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Unformatted text preview: PHYS 115A HW6 Solutions 1. In order to get the e—values and evectors 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 evalues are real this matrix is Herrnitian (we can also see this if we take the
conjugatetranspose, which in this case would just be the transpose). Now to solve for
the evectors. 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 ,andq=—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 evalues: —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 evectors 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/ ' *"
wax336: —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 Rowreducing: 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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 Spring '03
 Nelson
 mechanics

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