Normalisation and Expectation Values
Total probability must be one, which fixes the constant in front of any wavefunction, and averages of position or momentum are integrals weighted by the probability density. Expectation values are what a long run of measurements would average to.
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
The wave function is a chance map: |Psi| squared at x is the probability density there, so the chance of finding the particle between x and x plus dx is |Psi| squared times dx. Over a wider stretch you integrate the density, which is the area under its curve. The particle must be somewhere, so the integral over all space has to equal one.
Picture a ball equally likely anywhere in a tube of length 4 m. Its wave function is a constant C inside and zero outside, and normalising gives C squared times 4 equals one, so C is 0.5. The chance of finding it in the first half is 0.25 times 2, which is 0.5, exactly as you would guess.
An expectation value is the long-run average over many trials: for position, integrate psi-star times x times psi over the allowed range. A symmetric state centred at zero averages to zero, since left and right balance. The average need not be a value any single trial can return: a state split over two far spots can average to the middle, where the particle is never found.
Normalise before you average, or every prediction is off scale. Square first: it is |psi| squared, never psi itself, that weights the average. With complex functions use psi-star times psi for the density.
Normalising makes all chances sum to one, and expectation values average many trials with the squared wave function.
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