Physics Lecture 18

Physics Lecture 18 - <|>| 2 C....

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I. Quantum Mechanical “States” A. State – a complete description of a system 1. Even if this is less knowledge than in the classical world 2. |ψ> - quantum state II. Superposition A. |ψ> = c 1 |φ> + c 2 |φ> 1. Multiply two vectors together to get a new vector 2. Example: Ben’s Daughter is either under the bed or in the closet a. |b> = under bed = 640 out of 1000 b. |c> = in closet = 360 out of 1000 i. |ψ> = m b |b> + m c |c> ii. Probability = .64 under bed : |m b | 2 = .64 iii. .64 = (.8) 2 = (-.8) 2 = (.8i)(-.8i) B. Particle-in-a-box 1. E 1 = h 2 /8mL 2 |1> 2. E 2 = 4h 2 /8mL 2 |2> 3. Consider: |ψ> = 1/sqrt(2) |1> + 1/sqrt(2) |2> III. A. Suppose we measure E and find E 2 then |ψ> = |2> B. Working With State Vectors 1. given |ψ>, Probability of a measurement of finding a particle in state |φ> is |
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Unformatted text preview: <|>| 2 C. Expectation (Average) Value 1. <A> = Prob(| 1 >)A 1 + Prob(| 2 >)A 2 + a. when |> = a 1 | 1 > + a 2 | 2 > + <A> = |a 2 | 2 A 1 + |a 2 |A 2 | + b. Particle-in-a-box i. E 1 = 3eV; E 2 = 12eV ii. |> = .8|1> + .6|2> iii. Measurement should be either 3 or 12 eV iv. Prob(3eV) = .64 v. (Expectation Value) <E> = .64(3) + .36(12) = 6.24eV vi. After finding E = 3eV => |> = 1 2. Expectation Values apply to classical world 3. Example a. Given State: | 1 > = 3/5| 1 > + i4/5| 2 > b. | 2 > = 4/5| 1 > + i3/5| 2 > c. < 2 | 1 > Superposition: c = c 1 |> + c 2 |> <| = c 1 *<| + c 2 *<| If |> is normalized and orthogonal, Then Prob{|>} = |c| 2 Normalized: < 2 | 1 > = 1 Orthogonal: <|> = 0...
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