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Unformatted text preview: Physics 137B, Fall 2007, Moore Problem Set 11 Solutions 1. For ˆ ρ 1 = 1 / 2 1 / 2 1 / 2 1 / 2 we have h S z i = Tr ~ 2 1 / 2 1 / 2 1 / 2 1 / 2 1 1 = ~ 2 Tr 1 / 2 1 / 2 1 / 2 1 / 2 = 0 , h S x i = Tr ~ 2 1 / 2 1 / 2 1 / 2 1 / 2 1 1 = ~ 2 Tr 1 / 2 1 / 2 1 / 2 1 / 2 = ~ 2 . For ˆ ρ 2 = 1 / 2 1 / 2 we have h S z i = Tr ~ 2 1 / 2 1 / 2 1 1 = ~ 2 Tr 1 / 2 1 / 2 = 0 , h S x i = Tr ~ 2 1 / 2 1 / 2 1 1 = ~ 2 Tr 1 / 2 1 / 2 = 0 . The fact that h S x i = ~ / 2 for ˆ ρ 1 indicates that this density matrix represents a pure state with spin along the +ˆ x axis. Indeed, letting  x, ↑i = 1 √ 2 ( ↑i + ↓i ) , we see that  x, ↑ih x, ↑ = 1 2 ( ↑i + ↓i )( h↑ + h↓ ) = 1 2 ( ↑ih↑ + ↑ih↓ + ↓ih↑ + ↓ih↓ ) 1 = 1 / 2 1 / 2 1 / 2 1 / 2 = ˆ ρ 1 , (where ↑i ≡  z, ↑i are the usual eigenstates of S z ). Now if we were to start in the state  x, ↑i and measure S z , we know that the system will collapse either to ↑i or ↓i , with probabilities h x, ↑  ↑i 2 = 1 / 2 and h x, ↑  ↓i 2 = 1 / 2 . So if we measured S z but didn’t look at the outcome of the measurement , we would say that our system is in a mixed state with 50% probability of being ↑i and 50% probability of being ↓i . Alternatively, we could imagine creating a large number of identical copies of our system, all in the initial state  x, ↑i . Then if we measured S z on all of them, but didn’t look at the results of the measurements, we would know that half of the copies were in the state ↑i and half were in the state ↓i . In either case, this is precisely the mixed state that the density matrix ˆ ρ 2 represents. 4. Suppose we have a mixed state of made up of pure states  α 1 i , . . . ,  α m i with nonzero probabilities W 1 , . . . , W m , m > 1. We will assume these states are orthogonal, so h α i  α j i = δ ij . The density matrix representing this system is ˆ ρ = m X i =1 W i  α i ih α i  . Its square is ˆ ρ 2 = m X i =1 W i  α i ih α i  ! m X j =1 W j  α j ih α j  ! = m X i,j =1 W i W j  α i ih α i  α j ih α j  = m X i =1 W 2 i  α i ih α i  . Since each W i is nonzero and their sum is 1, each W i must be strictly less than 1. Therefore W 2 i 6 = W i , and hence we see that ˆ ρ 2 6 = ˆ ρ . 2 2,3. (Courtesy of Roger Mong) (General Method) In all four cases, the potential V ( r ) is spherically symmetric; V ( r ) is not a function of the angle. The first born approximation is dependent only on the the momentum transferred q : 1 f (1) B ( q ) = 1 4 π Z e i q · r U ( r ) d 3 r (Bransden 13.151) (1) = 1 q Z ∞ U ( r ) sin( qr ) r dr (Bransden 13.156) (2) where: U ( r ) = 2 m ~ 2 V ( r ) (Bransden 13.22) q = k k = 2 k sin θ 2 (3) The scattering amplitude can be computed by evaluating the integral, or by identi fying fourier transform pairs....
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This note was uploaded on 08/01/2008 for the course PHYSICS 137B taught by Professor Moore during the Fall '07 term at Berkeley.
 Fall '07
 MOORE
 Physics, mechanics

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