hws07 - HW 7 Solutions 1 T 00 1 fl w ’ 00 00...

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Unformatted text preview: HW 7 Solutions 1. T . 00 1 fl w ’. 00 00: . ) Uze‘QZSI—gan: 2g :ZQle:Z_lQ:e iQ:>U7-[j—I o n! 0 n! o n/ 0 n! 2. Because Q is hermitian we can work in a basis where Q is diagonal. Let V be a matrix that diagonalizes Q. m . h 00 -~ h VT U V=V7 2b Q V :[email protected] Q is the diagonal form of Q. on! on! dez(v7‘ U V):det(V*)det(U)dez(V):dez(Vl)det(v)det(U)=det(vlv)dez(U)=det(U) Since the exponent of a diagonal matrix is a diagonal matrix of the exponents we see that: det(U)=H eiq'where qi are the eigenvalues of Q. 3) a) If S and Q commute then they are simultaneously diagonalizable. 1o1211211101303l303 [Q,S]=00010 —1—10 —1000=000—000=0 1011—121—12‘101303303 b) The eigenspaces for Q and S are: 1 —.|. o 1 1 Q:qi=l0,0,2l lqi>= —- 01— 0 1 1 2 2 1 1 0 1 —1 —1 1 1 S..Sl,:r—17273] lsi>: —— 0 ’_ —1 ’L 2 2 1 \f3— 1 6 1 Because Q has degenerate e—values and the second two are linear combinations of those two degenerate vectors then the eigenvectors that diagonalize S diagonalize Q. C) 1 a a U:0 E l _ L 13 [3 This matrix diagonalizes both S and Q. The math here is rather tedious and is just matrix multiplication so in the interest of time I'm going to skip it. a» aw a; 4) £<Q>=£<W(O)IWQ(I)U|w(0)>=<w(0)li(U7)Q(t)U+UT1(Q(I))U+UTQ(I)1 U 0 dt dt dt NM )> —'Ht/h 'Ht/h U=e ‘ , U7L=el =i<Q>=<w(0)l%HU’Q(t)U|w(0)>+<w(0)|UTQ’t)( 101qu )>+<w(0 11301Q<z1ulzp<o1> =%<Y(t)IHQ(t)-Q(I)H|Y’()—>+< —>=—<1H Qrg—(1J>+< Q1 Where we used the fact that U and H commute since U can be written as a polynomial in H. 5) For this problem we assume H, X, P are not explicit functions of time. a) i<H>=‘—[H,H]+<a—H>=0+0=0 (it h at This is energy conservation. b) i<x>=imx1+o=i—L[p-p,xi+wfii= i (p[p.x]+[p,x]p) dt h fizm 2km = l (—ifip—ihp)=p— 271m m In classical mechanics velocity v = p/m. C) d z i 1 —<p>=-[H,pl+0=-(— 2 +[V(x),pl) dt 5 5 2m 2%—'fiv(x)aXY’+ihax(V(x)Y’))= —axv<x>=i<p> dt —> This is the same as: [—52—V U ...
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