Normalisation and Expectation Values · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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One total chance and honest averages

Science · Quantum & Modern Physics · ages 19-20
Name ______________________   Date ____________
  1. Where ψ = 2i, the probability density |ψ|² equals 4.

    Circle one:   True   False

  2. What is the normalisation condition?

    • The integral of ψ itself over space equals one
    • The integral of |ψ|² over space equals one
    • The value of ψ at every position equals one
  3. Using dimensionless x, ψ = 1/√10 on [0, 10] and zero elsewhere. What is P(0 < X < 3)?

    • 0.03
    • 3.00
    • 0.30
  4. For a normalised one-dimensional state with finite position expectation, how is E[X] computed?

    • Integrate x times |ψ|² over its range
    • Integrate x times ψ over its full range
    • Choose the position of the highest density
  5. Using dimensionless x, ψ(x) = 3C on [0, 1], 4C on (1, 2], and zero elsewhere. What positive real C normalises ψ?

    Answer: ______________

  6. A normalised position density is even about zero, and ∫ |x|ρ(x) dx is finite. What is E[X]?

    • Zero, from cancelling contributions
    • The largest positive position allowed
    • The distance between the two outer peaks
  7. Using dimensionless position x, a normalised wavefunction is ψ(x) = √(3/8)x on [0, 2] and zero elsewhere. Find the expectation of position.

    Answer: ______________

  8. A student says the average position must be a spot where the particle can be found.

    Circle one:   True   False

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Answer key

For grown-ups. Fold this page away before handing over the rest.

One total chance and honest averages W1-mt_6Y3DZ8P1qm-s1

  1. True · The conjugate of 2i is -2i. Their product is (-2i)(2i) = -4i² = 4.
  2. The integral of |ψ|² over space equals one · Total probability must be one, so |Psi| squared integrates to one.
  3. 0.30 · The density is 1/10. The interval has width 3, so its probability is 3/10 = 0.30.
  4. Integrate x times |ψ|² over its range · Multiply each position x by its density |ψ(x)|² and integrate over the full range.
  5. 0.2 · The two intervals each have width 1. Total probability is 9C² + 16C² = 25C², so the positive normalising constant is 1/5 = 0.2.
  6. Zero, from cancelling contributions · Even density gives equal weights to x and -x. Their position contributions cancel.
  7. 1.5 · The density is 3x²/8. Its position expectation is ∫₀² 3x³/8 dx = 1.5.
  8. False · An average over trials can land where no single trial lands.
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