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 ² ; χ - = ± 0 1 ² ; S z = ¯ h 2 ± 10 0 - 1 ² ; S x = ¯ h 2 ± 01 ² ; S y = ¯ h 2 ± 0 - i i 0 ² The spin is placed in a uniform magnetic Feld, ± 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 ± = 0 ¯ h/ 2. A small magnetic Feld ± b = b ˆ y is applied in addition to ± B , making the total Hamiltonian H = H (0) + H ² with H ² = - γbS y . Use perturbation theory to Fnd: 1.) (4 pts) The Frst 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 ² | χ ± ² . E (1) + = ± χ + |- y | χ + ² = - γb ¯ h 2 (1 0) ± 0 - i i 0 ²± 1 0 ² = - ¯ h 2 ± 0 i ² =0 E (1) - = ± χ - y | χ - ² = - ¯ h 2 (0 1) ± 0 - i i 0 0 1 ² = - ¯ h 2 ± - i 0 ² 2.) (5 pts) The second order energy corrections to E + and E - . Solution: Because there are only two states, the sum over states, m ³ = n , in the expression for E (2) n reduces to a single term: E (2) + = χ - | H ² | χ + ²| 2 E + - E - = γ 2 b 2 ¯ h 2 4 ³ ³ ³ ± 0 - i i 0 1 0 ² ³ ³ ³ 2 - 0 ¯ h/ 2 - 0 ¯ h/ 2 = γ 2 b 2 ¯ h 2 4 ³ ³ ³ ± 0 i ² ³ ³ ³ 2 - 0 ¯ h = - 2 ¯ h 4 B 0 E (2) - = χ + | H ² | χ - ²| 2 E - - E + = γ 2 b 2 ¯ h 2 4 ³ ³ ³ ± 0 - i i 0 0 1 ² ³ ³ ³ 2 0 ¯ h/ 2 - ( - 0 ¯ h/ 2) = γ 2 b 2 ¯ h 2 4 ³ ³ ³ ± - i 0 ² ³ ³ ³ 2 0 ¯ h = 2 ¯ h 4 B 0 3.) (4 pts) The Frst order corrections to the wavefunctions χ + and χ - .
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solmx2_04 - 1 Physics 325 Solution to Second Midterm Exam...

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