Assume that you have a light source, e.g. a laser diode, with a polarization extinction ratio (ER) of 100:1, and you need to improve the extinction ratio as much as possible by adding polarizers, e.g. Glan-Taylor prisms with an ER of 100000:1.

How can you mathematically prove, e.g. using Stokes parameters and Müller calculus, that stacking multiple polarizers in equal orientation will or will not improve the final ER?

The ER can be measured by adding a rotatable polarizer, called analyzer, and measuring the optical power $P$ when the analyzer is adjusted to parallel ($0°$) and crossed ($90°$) orientation:

$$ ER = \tfrac{P_{max}}{P_{min}} = \tfrac{P(0°)}{P(90°)} $$

The analyzer can be assumed to be ideal, i.e. a perfect polarizer with an infinite ER.

  []----->-----[]----->-----[]----->-----[]     => Extinction ratio?
Source      Polarizer    Analyzer    Powermeter
(100:1)     (100000:1)   (ideal)

  []----->-----[]----->-----[]----->-----[]----->-----[]     => Extinction ratio?
Source      Polarizer    Polarizer    Analyzer    Powermeter
(100:1)     (100000:1)   (100000:1)   (ideal)

  []----->-----[]----->-----[]----->-----[]----->-----[]----->-----[]     => Extinction ratio?
Source      Polarizer    Polarizer    Polarizer    Analyzer    Powermeter
(100:1)     (100000:1)   (100000:1)   (100000:1)   (ideal)

  • 1
    $\begingroup$ A formal proof will be based on an idealized model, but you seem to be asking about an actual experimental situation. In the real world, you have to account for every significant leakage and scattering path. It's not the straight line you've drawn. $\endgroup$
    – John Doty
    Aug 31, 2023 at 11:52
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    $\begingroup$ Why would you expect it to be any more complicated than multiplying the ratios? That's how cascaded filters (ideally) work. $\endgroup$
    – John Doty
    Aug 31, 2023 at 12:07
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    $\begingroup$ @EdV Thanks, I'm curious what result you will get! Is it possible to try your software for free? Btw, not that I'm mainly interested in the correct formalism...once the maths is clear, I could also write e.g. a small Python script to simulate different scenarios. $\endgroup$
    – srhslvmn
    Aug 31, 2023 at 12:20
  • 1
    $\begingroup$ In theory the response of cascaded linear filters is their product. You're overthinking this. $\endgroup$
    – John Doty
    Aug 31, 2023 at 12:39
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    $\begingroup$ @EdV Hey Ed, you posted an awesome link to your simulation, but it's gone now and I didn't have time to screenshot your calculations + explanations. It was super helpful, exactly what I was looking for...I just didn't have time yet to work through it carefully. Would you mind posting the results here or maybe send me the screenshot from your calculations? Thx a lot $\endgroup$
    – srhslvmn
    Sep 2, 2023 at 17:41

3 Answers 3


I will borrow the formalism of qubit states instead of using Mueller matrices, because it is a more convenient way to do calculations. Instead of describing the polarization using a four-component vector

$$ S = \begin{bmatrix} S_0\\S_1\\S_2\\S_3 \end{bmatrix} $$ one uses a two-by-two matrix:

$$ \rho_S = \frac{1}{2}(S_0 I + S_1\sigma_x + S_2\sigma_y + S_1\sigma_z) $$

where $\sigma_x,\sigma_y,\sigma_z$ are the Pauli matrices. Intensity-normalized horizontally and vertically polarized states can be written

$$ |H\rangle\langle H|, \quad |V\rangle\langle V| $$


$$ |H\rangle = \begin{bmatrix} 1\\0 \end{bmatrix}, \quad |V\rangle = \begin{bmatrix} 0\\1 \end{bmatrix} $$

and an unpolarized state can be written $$ \rho = \frac{1}{2}I = \frac{1}{2}(|H\rangle\langle H| + |V\rangle\langle V|). $$

An ideal polarizer corresponds to a projection onto a specific polarization, for example a vertical polarizer would simply be

$$ P_V = |V\rangle\langle V|, $$

and polarizer with a finite extinction ratio can be written as a projection onto a partially polarized state

$$ P = a|H\rangle\langle H| + b|V\rangle\langle V|. $$ If we initially have an unpolarized state, then the state after such a polarizer is

$$ P\rho = \frac{1}{2}(a|H\rangle\langle H| + b|V\rangle\langle V|) $$ and the polarization extinction ratio is $\left|\frac{a}{b}\right|$ (assuming $a>b$).

Passing the state through a second identical polarizer results in the state

$$ P^2\rho = \frac{1}{2}(a^2|H\rangle\langle H| + b^2|V\rangle\langle V|) $$ with extinction ratio $\left|\frac{a^2}{b^2}\right|$, which again assuming $a>b$ is greater than the ratio achieved using only one polarizer.

Using this model of a non-ideal polarizer we get the results that stacking polarizers does improve the extinction ratio. However, it is very possible for a physical device to introduce some amount of depolarization, and in that case the above model is not a good description of the polarizer anymore, and sequential polarizers could never improve the extinction ratio above the limit set by the induced depolarization. In my personal experience, combining two relatively low ER polarizers (say 1000:1) can improve the extinction ratio, but this ceases to be useful for polarizers with very high extinction ratios. You also have the practical problem of simply aligning the axes of these polarizers well enough for such an exercise to be useful even in an ideal case.

  • $\begingroup$ Over-thinking, too abstract. With only linear polarization in play, you only have two components. With the polarizers assumed to be aligned, those two are independent. $\endgroup$
    – John Doty
    Aug 31, 2023 at 14:33
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    $\begingroup$ Could you try to make a more coherent point? $\endgroup$
    – fulis
    Aug 31, 2023 at 14:40
  • $\begingroup$ What do you mean by more coherent? $\endgroup$
    – John Doty
    Aug 31, 2023 at 14:59
  • $\begingroup$ (+1) Good answer and I especially liked the final paragraph. The devil is always in the details. $\endgroup$
    – Ed V
    Sep 6, 2023 at 1:15

Everyone is ignoring the reality of the situation. The extinction ratio of a given polarizer is usually limited by stress induced birefringence, or some other defect.

So, if I have a polarizer (doesn't matter which) with an ER of 100:1, and transmit a light source with an ER of 100:1, the result will still be 100:1 because some of the light polarized in the "100" direction will be rotated into the "1" axis due to the defect.

  • $\begingroup$ Thanks for your input. But please consider that I already had precisely this discussion with basically different opinions and explanations...which is the motivation to post this question in the first place: To get a definite answer from theory (and then discuss limiting effects such as depolarization, scattering on defects, stress-induced birefringence etc. as needed) $\endgroup$
    – srhslvmn
    Sep 6, 2023 at 0:01
  • $\begingroup$ There's nothing fundamental about polarization that wouldn't let you achieve a perfect state of polarization. In fact, if you had a 100:1 ER represents a state of perfect polarization. Using a polarimeter you may achieve any other polarization state. $\endgroup$
    – JQK
    Sep 6, 2023 at 0:44
  • $\begingroup$ (+1) Your statement about stress induced birefringence brought back a not so pleasant memory of finding out the hard way that sapphire windows are not so good around 4 kelvin. $\endgroup$
    – Ed V
    Sep 6, 2023 at 1:19
  • 1
    $\begingroup$ Birefringence by itself doesn't affect the degree of polarization, it just rotates the state of polarization. Linear loss cannot induce depolarization either. Spatial non-uniformities, however, are a contributor to a lower degree of polarization, because in detection you average over the whole transverse profile of the beam. Ultimately, whether or not stacking polarizers improves the degree of polarization is situation dependent, but it is possible and I've personally observed it. $\endgroup$
    – fulis
    Sep 6, 2023 at 9:15
  • $\begingroup$ Birefringence, in general, will move the state of polarization from one point on the Poincare sphere to another. If you add a random variable, spatially or temporally, then, while there is an always a definite state of polarization, the description on the Poincare sphere may be a time-varying distribution. $\endgroup$
    – JQK
    Sep 6, 2023 at 11:27

The OP requests a formal proof, using Stokes vector and Mueller calculus formalism, as to whether stacking polarizers either does (or does not) theoretically improve the extinction ratio of a light source, e.g., a laser diode. Issues arising in practice, such as light scattering, reflection at interfaces, polarizer imperfections, depolarization, and so on, are to be neglected.

To begin, figure 1 shows the posited optical component train and the relevant Mueller calculus calculation that would be used to compute (manually) the output Stokes vector at G:

Model and Mueller calculus computation

The manual computation is straightforward, albeit a bit tedious, once the Mueller matrices for the linear polarizers have their requisite x and y transmittances specified. This is discussed below, in connection with the extinction ratio, $\epsilon$.

Now consider figure 2, which shows the assumed Mueller matrix for the linear polarizers and the dialog box for the polarizer block in my optical calculus software used to perform the non-manual computations:

Polarizer Mueller matrix and dialog box

The polarizer Mueller matrix is that of an ideal linear dichroism optical element. It is widely used to model ideal linear polarizer optical elements, as per Shurcliff, Kliger et al., and Jensen et al. As a convenience, two addendum figures, after the references, explicitly show the matrices used by the listed authors.

Note that the software polarizer block, that evaluates the polarizer’s Mueller matrix, allows for user-specification of the x and y transmittances. It also has a convenience checkbox that can be used to render the block transparent, i.e, the block then would simply transmit through whatever input it received. This facilitates trial removal of inline optical components.

Figure 3 shows how the laser light source is modeled as a unit intensity unpolarized Stokes vector followed by a non-ideal x-oriented linear polarizer having extinction ratio, $\epsilon$, defined as shown in the figure. This definition, which will be used throughout what follows, has long been in common usage and is found in, e.g., the reference by Kliger et al. on page 30. Note that $\epsilon$ ranges from 0, for an ideal polarizer, to 1, for non-polarizing. In the laser model, the non-ideal x polarizer was used to produce an output Stokes vector with $\epsilon = 0.01$, as explained below. The remaining blocks simply verify that the $\epsilon$ value is as specified. This is done by processing the laser output Stokes vector through orthogonal ideal analyzers and calculating the quotient of the light intensities they transmit.

Laser source model

It is also common to define $\epsilon$ as the reciprocal quantity, as per the link proved by the OP. Accordingly, such extinction ratios range from 1, for non-polarizing, to infinity, for an ideal polarizer. This alternative definition will not be used herein.

The figure shows how the model of the laser is formulated and processed via optical calculus modeling software. The output of the laser is shown in the software’s Stokes vector sink dialog box. The exact Stokes vector of the laser output is shown to the left of the dialog box. It is easily calculated manually, as shown in the Mueller calculus expressions in figure 4:

Manual Mueller calculus computation

The figure shows that obtaining a given $\epsilon$ is simply a matter of specifying a y component intensity that is $\epsilon$ times the intensity of the x component of the non-ideal x polarizer. With x + y = 1 and $\epsilon = 0.01$, this yields x = 100/101 and y = 1/101. The laser output Stokes vector is used in all subsequent calculations.

The degree of polarization, P, is a measure of polarization purity. It ranges from 0, for unpolarized light, to unity, for perfectly polarized light. The laser output Stokes vector has P = 99/101, as shown in the figure.

With the above preliminaries out of the way, the question is now this: what will happen to the degree of polarization if 1, 2, or 3 good polarizers (i.e., all with $\epsilon = 0.00001$) are added after the light source in an attempt to improve the polarization of the light source?

To answer this, consider figure 5:

Measurement protocol

To perform the optical calculus computation via the software, all that is necessary is to add another Stokes vector sink at the output of polarizer E. Then the simulation is run multiple times, with polarizer D (or both D and E) made transparent via their respective dialog box checkboxes.

The results are shown in figure 5. After polarizer C, the degree of polarization is high and the intensity of y polarized light is very low. Adding one or two more polarizers has only a very small positive effect. The effects are so small as to be negligible.

In a real system, a variety of non-idealities usually exist and these often cause significant departures from idealized modeled behavior. Mueller matrices (and Jones matrices) have demonstrated their utility, effectiveness and value over many decades, but they cannot perfectly model real optical elements such as polarizers. In particular, these calculuses assume individual light rays, not bundles of light rays. So they require homogeneous optical media. The typical tabulated matrices neglect scattering, reflections, depolarization, and so on, that real optical media exhibit. It is always possible to construct ever more complex matrices and models, in an attempt to achieve maximum verisimilitude, but this is ultimately futile, for two reasons:

  1. As statistician George Box famously stated, in slightly variant ways over the years, “all models are wrong, but some models are useful.”

  2. The Bonini paradox arises: the more complicated the model is made, the less understandable it becomes.

So it is futile, in general, to try to make a grand “everything plus the kitchen sink” optical calculus matrix. A more productive route is to size up the various complexities and deal with the most relevant of them, neglecting the rest.

References for the Mueller matrices for the ideal and non-ideal x polarizers:

  1. H.P. Jensen, J.A. Schellman, T. Troxell, “Modulation Techniques in Polarization Spectroscopy”, Applied Spectroscopy, 32 (1978) 192-200. Table IV with only LD > 0. (LD stands for linear dichroism, as used in the paper.)

  2. W.A. Shurcliff, Polarized Light, Harvard University Press, Cambridge, MA, 1962. Page 166 for ideal polarizers and page 168 for “Other polarizers”.

  3. D.S. Kliger, J.W. Lewis, C.E. Randall, “Polarized Light in Optics and Spectroscopy”, 1st ed., Academic Press, Boston, MA, 1990. Page 285 for ideal polarizers and page 290 for Linear Dichroism matrices.

Addendum figures:

Shurcliff and Kliger et al. matrices


Jensen et al. matrices

  • $\begingroup$ This is for questions but it also applies to answers and your post in particular. $\endgroup$ Sep 6, 2023 at 0:18
  • $\begingroup$ @ZeroTheHero In this case, I would accept the text-containing figures. The only figure where typesetting wouldn't involve too much work are the two bullet points in fig. 5. The rest is either mostly equations/formulae (which IMHO nobody is going to search for), figures reused from older answers, or screenshots from publications, which are already referenced and can be downloaded and searched. $\endgroup$
    – srhslvmn
    Sep 6, 2023 at 0:31
  • $\begingroup$ @srhslvmn with due respect: you are free to accept what you like but for the benefit of the community and users at large, having a searchable website and taking time to reflect on a question or an answer so it is something more than what can be cut-and-pasted from another source are considered important for this community. Clearly the vote outcomes of both the linked questions and multiple answers leaves no doubt about this. $\endgroup$ Sep 6, 2023 at 0:43
  • $\begingroup$ @srhslvmn You might want to copy the whole answer, just in case it vanishes away, softly and suddenly, like the snark hunter. (The P value was actually 1, not 0. Just a typo when I copied the numbers.) $\endgroup$
    – Ed V
    Sep 6, 2023 at 0:47
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    $\begingroup$ @EdV Do you think it's time for an Optics and Photonics SE proposal in Area 51? Asking and answering those questions here often feels like an uphill battle, and in my opinion Physics SE is too big! 1, 2. $\endgroup$
    – uhoh
    Sep 8, 2023 at 4:11

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