hw201a - Physics 512 Homework Set #1 Solutions 1. The Pauli...

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Physics 512 Winter 2003 Homework Set #1 – Solutions 1. The Pauli Hamiltonian (for g s = 2) was presented as H = 1 2 m ( ~p q c ~ A ) 2 q mc ~ S · ~ B + Using the properties of the Pauli matrices (and being careful with operators), show that this may be rewritten in the equivalent form H = 1 2 m h · ( q c ~ A ) i 2 + It is actually easier to work back from the second form of the Pauli Hamiltonian to the Frst. ±or the Pauli matrices, we may use ( · ~ A )( · ~ B )= ~ ~ B + i~σ · ( ~ ~ B ) Using ~ A = ~ B =( q c ~ A ) , we obtain h · ( q c ~ A ) i 2 q c ~ A ) 2 + · ( q c ~ A ) × ( q c ~ A ) Note that, as counterintuitive as it may be, ~ ~ A does not necessarily vanish for a vector operator A since its various components may not commute with each other. We have to be a bit careful whenever things do not commute. Nevertheless, commutes with itself and likewise ~ A commutes with itself. So the only terms we need to worry about in the cross product involve with ~ A . h · ( q c ~ A ) i 2 q c ~ A ) 2 i q c · × ~ A + ~ A × (1) We may use the coordinate representation = i ¯ h ~ and imagine acting on a function f to Fnd × ~ A + ~ A × f = i ¯ h ~ ∇× ( ~ Af )+ ~ A × ( ~ f ) = i ¯ h ( ~ ~ A ) f +( ~ f ) × ~ A + ~ A × ( ~ f ) Now the last two terms vanish due to antisymmetry of the cross product. So we have simply × ~ A + ~ A × = i ¯ h ~ ~ A = i ¯ h ~ B 1
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(Alternatively, we could use the commutation relation [ f ( ~x ) ,~p ]= i ¯ h ~ f ( ) applied to ~ A ( ) .) Inserting this back into (1) yields h · ( ~p q c ~ A ) i 2 =( q c ~ A ) 2 q ¯ h c · ~ B so that H = 1 2 m h · ( q c ~ A ) i 2 + = 1 2 m ( q c ~ A ) 2 q ¯ h 2 mc · ~ B + Finally, we identify ~ S = 1 2 ¯ h~σ for spin- 1 2 . 2. We model the reflection of a neutron beam perpendicularly incident on a block of a ferromagnetic material =0 x x n with the Hamiltonian H = p 2 2 m + V ( x ) ,V ( x )= ± 0 ,x < 0; V 0 + ω 0 S z ,x> 0 where V 0 and ω 0 are constant, and satisfy 0 < 1 2 ¯ 0 <V 0 . a ) Assume the incident neutron beam is unpolarized. Calculate the degree of po- larization of the reflected beam. Note that you may have to consider diFerent incident beam energies. We ±rst note that the potential V ( x ) is a spin dependent potential. However, since it only depends on S z , we naturally choose to quantize spin along the z - axis. Since spin-up |↑i and spin-down |↓i are eigenstates of S z , it is simplest to describe the unpolarized beam as equal probability spin-up and spin-down (this is a mixed state, or an ensemble). For x< 0 , the potential vanishes, while for x> 0 up-spins see V = V 0 + 1 2 ¯ 0 and down-spins see V = V 0 1 2 ¯ 0 . This problem then reduces to that of reflection from a one-dimensional potential barrier V ( x ) V = 0 1 2 ω h + V Be -ikx Ae ikx Ce ik’x V = 0 1 2 ω h - V x 2
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Recall that the reflection and transmission coefficients (probabilities) are given by R = ± ± ± ± B A ± ± ± ± ,T = Re k 0 k ± ± ± ± C A ± ± ± ± where the amplitudes are given by B A = k k 0 k + k 0 , C A = 2 k k + k 0 where ¯ hk = 2 mE, ¯ hk 0 = p 2 m ( E V ef ) Here V ef is either V or V , depending on the spin state of the neutron. Note that for E<V ef we let k 0 be pure imaginary. In this case,
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This note was uploaded on 02/15/2011 for the course PHYS 512 taught by Professor Unknow during the Winter '03 term at Cornell College.

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hw201a - Physics 512 Homework Set #1 Solutions 1. The Pauli...

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