HOMEWORK 08

For a thermal electron beam, you need to consider the average of the probabilities over all possible orientations. You can use the average of the probabilities calculated in steps 1 to 6: Average Probability = (P1 + P2 + P3 + P4 + P5 + P6) / 6 Where P1 to P6 are the probabilities calculated for each case. If delta(θ) changes to 100, you can repeat the calculations using the new delta(θ) value in the probability calculations for steps 7 and 8. The probabilities will change accordingly. 2. 1. Express |up arrow ⟩ and |down arrow ⟩ in the y-polarized basis: |up arrow ⟩ = (1/√2)(|↑ ⟩ - i|↓ ⟩ ) |down arrow ⟩ = (1/√2)(|↑ ⟩ + i|↓ ⟩ ) 2. Rewrite the |S=0 ⟩ state: |S=0 ⟩ = (1/√2)(|up arrow down arrow ⟩ - |down arrow up arrow ⟩ ) 3. Substitute the expressions for |up arrow ⟩ and |down arrow ⟩ : |S=0 ⟩ = (1/√2)[(1/√2)(|↑ ⟩ - i|↓ ⟩ )(1/√2)(|↑ ⟩ + i|↓ ⟩ ) - (1/√2)(|↑ ⟩ + i|↓ ⟩ )(1/√2)(|↑ ⟩ - i|↓ ⟩ )] 4. Now, distribute and simplify: |S=0 ⟩ = (1/√2)[(1/2)(|↑ ⟩⋅ |↑ ⟩ - i|↑ ⟩⋅ |↓ ⟩ - i|↓ ⟩⋅ |↑ ⟩ + i²|↓ ⟩⋅ |↓ ⟩ ) - (1/2)(|↑ ⟩⋅ |↑ ⟩ - i|↑ ⟩⋅ |↓ ⟩ - i|↓ ⟩⋅ |↑ ⟩ + i²|↓ ⟩⋅ |↓ ⟩ )] 5. Use the fact that |↑ ⟩⋅ |↑ ⟩ = 1, |↓ ⟩⋅ |↓ ⟩ = 1, and |↑ ⟩⋅ |↓ ⟩ = 0: |S=0 ⟩ = (1/√2)[(1/2)(1 - i(0) - i(0) + i²(1)) - (1/2)(1 - i(0) - i(0) + i²(1))] 6. Further simplify: |S=0 ⟩ = (1/√2)[(1/2)(1 - 0 - 0 - 1) - (1/2)(1 - 0 - 0 - 1)] 7. Calculate the final expression: |S=0 ⟩ = (1/√2)[(1/2)(-1) - (1/2)(-1)] |S=0 ⟩ = (1/√2)(-1/2 + 1/2)