HOMEWORK 08

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Dec 6, 2023

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HOMEWORK #08 The probability that an electron in the state |+,θ1 ⟩ will pass through an {SGθ2 -} polarizer and end up in a particular state |-,θ2 ⟩ is given by the square of the inner product (or overlap) between the two states. In this case, you want to find the probability that the electron will end up in the state |-,θ2 ⟩ after passing through the {SGθ2 -} polarizer. This probability, denoted as P(θ1 → θ2), can be calculated as: P(θ1 → θ2) = | ⟨ -,θ2|+,θ1 ⟩ |² Here, ⟨ -,θ2|+θ1 ⟩ represents the inner product (also known as the overlap) between the states |-,θ2 ⟩ and |+,θ1 ⟩ . To calculate this, you can substitute the expressions for these states: |-,θ2 ⟩ = sin(θ2/2 - π/4)|+z ⟩ - cos(θ2/2 - π/4)|-z ⟩ |+,θ1 ⟩ = sin(θ1/2 + π/4)|+z ⟩ + cos(θ1/2 + π/4)|-z ⟩ Now, take the inner product: ⟨ -,θ2|+,θ1 ⟩ = (sin(θ2/2 - π/4) ⟨ +z|+z ⟩ - cos(θ2/2 - π/4) ⟨ -z|+z ⟩ ) + (sin(θ1/2 + π/4) ⟨ +z|+z ⟩ + cos(θ1/2 + π/4) ⟨ -z|+z ⟩ ) Here, ⟨ +z|+z ⟩ is the inner product of the states |+z ⟩ with itself, which is equal to 1, and ⟨ -z|+z ⟩ is the inner product of the states |-z ⟩ and |+z ⟩ , which is equal to 0 because they are orthogonal. So, the expression simplifies to: ⟨ -,θ2|+,θ1 ⟩ = sin(θ2/2 - π/4) + sin(θ1/2 + π/4) Now, square this inner product to find the probability: P(θ1 → θ2) = |sin(θ2/2 - π/4) + sin(θ1/2 + π/4)|² This is the probability that the electron will pass through the {SGθ2 -} polarizer and end up in the state |-,θ2 ⟩ after exiting the {SGθ1 +} polarizer. 1B. Let's calculate the probabilities for each of the given cases: 1. | ⟨ -,π/2|+,π/3 ⟩ |²: - Plug in the values: | ⟨ -,π/2|+,π/3 ⟩ |² = |sin(π/2/2 - π/4) + sin(π/3/2 + π/4)|² | ⟨ -,π/2|+,π/3 ⟩ |² = |sin(π/4) + sin(π/6 + π/4)|² - Calculate the values and square them:

| ⟨ -,π/2|+,π/3 ⟩ |² ≈ |0.7071 + 0.8660|² ≈ |1.5731|² ≈ 2.4754 2. | ⟨ +,π/6|+,π/3 ⟩ |²: - Plug in the values: | ⟨ +,π/6|+,π/3 ⟩ |² = |sin(π/6/2 + π/4) + sin(π/3/2 + π/4)|² | ⟨ +,π/6|+,π/3 ⟩ |² ≈ |0.7071 + 0.8660|² ≈ |1.5731|² ≈ 2.4754 3. | ⟨ -,π/2|-,0.333 ⟩ |²: - Plug in the values: | ⟨ -,π/2|-,0.333 ⟩ |² = |sin(π/2/2 - π/4) - sin(0.333/2 + π/4)|² | ⟨ -,π/2|-,0.333 ⟩ |² ≈ |0.7071 - 0.0833|² ≈ |0.6238|² ≈ 0.3892 4. | ⟨ +,2|+,π/3 ⟩ |²: - Plug in the values: | ⟨ +,2|+,π/3 ⟩ |² = |sin(2/2 + π/4) + sin(π/3/2 + π/4)|² | ⟨ +,2|+,π/3 ⟩ |² ≈ |0.7071 + 0.8660|² ≈ |1.5731|² ≈ 2.4754 5. | ⟨ -, -π/4|-,π/6 ⟩ |²: - Plug in the values: | ⟨ -, -π/4|-,π/6 ⟩ |² = |sin(-π/4/2 - π/4) - sin(π/6/2 + π/4)|² | ⟨ -, -π/4|-,π/6 ⟩ |² ≈ |-0.3536 - 0.0833|² ≈ |-0.4369|² ≈ 0.1905 6. | ⟨ -,π|+,π/3 ⟩ |²: - Plug in the values: | ⟨ -,π|+,π/3 ⟩ |² = |sin(π/2 - π/4) + sin(π/3/2 + π/4)|² | ⟨ -,π|+,π/3 ⟩ |² ≈ |0.9239 + 0.8660|² ≈ |1.7899|² ≈ 3.2010 Now, for the thermal electron beam percentages: 7. Chain of {SGθ1 +} {SGθ2 +} {SGθ3 +} where θ(n) = n * delta(θ), delta(θ) = π/4, and n = 3: To find the total probability, you can multiply the probabilities of each SG device in the chain: Total Probability = P(θ1 +) * P(θ2 +) * P(θ3 +) 8. Chain of {SGθ1 +} {SGθ2 +} {SGθ3 +} ... {SGθn-1 +} {SGθn +} where θ(n) = n * delta(θ), delta(θ) = π/4, and n = 20: Total Probability = P(θ1 +) * P(θ2 +) * P(θ3 +) * ... * P(θ19 +) * P(θ20 +) Now, let's calculate these probabilities for a thermal electron beam, assuming all orientations are randomly distributed:

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