Normalisation and Expectation Values
A finite, nonzero wavefunction norm determines the magnitude of a normalising constant, but leaves its overall phase free. Position averages integrate x against the probability density. Other observables use their operators.
What a learner can do afterwards
- Normalises a given wavefunction over its allowed range
- Computes the expectation value of position for a stated state
- Explains why the expectation value need not be a value the measurement can return
1 · Read
Start with a candidate function u(x). Its squared norm is N = ∫ |u(x)|² dx over the full range. If 0 < N < ∞, set ψ(x) = u(x)/√N. Then ρ(x) = |ψ(x)|² has integral one. Interval probability is ∫ₐᵇ ρ(x) dx. Multiplying ψ by a constant of magnitude one changes its overall phase but leaves these probabilities unchanged.
Use dimensionless x. Let u(x) be 1 on [-1, 0], 2 on (0, 1], and zero elsewhere. Its squared norm is 1² × 1 + 2² × 1 = 5, so ψ = u/√5. The two interval probabilities are 1/5 and 4/5. On a constant-density interval, multiply its width by the density. For ψ = Cu, normalisation requires |C|²N = 1; choosing C positive gives C = 1/√N.
For a normalised one-dimensional state, position expectation is E[X] = ∫ xρ(x) dx = ∫ x|ψ(x)|² dx. Require ∫ |x||ψ(x)|² dx < ∞ for a finite mean. Multiply position by its density before integrating. This predicts the average across freshly prepared copies of the same state, not the result of one trial.
Complex conjugation reverses the sign of the imaginary part. The conjugate of ψ = a + ib is a - ib, so |ψ|² = (a - ib)(a + ib) = a² + b². For ψ = 3 + 4i, the density is 9 + 16 = 25. Using ψ² instead can give a complex or negative result. A density may exceed one; its integral over an interval, not its height, is the probability.
For a normalised density with finite ∫ |x|ρ(x) dx, E[X] = ∫ xρ(x) dx. If ρ(-x) = ρ(x), the contributions xρ(x) and -xρ(-x) cancel, so E[X] = 0. The mean need not be a possible location: equal probability in separated symmetric regions can balance in a gap where density is zero.
Normalisation makes ∫ |ψ|² dx = 1. For a finite position mean, integrate x|ψ|². Squared magnitude, not ψ², is the probability density.
Lesson and questions written for this dataset using standard probability-density and wavefunction-normalisation methods.
Readings and the overall-phase flashcard written for this dataset. The other flashcards were drafted from collected fact candidates whose individual sources were not recorded.
2 · Watch
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Where it sits
Optional connections
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.