When a single photon is emitted as a result of relaxation of an electronic energy level, it (the field) is clearly in a Fock state because it has a well-defined number of quanta. By the uncertainty relation $\Delta n \Delta \phi \ge 1$ it has maximum uncertainty in its phase. How can something wave-like have a frequency but no phase?

  • 1
    $\begingroup$ A phase is relative to a reference. What reference would that be? $\endgroup$
    – ACuriousMind
    Feb 11, 2015 at 13:54
  • 2
    $\begingroup$ The reference would be the time that it emitted. It has a frequency and a time-zero. Shouldn't that mean that its phase should also be defined? Or does the uncertain phase mean that it's emission time is equally undefined? $\endgroup$
    – mactud
    Feb 11, 2015 at 14:23
  • 1
    $\begingroup$ Also, why the downvote? Is it a stupid question? $\endgroup$
    – mactud
    Feb 11, 2015 at 14:24
  • 1
    $\begingroup$ This is not a stupid question. @ACuriousMind, the reference could be my pocket watch. $\endgroup$
    – DanielSank
    Feb 11, 2015 at 16:47
  • 1
    $\begingroup$ @ACuriousMind the reference, is itself at a previous time. $\endgroup$
    – Sofia
    Feb 11, 2015 at 17:16

4 Answers 4


Photons indeed have phase (or phases, in the plural) as folloows: one can turn Hossein's answer on its head in the following way. A "lone photon" is simply a particular pure state - a so called Fock State - of the quantum electromagnetic field (see the "Quantization of the Electromagnetic Field" Wikipedia article) where the number observable is certain to return a measurement of $n=1$.

One can form the following quantities related to a one-photon Fock state $\psi$ of the quantum light field as follows:

$$\begin{array}{lcl}\vec{\phi}_E(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{E}}^+(\vec{r},t)\left|\left.\psi\right>\right.\\ \vec{\phi}_B(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{B}}^+(\vec{r},t)\left|\left.\psi\right>\right. \end{array}\tag{1}$$

where $\psi$ is the (Heisenberg picture) light field quantum state, $\mathbf{\hat{B}}^+,\,\mathbf{\hat{E}}^+$ are the positive frequency parts of the (vector valued) electric and magnetic field observables and, of course, $\left<\left.0\right.\right|$ is the unique ground state of the quantum light field.

These quantities have two interesting properties:

  1. They fulfill Maxwell's equations (if one replaces the classical $\vec{E}$ and $\vec{B}$ vectors by $\vec{\phi}_E(\vec{r},\,t)$ and $\vec{\phi}_B(\vec{r},\,t)$ respectively and
  2. They represent a probability amplitude to absorb a photon at a particular point in space and time (e.g. by observing a detection event in a CCD array or PMT as the quantized EM field drops to its ground state) because the analogue of the classical energy density $p(\vec{r},\,t) = \frac{1}{2}\,\epsilon_0\,|\vec{\phi}_E|^2 + \frac{1}{2\,\mu_0}\,|\vec{\phi}_B|^2$ is the probability in question.

Thus we see that we can define probability amplitude functions of space and time from the quantized electromagnetic field that fulfill Maxwell's equations, which can thus be thought of as the analogue of the Dirac equation for the lone photon. In the other direction, (1) is invertible so that, given any classical solution of Maxwell's equations, one can find a unique one photon Fock state $\psi$ that would yield the quantities $\vec{\phi}_E(\vec{r},\,t)$ and $\vec{\phi}_B(\vec{r},\,t)$ in (1), to within a normalization constant. So, every classical solution of ME defines a corresponding one photon state and contrariwise.

Naturally there are phases associated with these quantities, being solutions of Maxwell's equations. The one photon state may, however, be a quantum superposition of energy eigenstates, in which case one needs to specify the phase of all the superposition weights to define it fully. In theory, one could even have a "white" photon, one with roughly equal superposition weights for every energy eigenstate over a broad wavelength range, although I don't know of any physical process that can emit a lone photon spread over more than a few MHz. Such a lone "white" photon is nonetheless a pure quantum state, with definite phases for each superposition weight, so is an exotic object quite different from white light comprising many photons all at different frequencies.

As in CuriousOne's comment:

A phase is relative to a reference. What reference would that be?

one can impart a global phase change $e^{i\,\phi}$ to the whole superposition and the one photon state will be unchanged. If the photon, however, is a superposition of many energy eigenstates, the relative phase between them has a real, physical meaning. It will influence, for example, interference patterns observed by the absorption of a large number of photons of the same state on a CCD array.


By the uncertainty relation between energy and time, when the energy is known with precision, the time at which the photon was emitted is totally unknown. So, in the Fock state $|1\rangle$ we cannot know the phase because we don't know when it was emitted. Also, since we cannot measure a photon without disturbing it, if we measure it once at a time $t_0$ we cannot do on it a second measurement to see whether at a time $t_1$ it evolved in phase with respect to the time $t_0$ as expected.


Please recall how the concept of a photon came to life. 1.You have Maxwell's equations, 2. you look for wavy solutions (all classically),3. you get a classical Lagrangian for EM fields and the classical action accordingly. You quantize this classical action. Frequency in this case is an energy (Planck constant is unity in selected system of units). The uncertainty relation is between energy and time. Your n is related to a NUMBER of quanta while your (delta fee) is related to a specific phase for a specific photon. This is like comparing apples with oranges...Sorry...


Since you asked how to interfere when there is no phase, the answer is that single photon interferes itself. This is because it is the smallest unit of energy that cannot be further divided. In a Mach-zehnder interferometer, single photon cannot transmits and reflects through symmetric beam splitter simultaneously, and in a double-slit interferometer, single photon cannot pass two slits simultaneously. However, the path superposition effect must be mentioned, where a single photon goes through two paths to explain interference effect. Thus, the relative phase difference of a single photon passing through the two paths must be taken into account. The individual phases cannot be known by the uncertainty principle, but the relative phase difference can be known. It is to find out the phase difference by looking at the interference effect. A single photon is not phaseless but we just don't know because of uncertainty. However, the relative phase difference is deterministic, not probabilistic. Mentioning a phase in a single photon is a very insightful judgment.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.