Wavefunctions and position representation
Treat a wavefunction as coordinates of a pure state, with normalization, phase, and continuous probabilities interpreted precisely.
On this page
The question for today
What you’ll understand
- Distinguish an abstract state ray from its position wavefunction.
- Normalize a continuous state and integrate detector probabilities.
- Separate normalizable wavepackets from generalized position eigenvectors.
A wavefunction is a representation of a quantum state, not an additional physical object chosen independently of that state. For a spinless particle on a line, position labels turn the abstract state into a complex function. Its squared magnitude supplies a probability density only after the state is normalized. This representation makes spatial questions convenient while leaving room for other descriptions, such as momentum coordinates, that encode the same preparation.
A state and its coordinates are different objects
A nonzero Hilbert-space vector specifies a pure state only up to multiplication by a nonzero complex number; after normalization the remaining freedom is an overall phase. The physical pure state is this ray. Choosing a position representation gives the amplitude density . Multiplying the entire function by one constant phase leaves every prediction unchanged, while position-dependent phase changes generally alter the state and can change its momentum statistics.
The Hilbert space for a spinless particle on the full line is the space of square-integrable functions, with functions equal almost everywhere identified. Its inner product is an integral rather than a finite component sum. This means changing a function's value at one isolated point does not change its Hilbert-space vector. Internal spin would add component indices, and many-particle positions would label configuration space; neither extension turns the wavefunction into an ordinary probability distribution over classical trajectories.
Density becomes probability only after integration
The Born rule for an ideal position measurement assigns a detector region the integral of the squared amplitude over that region. A density can exceed one because its units are inverse length; the integral, not the height at a point, must lie between zero and one. For a continuous position distribution, the probability of any one exact position is zero even though intervals around it may have large probability. This does not mean that a detector cannot produce a finite-resolution reading.
Normalization fixes the magnitude of a multiplicative constant but leaves its global phase free. Change variables to a dimensionless coordinate before integrating when a length scale appears repeatedly. This exposes the units of the normalization constant and prevents a numerical answer from depending on whether length was entered in metres or nanometres. A probability calculated from an unnormalized trial function must be divided by the total norm, or the function must first be normalized.
An exact position ket is not a prepared wavepacket
The formal position ket corresponds to a delta distribution, not a square-integrable function. Its overlap with another position ket is delta normalized. These generalized eigenvectors are useful in integral resolutions of the identity and in defining representations, but they do not have ordinary unit norm. One must not square a delta distribution and announce a normalized position probability. Physical localized preparations use finite-width normalizable packets with finite-resolution measurement predictions.
Square integrability alone also does not place a function in every operator's domain. A state can be normalized while a derivative observable or an energy moment is undefined or divergent. Boundary regularity becomes essential when applying an unbounded momentum or Hamiltonian operator. This lesson uses normalization and position integrals only; additional dynamical assumptions must be checked before treating an arbitrary normalized shape as a finite-energy eigenstate of a particular potential.
Work it through
Normalize a confined spatial shape
On the interval 0<x<L, a particle has and no other position support. Choose C real and positive. Find C, the probability in the left quarter, and the mean position.
Put u=x/L. The squared norm is . Unit norm therefore gives ; multiplying by two powers of length leaves the required wavefunction units.
The cumulative probability to x=tL is . At t=1/4 this becomes . The interval is one quarter of the box but does not contain one quarter of the probability.
The density is invariant under x replaced by L−x, so its mean is L/2. This symmetry result agrees with integrating x times the normalized density and avoids unnecessary polynomial arithmetic.
The normalization is , the left-quarter probability is 53/512, and the mean position is L/2. No energy eigenstate claim was needed for these position predictions.
The living notebook
Explore the model
Interactive lab · vectors
Coordinates of a normalized real state
Use the vector as two real coefficients in a fixed orthonormal spatial-mode basis. Reset and set both controls to Angle 0° and Magnitude 1; the squared coefficients are 1 and 0. For a fresh comparison, Reset and set both controls to Angle 45° and Magnitude 1; each squared coefficient is 1/2. The norm stays one although its coordinates change.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forBasis coefficients describe a state relative to chosen modes. This real two-component geometry does not display a continuous position density, a detector region, or a complex wavefunction.
Pause & reason
Check your understanding
Your turn
Try it yourself
On the full line take with a>0 and C real positive. Normalize it and find the probability that |x|<a.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The density is even. Evaluate twice the integral over positive x, using the exponent −2x/a after squaring the amplitude.
Compare with the solution
The total norm is , giving . The desired probability is . Its dimensionless value does not depend on a because the detector interval scales with the packet width. The cusp at the origin is irrelevant to these position integrals but would need attention in a Hamiltonian-domain calculation.
Keep the idea
Take it with you
- A position wavefunction represents a pure-state ray after choosing position coordinates.
- Integrate squared amplitude over a region to obtain a probability.
- Delta-normalized position kets are generalized eigenvectors rather than normalizable preparations.
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