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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)Uw(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)Uw(0)>+<w(0)UTQ’t)( 101qu )>+<w(0 11301Q<z1ulzp<o1> =%<Y(t)IHQ(t)Q(I)HY’()—>+< —>=—<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[pp,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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 Winter '03
 Nelson
 mechanics

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