{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

p137bp11sol

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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
= 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
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}