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What does the electric field of unpolarized light look like when measured?

+ 6 like - 0 dislike
355 views

When we talk about fermions in mixed state, we say that their state can't be described by a wavefunction and just compute all the probabilities using density matrix. That's OK because the wavefunctions aren't observable. But the analogous function for photons (which is admittedly not quite a wavefunction, but still has many properties of it) is observable — electromagnetic tensor or even just electric and magnetic fields.

Now I might just say that we can't use electric and magnetic fields to describe unpolarized light, but observability of these quantities makes me feel that this is wrong.

So my question is, what would electric field of unpolarized EM wave look like when measured (assuming low enough frequencies to be measurable)? Will it just fluctuate wildly in direction, remaining regularly changing in amplitude?


This post imported from StackExchange Physics at 2016-02-11 14:28 (UTC), posted by SE-user Ruslan

asked Apr 21, 2015 in Theoretical Physics by Ruslan (85 points) [ revision history ]
edited Feb 11, 2016 by Dilaton
The photon has a complex wave function, not measurable, it is the solution of the potential form of maxwell's equation turned into an operator on psi equation. Light is built up by a confluence of innumerable photons

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user anna v
continued see this blog on photons building up classical em wave motls.blogspot.com/2011/11/…

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user anna v
@annav: I don't think his question is about quantization of the field or how to obtain photon solutions. He is asking about how unpolarized light behaves and how it makes sense.

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user Numrok
@Numrok he is talking of "the analogous function for photons" does not exist, and that is wrong. It does and is unobservable and in that sense the link I gave describes the analogous build up to a density matrix. And I think the linked answer is correct .

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user anna v
@Ruslan - Are you asking what a hodogram showing two components of a vector field plotted against one another would look like?

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user honeste_vivere
@honeste_vivere if I understand, what a hodogram is, correctly (a trace of vector endpoint as it changed in time), then yes.

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user Ruslan
@Ruslan - Then an unpolarized signal will look like a "blob" when plotted (i.e., if enough points are used, it will effectively be a disk but no subinterval will show circular or consecutive linear polarizations).

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user honeste_vivere

4 Answers

+ 8 like - 0 dislike

Polarized, partially polarized, and unpolarized light are a phenomenon of classical optics known for a long time. The first successful theory dates back to 1809, and a complete theoretical description was given by Stokes in 1852, though without reference to the electromagnetic field.

The correct way to model classical unpolarized light using electromagnetic language is by means of the stochastic Maxwell equations. This is discussed at length in the book Optical Coherence and Quantum Optics by Mandel and Wolf.

In particular, unpolarized light is discussed on p.350. It is modelled in (6.3-4) by a coherence matrix $J$ that is a multiple of the identity matrix. The coherence matrix is defined on p.343 in (6.2-6) as the time-independent equal-time covariance matrix of the part of the fluctuating electric field transversal to the light beam.

Here ''fluctuating'' refers to the fact that one has to use a stochastic description. This means that the electromagnetic field is considered in a stochastic framework, where the Maxwell equations propagate stochastic initial conditions. How this works is describe in the simpler case of a scalar field on p.147. The necessary background on stochastic processes is described in Chapter 2 (p.41ff).

Mandel and Wolf work in a slab geometry along an optical axis aligned with the $z$-axis ($z=x_3$). The solution is of interest for $z\in[0,z_{\max}$. The initial values (at $z=0$, for arbitrary $x,y,t$) are stochastic, all field expectation values are zero. The dynamics (in $z$, representing a propagating wave) is completely deterministic, given by the free Maxwell equations. Because of linearity one doesn't need to specify the initial probability distribution exactly. If one specifies the first $k$ moments one obtains exact, deterministic, linear equations for these moments. For most effects (including polarization effects), it is sufficient to consider second order moments ($k=2$). In this case one gets linear deterministic equations for the covariance matrices. To model the behavior of a ray of light in $z$-direction, one uses in addition the paraxial approximation, valid for fields that have an appreciable energy density only close to the $z$-axis.

On the other hand, polarized, partially polarized, and unpolarized light is also the simplest instance of quantum mechanics. In the monochromatic plane wave case it gives a classical model for a qubit, faithful in every respect. This is discussed in my lecture Optical models for quantum mechanics.

answered Feb 5, 2016 by Arnold Neumaier (12,355 points) [ revision history ]
edited Feb 11, 2016 by Arnold Neumaier
+ 2 like - 0 dislike

I will try to answer this question in two parts: firstly I disagree with the answer in the question you linked. There is "classical unpolarized light" (in some sense) and it is instructive to look at it. Then I will show how this relates to the quantum version.

Classical unpolarized light: classically, all the information about the wavefield is contained in the time varying electric field. The mathematical object we would use is the general state of coherence

$ \langle \vec{E}(r_1) \vec{E}(r_2) \rangle $

Here the brackets denote an ensemble average. You can also think of it in terms of a time average, so it is exactly like you say: the field fluctuates wildly. The product of the two fields is the dyadic product (so that the overall object is a 2x2 matrix for the fields perpendicular to the travel direction). For simplicity we can consider coherent light, i.e. there is no correlations between different positions $r_1$ and $r_2$. For unpolarized light travelling in the z-direction we then get:

$ \langle \vec{E}(r_1) \vec{E}(r_2) \rangle \propto \mathbb{1} \delta^3(r_1 - r_2) $

where $ \mathbb{1} $ is the identity matrix. What can we get from this object? All important time averages are contained, e.g. the power can be obtained by tracing over polarization and direction degrees of freedom: $ P = 1/Z_0 \int \int d^3r_1 d^3r_2 tr(\langle \vec{E}(r_1) \vec{E}(r_2) \rangle) $ Note that this for example yields the power addition law for unpolarized light, since the trace adds the diagonal entries of the matrix (ie. the individual powers get added).

I'll stop boring you with formalism and get to your question: what about the fields? They fluctuate, so instantaneously we could still say the light is polarized. It is unpolarized in the average over a certain characteristic time scale though. This time scale has to be large enough to contain a "representative fluctuation". The field at a certain time is still well defined though and you can measure it.

Quantum view: In the quantum case we don't need this concept of a fluctuation time scale, since we can instantaneously have a state that has a probability to be either polarization. If you measure the field at multiple times then it would still fluctuate.

Comparison: I personally think that there is almost no difference between the classical probabilistic and the quantum picture. In fact the general state of coherence I presented above IS the density matrix if the characteristic time scale I mentioned is zero. In practice you don't get fields that classically fluctuate very fast since sources can't produce that light. You can get the quantum unpolarized light though (e.g. in a laser or something, doesnt really matter for the argument).

For more information on the mathematical background google e.g. "partially coherent wavefield modelling"

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user Numrok
answered Feb 3, 2016 by Numrok (20 points) [ no revision ]
''In practice you don't get fields that classically fluctuate very fast since sources can't produce that light.'' This is incorrect. Most ordinary light is unpolarized, which is possible only with fast fluctuations. Nonfluctuating sources can produce completely polarized light only.

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user Arnold Neumaier

@Arnold Neumaier: I agree, sorry for the bad formulation

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user Numrok

+ 2 like - 0 dislike

Answer

So my question is, what would electric field of unpolarized EM wave look like when measured (assuming low enough frequencies to be measurable)?

To create an unpolarized signal we can start by creating some random noise, allowing it to constructively and destructively interfere with itself. The following is some Mathematica code but written in TeX for visualization purposes:

xx = RandomReal[{0, 2$\pi$}, 20]

mysin[t_] := $\sum_{i=1}^{20}$ Sin[t + xx[[i]]]

mycos[t_] := $\sum_{i=1}^{20}$ Cos[t + xx[[i]]]

ParametricPlot[{mycos2[t], mysin[t]}, {t, 0, 2$\pi$}]

The output can be seen below:
hodogram of random noise

As you can see, there is no clear polarization and no filtering will help clear this up unless there is a clear frequency peak in the power spectrum.

Will it just fluctuate wildly in direction, remaining regularly changing in amplitude?

That depends upon the signal. One could imagine a scenario where the amplitude was modulated but the signal remained unpolarized. For instance, the signal produced by the mysin function above looks like the following:
enter image description here

You can see that if one were to overplot the envelope of this signal, the amplitude of the envelope would change in time but the mixture of the two components is still unpolarized.

Mathematical Description

We start by assuming we are working in $n$-dimensional unitary space. If a wave/fluctuation is in a pure state or is totally polarized, we can determine all of its polarization information from a single state vector, $\mathbf{u}$, in a an orthonormal basis.

We can construct a spectral matrix from an $n$-variate time series, $\mathbf{x}^{T} = \left[ x_{1}\left( t \right), ..., x_{n}\left( t \right) \right]$, given by: $$ \mathbb{S}\left( \omega, \Delta \omega \right) = \int_{\omega - \Delta \omega}^{\omega + \Delta \omega} \int_{-\infty}^{\infty} \ d\tau \ d\omega \mathbb{C}\left( \tau \right) e^{-i \ \omega \tau} \\ \mathbb{C}\left( \tau \right) = \langle \mathbf{x}\left( t \right) \ \mathbf{x}^{T}\left( t + \tau \right) \rangle $$ where $\omega$ is a frequency, $\Delta \omega$ is the bandwidth about $\omega$, $\tau$ is an effective time delay or time lag, $T$ is the transpose, and $\langle \rangle$ is expectation value or ensemble average.

We can expand $\mathbb{S}$ into a set of $n^{2}$ outer products, $\mathbf{u}_{j} \ \mathbf{u}_{k}^{\dagger}$, where the vectors $\mathbf{u}_{j}$ form a complete orthonormal basis. In other words, we have $\mathbf{u}_{j}^{\dagger} \ \mathbf{u}_{k} = \delta_{j,k}$, where $\delta_{j,k}$ is the Kronecker delta. The expansion yields the following, if $\mathbf{u}_{j}$ are eigenvectors (thus, the expansion will only have $n$ terms instead of $n^{2}$): $$ \mathbb{S} = \sum_{j=1}^{n} \ \lambda_{j} \ \mathbf{u}_{j} \ \mathbf{u}_{j}^{\dagger} $$ where $\lambda_{j}$ are eigenvalues, $\lambda_{j}^{1/2} \ \mathbf{u}_{j}$ are state vectors of $\mathbb{S}$, $\dagger$ is the Hermitian adjoint, and $n$ is the number of degrees of freedom (e.g., $n$ = 3 for many cases one is likely to encounter, thus, $j$ corresponds to the components in a 3-vector).

Note that in the event that $\mathbb{S}$ describes a purely polarized wave, then it can be described by one state vector because only one of the eigenvalues is non-zero. We should also note that any given state vector can be multiplied by an arbitrary phase factor, $e^{i \ \phi}$, and the result will still be an eigenvector of $\mathbb{S}$. This allows us to define the following: $$ e^{i \ \phi} \ \mathbf{u} = \mathbf{r}_{1} + i \ \mathbf{r}_{2} $$ where $\mathbf{r}_{1,2}$ are vectors in real space satisfying $\mathbf{r}_{1}^{T} \ \mathbf{r}_{2} = 0$ (i.e., they are orthogonal). This allows us to rewrite $\mathbb{S}$ as: $$ \mathbb{S} = \lambda_{j} \ \mathbf{u}_{j} \ \mathbf{u}_{j}^{\dagger} = \left( \mathbf{r}_{1,j} + i \ \mathbf{r}_{2,j} \right) \left( \mathbf{r}_{1,j}^{T} - i \ \mathbf{r}_{2,j}^{T} \right) $$ This form is useful since is gives way to an easy construction of projection operators, where the power, $D$, in the $\mathbf{v}$ direction is given by: $$ D = \mathbf{v}^{\dagger} \ \mathbb{S} \ \mathbf{v} = Tr\left[ \mathbf{v} \ \mathbf{v}^{\dagger} \ \mathbb{S} \right] $$

We can write the state vectors in the following form: $$ \lambda_{j}^{1/2} \ \mathbf{u}_{j}^{T} = \left[ a_{1}, a_{2} \ e^{i \ \phi_{2}}, ..., a_{n} \ e^{i \ \phi_{n}} \right] $$ where $a_{i}$ are amplitudes associated with each phase, $\phi_{i}$. If we assume a pure state, then we can rewrite $\mathbb{S}$ as: $$ S_{j k} = a_{j} \ a_{k} e^{i \left( \phi_{j} - \phi_{k} \right)} $$ and from our assumption of a pure state we know that: $$ \left( Tr\left[ \mathbb{S} \right] \right)^{2} - Tr\left[ \mathbb{S}^{2} \right] = 0 $$ From this we can define the degree of polarization as: $$ P^{2} = \frac{ n \ Tr\left[ \mathbb{S}^{2} \right] - \left( Tr\left[ \mathbb{S} \right] \right)^{2} }{ \left( n - 1 \right) \ \left( Tr\left[ \mathbb{S} \right] \right)^{2} } $$ where in the limit that $P = 1$ we have a pure state (e.g., circularly polarized wave) or purely polarized wave. Given that we restrict our systems such that $n$ > 1, then we can see that the degree of polarization is limited to $0 \leq P \leq 1$.

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user honeste_vivere
answered Feb 4, 2016 by honeste_vivere (20 points) [ no revision ]
Why did you make x an array of delayed-initialized random values if you could just as well make it a single value like x:=RandomReal[{0,2Pi}]? In your plots every point was random, so there was no need in the array (you can see this by checking that there're much more than 20 peaks in the second plot).

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user Ruslan
@Ruslan - 20 is not the number of points in each plot. 20 is the number of different sine/cosine waves being self-interfered to produce the noise estimates. The plots are created using something like: Plot[mysin[t], {t, 0, 2$\pi$}]. The delayed initialization is probably not important.

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user honeste_vivere
Due to delayed initialization every evaluation of mysin[t] yields a new random variable.

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user Ruslan
Hmm, why the downvote? Is the answer incorrect in principle?

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user Ruslan
Yes, I was curious about that as well... Perhaps people were expecting some complicated discussion of the properties of the spectral matrix: $$\mathbb{S} = \sum_{j=1}^{n} \lambda_{j} \ \mathbf{u}_{j} \ \mathbf{u}_{j}^{\dagger}$$, where $\lambda_{j}$ are eigenvalues and $\mathbf{u}_{j}$ are state vectors in an orthonormal basis...

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user honeste_vivere
Then the degree of polarization is formally defined as $$P^{2} = \frac{ n Tr\left[ \mathbb{S}^{2} \right] - \left( Tr\left[ \mathbb{S} \right] \right)^{2} }{ \left( n - 1 \right) \left( Tr\left[ \mathbb{S} \right] \right)^{2} }$$. Then one can see that $0 \leq P \leq 1$ such that for $P = 1$ we have a pure state (e.g., circularly polarized).

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user honeste_vivere
The problem is, you did not ask about the math, you asked what it looked like, which is what I provided in my answer. So there really is no reason for the down vote.

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user honeste_vivere
+ 0 like - 2 dislike

In a photon electic field vibrate only along a single direction and does not change along its axis.

When you see an unpolarised wave you actually see a number of photons. Vibrational axis of each photon is independent of each other. In unpolarized wave we see all these random directions. And talk about changing electric field which is misconception.

In polarised light all photons vibrate in particular direction and appear to be same direction.

This post imported from StackExchange Physics at 2016-02-11 14:29 (UTC), posted by SE-user Anubhav Goel
answered Feb 9, 2016 by Anubhav Goel (-20 points) [ no revision ]

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