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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?

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related: physics.stackexchange.com/q/437/4552 –  Ben Crowell Jun 3 '13 at 15:41

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up vote 13 down vote accepted

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.

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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.

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a link (to the electric and magnetic field operators) would be good to have for us who are refreshing our knowledge –  anna v Dec 18 '12 at 4:15
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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? –  John Rennie Dec 18 '12 at 7:14
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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. –  emarti Dec 18 '12 at 8:34
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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). –  Vladimir Kalitvianski Dec 18 '12 at 14:19
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@andrey.baj: I'm glad to hear that my guess was wrong. And yes, you're basically correct. If you have a photon which is roughly localized in both position and momentum space, you can estimate the largest expected field values. But you'll get into trouble if you assume the photon's energy is perfectly known. –  user1504 Dec 18 '12 at 21:55

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.

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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). –  Peter Shor Jan 27 '13 at 15:46

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