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p137bp11sol - Physics 137B Fall 2007 Moore Problem Set 11...

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Physics 137B, Fall 2007, Moore Problem Set 11 Solutions 1. For ˆ ρ 1 = 1 / 2 1 / 2 1 / 2 1 / 2 we have S z = Tr 2 1 / 2 1 / 2 1 / 2 1 / 2 1 0 0 - 1 = 2 Tr 1 / 2 - 1 / 2 1 / 2 - 1 / 2 = 0 , S x = Tr 2 1 / 2 1 / 2 1 / 2 1 / 2 0 1 1 0 = 2 Tr 1 / 2 1 / 2 1 / 2 1 / 2 = 2 . For ˆ ρ 2 = 1 / 2 0 0 1 / 2 we have S z = Tr 2 1 / 2 0 0 1 / 2 1 0 0 - 1 = 2 Tr 1 / 2 0 0 - 1 / 2 = 0 , S x = Tr 2 1 / 2 0 0 1 / 2 0 1 1 0 = 2 Tr 0 1 / 2 1 / 2 0 = 0 . The fact that S x = / 2 for ˆ ρ 1 indicates that this density matrix represents a pure state with spin along the +ˆ x axis. Indeed, letting | x, = 1 2 ( |↑ + |↓ ) , we see that | x, x, ↑| = 1 2 ( |↑ + |↓ )( ↑| + ↓| ) = 1 2 ( |↑ ↑| + |↑ ↓| + |↓ ↑| + |↓ ↓| ) 1
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= 1 / 2 1 / 2 1 / 2 1 / 2 = ˆ ρ 1 , (where |↑ ≡ | z, are the usual eigenstates of S z ). Now if we were to start in the state | x, and measure S z , we know that the system will collapse either to |↑ or |↓ , with probabilities | x, ↑ | ↑ | 2 = 1 / 2 and | x, ↑ | ↓ | 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 |↑ and 50% probability of being |↓ . Alternatively, we could imagine creating a large number of identical copies of our system, all in the initial state | x, . 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 |↑ and half were in the state |↓ . 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 , . . . , | α m with nonzero probabilities W 1 , . . . , W m , m > 1. We will assume these states are orthogonal, so α i | α j = δ ij . The density matrix representing this system is ˆ ρ = m i =1 W i | α i α i | . Its square is ˆ ρ 2 = m i =1 W i | α i α i | m j =1 W j | α j α j | = m i,j =1 W i W j | α i α i | α j α j | = m i =1 W 2 i | α i 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 = W i , and hence we see that ˆ ρ 2 = ˆ ρ . 2
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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 e - i q · r U ( r ) d 3 r (Bransden 13.151) (1) = - 1 q 0 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. The differential cross section is simply the square modulus of the scattering am- plitude: d Ω = f B 2 The total cross section is computed by integrating the differential cross section over the entire sphere, where k = k = k . σ tot ( k ) = d Ω ( k ) sin θ dθ dφ = 2 π k 2 2 k 0 f (1) B ( q ) 2 q dq (Bransden 13.157) (4) The latter expression is a result of substituting for dq integral, applying azimuthal symmetry, and restricting the Born series to the 1 st order.
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