Given a light pulse in vacuum containing a single photon with an energy $E=h\nu$, what is the peak value of the electric / magnetic field?

  • 1
    $\begingroup$ related: physics.stackexchange.com/q/437/4552 $\endgroup$
    – user4552
    Commented Jun 3, 2013 at 15:41
  • $\begingroup$ @Ben Crowell Is wavefunction of a single photon the same thing as EM wave corresponging the single photon? $\endgroup$
    – jw_
    Commented Dec 12, 2019 at 12:18

8 Answers 8


The electric and magnetic fields of a single photon in a box are in fact very important and interesting. If you fix the size of the box, then yes, you can define the peak magnetic or electric field value. It's a concept that comes up in cavity QED, and was important to Serge Haroche's Nobel Prize this year (along with a number of other researchers). In that experiment, his group measured the electric field of single and a few photons trapped in a cavity. It's a very popular field right now.

However, to have a well defined energy, you need to specify a volume. In a laser, you find an electric field for a flux of photons (n photons per unit time), but if you confine the photon to a box you get an electric field per photon. I'll show you the second calculations because it's more interesting.

Put a single photon in a box of volume $V$. The energy of the photon is $\hbar \omega$ (or $\frac{3}{2} \hbar \omega$, if you count the zero-point energy, but for this rough calculation let's ignore that). Now, equate that to the classical energy of a magnetic and electric field in a box of volume $V$:

$$\hbar \omega = \frac{\epsilon_0}{2} |\vec E|^2 V + \frac{1}{2\mu_0} |\vec B|^2 V = \frac{1}{2} \epsilon_0 E_\textrm{peak}^2 V$$

There is an extra factor of $1/2$ because, typically, we're considering a standing wave. Also, I've set the magnetic and electric contributions to be equal, as should be true for light in vacuum. An interesting and related problem is the effect of a single photon on a single atom contained in the box, where the energy of the atom is $U = -\vec d \cdot \vec E$. If this sounds interesting, look up strong coupling regime, vacuum Rabi splitting, or cavity quantum electrodynamics. Incidentally, the electric field fluctuations of photons (or lack thereof!) in vacuum are responsible for the Lamb shift, a small but measureable shift in energies of the hydrogen atom.

  • $\begingroup$ how do you define the volume? would not the size of the volume change the total energy of the photon if it is equal to E=hf? $\endgroup$
    – torgny
    Commented Dec 26, 2019 at 17:46

This is a reasonable question to ask, but the answer is probably not what you're expecting: the electric and magnetic fields don't have well-defined values in a state with a fixed number of photons. The electric and magnetic field operators do not commute with the number operator which counts photons. (They can't, because they are components of the exterior derivative of the field potential operator, which creates/annihilates photons.) The lack of commutativity implies via Heisenberg's uncertainty principle that the field might have arbitrarily large values.

  • 1
    $\begingroup$ Classically, any wavepacket won't have a single frequency. You need an infinitely long wave to get a single frequency. @user1504 is this related to your answer? $\endgroup$ Commented Dec 18, 2012 at 7:14
  • 5
    $\begingroup$ Anna V: the quantum electromagnetic field has operators analogous to the harmonic oscillator. Instead of the Hamiltonian $H = p^2 + x^2$ (I'm dropping the units here) and energy $(\frac{1}{2} + n) \hbar \omega$, we write a Hamiltonian for each frequency $\omega$: $H = \frac{\epsilon_0}{2} E^2 + \frac{1}{2\mu_0} B^2$, where $E$ and $B$ are treated as conjugate quantum operators (just like $x$ and $p$). We find creation and annihilation operators such that $E = a^\dagger + a$ and $B = a^\dagger - a$, so the energy must be $(\frac{1}{2} + n) \hbar \omega$. $\langle E \rangle$ is well-defined. $\endgroup$
    – emarti
    Commented Dec 18, 2012 at 8:34
  • 1
    $\begingroup$ The peak value depends on the wave packet size. Assuming a very large size in vacuum, you get a very low amplitude, but such a photon is absorbed anyway with a resonator. It takes just more time to pump a resonator (an atom, for example). $\endgroup$ Commented Dec 18, 2012 at 14:19
  • 1
    $\begingroup$ @emarti: Thanks for adding those details. Andrey.Baj: You should accept emarti's answer. I was having fun being pedantic, but emarti's answer is better physics. $\endgroup$
    – user1504
    Commented Dec 18, 2012 at 14:37
  • 2
    $\begingroup$ @user1504 This is not a homework question, I thought about it myself. Unfortunately I have only a little background in quantum mechanics so most of the technical detail of the answers is unclear to me at this moment. I hope later on my studies I'll catch up. Meanwhile, I think I've understood the general idea. When the photon is localized in a box we can calculate the peak values of the electromagnetic field from the box's volume and the photon's energy. When it's not localized, given only the energy, we can't do so. It can take arbitrarily large values, with some probability. Am I correct? $\endgroup$
    – Andrey S
    Commented Dec 18, 2012 at 21:26

@charles Fransis correctly points out that the expected value of the electric field is zero. (that is, $\langle 1|E|1 \rangle = 0$)

And to quote @user1504 (also stated by @vadim):

The electric and magnetic field operators do not commute with the number operator which counts photons. (They can't, because they are components of the exterior derivative of the field potential operator, which creates/annihilates photons.) The lack of commutativity implies via Heisenberg's uncertainty principle that the field might have arbitrarily large values.

So yes, we know that since the photon number operator doesn't commute with the E-field operator, so we know the E is uncertain. But realistically speaking, can a really photon have any arbitrarily high E-field amplitude? It could in the same sense that a bounded electron might appear on the moon because it's wavefunction has some tiny component.

Putting things more precisely: what is the probability distribution associated with measuring a particular E-field amplitude for a single photon (that is what is $|\langle E|1\rangle|^2 = |\psi_1(E)|^2$).

The answer is that a photon has a very unique probability distribution that looks like this:

enter image description here

Which, for comparison to if there was only vacuum:

enter image description here

So here we can visually see how much of an electric field amplitude we really expect to see, with the vacuum as a reference. While the average value is zero, the absolute value of that amplitude is certainly larger than the absolute value of the vacuum and is around the FWHM (full width half max) of the vacuum distribution. (or eyeballing it, roughly double the value of the vacuum on average)

This is a very commonly measured feature of single photons (check out this paper for a classic take on it, and this for something more modern), and this probability distribution (when measuring a photon's electric field) is often used to identify if the quantum state of light is (or isn't) a "pure" single photon.

To see this more clearly, a coherent state, when looking at the statistics as the "phase" of the light changes, it would look like this:

enter image description here

(which is basically $\cos(\omega t)$ with gaussian noise). By comparison a single photon state looks like this:

enter image description here

And here we see that changing a "phase" doesn't do anything to a "single photon" (which may have to do with single photons not having a well defined phase.)

Here's an example in the linked experiment:

enter image description here

(in this case the x-axis is TIME, not phase!) Pulses with a specific time delay are sent into a homodyne detector (which measures the E-field). Here you can see that for a specific temporal value associated with when the photon hits the detector (repeated for multiple measurements). The value of the electric field has a new distribution, and you see the dip at zero.

  • $\begingroup$ Is $\langle E |$ an eigenstate of the electric field operator? $\endgroup$
    – A. P.
    Commented Jan 17, 2021 at 20:11
  • 2
    $\begingroup$ yup. I am referring to the operator $\propto a^\dagger - a$. In textbooks this is often refered to as the "position quadrature" and I think a lot of people (because of the abstract language) don't realize that this operator really is the amplitude of the electric field. $\endgroup$ Commented Jan 17, 2021 at 20:33
  • $\begingroup$ While this is a good answer (+1), I don't think there's anything wrong with the existing answers. The accepted answer describes how the amplitude depends on the cavity volume, while this one describes the distributions of measured amplitudes at a single point. So really it's the combination of them that gives the full picture. $\endgroup$
    – knzhou
    Commented Jan 21, 2021 at 6:08
  • $\begingroup$ @knzhou, while its true that you can effectively increase the interaction strength by confining a photon in a cavity - that doesn't to me seem like a natural way of describing that the E-field of a photon is. Photons are rarely in nature confined to high finesse cavities, and only answering the question in this context is not helpful for people looking to understand what photons actually are. $\endgroup$ Commented Jan 22, 2021 at 19:15
  • $\begingroup$ No, my point is that your answer doesn’t mention at all what the typical electric field amplitudes are. How many volts per meter is it? That’s what I would imagine most people care about, and that’s what the accepted answer describes. $\endgroup$
    – knzhou
    Commented Jan 22, 2021 at 23:32

The electromagnetic field can be understood as the expectation $\langle A(x) \rangle$ of the photon field operator, $A(x)$, which annihilates or creates a photon in an interaction with an electron (or other charged particle), provided of course that a charged particle is present for the interaction to take place. For a single photon state $|\phi \rangle$ the action of $A$ will be to annihilate the photon, or create another one, meaning that the resulting state is a superposition of states with two or no photons. The inner product with a one photon state is zero and you necessarily have $$ \langle A(x) \rangle = \langle \phi | A(x) |\phi \rangle = 0. $$

Which answers the question. The amplitude of such a state is necessarily $0$. You only get a non-zero classical electromagnetic field $\langle A(x) \rangle$ (including a classical electromagnetic wave) from states which contain an indeterminate number of photons.

Equivalently, as @user1504 put it

The electric and magnetic field operators do not commute with the number operator which counts photons.

In other words, for a state with a definite number of photons, it is not meaningful to talk of the classical electric or magnetic field, or of a classical electromagnetic wave.

  • 2
    $\begingroup$ You've found the expected value that's zero even classically. If you instead find expected value of actual amplitude of the field, as requested in the question title, rather than its instantaneous value—i.e. find $\langle \sqrt{A^2}\rangle$ or just $\langle A^2\rangle$—you should get non-zero. $\endgroup$
    – Ruslan
    Commented Jan 16, 2021 at 20:04
  • $\begingroup$ @Ruslan, you are confusing the photon field operator, which does not have an amplitude, with the classical e.m. field which is the expectation of the photon field operator. Classically you would need to calculate the expectation of the expectation of the photon field operator! $\endgroup$ Commented Jan 16, 2021 at 22:24
  • $\begingroup$ What quantity does this operator correspond to? Isn't it electric field/magnetic field/4-potential? $\endgroup$
    – Ruslan
    Commented Jan 16, 2021 at 22:27
  • $\begingroup$ @Ruslan, an operator is not a quantity. It acts on the photon Fock space. To get a classical quantity, we can form the expectation as in the answer. This expectation is the classical $A$ field. $\endgroup$ Commented Jan 16, 2021 at 22:31
  • $\begingroup$ OK, don't see any contradiction with my comments so far. In ordinary non-relativistic QM we can e.g. form the kinetic energy operator by multiplying the momentum operator by itself (and scaling by a constant). Then its diagonal matrix element will give the average value of corresponding quantity. Why can't we do the same with the $A$-field operator? $\endgroup$
    – Ruslan
    Commented Jan 16, 2021 at 22:35

The amplitude of an electromagnetic wave does not commute with the photon number. A state with a single photon is an eigenstate of the photon number operator: $$ n_{\mathbf{k},\lambda}|1_{\mathbf{k},\lambda}\rangle=a^\dagger_{\mathbf{k},\lambda}a_{\mathbf{k},\lambda}|1_{\mathbf{k},\lambda}\rangle=|1_{\mathbf{k},\lambda}\rangle. $$ On the other hand, the states having a definite value of an amplitude (whether one talks about electric field, magnetic field or vector potential) are more like position and momentum operators of the quantum oscillator, corresponding to the given quantum mode, i.e., they do not commute with the number operator: $$ \hat{E} = E_0a_{\mathbf{k},\lambda} + E_0^* a_{\mathbf{k},\lambda}^\dagger $$ (see, e.g., the Wikipedia for the expressions for the coefficients $E_0$.) Thus, the number of photons and the amplitude of the field are related via the uncertainty relation: if one of them is measured, the other can have any value. More specifically: $$ \langle 1_{\mathbf{k},\lambda}|\hat{E}|1_{\mathbf{k},\lambda}\rangle=0,\\ \langle 1_{\mathbf{k},\lambda}|\hat{E}^2|1_{\mathbf{k},\lambda}\rangle= |E_0|^2 \langle 1_{\mathbf{k},\lambda}|2a^\dagger_{\mathbf{k},\lambda}a_{\mathbf{k},\lambda}+1|1_{\mathbf{k},\lambda}\rangle = 3|E_0|^2. $$

  • 1
    $\begingroup$ Is that statement about coherent states true? The position basis wave function of any coherent state is just the ground state Gaussian translated by $x+ip$. Gaussians have tails out to infinity. $\endgroup$
    – kaylimekay
    Commented Jan 15, 2021 at 9:15
  • $\begingroup$ @kaylimekay you are right, these are more like position and momentum eignestates. $\endgroup$
    – Roger V.
    Commented Jan 15, 2021 at 9:21

A single photon's wave could have different shapes, really, so the maximum of the Electric field would be impossible to compute given the parameters above. It could be really short with a very high electric field or really long with a low electric field. Or, whatever. That light comes in "chunks" as photons doesn't restrict you.

Suppose the energy of an ocean wave were hv... then how high is it? Well, it would depend on the width and other factors... as surf comes in waves so does light come in photons, but we wouldn't know the exact shape from the question.


If an atom emits energy hf, it emits also an angular momentum (spin). That combination is called "photon" or "wave packet". Linking the appropriate formulas from QM and E&M waves, you get the diameter of the wave packet (about λ/2) but not the length. The radius and the direction of propagation do not change as long as the wave packet is not disturbed. It is not locked in a box but propagates in vacuum.

If the coherence length L is accepted as the length of the cylindrical wave packet, you can calculate the energy density u~f³/L and the electric field strength E~sqrt(f³/L), which is constant inside the cylinder.

I got the following results: a) The Hydrogen line at 1420 MHz has FWHM≈5 kHz, L≈60,000 m, E≈1e-8 V/m

b) The Sodium D-Line has FWHM≈10 MHz, L≈6 m, E≈220 V/m

c) X-ray, λ≈1e-12 m, L≈1000λ, E≈1e16 V/m

If you choose a different shape, perhaps like a cigar, those values differ

  • $\begingroup$ Where have you read that Energy+Angular momentum=Photon? $\endgroup$
    – jinawee
    Commented Aug 24, 2014 at 11:00
  • 1
    $\begingroup$ users.df.uba.ar/schmiegelow/materias/FT2_2010_1C/extra/… Beth called it light. Some people insist that light consists of photons. Some say, light = electromagnetic waves. $\endgroup$
    – 9herbert9
    Commented Aug 24, 2014 at 14:45
  • $\begingroup$ You may be making a mistake relying on a 1936, experimental paper as your source for the semantics of photons. Firstly the paper predates QED and therefore misses out on a lot of important work on how a photon should be understood and secondly experimental papers tend to give only as much theory as needed for that measurement and only in the interpretation in which the measurement is clearest (I speak as an experimenter). All the qualified theorists I've talked to about this subject suggest caution in trying to impose a simple wave-packet interpretation on a photon. $\endgroup$ Commented Aug 26, 2014 at 18:34

In a box of defined, thus finite, volume an infinitely long wave is by definition impossible. Positing an infinitely long wave would also deny the physical reality of the photon having a wavelength, as wavelength is never infinite; measured wavelengths, of visible light for instance, are extremely short, not infinite.

By defining the volume of the box, i.e. by setting a volume arbitrarily, one is in effect setting an arbitrary upper limit on wavelength. But a single photon cannot yield a value for wavelength, since there is no possibility of measuring a peak-to-peak distance between adjacent peaks in the waveform, when there is no second peak to measure to.

Energy is a derivative of amplitude, but only in a statistical sense, as an average of many photons per second, since the uncertainty principle makes measuring a single photon problematic. Its electric and magnetic field values are only a statistical average; individual photons may deviate widely from that average. Equations derived from these group averages are likewise valid only for the group, not for individual photons.

  • 3
    $\begingroup$ An infinitely long wave can have a perfectly well-defined wavelength in quantum mechanics (and in fact, only infinitely long waves have well-defined wavelength, both classically and quantum mechanically). $\endgroup$ Commented Jan 27, 2013 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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