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Quantum Foundations

Here

Use the photoelectric effect and single-particle interference to motivate quantum ideas.

Wavefunctions

Next

Interpret amplitudes, probabilities, and measurement in a one-dimensional quantum state.

Quantization

Finish

Use bound-state potentials to build intuition for discrete energies in quantum systems.

Quantum Foundations

Classical physics explains an enormous range of phenomena, but early twentieth-century experiments revealed sharp limits to that picture. Light sometimes behaves as if it arrives in packets of energy. Electrons sometimes produce interference patterns that look unmistakably wave-like. Quantum mechanics provides a framework for making sense of these unexpected results.

Where the Classical Picture Breaks

At the turn of the 20th century, Max Planck found that the spectrum of blackbody radiation could be explained if energy was exchanged in discrete packets. This quantization of energy set a theoretical relationship between the frequency of light and its corresponding energy:

E=hf.E = hf.

At the time, this resolved the “ultraviolet catastrophe”, a failure of combining statistical mechanics with classical electromagnetism. More importantly, the quantized treatment of light would eventually lead to a revolution in physics. Planck’s constant h=6.6×10−34 J⋅sh= 6.6 \times 10^{-34} \ \text{J} \cdot \text{s} sets the scale for this new quantum theory.

To better appreciate the failings of classical physics, we’ll consider two important experimental results:

  • In the photoelectric effect, light ejects electrons from a metal only when the light frequency is high enough.
  • In the double-slit experiment, wave-like interference patterns are formed one particle at a time.

A classical view of physics fails in both cases. If light were only a wave, then a sufficiently bright low-frequency beam should eventually shake electrons loose. If particles behaved classically, then opening two slits should produce nothing more than the sum of two piles on the screen. It’s as if nature requires some blend of a particle-like and wave-like description.

Photons and the Photoelectric Effect

Einstein’s explanation of the photoelectric effect treats light as photons. Following Planck’s lead, each photon carries energy E=hfE = hf, where ff is the frequency of the corresponding electromagnetic wave (it’s not clear yet what it means for a particle to have a frequency). A metal surface needs a minimum energy called the work function Φ\Phi to remove one electron. Any leftover energy appears as the emitted electron’s kinetic energy:

Kmax⁡=hf−Φ.K_{\max} = hf - \Phi.

A classical wave account would suggest that brighter light should deliver energy more strongly to the surface, so one might expect low-frequency light to work if the beam is intense enough or if the electrons have time to “soak up” the energy.

In the actual experiment we observe a threshold frequency. Below it, no electrons are emitted no matter how bright the beam becomes. Above threshold, emission begins essentially without delay. Increasing the intensity mainly increases how many electrons are emitted per second, while increasing the frequency raises the maximum kinetic energy of each emitted electron.

A further check comes from the stopping potential. A negative collector voltage is used to slow and stop incoming electrons. When it’s just strong enough to stop even the fastest emitted electrons (those with maximum kinetic energy), the stopping potential satisfies

eVstop=Kmax⁡=hf−Φ.eV_{\text{stop}} = K_{\max} = hf - \Phi.
+-CURRENTCycle cathode material. Current material Sodium.LIGHT SOURCECATHODE [Na]COLLECTOR

Light

Collector Voltage0.0 V
0.0 V-6.0 V

You are just above threshold. Electrons appear right away because a single photon can already free one electron, but their maximum kinetic energy is still modest.

Photon energy: 2.98 eVWork function: 2.28 eVMax kinetic energy: 0.70 eVStopping potential: 0.70 V

Matter Waves and the Double Slit

When a wave passes through a single opening in a barrier, the waveform will appear to spread out, or diffract. When coherent waves pass through two nearby openings, the resultant waveforms interfere constructively or destructively, forming an interference pattern. (A single opening also produces a similar diffraction pattern.)

It’s been known since 1801 that a pattern of bright and dark fringes can be formed by passing light through two closely spaced slits. If you are familiar with wave phenomena, there’s nothing particularly surprising here. However, the explanation of the photoelectric effect indicates that light sometimes requires a particle-like description.

Since a previously wave-like phenomenon has particle-like behavior, de Broglie hypothesized that massive particles also have associated wavelengths. For a particle, like an electron, carrying momentum pp, its de Broglie wavelength is

λ=hp.\lambda = \frac{h}{p}.

Careful analysis of double-slit experiments over the 20th century shows that this mixed description is closer to the truth: very dim light sources produce the interference pattern one “hit” at a time, and electrons also form the same double-slit interference pattern.

predictionSOURCELIGHTBARRIERSCREEN

Light Source

Barrier

For light, classical wave theory and quantum theory agree on the interference pattern. What this display adds is the one-by-one build-up from individual photon detections.

Source: LightMode: Coherent two slitsDetections: 0Fringe spacing: 0.48 units

The screen is therefore telling you something subtle but decisive. Quantum mechanics keeps the one-at-a-time, particle-like detection events, yet it predicts their distribution by adding amplitudes as if a wave-like description of the possible paths still matters.

Classical vs. Quantum Interference

A helpful classical comparison comes from looking at the electric field component of classical electromagnetic waves. For two coherent sources (like passing a plane wave through a double-slit barrier), the intensity on the screen can be written as

I∝∣E1+E2∣2=E12+E22+2E1E2cos⁡(Δϕ).I \propto \left| E_1 + E_2 \right|^2 = E_1^2 + E_2^2 + 2E_1E_2\cos(\Delta \phi).

The interference ultimately comes from the difference in phase Δϕ\Delta \phi at a particular location on the screen. In quantum mechanics, we use wavefunctions ψ\psi, and so called “probability amplitudes” take the place of electric-field amplitudes:

P∝∣ψ1+ψ2∣2=∣ψ1∣2+∣ψ2∣2+2 Re ⁣(ψ1∗ψ2).P \propto \left| \psi_1 + \psi_2 \right|^2 = \left| \psi_1 \right|^2 + \left| \psi_2 \right|^2 + 2\,\mathrm{Re}\!\left(\psi_1^* \psi_2\right).

Once again, the last term is the interference term. In both equations, it is the cross term that can reinforce the pattern in some places and cancel it in others. The important difference is interpretive, not algebraic: EE describes a classical electric field, while ψ\psi is a quantum probability amplitude.

Problem Solving

Photoelectric Effect

Problem

A metal has work function Φ=2.2 eV\Phi = 2.2\,\text{eV}. Light of frequency f=7.5×1014 Hzf = 7.5 \times 10^{14}\,\text{Hz} shines on the surface. Using h=4.14×10−15 eV⋅sh = 4.14 \times 10^{-15}\,\text{eV}\cdot\text{s}, determine whether electrons are emitted and find the maximum kinetic energy if they are.

Given

  • Φ=2.2 eV\Phi = 2.2\,\text{eV}
  • f=7.5×1014 Hzf = 7.5 \times 10^{14}\,\text{Hz}
  • h=4.14×10−15 eV⋅sh = 4.14 \times 10^{-15}\,\text{eV}\cdot\text{s}

Use

Kmax⁡=hf−Φ.K_{\max} = hf - \Phi.

Solve

First compute the photon energy:

hf=(4.14×10−15)(7.5×1014)≈3.11 eV.hf = (4.14 \times 10^{-15})(7.5 \times 10^{14}) \approx 3.11\,\text{eV}.

Then subtract the work function:

Kmax⁡=3.11−2.2=0.91 eV.K_{\max} = 3.11 - 2.2 = 0.91\,\text{eV}.

Since the photon energy exceeds the work function, electrons are emitted.

Check

The result is positive, so the photons have enough energy to free electrons and still leave some kinetic energy behind.

De Broglie Wavelength

Problem

An electron has momentum p=6.6×10−24 kg⋅m/sp = 6.6 \times 10^{-24}\,\text{kg}\cdot\text{m/s}. Estimate its de Broglie wavelength using h=6.63×10−34 J⋅sh = 6.63 \times 10^{-34}\,\text{J}\cdot\text{s}.

Given

  • p=6.6×10−24 kg⋅m/sp = 6.6 \times 10^{-24}\,\text{kg}\cdot\text{m/s}
  • h=6.63×10−34 J⋅sh = 6.63 \times 10^{-34}\,\text{J}\cdot\text{s}

Use

λ=hp.\lambda = \frac{h}{p}.

Solve

λ=6.63×10−346.6×10−24≈1.00×10−10 m.\lambda = \frac{6.63 \times 10^{-34}}{6.6 \times 10^{-24}} \approx 1.00 \times 10^{-10}\,\text{m}.

So the wavelength is about

0.10 nm.0.10\,\text{nm}.

Check

A sub-nanometer wavelength is microscopic but not absurdly tiny, which is why electron diffraction can be observed in atomic-scale experiments.

Quantum Foundations Checkpoint

Question 1 of 3

In the photoelectric effect, what happens if the light intensity is increased while the light frequency stays below the threshold frequency?

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