How to predict the photoelectric effect with modern quantum theory?

Question

In introduction class to quantum mechanics, the example of the photoelectric effect is often shown to the students to explain how the classical physics fails to explain it. We are told that one can solve the problem by only allowing the light to have discrete energies, proportional to the frequency of the light.

But somehow I can't see what is the connexion between these discrete energies of the photons and the rest of the stuff we learn (wave function, Schrödinger equation). Is it possible to predict the photoelectric effect using Schrödinger's equation on a wave function? If this is the case, I would be very happy to know how.

EDIT: If I understood it well, the weird thing in the photoelectric effect is that even at very high intensity (classically proportional to energy squared) the light can't overcome the work function if the light has a frequency which is too low. So my question is rather : How can one see that light is quantised and that the energy of a quanta depends on its frequency using quantum mechanics?

Answer 1

The photoelectric effect indicates that either energy levels in matter or light is quantized. Willis Lamb and Marlan Scully authored a relevant paper in 1968, entitled "The Photoelectric Effect without Photons.".

There are several phenomenon, particularly spontaneous emission (which is used in lasers), that require quantization of the electromagnetic field in order to be described.

Answer 2

Photoelectric effect

• The energy (density) of a classical electromagnetic wave is proportional to its intensity, i.e., the squared amplitude of the intensity (see Poynting vector), whereas in quantum theory all particles carry energy proportional to their frequency, $E=\hbar\Omega$.
• Ejecting an electron from a material requires energy known as work function, $W$, (whether classical or quantum does not matter).

It is now easy to check experimentally, by varying independently the EM wave intensity and frequency, that for $W<\hbar\Omega$ no electrons are extracted, whatever is the field intensity. On the other hand, for $\hbar\Omega>W$ the photoelectric effect takes place even at low intensities.

Photoelectric effect in the introductory QM
The complete description of photoelectric effect, as absorption of single photons, requires quantization of electromagnetic field, which is not done in the introductory QM courses. However, all introductory quantum mechanics courses treat absorption of time-dependent electromagnetic field, e.g., in the form of the Fermi Golden Rule, where the key expression relating the change of the electron energy and the frequency of the EM wave appears: $$w_{i\rightarrow f}=\frac{2\pi}{\hbar}|V_{fi}|^2\delta(E_f-E_i-\hbar\Omega).$$ It is in this sense that one can interpret photoelectric effect without invoking the quantization of field - as mentioned in the other answers and this thread. Fermi Golden rule however rests on many assumptions/hand-waving-arguments that are fully resolved only using the quantized EM field, e.g., see here.

Yet most discussions are done as a warm-up/motivation, and therefore carried out on the level of De Broglie relations $$E=\hbar\omega, \mathbf{p}=\hbar\mathbf{k}$$.

Answer 3

The photoelectric effect is evidence more for the quantum nature of matter, rather than quantum nature of light.

Applying the Schrodinger equation in a bound system, such as a crystalline solid or just a box potential, gives you a set of discrete energy levels. Being bound, the electrons in the material have negative energy, that is $0$ (where they are free) minus whatever the energy level they are in. This energy is the one that needs to be supplied in order to knock the electron out of the metal, that is to make its energy go from negative to $\geq 0$. It is referred to as the work function, which you will recognise from the photoelectric effect.

So: the photoelectric effects proves that electrons are knocked off a material only when certain (discrete) energies are supplied to them. These energies can be derived from the Schrodinger equation applied to the binding potential in question.

The effect also provides a necessary (but not sufficient) condition for light itself to be quantised in packets i.e. photons.

Answer 4

How can one see that light is quantised and that the energy of a quanta depends on its frequency using quantum mechanics?

The photoelectric effect is one of the physical phenomena that made necessary the introduction of quantum mechanics, a new theory. The other two are the black body radiation, and the atomic spectra.

The photoelectric effect is the puzzle of why below a certain frequency the intensity of light cannot raise photo electrons.

The black body radiation needed to have radiation in energy increments $hν$, $h$ Plancks constant and $ν$ the frequency of the classical electromagnetic wave in order to solve the ultraviolet catastrophe of classical black body radiation.

The atomic spectra could be fitted by Bohr assuming the electrons were in orbits about the nucleus, in quantized angular momentum states, and could only change them when the atom absorbed a photon of the appropriate energy interval.

All three are needed to explain the observations and come to understand that " light is quantized and the energy of a quantum of light depends on its frequency".

The ad hoc Bohr model was replaced by the Schrodinger equation and the postulates of quantum mechanics which give a coherent theoretical model to be able to calculate interactions at the level of particles, atoms,molecules.

Answer 5

Trying to understand the Photoelectric Effect from a QFT perspective has the potential to helpfully shatter a lot of illusions one may have built up from studying QM/QFT.

The cross-section can be calculated explicitly for the K-shell electrons, i.e. for a hydrogen atom - it's sometimes called the 'Stobbe' cross section, and it's not simple to derive (it's done in a 'modern' fashion in [1] and [2], it's done in an older though still similar manner in [3]).

Notice we are talking about an electron moving in the external electromagnetic field of the nucleus (the Coulomb potential), while also allowing a single quantized photon to interact with the electron moving in that field in an absorption process.

In other words, we are already talking about 'QFT' for a 'bound state' in an 'external field'.

Further we are talking about wave functions of single Hydrogen atom electrons in the initial state, and wave functions of a single electron in the continuous spectrum in the final state, where the process of calculating the cross sections involves using normalization factors for those continuous spectrum wave functions, a statement that should ring alarm bells since continuous spectrum wave functions are supposed to be non-normalizable and non-physical right?

Even worse, since we are talking about a transition involving a single photon, it means we are doing 'first order QED', which is 'proven' to be trivial at first order in QFT books.

First order QED? Bound states? External potentials? Everything is time-independent? Continuous spectrum wave functions which are normalized? Electron wave functions which are solutions of the Dirac/Schrodinger equation and not expanded in terms of creation and annihilation operators?

The applicability of some of these statements to 'QFT' are commonly denied/seriously-questioned even on this site, yet one of the most important problems in physics uses all of them.

But we are doing 'First order QED' using external fields, so that by-passes the usual 'first order' proof, and we are using wave function solutions of the Dirac equation (or Schrodinger equation, in the non-relativistic limit) to capture the effect of the external field, and we are using quantized EM fields to explain why the photon gets annihilated, and the single particle Dirac equation and it's stationary state solutions obviously apply because we can obviously measure the process, and if we've studied the continuous spectrum properly we know there's obviously no issue with normalizing them and physically interpreting the result. Further everything is time-independent, so in the derivation of the Fermi Golden Rule one does not even touch the matrix element of the interaction term, the interesting thing (given the unbelievable comments in the links above) about the Fermi Golden Rule is that since it involves the continuous spectrum one may find extra factors in the formula as one does in [1].

References:

1. Landau and Lifshitz, "Quantum Electrodynamics"
2. Akhiezer and Berestetskii, "Quantum Electrodynamics"
3. Heitler, "Quantum Theory of Radiation".