#### [SOLVED] How can we interpret polarization and frequency when we are dealing with one single photon?

2013-09-28 23:52:07 8 Comments

If polarization is interpreted as a pattern/direction of the electric-field in an electromagnetic wave and the frequency as the frequency of oscillation, how can we interpret polarization and frequency when we are dealing with one single photon?

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## 3 comments

## @WetSavannaAnimal 2013-09-29 02:44:30

Maxwell's equations

exactlydefine the propagation of a lone photon in free space. The state of a photon can be defined by a vector valued state in Hilbert space and this vector valued state is a precise mathematical analogy of the $\vec{E}$ and $\vec{H}$ fields of a macroscopic, classical field. That's not to say that, for a single photon, the $\vec{E}$ and $\vec{H}$ are to be interpreted as electric and magnetic field: the vector valued $\vec{E}$ and $\vec{H}$ state is the unitarily evolving quantum state before any measurement is made. But:This is the

first quantizeddescription of the photon. To understand what measurements a photon state implies, one has to shift to a second quantized description where we have electic and magnetic field observables, whose measurements behave more and more like classical measurements as the number of photons gets bigger. A classical state is acoherent stateof the second quantised field. But, given a photon can be described by a vector valued quantum state, it should be clear that polarization and all like "classical" attributes are meaningful for a lone photon.In particular, a photon can be a

quantum superpositionof eigenstates, so:One can even broaden this concept to propagation through dielectric mediums: the light becomes a quantum superposition of free photons and excited matter states, and the lone, first quantized quasiparticle that results from this superposition (strictly speaking a "polariton" rather than a true, fundamental, photon) has a quantum state which evolves following Maxwell's equations solved for the medium. Thus, for example, we talk about lone photons propagating in the bound modes of optical fibres.

Another take on the one photon state is given in the first chapter of Scully and Zubairy "Quantum Optics". The one photon state $\psi$ can be defined by the

ensemblestatistics derived from the second quantized electric and magnetic field observables:$$\vec{E} = \left(\begin{array}{c}\left<0 | \hat{E}_x | \psi\right>\\\left<0 | \hat{E}_y | \psi\right>\\\left<0 | \hat{E}_z | \psi\right>\end{array}\right);\quad\quad \vec{B} = \left(\begin{array}{c}\left<0 | \hat{B}_x | \psi\right>\\\left<0 | \hat{B}_y | \psi\right>\\\left<0 | \hat{B}_z | \psi\right>\end{array}\right)$$

where $\hat{E}_j$ is the $j^{th}$ component of the vector valued electric field observable and $\hat{B}_j$ that of the magnetic induction observables. ($[\hat{E}_j, \hat{B}_j]=0$ for $j\neq k$ and, in the right units, $[\hat{E}_j, \hat{B}_j]=i\,\hbar\,I$). For a one-photon state $\psi$, these statistics:

exactlyfollowing Maxwell's equations;Things are much more complicated for general, $N$ photon states so we need much more information than simple means to fully define the quantum state particularly with entangled states. Going back to our classical probability distribution analogy, the normal distribution needs

twoindependent parameters, mean and variance, to wholly specify it, so it's a more complicated thing than the Poisson distribution, which is defined by only its mean (which equals the variance).So quantum fields are hugely more complicated things than classical ones. But acoherentstate of any photon is again uniquely defined by the mean values of the field observables, which means again propagate following the same Maxwell equations as the one-photon means: hence the one-to-one, onto correspondence between classical and one-photon states I spoke of - I like to call this the one photon correspondence principle ("OpCoP"). Why our macroscopic EM fields seem to behave like coherent quantum states rather than hugely more general, entangled ones (unless one goes to considerable experimental effort to observe entanglement) is still an open question. It is interesting to note, though, that the class of coherent states is the unique class of quantum harmonic oscillator states that achieve the lower bound of the Heisenberg uncertainty inequality.Also see my answers to:

Incidentally, even though general, entangled light states are hugely more complicated than one-photon (and, equivalently through the OpCoP, classical) light states, in principle we can still decompose them into a quantum superposition of tensor products of coherent states and so represent a general state by a set of field observable means. This was one of the contributions of 2005 Nobel Laureate Roy Glauber, who showed the above in 1963 in:

R. Glauber, "Coherent and Incoherent States of the Radiation Field", Phys. Rev. 131, 2766â€“2788 (1963)

The coherent state tensor products are, however,

overcomplete so the decomposition of a general quantum state into coherent states is highly not unique. Nonetheless, such a decomposition allows classical-like techniques to be brought to bear on entangled quantum states (in principle - in practice it is still complicated!).If you google Iwo Bialynicki-Birula and his work on the photon wave function, he has heaps more to say about the one-photon wave function. He defines the photon wave function as the positive frequency part of left and right circularly polarized eigenfunctions $\vec{F}_\pm = \sqrt{\epsilon} \vec{E} \pm i \sqrt{\mu} \vec{H}$. Iwo Bialynicki-Birula's personal website is http://cft.edu.pl/~birula and all his publications are downloadable therefrom. $|\vec{F}_+|^2 + |\vec{F}_-|^2$ is the electromagnetic energy density. He defines the pair $(\vec{F}_+, \vec{F}_-)$, normalised so that $|\vec{F}_+|^2 + |\vec{F}_-|^2$ becomes a probability density to absorb the photon at a particular point, to be a first quantized photon wave function (without a position observable). There is special, nonlocal inner product to define the Hilbert space and in such a formalism the general Hamiltonian observable is $\hbar\, c\, \mathrm{diag}\left(\nabla\wedge, -\nabla\wedge\right)$. Please also see Arnold Neumaier's pithy summary (here) of a key result in section 7 of Bialynicki-Birula's "Photon wave function" in

Progress in Optics36V (1996), pp. 245-294 also downloadable from arXiv:quant-ph/0508202. The Hilbert space of Riemann Silberstein vector pairs that Bialynicki-Birula defines is acted on by an irreducible unitary representation, defined by Bialynicki-Birula's observables $\hat{H}$, $\hat{\mathbf{P}}$, $\hat{\mathbf{K}}$ and $\hat{\mathbf{J}}$, of the full PoincarĂ© group presented in the paper.## @WetSavannaAnimal 2013-10-18 23:14:31

@Programmer Thanks. I've spent quite a bit of time over the years thinking about what a photon "looks like" and I like the idea that one can look at a classical state (e.g. in a microwave lab) and think "this is what one photon "looks like"". (even though it's not one photon, of course)

## @user26165 2013-09-29 02:25:26

Well the energy of the photon is simply h.f , so if you can determine the energy of a single photon, you can determine its frequency. One way to determine the energy of a photon; assuming that you can generate one photon at a time, all of the same energy, would be to use the photo-electric effect, with adjustable band gap photo-cathode materials in a PMT, which can detect single photons. Variable bandgap photo-cathodes, can be made over limited ranges from III-V Ternary or quaternary compounds, such as GaAsP or InGaAsP. Smaller bandgap cathodes will emit a photo-electron; higher bandgap ones will not.

Now you didn't say you wanted to know a practical way of doing that, but if you have a need for single photons of known frequency and polarization, making the PMTs should be no problem for you.

## @electronpusher 2017-03-23 07:43:01

"The energy of the photon is simply hf..." I'm pretty sure that "simple" expression won a Nobel prize and sparked a revolution in both science and philosophy that still echoes over a century later. ;)

## @anna v 2013-09-29 00:53:45

The classical wave is composed by a large ensemble of photons. Both the photon/particle equations and Maxwell's equations contain the state of the electric field in their solutions. Thus, it is not a matter of interpretation but a matter of showing how from single individual photons mathematically described by the equation of second quantization as such, one can derive for an ensemble of photons the electromagnetic wave.

This is not simple but it has been done. A demonstration is given in the article in this blog.

Hand waving an answer: the functions describing the photons have to be coherent (in phase), then the constants in their mathematical description which pertain to the electric and magnetic field "miraculously" build up a classical electromagnetic field which carries the frequency which is contained in the particle description in E=h*nu.

## @electronpusher 2017-03-23 07:40:04

Thank you. While I have great respect for WetSavannah's explanation, your answer was much more helpful to me as a layman.