9478 / 2027
Quantum Physics overview

Lesson 4 of 7

Adding probability amplitudes

  • I can apply superposition to position wavefunctions to explain single-particle interference and standing solutions in a box.Syllabus

    Syllabus 9478, 19(g). Apply superposition to position wavefunctions to explain single-particle interference and standing solutions in a box.

Make a guess. You are not marked.

Single photons pass one at a time through a double slit to a sensitive screen. Where do the first few photons land?

  1. At scattered single points; a fringe pattern builds up after many detections.
  2. As two faint bright bands, one behind each slit, from the very start.
  3. As a faint copy of the complete fringe pattern, spread over the screen at once.
Show the answer

At scattered single points; a fringe pattern builds up after many detections.

Each photon is detected at one point. Where it lands is random, but the probability is greatest where the two amplitudes add. Many detections together show the bright and dark fringes.

Coherent contributions combine as probability amplitudes. Add them first, then square the result to find density. Adding the separate densities omits interference.

A wavefunction describes one particle's state. In a double-slit arrangement, the coherent alternatives contribute to the amplitude at a detector position. They are not two independently detected fractions of the electron.

For locally real amplitude contributions a and b:

Resulting density = (a + b)2
= a2 + b2 + 2ab

The 2ab term depends on the relative signs. The same separate squares can therefore produce different combined densities. For general amplitudes the rule is the modulus squared of their sum.

Worked local cancellation

A zero amplitude at one point

At one detector position, suppose the already weighted contributions are +0.300 and -0.300 mm-1/2:

a + b = 0
|a + b|2 = 0 mm-1

Adding their separate squares would give 0.180 mm-1 and miss the cancellation. These are local contributions, not separately normalised complete states. Zero density at an ideal point does not establish zero probability throughout a finite detector bin.

Check your understanding

At one detector position, two coherent, already weighted real amplitude contributions are +0.300 and -0.300 mm^(-1/2). What is the resulting probability density there?

  1. -0.090 mm-1.
  2. 0.180 mm-1, the sum of the separate squares.
  3. Zero at that point.
Show the answer

Zero at that point.

The total amplitude is +0.300 - 0.300 = 0, so its square is zero. This local statement does not establish zero probability throughout a finite detector bin.

Normalise a complete combined state

For a separate 4.00 nm interval, let L = 4.00 nm and define two normalised profiles, zero outside that interval:

u = √(2/L) sin(πx/L)
v = √(2/L) sin(2πx/L)

Each squared profile has area 1. The product uv has equal positive and negative areas, which cancel over the whole interval. Consequently the total squared area of u + v, or of u - v, is 2. Dividing the amplitude by √2 normalises either combined state:

ψ+ = (u + v)/√2
ψ- = (u - v)/√2

The factor works because of these particular normalised components and their cancelling product area. It is not an automatic coefficient for every pair of overlapping wavefunctions.

Start with two signed component amplitudes

Start with two signed component amplitudesTwo real normalized component amplitudes u and v share the interval zero to four nanometres and the same amplitude axis from minus one to plus one per square root nanometre. The solid first sine has one positive lobe. The dashed second sine has a positive first lobe, a node at two nanometres and a negative second lobe. Both coefficient magnitudes are square root of one half. Neither trace is a particle path.-1-0.500.5101234Amplitude / nm-1/2Position x / nmuv

The component profiles u and v each have squared area 1 and are orthogonal over this interval. Their relative signs matter when amplitudes are combined.

Combine amplitudes and normalize the state

Combine amplitudes and normalize the stateSolid psi-plus is u plus v divided by square root two; dashed psi-minus is u minus v divided by square root two. Their common real-amplitude axis runs from minus one to plus one per square root nanometre. The plus profile has an interior zero at eight thirds nanometres, and the minus profile at four thirds. Plus amplitude at three nanometres is negative, approximately minus0.146447. These are normalized state profiles at one specified instant, not stationary trajectories.-1-0.500.5101234Amplitude / nm-1/2Position x / nmψ+ψ-

The factor 1/sqrt(2) normalizes these particular orthogonal equal-weight combinations. Changing the relative sign changes the spatial density. A mixed-energy superposition generally evolves; this is one instant.

Square the combined amplitude to obtain density

Square the combined amplitude to obtain densityThe separate density axis runs from zero to one per nanometre. The solid plus-state and dashed minus-state densities are nonnegative and mirror images about two nanometres. Each has total area one. At one nanometre their values are 0.728553 and 0.0214466; at three nanometres these values are exchanged. Squaring the negative plus-amplitude lobe gives positive density. The quantity plotted is the square of the combined amplitude, not the sum of component squares.00.250.50.75101234Density / nm-1Position x / nm|ψ+|2|ψ-|2

Both full densities have area 1. Their relative sign information came from amplitude addition before squaring. A negative amplitude is not a negative probability.

The first panel shows u and v, the next their two normalised combinations, and the last the corresponding densities. All use x from 0 to 4 nm. Signed amplitude and nonnegative density have separate units and scales; the two final density areas are each 1.

At x = 1.00 nm, the plus-state density is about 0.728553 nm-1, while the minus-state density is about 0.0214466 nm-1. At 3.00 nm they swap. In particular, ψ+ there is -0.146447 nm-1/2, whose square is positive. At the midpoint both densities are 0.250 nm-1.

These are state profiles at a specified instant. A superposition of different energy states generally has a density that changes with time; it is not a stationary particle tracing the plotted curve.

Connect interference and standing shapes

Coherent alternative paths produce interference because their amplitudes combine. In a confined region, combining appropriate oppositely travelling amplitude components can produce standing spatial shapes that meet the boundaries. Both use amplitude superposition, but neither is obtained by adding probability densities.

The infinite-well model makes the boundary restriction explicit. Its standing wavefunction does not require a particle to have a definite classical leftward or rightward trajectory.