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solmx2_04

# solmx2_04 - 1 Physics 325 Solution to Second Midterm Exam...

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1 Physics 325 Solution to Second Midterm Exam Winter 2004 PART I (19 pts) Spin 1 / 2: This problem is about a spin 1 / 2 particle. The eigenstates of S 2 and S z are labelled χ ± with: χ + = 1 0 ; χ - = 0 1 ; S z = ¯ h 2 1 0 0 - 1 ; S x = ¯ h 2 0 1 1 0 ; S y = ¯ h 2 0 - i i 0 The spin is placed in a uniform magnetic field, B = B 0 ˆ z , giving a Hamiltonian ˆ H = H (0) = - γB 0 S z . The eigenstates of H (0) are then χ ± with eigenvalues E ± = γB 0 ¯ h/ 2. A small magnetic field b = b ˆ y is applied in addition to B , making the total Hamiltonian H = H (0) + H with H = - γbS y . Use perturbation theory to find: 1.) (4 pts) The first order energy corrections to E + and E - . Solution: Because the two states, χ ± , have different energies, we use non-degenerate perturbation theory: E (1) ± = χ ± | H | χ ± . E (1) + = χ + | - γbS y | χ + = - γb ¯ h 2 ( 1 0 ) 0 - i i 0 1 0 = - γb ¯ h 2 ( 1 0 ) 0 i = 0 E (1) - = χ - | - γbS y | χ - = - γb ¯ h 2 ( 0 1 ) 0 - i i 0 0 1 = - γb ¯ h 2 ( 0 1 ) - i 0 = 0 2.) (5 pts) The second order energy corrections to E + and E - . Solution: Because there are only two states, the sum over states, m = n , in the expression for E (2) n reduces to a single term: E (2) + = | χ - | H | χ + | 2 E + - E - = γ 2 b 2 ¯ h 2 4 ( 0 1 ) 0 - i i 0 1 0 2 - γB 0 ¯ h/ 2 - γB 0 ¯ h/ 2 = γ 2 b 2 ¯ h 2 4 ( 0 1 ) 0 i 2 - γB 0 ¯ h = - γb 2 ¯ h 4 B 0 E (2) - = | χ + | H | χ - | 2 E - - E + = γ 2 b 2 ¯ h 2 4 ( 1 0 ) 0 - i i 0 0 1 2 γB 0 ¯ h/ 2 - ( - γB 0 ¯ h/ 2) = γ 2 b 2 ¯ h 2 4 ( 1 0 ) - i 0 2 γB 0 ¯ h = γb 2 ¯ h 4 B 0 3.) (4 pts) The first order corrections to the wavefunctions χ + and χ - .

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