0 0 0 1 0 0 0 e-i Rv =

  11. Block diagonal form means NO MIXINGof components! e+i0 0 0 1 0 0 0 e-i Rv = Reduces to new “1-dim” representation of the operator acting on a new “1-dim” basis: e+i + e-i -

  12. R(1) R(2)= R(1+2) UNITARYnow! (not orthogonal…) ei  is the entire set of all 1-dim UNITARYmatrices, U(1) obeying exactly the same algebra as SO(2) SO(2)is ISOMORPHIC toU(1)

  13. SO(2) is supposed to be the group of allORTHOGONAL 22 matrices withdet(U) = 1 a b c d a c b d a2+b2 ab+bd ac+bd c2+d2 = a2 + b2= 1 ac = -bd c2 + d2= 1 and det(U) = ad – bc = 1 along with: abd – b2c = b -a2c – b2c = b -c(a2 + b2) = b -c = b which means: ac = -(-c)d a = d

  14. So all matrices have the SAME form: SO(2) a b -b a a2 + b2= 1 with i.e., the set of all rotations in the space of 2-dimensions is the complete SO(2) group!

  15. det(A)  n1 n2 n3···nN An11 An22 An33 …AnNN N n1,n2,n3…nN completely antisymmetric tensor (generalized Kroenicker  ) since these are just numbers some properties det(AB) = (detA)(detB) = (detB)(detA) = det(BA) which means Determinant values do not change under similarity transformations! det(UAU†) = det(AU†U) = det(A) So if A is HERMITIAN it can be diagonalized by a similarity transformation (and if diagonal) det(A)  …(n1 n2 n3···nN An11)An22 An33 …AnNN N N N N nN n3 n2 n1 Only A11 term0 only diagonal terms survive, here that’s A22 det(A)  123…NA11 A22 A33 …ANN = l1l2l3...lN

More Related