hws06 - PHYS 115A HW6 Solutions 1. In order to get the...

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Unformatted text preview: PHYS 115A HW6 Solutions 1. In order to get the e—values and e-vectors we must solve for the characteristic equation. 1 1 —1 l—q 1 —1 l 1 —1 1 =>det(Q—ql)=det 1 —1——q 1 =0 —1 1 1 —1 1 1—qJ =—q3+q2+4q—4=0=>q={1,i2] Since the e-values are real this matrix is Herrnitian (we can also see this if we take the conjugate-transpose, which in this case would just be the transpose). Now to solve for the e-vectors. There are a few ways to do this. This is the way I learned in college so this is the way I still do it (when Mathematica is not an option). We solve for the kernel of the matrix Q — qI, where we plug in the different q values. E1: 1—(1) 1 —1 0 1 —1 0 1 ~1—(1) 1 =1 —2 1 0 —1 l 1—(1) -l 1 0 0 1 0 —1 0 Row reducing we get: 0 1 —1 0 0 O O O Letting x3 = t we get that: =1 x2 =t 1 =lq=1>fi 1 x3] 1 1 Similarly we can solve for the eigenspaces of +2 and — 2 and we get that: 1 ‘1 1 1 Iq=+2)=— 0 ,and|q=—2>=— _2 J3 6 1 1 2) a) 0 0 1 A 0 0 1 Q=0 0 0,Q*=0 0 O=Q=>QisHermitian. 1 0 0 1 0 0 b) First, finding the e-values: —q 0 1 det(Q—611)=det 0 —-q 0 =0=~q3+q=>q=[0,il} 1 0 —q Plugging q = 0 back into the eigenvalue equation: 0010 1000 0000—>0000 1000 0010 Setting x2 = t we get for a normalized e—vector: 0 |0>= 1 0 This vector is obviously normalized. Moving on to the next e—value, +1. —10 0—1 10—1 Finally for q = -1: 10101010 1_1 0100—10100=|—1>=-— \/'2'0 10100000 1 0) - Let U be .the matrix compose of columns of e-vectors of Q. $ 01/5 4/5 0 1 0 U=1 =01: 1/5 0 1/15 0 0 0 1/6 1/15 4/5 0 W3 010 001005—105 UTQU= 1N5 01/50001 0 0 —1/\/§ 0 1/6 1 0 0 0 1/15 1/1/5 010 01/15 1/15 21/1/501/50 0 0 —1/\/§ 0 WE 0 1/5 —1/\/3 0 0 0 611 0 0 =0 1 0 —0 q2 0 3) a) [A,BC]=ABC—BCA [A,B]C+B[A,C]=(AB—BA)C+B(AC—CA)=ABC—BAe+—BAG—BCA =ABC—BCA=[A,BC] b) 1.[A, [B, C]]=[A, BC—CB]=[A, BC]—[A, CB]=ABC—BCA—ACB+CBA 2.[B, [C, A]]=[B, CA]—[B, AC]=BCA—CAB—BAC+ACB 3.[C,[A, B]]=[C, AB]—[C, BA]=CAB—ABC—CBA+BAC Adding 1+2: 1+2+3=ABe—BGA—AGB+GBA+BeA—e43—BAG+AGB+643—Age—GBA+BAG [AJBxflH{BJC~H}HCiA,BH=0 {This [5 knflu/n 4.5 #16 Jacobi Idea-’15)): [A,B]T=(AB—BA)T=(AB)T—(BA)’=BTAT—A’B7‘=[B’,Af] 4) a) Q=AB Q =AB => Q =( AB)’ 23* AT =BA 2 Q is not Hermitian but somthing else. b) Q=AB+BA Q=AB+BA=>QT=(AB+BA)T=(AB)T+(BA)T=BTAT+ATBT=BA+AB=>Q is Hermitian. 0) Q=[A,B] Q=[A,B]=AB—BA:>Q’=[A,B]"=[B*,A’] (from 3c) =[B,A]=—[A,B]=—Q=>Q is anti—Hermitian. d) Q=A+iB Q=A+iB=>Q’=(A+iB)T=A*—iB"=A—iB=>Q is not Hermitian. cos 9 sin 6 cos 6 —sin 6 Q: 97‘: —sin 6 cos 6 sin 9 cos 9 C0520+sin29 cosOsinQ—sinacose _ 1 0 _I -sinecose+cos(9sin6 cos29+sin26 0 1 This means that is unitary. b) cose—a sine det =0=(cos€)—a)2+sin26=cos29—2acose+a2+sin26 -—sin 9 c036 —a 2 0:V4 20—4 1/ ' *" wax-336: —sin20=cos(9:is1n9=e‘9 2 0=a2—2acose+1=>a= 0 0 cos9—(cos(9+isin6) sine —iS s ‘0 —sin6 cos0—(cos(9+isin0) —S —iS 0 Row-reducing: 1 1 0 1 _l- . . 1 ,- =>| >=— $1m11ar1y |—>=— 0 0 0 1/3 1 1/31 <—|+>:%(—i,1) _i =%(—1+1)=0=> they are orthogonal. 1 =i _l l and Ufz—l— l _l \/E 1 1 J51 1 UTQUZLI —t C S —l z_lz —z —zC+S zC+S 21 1 —S C 1 1 21 1 iS+C —iS+C __1_C+iS+iS+C 0 _e"" 0 2 0 C—iS+C—iS 0 e‘” ...
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hws06 - PHYS 115A HW6 Solutions 1. In order to get the...

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