9478 / 2027
Quantum Physics overview

Full chapter

Quantum Physics

All 7 topics and the revision summary on one page.

01

Evidence for photons

Light produces interference and diffraction patterns, but it also transfers energy in individual quanta. A successful explanation must account for both kinds of evidence.

In the photoelectric effect, electrons can leave a material when it absorbs suitable electromagnetic radiation. For a given surface there is a threshold frequency: below it, increasing ordinary illumination does not cause emission. Above it, emission is possible even with weak illumination.

A photon is one quantum of electromagnetic radiation. Its energy depends on frequency:

E = hf

Here h is the Planck constant, 6.63 × 10-34 J s. At fixed frequency, higher intensity means more incident photons per second for a fixed illuminated area. It does not increase each photon's energy.

Compare frequency and intensity separately

Consider a surface with supplied threshold f0 = 5.00 × 1014 Hz. Keep the surface, geometry and collection conditions unchanged. The comparisons below use the ordinary single-photon model and, above threshold, conditions without saturation.

Controlled photoelectric comparisons for the supplied surface
FrequencyIlluminationPhotoelectric result
4.00 × 1014 HzLower intensityNo emission: insufficient energy per photon.
4.00 × 1014 HzHigher intensityStill no emission in this model.
6.00 × 1014 HzLower intensityEmission is possible.
6.00 × 1014 HzHigher intensityThe emission rate can increase.

Below the threshold: greater intensity is insufficient

Below the threshold: greater intensity is insufficientThe same metal surface has threshold frequency five times ten to the fourteen hertz. Incident photons move right towards its exposed left surface at frequency four times ten to the fourteen hertz. No emitted electron or outgoing electron arrow is drawn. Increasing ordinary single-photon illumination at this unchanged frequency does not supply enough energy per photon. Photon and electron symbols and arrow lengths are schematic, not measured positions, sizes or rates.Incident f = 4.00 × 1014 HzSurface f0 = 5.00 × 1014 HzPhotonPhotonMetalNo photoelectric emission

Compare frequencies while holding the material and collection conditions fixed. The arrows show incoming photons and, where possible, outgoing electrons. Icon counts do not specify an efficiency or a measured rate.

Above the threshold: emission is possible

Above the threshold: emission is possibleThe same metal surface has threshold frequency five times ten to the fourteen hertz. Incident photons move right towards its exposed left surface at frequency six times ten to the fourteen hertz. One schematic electron is shown leaving the surface, moving up and left. Emission is possible even with weak ordinary illumination, but this icon count does not state an emission or collection efficiency. Photon and electron symbols and arrow lengths are schematic, not measured positions, sizes or rates.Incident f = 6.00 × 1014 HzSurface f0 = 5.00 × 1014 HzPhotonPhotonElectronMetalEmission is possible

Compare frequencies while holding the material and collection conditions fixed. The arrows show incoming photons and, where possible, outgoing electrons. Icon counts do not specify an efficiency or a measured rate.

Incoming photons and possible emitted electrons have different arrows. The supplied frequency determines whether the threshold condition is met; icon size, colour and number are schematic and do not specify an emission efficiency.

The threshold supports quantised energy exchange: adding more photons whose individual energies are too small does not meet the single-photon condition. This does not remove the wave evidence from interference and diffraction. Nor does it claim that one ordinary single-photon model covers arbitrarily intense fields.

The work function is an energy

The work function, written here as Φwork, is the minimum energy needed to remove an electron from the specified surface. At threshold:

Φwork = hf0
= (6.63 × 10-34)(5.00 × 1014)
= 3.315 × 10-19 J
= 2.07 eV

One electronvolt is the energy gained by a particle of charge magnitude e accelerated through a potential difference of 1 V. Thus 1 eV = e × 1 V = 1.60 × 10-19 J using the supplied elementary charge. An eV is an energy unit, not a voltage. Φwork is measured in joules; it is distinct from magnetic flux Φ measured in webers.

Worked photon calculation

Energy, wavelength and momentum

An optical frequency of order 5 × 1014 Hz with h of order 7 × 10-34 J s suggests a few times 10-19 J per photon: a few eV. Expect a wavelength of a few hundred nanometres and momentum of order 10-27 kg m/s.

For f = 6.00 × 1014 Hz, use c = 3.00 × 108 m/s in vacuum:

λ = c/f = 5.00 × 10-7 m = 500 nm
E = hf = 3.978 × 10-19 J
E = (3.978 × 10-19)/(1.60 × 10-19)
= 2.48625 eV ≈ 2.49 eV

A photon has zero rest mass but nonzero momentum. Its momentum points along propagation, with magnitude:

p = E/c = h/λ
= 1.326 × 10-27 kg m/s
≈ 1.33 × 10-27 N s

The non-relativistic formula p = mv for a massive particle is not a way to set photon momentum to zero. Check that energy divided by speed has momentum units.

A monochromatic beam with supplied power 1.989 µW at this frequency carries P/E = 5.00 × 1012 photons/s. That incident photon rate is not automatically a recorded electron rate: absorption, emission and collection efficiencies matter.

Optional check A surface has threshold frequency 5.00 x 10^14 Hz. In the ordinary single-photon model, what happens if the intensity at 4.00 x 10^14 Hz increases while frequency and other conditions stay fixed?
A surface has threshold frequency 5.00 x 10^14 Hz. In the ordinary single-photon model, what happens if the intensity at 4.00 x 10^14 Hz increases while frequency and other conditions stay fixed?

02

Matter waves and detections

An electron can produce a localised detection while an accumulated pattern displays wave behaviour. The detection position and the state used to predict its distribution are different parts of the explanation.

Two distinct observations

In an electron-diffraction demonstration, a beam encounters a thin crystalline specimen and produces a diffraction pattern at a detector. Its ordered structure supplies a suitable spatial scale. Thin polycrystalline graphite can produce rings; a ring pattern is not universal to every crystal and orientation.

In a single-particle double-slit experiment, individual detections accumulate into an interference pattern when the alternatives remain coherent. A sufficiently weak beam still builds the pattern one detection at a time. Collisions between neighbouring electrons are not required to explain it.

Electron diffraction from polycrystalline graphite

Electron diffraction from polycrystalline graphiteThe apparatus arrangement has an electron source, a thin polycrystalline graphite specimen and a detector. A beam travels from the source towards the specimen. Pale dashed guides locate the detector region; they are not measured single-electron trajectories. A separate face-on detector view below shows a central spot and two schematic rings. The specimen is explicitly polycrystalline graphite; rings are not the universal result for every crystal. The ring radii are qualitative, with no numerical diffraction calculation or claimed experimental data.SourceDetectorThin polycrystallinegraphiteDetector viewed face-onRing radii are schematic.

The detector pattern is evidence of electron diffraction. This is a schematic of the arrangement and pattern type; the ring spacing is not supplied data. The enlarged face view uses its own drawing scale.

Two openings and a position-sensitive detector

Two openings and a position-sensitive detectorA low-rate single-particle source faces two distinct openings in an opaque barrier. The openings are between y110 and126 and between y149 and165; solid barrier segments separate them. Pale dashed arrangement guides run from the source to the openings and from the openings towards the detector; no drawn curve is a measured particle path. A single filled point on the detector represents one localised detection. It does not indicate which opening was used, a half-charge particle at each opening, or a classical path from the source. The following count maps enlarge detector positions independently of this schematic.SourceLow rateTwo openingsDetector12One localised detectionGuides show the arrangement,not measured particle trajectories.

A detection records a position. Repeating the coherent two-opening experiment builds a spatial distribution. Sparse particles can still produce an interference pattern; neighbouring particles do not need to collide.

The crystal diffraction arrangement and the single-particle two-slit arrangement show different experiments. The ring positions and apparatus geometry are schematic. Guides locate source, specimen or openings, and detector; they are not measured electron trajectories.

A detection dot tells us where an interaction was recorded. It does not reveal a definite classical path between the source and detector. An electron is not a tiny ball following a sine curve or two half-charge balls passing separately through the slits. Obtaining which-path information changes the experimental conditions; simply looking at an already recorded screen does not cause that change.

Worked de Broglie wavelength

Choose the momentum model first

For a particle with momentum magnitude p, the de Broglie wavelength is:

λ = h/p

An electron of mass about 10-30 kg moving at about 106 m/s has momentum of order 10-24 kg m/s and wavelength of order 1 nm. Its speed is well below c.

Use me = 9.11 × 10-31 kg, h = 6.63 × 10-34 J s and v = 1.50 × 106 m/s. Here v/c = 0.00500, so the non-relativistic p = mv approximation is suitable:

p = mev = 1.3665 × 10-24 kg m/s
λ = (6.63 × 10-34)/(1.3665 × 10-24)
= 4.85181 × 10-10 m ≈ 0.485 nm

Doubling v doubles p and halves λ to 0.243 nm. Kinetic energy instead scales as v2 and quadruples. Initially it is (1/2)mev2 = 1.024875 × 10-18 J, or about 6.41 eV. Do not use the photon's E = pc as this electron's kinetic-energy equation.

Optional check An electron at 1.50 x 10^6 m/s has de Broglie wavelength 0.485 nm. Its speed doubles and the non-relativistic approximation remains valid. What is the new wavelength?
An electron at 1.50 x 10^6 m/s has de Broglie wavelength 0.485 nm. Its speed doubles and the non-relativistic approximation remains valid. What is the new wavelength?

Interpret a finite collection of detections

The following constructed illustrative counts use nine detector-position bins of width 1.0 mm. Centres run from -4 to +4 mm, with edges from -4.5 to +4.5 mm. They illustrate accumulation; they are not a reported experiment or an exact diffraction calculation.

First 20 detections

First 20 detectionsThis is a constructed count illustration, not experimental observations. Nine equal one-millimetre bins have centres minus4 to plus4 millimetres and edges minus4.5 to plus4.5. The counts from left to right are 1, 1, 3, 0, 10, 1, 3, 0, 1, totalling 20. Every dot lies inside its specified bin. All twenty positions recur exactly in the later panel. Vertical positions are schematic; only the horizontal detector coordinate has a physical scale.Constructed illustrative detections-4-3-2-101234Detector position / mmBin edges run from -4.5 to +4.5 mm.Vertical placement is schematic.

The small sample gives limited information about the distribution. An empty early bin does not establish zero underlying probability.

The same constructed run after 200 detections

The same constructed run after 200 detectionsThis is a constructed count illustration, not experimental observations. Nine equal one-millimetre bins have centres minus4 to plus4 millimetres and edges minus4.5 to plus4.5. The counts from left to right are 10, 4, 34, 8, 88, 8, 34, 4, 10, totalling 200. Every dot lies inside its specified bin. The twenty earlier dots remain at the exact same coordinates as a subset of these two hundred. Vertical positions are schematic; only the horizontal detector coordinate has a physical scale.Constructed illustrative detections-4-3-2-101234Detector position / mmBin edges run from -4.5 to +4.5 mm.Vertical placement is schematic.

The count distribution becomes clearer after more detections. These are constructed illustrative counts with fixed positions, not a fitted diffraction model or authentic measurement record.

Count per equal-width bin after 200 detections

Count per equal-width bin after 200 detectionsThe histogram uses the same nine one-millimetre bin boundaries as the detector maps. Bar heights are 10,4,34,8,88,8,34,4,10 counts on a linear count axis from zero to one hundred. Adjacent bars have equal width and no inferred smooth curve. The central bar covers minus0.5 to plus0.5 millimetre and has eighty-eight counts, giving relative frequency 88/200 equals0.44 for that finite interval. It is not the probability of one exact detector coordinate.02040608010010-44-334-28-18808134243104Count in each 1.0 mm binDetector position / mm

The middle bin contains 88 of 200 detections: relative frequency 0.44 over -0.5 to +0.5 mm. Histogram height here is a count, not probability density. A density estimate would also divide the relative frequency by the bin width.

The first 20 positions remain a subset of the same constructed run after 200 detections. Horizontal positions lie in the specified bins; vertical placement is schematic. Histogram height is count per equal-width bin, and the geometry does not supply a wavelength measurement.
Constructed counts in the same nine detector bins
Centre / mmFirst 20After 200
-4110
-314
-2334
-108
01088
118
2334
304
4110

The middle bin contains 88/200 = 0.44 of the later detections. This estimates probability over the finite interval -0.5 to +0.5 mm, not probability at exactly x = 0. An empty bin in the smaller sample does not establish zero underlying probability there.

Count-data exercise

Keep bin probability separate from density

Download the constructed illustrative count CSV. Import it as comma-delimited numeric columns, preserving raw A2:E10. The headers are centre, left edge and right edge in mm, then counts after 20 and 200 detections. The spreadsheet workflow gives the entry and import steps.

Add F1 = width / mm, G1 = early relative frequency, H1 = later relative frequency and I1 = later density estimate / mm^-1. Enter:

First formulas for the count CSV after import
CellFormula
F2=C2-B2
G2=D2/SUM($D$2:$D$10)
H2=E2/SUM($E$2:$E$10)
I2=H2/F2

Fill F2:I2 through row 10. Check that the count totals are 20 and 200, both relative-frequency columns sum to 1, and every later count is at least its earlier count. Use bars over the actual bin edges. These supplied values do not determine a unique fitted sine curve.

Interpret the middle bin

Its later relative frequency is 0.44. Dividing by its 1.0 mm width gives a bin-average density estimate of 0.44 mm-1. Multiplying that density estimate by the same width returns the dimensionless bin probability estimate. The coordinate unit here is mm, separate from the nm model on the probability page.

03

Wavefunctions and probability

A wavefunction represents a particle's state. Its amplitude is not a displacement of matter. Squared amplitude gives a probability density; area under that density gives an interval probability.

For a real wavefunction ψ, density is ψ2. More generally it is the modulus squared, |ψ|2. A negative real amplitude is allowed, but its squared density is nonnegative. Reversing the sign of the whole real wavefunction leaves its density unchanged; relative signs matter when combining contributions.

In one dimension, the probability of a position interval is the area under |ψ|2 over that interval. Normalisation sets the total area over all possible positions to 1. A density value alone is not a probability: it has an inverse-length unit and can be numerically greater than 1 without contradiction.

Matching coordinate, amplitude and density units in one dimension
Coordinate xAmplitude ψDensity |ψ|2
mm-1/2m-1
nmnm-1/2nm-1

Density multiplied by a width in the matching unit is dimensionless. An atomic electron cloud instead represents a three-dimensional spatial probability distribution, not a classical orbit. Its full volume density has unit m-3, and probability comes from a volume, rather than a one-dimensional area under a curve.

Square-profile normalisation

Determine the height before finding an interval area

Let ψ = C for 0 < x < 4.00 nm, with zero amplitude outside. For this constant profile:

Total probability = C2(4.00 nm) = 1
C = 1/√(4.00 nm) = 0.500 nm-1/2
|ψ|2 = 0.250 nm-1

Between 1.00 and 2.50 nm the width is 1.50 nm, giving:

P = (0.250 nm-1)(1.50 nm) = 0.375

Square the amplitude first, then find the density area. Squaring the area under ψ gives a different result with the wrong units. Choosing C = -0.500 nm-1/2 would give the same probabilities.

In metre units the same amplitude is 1.58114 × 104 m-1/2, and density is 2.50 × 108 m-1. The interval width is then 1.50 × 10-9 m, so its probability remains 0.375.

This abrupt square profile is an ideal normalisation exercise. The values at its isolated endpoints do not change its area. It is not an infinite-well stationary state.

Optional check A normalised real wavefunction is 0.500 nm^(-1/2) for 0 < x < 4.00 nm and zero outside. What is the probability of finding the particle between 1.00 and 2.50 nm?
A normalised real wavefunction is 0.500 nm^(-1/2) for 0 < x < 4.00 nm and zero outside. What is the probability of finding the particle between 1.00 and 2.50 nm?

Sinusoidal-profile normalisation

Use the mean square over the whole interval

Now let ψ = C sin(πx/L) within 0 < x < L, where L = 4.00 nm, and zero outside. The mean of sin2 over this half-wave is 1/2:

Total probability = C2L/2 = 1
C = √(2/L) = 0.707107 nm-1/2
|ψ|2 = 0.500 sin2(πx/4.00) nm-1

The last expression uses x in nm, making πx/4.00 a dimensionless angle in radians. Symmetry gives probability 0.500 in each half of the interval. The central half, 1.00 to 3.00 nm, contains more probability than the combined two outer quarters because its density is higher on average.

The different coefficient from the square profile is essential. Using 1/√L here would leave total probability 1/2, not 1.

Optional check For 0 < x < L, psi = C sin(pi x/L), with L = 4.00 nm and zero amplitude outside. The mean of sin^2 over this interval is 1/2. Which positive C normalises the state?
For 0 < x < L, psi = C sin(pi x/L), with L = 4.00 nm and zero amplitude outside. The mean of sin^2 over this interval is 1/2. Which positive C normalises the state?

Square profile: signed amplitude

Square profile: signed amplitudeThe amplitude is 0.500 per square root nanometre strictly between zero and four nanometres, and zero outside. This panel is amplitude in inverse square root nanometres, with maximum 0.500. No region under the amplitude is shaded as probability. All four profile plots have the same x mapping, with zero and four nanometres at eighty and three hundred drawing units. Pale outside regions show the zero profile beyond this support. Separate constant segments and dashed jump guides show the discontinuities; the vertical guides are not additional function values.00.250.500.7501234ψ / nm-1/2Position x / nmZero outside 0 < x < 4 nm

This discontinuous square profile is an ideal normalization exercise, not an infinite-well stationary state. Values at single endpoints do not change an interval area.

Square profile: density and interval probability

Square profile: density and interval probabilityThe amplitude is 0.500 per square root nanometre strictly between zero and four nanometres, and zero outside. This panel squares the real amplitude to obtain density in inverse nanometres. Its maximum is 0.250 per nanometre. The shaded interval is one to two and a half nanometres, with dimensionless probability 0.375. All four profile plots have the same x mapping, with zero and four nanometres at eighty and three hundred drawing units. Pale outside regions show the zero profile beyond this support. Separate constant segments and dashed jump guides show the discontinuities; the vertical guides are not additional function values.00.250.5001234|ψ|2 / nm-1Position x / nmShaded: 1 to 2.5 nm

Shaded width: 1.50 nm. Density: 0.250 per nm. Their product is probability 0.375, with no unit. The total width of 4.00 nm gives total probability 1.

Sinusoidal profile: signed amplitude

Sinusoidal profile: signed amplitudeThe amplitude is square root of one half times sine of pi x over four, with x in nanometres. This panel is amplitude in inverse square root nanometres, with maximum 0.70710678. No region under the amplitude is shaded as probability. All four profile plots have the same x mapping, with zero and four nanometres at eighty and three hundred drawing units. Pale outside regions show the zero profile beyond this support. The plotted curve is sampled densely from the exact sine or squared-sine function.00.250.500.7501234ψ / nm-1/2Position x / nmZero outside 0 < x < 4 nm

The coefficient is sqrt(2/L), with L = 4.00 nm. Amplitude is not a material displacement, and area under this amplitude graph is not probability.

Sinusoidal profile: density and interval probability

Sinusoidal profile: density and interval probabilityThe amplitude is square root of one half times sine of pi x over four, with x in nanometres. This panel squares the real amplitude to obtain density in inverse nanometres. Its maximum is 0.500 per nanometre. The shaded interval is one to three nanometres, with dimensionless probability 0.818309886. All four profile plots have the same x mapping, with zero and four nanometres at eighty and three hundred drawing units. Pale outside regions show the zero profile beyond this support. The plotted curve is sampled densely from the exact sine or squared-sine function.00.250.5001234|ψ|2 / nm-1Position x / nmShaded: 1 to 3 nm

Only area under density gives an interval probability. The central 1-3 nm area is 0.818309886...; the total density area is 1. The shaded area is not the square of the area under amplitude.

Amplitude and density use the same position scale but different vertical quantities and units. Only density areas are shaded: 1.00-2.50 nm for the square profile and 1.00-3.00 nm for the sine profile. Both complete density areas are 1.

Probability-area workbook

Generate the sine model and refine its area estimate

Download the probability starter workbook. Its Coarse, Fine and Finer sheets contain coordinates for this generated 4.00 nm model, not experimental observations. The calculation cells are blank for you to complete.

All three sheets use A for x / nm, B for ψ / nm-1/2, C for density / nm-1 and D for interval probability. G2 contains L = 4; I2 contains the coordinate step. H2 is the blank normalisation coefficient. Enter these first formulas on each sheet:

First formulas for the supplied sine-profile sheets
CellFormula
H2=SQRT(2/$G$2)
B2=$H$2*SIN(PI()*A2/$G$2)
C2=B2^2
D2=(A3-A2)*(C2+C3)/2

Keep the fixed references when copying. B and C have a value at every coordinate. D is the trapezium area between two consecutive coordinates, so its last formula belongs one row earlier:

Coordinate and formula boundaries in the three starter sheets
Sheet and stepCoordinates; B/C fillD interval fill
Coarse: 0.5 nmA2:A10; B2:C10D2:D9
Fine: 0.25 nmA2:A18; B2:C18D2:D17
Finer: 0.125 nmA2:A34; B2:C34D2:D33

Plot B against numeric A and C against numeric A in separate XY graphs, with both vertical quantities and units labelled. The sine argument is in radians. For area, density in nm-1 must multiply widths in nm; mixing an unchanged nm-1 density with metre widths would change the answer incorrectly.

Use a blank cell for each of the following sums. Preserve the stored values while formatting display precision.

Full and central-interval probability sums
SheetProbability sums
CoarseFull: =SUM(D2:D9)
1-3 nm: =SUM(D4:D7)
FineFull: =SUM(D2:D17)
1-3 nm: =SUM(D6:D13)
FinerFull: =SUM(D2:D33)
1-3 nm: =SUM(D10:D25)
Compare the three area estimates

The central-interval estimates are 0.8017766953, 0.8142087183 and 0.8172865746, respectively. Compare them with the supplied exact model area 1/2 + 1/π = 0.8183098862. The shortfalls are about 0.0165332, 0.00410117 and 0.00102331.

The full-range sums happen to equal 1, apart from numerical round-off, on all three unrounded regular grids. That coincidence does not make each partial area exact. The exact smooth density and the straight-sided trapezia are different representations.

Refining the step reduces a numerical approximation error here. It is distinct from experimental uncertainty and from the quantum position-momentum width relation. Every interval probability must stay between 0 and 1; a dimensional density has no such numerical upper limit.

04

Adding probability amplitudes

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.

Optional check 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?
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?

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.

05

Localisation and momentum spread

A sharply localised state needs a spread of momentum components. This is a property of the state, not merely poor measuring technique or an uncertainty about one definite classical speed.

A single perfectly defined travelling-wave momentum does not form a localised packet. Combining momentum components can localise the state, with a narrower position distribution requiring a broader momentum distribution along the same axis.

Δx Δp ≳ h

Here Δx and Δp are characteristic width scales, and the relation means greater than or of the order of h. Use it for an order-of-magnitude estimate. It is not an exact equality and does not require a state to attain a minimum product. It is also not an instrument specification.

Compare position widths on one common scale

Compare position widths on one common scaleTwo normalized Gaussian position densities have the same mean of zero. The horizontal axis is X equals position divided by reference width a, from minus four to plus four. The vertical axis is a times position density, so both axes are dimensionless. Solid state A has unit width and peak 0.398942 on these relative axes. Dashed state B has width 0.5 and peak 0.797885. Both curves use exactly the same horizontal and vertical scales within the panel. Their full tails continue beyond the displayed window and their full areas are one. The corresponding states use widths a and a over two in position, and b and two b in momentum.00.20.40.60.8-4-2024a × position density(dimensionless)Scaled position X = x/aState AState B

State B has half the position width of A and twice its peak density. Both mean positions are zero. These relative Gaussian shapes illustrate corresponding spreads; they do not replace the separate syllabus order-of-magnitude estimate.

The narrower-position state has a broader momentum distribution

The narrower-position state has a broader momentum distributionTwo normalized Gaussian momentum densities have the same mean of zero. The horizontal axis is P equals momentum divided by reference width b, from minus six to plus six. The vertical axis is b times momentum density, so both axes are dimensionless. Solid state A has unit width and peak 0.398942 on these relative axes. Dashed state B has width 2 and peak 0.199471. Both curves use exactly the same horizontal and vertical scales within the panel. Their full tails continue beyond the displayed window and their full areas are one. The corresponding states use widths a and a over two in position, and b and two b in momentum.00.20.40.60.8-6-3036b × momentum density(dimensionless)Scaled momentum P = p/bState AState B

Follow the same state labels: B is twice as broad in momentum, although both mean momenta remain zero. Its peak is half as high so the full density area remains 1. The tails continue beyond this crop.

Both states have the same mean position and the same mean momentum. On a common scale, halving the position width accompanies doubling the momentum width. Each complete density has area 1 and continuing tails; the vertical axes are scaled densities, not probabilities at a single point.

The momentum spread describes the distribution of possible results for similarly prepared states. Its mean can remain zero while its width is large. Reducing ruler or detector errors does not remove this state-level relationship.

Worked width-scale estimate

Estimate momentum spread before a velocity scale

Take an electron localised on the scale Δx = 2.0 × 10-10 m. With h = 6.63 × 10-34 J s:

Characteristic required Δp is of order h/Δx
= (6.63 × 10-34)/(2.0 × 10-10)
= 3.315 × 10-24 kg m/s
≈ 3.3 × 10-24 kg m/s

Using the non-relativistic electron mass me = 9.11 × 10-31 kg gives a corresponding velocity-spread scale:

Δv of order Δp/me
= 3.64 × 106 m/s

This is a spread scale, not a measured speed or a required mean velocity. Halving Δx to 1.0 × 10-10 m doubles the characteristic required momentum and velocity spreads under the same convention. It does not require the mean momentum to double.

Why the width convention matters

The estimate above uses characteristic widths and suppresses numerical factors. If position and momentum widths are defined specifically as standard deviations σx and σp, their precise lower bound is σxσp ≥ h/(4π). Do not substitute standard-deviation numbers into the rough-width form while treating it as an exact theorem.

Optional check Using the same order-of-magnitude position-momentum width convention, a particle is localised to half its previous position-width scale. How does the characteristic required momentum-spread scale change?
Using the same order-of-magnitude position-momentum width convention, a particle is localised to half its previous position-width scale. How does the characteristic required momentum-spread scale change?

06

A particle in a box

Confinement restricts the wavefunctions that can satisfy the boundaries. In a one-dimensional infinite well, only certain standing shapes and corresponding energies are allowed.

Define the well and its boundaries

The potential energy V is zero inside 0 < x < L and is an ideal infinite barrier outside. Here V means potential energy, not voltage. The particle cannot penetrate the barrier, so its wavefunction vanishes outside and at x = 0 and x = L.

A nonzero standing shape must fit n half-wavelengths between the walls:

L = nλn/2
λn = 2L/n
ψn = √(2/L) sin(nπx/L)
n = 1, 2, 3, ...

The coefficient normalises each state. An n = 0 sine profile would be zero everywhere and could not have total probability 1. Each state has n - 1 interior nodes, in addition to its two boundary nodes.

n = 1: no interior node

n = 1: no interior nodeTwo aligned plots use the same dimensionless position X equals x over L, with walls at zero and one. The upper dimensionless amplitude is square root L times psi-n, equal to square root two sine of 1 pi X. Its full-scale extrema are plus and minus square root two. The lower dimensionless density is L times modulus psi-n squared, equal to two sine squared of 1 pi X; its maxima are two and it is never negative. Both vanish at the boundaries and outside. There is no interior node. Filled markers at interior nodes mark graph zeros, not particle detections. Each lower curve has area one against X.√L ψ1 (dimensionless)-√20+√201/21X = x/LL|ψ1|2 (dimensionless)01201/21X = x/L

The upper curve is a signed state amplitude, not a moving particle or a material displacement. Its square gives the lower density. Each stationary-state density has total area 1 and remains fixed in time; it is not multiplied by an oscillating probability factor.

n = 2: 1 interior node

n = 2: 1 interior nodeTwo aligned plots use the same dimensionless position X equals x over L, with walls at zero and one. The upper dimensionless amplitude is square root L times psi-n, equal to square root two sine of 2 pi X. Its full-scale extrema are plus and minus square root two. The lower dimensionless density is L times modulus psi-n squared, equal to two sine squared of 2 pi X; its maxima are two and it is never negative. Both vanish at the boundaries and outside. An interior node at X one half is zero in both plots. Filled markers at interior nodes mark graph zeros, not particle detections. Each lower curve has area one against X.√L ψ2 (dimensionless)-√20+√201/21X = x/LL|ψ2|2 (dimensionless)01201/21X = x/L

The upper curve is a signed state amplitude, not a moving particle or a material displacement. Its square gives the lower density. Each stationary-state density has total area 1 and remains fixed in time; it is not multiplied by an oscillating probability factor.

n = 3: 2 interior nodes

n = 3: 2 interior nodesTwo aligned plots use the same dimensionless position X equals x over L, with walls at zero and one. The upper dimensionless amplitude is square root L times psi-n, equal to square root two sine of 3 pi X. Its full-scale extrema are plus and minus square root two. The lower dimensionless density is L times modulus psi-n squared, equal to two sine squared of 3 pi X; its maxima are two and it is never negative. Both vanish at the boundaries and outside. Interior nodes at X one third and two thirds are zero in both plots. Filled markers at interior nodes mark graph zeros, not particle detections. Each lower curve has area one against X.√L ψ3 (dimensionless)-√20+√201/32/31X = x/LL|ψ3|2 (dimensionless)01201/32/31X = x/L

The upper curve is a signed state amplitude, not a moving particle or a material displacement. Its square gives the lower density. Each stationary-state density has total area 1 and remains fixed in time; it is not multiplied by an oscillating probability factor.

All three states use X = x/L, with the same walls at 0 and 1. The upper axis is the scaled amplitude √L ψ; the lower is scaled density L|ψ|2. Their separate vertical labels matter. Each density has unit area against X, and all amplitude nodes are also density zeros.

The second state has an interior node at x/L = 1/2; the third has nodes at 1/3 and 2/3. Negative amplitude lobes still give positive density. These stationary states have time-independent position densities. Their full wavefunctions can change phase, but a time factor must not make total probability oscillate between zero and one.

Connect the allowed wavelength to energy

The spatial wavelength sets a momentum scale h/λn = hn/(2L). Combining this scale with the non-relativistic energy relationship gives the allowed well energies:

En = (hn/2L)2/(2m)
= h2n2/(8mL2)

This is an energy calculation for a confined standing state. It does not give one definite measured classical momentum, or restrict all momentum measurements to just two values. Confinement requires a momentum spread.

An infinite-well potential is an ideal energy model

An infinite-well potential is an ideal energy modelThe horizontal coordinate is X equals x over L. The impenetrable boundaries are X zero at drawing coordinate eighty and X one at three hundred, exactly the positions used in all subsequent state plots. Potential energy V is zero inside. Vertical wall arrows point upwards towards infinity; their drawn height is symbolic, not a finite barrier value. Shaded outside regions are inaccessible in this ideal model. No wavefunction penetration or tunnelling tail is shown. V denotes potential energy in joules, not electric potential in volts.Potential energy V / J∞∞0V = 001Scaled position X = x/LInfinite arrows are symbolic.

Inside: V = 0. Outside: the ideal potential energy is infinite. The wavefunction vanishes at both boundaries and outside. The electron example uses L = 1.00 nm, independently of the earlier 4.00 nm normalization profiles.

Allowed energies have the ratio 1 : 4 : 9

Allowed energies have the ratio 1 : 4 : 9The vertical axis is dimensionless E over E1 and uses one linear scale: y equals three hundred and sixty minus twenty-eight times E over E1. The n equals one, two and three levels are at relative energies one, four and nine, so their y coordinates are three hundred and thirty-two, two hundred and forty-eight, and one hundred and eight. The zero dashed baseline is a reference only, not an allowed n equals zero state. The n equals two to one gap is three E1 and the n equals three to two gap is five E1. Horizontal level lengths are not position or energy magnitudes.1n = 14n = 29n = 30E / E1 (dimensionless)Zero is a reference,not an allowed n = 0 state.

For the supplied 1.00 nm electron box, E1 is about 0.377 eV. The first three energies are about 0.377, 1.51 and 3.39 eV. The increasing gaps are genuine; the levels have not been evenly spaced for convenience.

The infinite barriers are symbolic potential-energy walls, with no tunnelling tail. The separate energy plot has a linear scale: allowed levels are E/E1 = 1, 4 and 9. Zero is the energy reference, not an allowed n = 0 state.

Worked electron well

Keep the width distinct from the earlier profiles

An electron confined across about 1 nm has a ground-energy scale of order 10-19 J, below a few eV. Doubling the width should reduce that energy scale by four.

Use L = 1.00 nm = 1.00 × 10-9 m, me = 9.11 × 10-31 kg and h = 6.63 × 10-34 J s. This is a new width, distinct from the 4.00 nm normalisation examples.

E1 = (6.63 × 10-34)2
/ [8(9.11 × 10-31)(1.00 × 10-9)2]
= 6.03141 × 10-20 J
= 0.376963 eV ≈ 0.377 eV
Allowed wavelengths and energies in the supplied 1.00 nm electron well
nλn / nmEn / eV
12.000.377
21.001.51
30.6673.39

The corresponding unrounded energy calculations for n = 2 and 3 give 2.41256 × 10-19 J and 5.42827 × 10-19 J. The ratio 1:4:9 is exact in the model; displayed energies are rounded.

E2 - E1 = 3E1 ≈ 1.13 eV
E3 - E2 = 5E1

The increasing gaps are not equally spaced. At fixed n and mass, doubling L quarters En. At fixed n and L, doubling m halves it.

The nonzero ground energy is consistent with the momentum spread needed for confinement. It does not mean that a stationary density graph depicts a ball travelling back and forth at a definite speed. The infinite well is also not the exact potential or energy sequence of an atom.

Optional check For the same particle mass, compare n = 2 in an infinite well of width 2L with n = 1 in a well of width L. How do their energies compare?
For the same particle mass, compare n = 2 in an infinite well of width 2L with n = 1 in a well of width L. How do their energies compare?

07

Atomic transitions and spectra

An isolated atom has discrete electronic energy levels. Emission and absorption involve differences between those levels, with the actual initial state determining which transitions are available.

Atomic hydrogen is a real example of discrete bound electronic states with spatial wavefunctions. Their energies need not be equally spaced or follow the infinite well's n2 pattern. The box is a model of confinement, not an exact model of hydrogen.

For a transition that emits a photon, the photon carries the decrease in atomic energy. Absorption requires a photon that supplies an available upward gap. Photon energy is positive:

Ephoton = |Efinal - Einitial|
= hf = hc/λ

Read a supplied three-level atom

Consider levels at -6.00, -3.00 and -1.00 eV relative to a zero ionisation reference. All three pairwise transitions are permitted in this supplied model. It is not a labelled hydrogen model. The following comparison considers only these bound-bound transitions.

Use the actual starting level and the allowed energy gap

Use the actual starting level and the allowed energy gapThis supplied atom has bound levels minus one, minus three and minus six electronvolts relative to the zero ionisation reference. It is not labelled hydrogen. The energy axis is linear: y equals sixty minus forty times energy in electronvolts, so zero and the three levels are at sixty, one hundred, one hundred and eighty, and three hundred. All three pairwise bound transitions are permitted in this model. Three brown downward emission lanes show gaps two, three and five electronvolts. Separate blue upward lanes start at the ground level and end at minus three and minus one electronvolts, requiring three and five electronvolt absorption. The five-electronvolt arrows pass the intermediate energy without ending there. Arrow width is schematic; photon energy is the positive magnitude of each level change.Energy / eV0Ionisation reference-1-3-62 eV3 eV5 eV3 eV5 eVAll three pairs are permitted here.EmissionAbsorption

The three downward transitions can emit photons when their upper states are occupied. A ground-state absorber can use the upward 3 eV or 5 eV transition; it cannot use the 2 eV gap from an unoccupied excited state. Zero is the ionisation reference, not another bound level.

The level positions use one linear energy scale, with zero as the ionisation reference. Separate arrow lanes identify the 2, 3 and 5 eV gaps without making a direct transition appear to stop at an intermediate level.

Worked emitted photon

Subtract the levels before using hf

For emission from -1.00 to -3.00 eV, the atom loses 2.00 eV. Using 1 eV = 1.60 × 10-19 J, h = 6.63 × 10-34 J s and c = 3.00 × 108 m/s:

Ephoton = 3.20 × 10-19 J
f = Ephoton/h = 4.82655 × 1014 Hz
λ = c/f = 6.215625 × 10-7 m
≈ 622 nm
Emitted photons for the three permitted downward transitions
Levels / eVPhoton energy / eVλ / nm
-1 to -32.00622
-3 to -63.00414
-1 to -65.00249

For the 3.00 and 5.00 eV photons, the frequencies are about 7.24 × 1014 and 1.21 × 1015 Hz. With the supplied constants, their wavelength calculations give 414.375 and 248.625 nm before rounding.

A direct upper-to-ground transition emits one 5.00 eV photon. A two-step cascade emits one 2.00 eV photon and one 3.00 eV photon, with the same total energy. It does not emit one photon at an average of their wavelengths.

Distinguish emission and absorption spectra

An excited low-density gas can produce separate emission lines against a dark background. Their wavelengths correspond to permitted downward gaps, with suitable excited-state populations. The level diagram alone does not determine each line's strength.

A cooler gas in front of a continuous source can remove selected wavelengths from the transmitted beam, producing absorption lines. The relevant lower levels must be occupied. Absorbed energy has not disappeared; later emission can send it in other directions.

An excited population can show all three emission lines

An excited population can show all three emission linesBoth spectra use the same numerical wavelength axis from two hundred to seven hundred nanometres, with x equal to fifty plus one half times wavelength minus two hundred. The exact three model wavelengths are 248.625,414.375 and621.5625 nanometres, at x74.3125,157.1875 and260.78125. The shortest line is ultraviolet, not visible violet. Three narrow bright markers stand on a dark background when suitable excited levels are occupied. The ultraviolet marker is neutral rather than coloured violet. The visible shorter and longer markers denote approximately violet and red wavelengths, without asserting relative line intensity. Marker widths and any displayed brightness are schematic, not supplied spectral-resolution or population data.Suitable excited-state populations249 nmUV414 nm622 nm200300400500600700Vacuum wavelength / nmLine widths and strengths are schematic.

An appropriately excited population can supply the 2 eV, 3 eV and 5 eV downward transitions. The numerical axis includes ultraviolet. Equal marker height or width does not mean equal emitted intensity.

Ground-state absorption removes only the available gaps

Ground-state absorption removes only the available gapsBoth spectra use the same numerical wavelength axis from two hundred to seven hundred nanometres, with x equal to fifty plus one half times wavelength minus two hundred. The exact three model wavelengths are 248.625,414.375 and621.5625 nanometres, at x74.3125,157.1875 and260.78125. The shortest line is ultraviolet, not visible violet. A neutral light band represents the supplied incident continuum; dark notches occur only at 248.625 and414.375 nanometres for a population entirely in the ground state. The 621.5625-nanometre position is unnotched because the lower state of that two-electronvolt transition is unoccupied. The band is not a literal colour display of ultraviolet. Marker widths and any displayed brightness are schematic, not supplied spectral-resolution or population data.All atoms initially in the ground state249 nmUV414 nm622 nmpasses200300400500600700Vacuum wavelength / nmNeutral band represents a continuum.

For this all-ground-state bound-transition model, 249 nm and 414 nm are removed from the beam; 622 nm is not. Absorbed energy is not destroyed, and subsequent emission can be in other directions. A different lower-state population changes the possible absorption lines.

The two spectra share a numerical wavelength axis. Suitable excited populations can produce all three emission lines. The stated ground-state absorber removes only the 249 and 414 nm components in this model. The 249 nm line is ultraviolet, not visible violet; line widths and strengths are schematic.

If all atoms start at -6.00 eV, photons of 3.00 or 5.00 eV can excite the listed upper states. A 2.00 eV photon cannot make either of those ground-state transitions. It could produce the -3.00 to -1.00 eV transition if that lower excited state were occupied.

These photons are below the 6.00 eV ground-state ionisation threshold. Start from the actual occupied level and the transitions permitted in the question. A mismatch with these specified gaps is not a universal claim that the photon cannot interact with matter in any other way.

Optional check A supplied atom has levels -6.00, -3.00 and -1.00 eV, with the three pairwise transitions permitted. It starts in the -6.00 eV ground state. Can a 2.00 eV photon produce one of these bound-bound absorptions?
A supplied atom has levels -6.00, -3.00 and -1.00 eV, with the three pairwise transitions permitted. It starts in the -6.00 eV ground state. Can a 2.00 eV photon produce one of these bound-bound absorptions?

Revision

Quantum physics at a glance

Identify the evidence and the quantity being predicted. A photon energy, a probability density, a momentum spread and an atomic energy gap require different calculations.

Choose the model before the equation

Decisions that prevent common quantum-model errors
TaskKeep this distinction
Light evidenceThreshold frequency supports quantised energy exchange; interference and diffraction support wave behaviour. Frequency changes energy per photon, while intensity at fixed frequency changes photon rate.
Momentum and wavelengthA photon has p = E/c = h/λ despite zero rest mass. A non-relativistic electron uses p = mv and λ = h/p. Do not use E = pc for its kinetic energy.
Position probabilityAmplitude can be signed; density |ψ|2 is nonnegative. An interval probability is density area, not density alone or the square of amplitude area.
Coherent alternativesAdd amplitudes before squaring. Relative signs or phases affect interference. A global sign reversal of one complete real state does not change its density.
LocalisationA narrower position distribution needs a broader momentum distribution. The width relation concerns spread, not a required mean momentum or apparatus error.
ConfinementBoundary conditions select standing shapes and quantised energies. A wavefunction plot is not a particle trajectory, and the box's level sequence is not an atom's exact sequence.
Spectral linesUse an allowed gap from the actual occupied level. Photon energy is the magnitude of the energy change, not the absolute value of one level.

Photon and matter-wave calculations

Photon: E = hf = hc/λ, p = E/c = h/λ
Threshold: Φwork = hf0
Matter wave: λ = h/p
Non-relativistic massive particle: p = mv

For the 6.00 × 1014 Hz photon, λ = 500 nm, E ≈ 2.49 eV and p ≈ 1.33 × 10-27 kg m/s. The 1.50 × 106 m/s electron instead has λ ≈ 0.485 nm. Doubling its speed halves wavelength while quadrupling kinetic energy.

Normalisation and area

For a profile that vanishes outside 0 < x < L:

Square profile ψ = C:
C2L = 1, so C = 1/√L

Sine profile ψ = C sin(πx/L):
C2L/2 = 1, so C = √(2/L)

Choose a global sign without changing probabilities. At L = 4.00 nm, the positive coefficients are 0.500 and 0.707107 nm-1/2. The square profile gives probability 0.375 over 1.00-2.50 nm. The sine profile's central 1.00-3.00 nm probability is about 0.818310; a coarse trapezium estimate need not equal that smooth-model area.

The coordinate and density units must match: nm-1 times nm gives a dimensionless probability. For coherent real contributions, (a + b)2 includes 2ab. Do not replace it with a2 + b2. Detection-bin frequencies estimate finite-interval probabilities and do not establish a classical path or exact point probability.

Width scales, box states and atomic gaps

Characteristic widths: ΔxΔp ≳ h
Infinite well: ψn = √(2/L) sin(nπx/L)
λn = 2L/n
En = h2n2/(8mL2)

The width relation is an order-of-magnitude convention, not an exact equality. Infinite-well states have n = 1, 2, 3, ... and n - 1 interior nodes. Their first energies are in the ratio 1:4:9. Doubling width quarters energies at fixed n and mass; n = 2 at width 2L has the same energy as n = 1 at width L.

In the supplied atom, levels -6, -3 and -1 eV give permitted gaps of 3, 2 and 5 eV. The emitted wavelengths are about 414, 622 and 249 nm. A ground-state absorber can use the 3 and 5 eV upward gaps; the 2 eV absorption requires the -3 eV state to be occupied. The 249 nm line is ultraviolet.

Quantities and units

Quantum quantities, unit conversions and local symbol meanings
Quantity and symbolUnitMeaning
Energy E, EnJ or eV1 eV = 1.60 × 10-19 J.
Work function ΦworkJMinimum surface-removal energy, distinct from magnetic flux.
Planck constant hJ s6.63 × 10-34 J s in these examples.
Momentum pkg m/s = N sPhoton and massive-particle relations have different conditions.
Frequency fHzSets energy per photon.
Vacuum light speed cm/s3.00 × 108 m/s.
Wavelength λ; width L; position xm or nm1 nm = 10-9 m. Match density units to the position coordinate.
Mass m, mekgme = 9.11 × 10-31 kg.
Elementary charge eC1.60 × 10-19 C; eV itself is an energy unit.
One-dimensional amplitude ψm-1/2 or nm-1/2Its modulus squared is the corresponding inverse-length density.
Probability; quantum number nNo unitProbability is between 0 and 1. Here n labels a box state, not amount of substance.
Box potential energy VJ or eVThe local symbol V here does not mean electric potential.

A three-dimensional electron cloud instead uses probability per volume, with SI density unit m-3. Always identify what the graph's vertical axis represents before reading a height or calculating an area.