2
$\begingroup$

I'd like to help use the following exercise to reinforce my understanding of some basic concepts of quantum states. Here's a picture of the setup:

enter image description here

Now, I'll try and make my series of questions about the insights taken from this exercise as clear as possible.

First question - why is the intensity after going through the first slit $1/2 \ I_0$ regardless of slit orientation?

Firstly, I attempted the following, which was flat out wrong (also I don't know how to use Bra-ket notation here, I'm going to use unit vectors instead.. hopefully that's okay? Please excuse this hideous 'notation', just trying to illustrate my confusions):

$$V' = \cos{(7\pi/12)} \hat V + \sin{(7\pi/12)}\hat H$$

Where $V'$ is the new vertical polarization state, and $\hat V$ and $\hat H$ are the vertical and horizontal states respectively.

I posited this because $\hat V$ and $\hat H$ represent the basis vectors $(0,1)$, $(1,0)$ that represent the vertical and horizontal aperture arrangements of the filters. We are instructed, as far as I could understand, to construct a new basis vector when there is an arbitrary aperture angle as a linear combination of the typical horizontal and vertical arrangements.. which led to something like this. This led me to state that the probability for absorption of the vertical filter to be..

$$cos^2(7\pi/12) \approx 0.06$$

It's clear that this stuff is fuzzy to me, but hopefully my confusion is laid relatively bare enough. $7\pi/12$ is merely the angle I took from $\theta_0 = 0$, so $\alpha = \pi/12 + \pi/2$.

This was an attempt to answer the first part of the question, which I'm aware was an answer of probability and not intensity, but I wasn't sure how to do this any other way. Anyway, that garbage attempt for physics aside, I'm told that the answer to this is that the intensity is halved from the first aperture, regardless of orientation of the slit. I don't really know why - my only guess is that it has something to do with the light being unpolarized. But, I mean, thinking off the cuff, if the light is a beam with a diameter roughly equal to the filter, then the filter would have to be a gap half the size of the circle in order for only half the light to make it through! I don't have any good justification as to why this doesn't make sense.

Second question - what is the sense behind the approach to find the probability for a photon to be detected by a PMT placed after the final filter?

My lecturer's answer states that after the light goes through the first slit, it is polarized along the $\alpha$ axis, which I think means it basically forms a line parallel and on top of the slightly diagonal, dotted line in the first filter. He contends that, since it makes an angle $\alpha + \beta$ with the transmission axis of the second filter (which is not apparent to me visually), the beam is reduced by a factor

$$cos^2(\alpha + \beta)$$

and by a similar argument for the third filter:

$$cos^2(\beta + \gamma)=cos^2(55^0) \approx 0.33$$

Which leads to a final intensity of $I \approx 0.068 I_0$.

So my specific confusions I need addressed are:

  • Why the intensity after going through the first slit is $1/2 \ I_0$ regardless of slit orientation?

  • Why we use a cosine squared term. I thought I knew why in my screw up above but I don't think I do anymore.

  • Why, in order to find the proper new orientation of polarization, we add the previous angle with the new one as our cosine argument

$\endgroup$
  • $\begingroup$ try not to confuse light , classical electromagnetic waves, where what is waving is the energy transferred by the beam, and photons. Photons are not ligth, they are the building blocks of light. Look at this answer of mine physics.stackexchange.com/q/154468 $\endgroup$ – anna v Apr 3 '18 at 17:08
2
$\begingroup$

Your question is interesting as the answer involves more than basic quantum mechanics.

Unpolarized light is actually best described as a mixed state, rather than a pure state. Basically, unpolarized light is a statistical superposition of two perpendicular polarization in the sense that half of the photons have one polarization, and half of the photon the other. Mixed states are described by a density matrix, which here would take the form \begin{align} \rho &=\frac{1}{2}\vert H\rangle\langle H\vert +\frac{1}{2}\vert V\rangle\langle V\vert\, ,\\ &=\frac{1}{2}\vert \updownarrow \rangle\langle \updownarrow\vert +\frac{1}{2}\vert \leftrightarrow \rangle\langle \leftrightarrow\vert \tag{1} \end{align} In mixed states, there is no possibility for the vertically polarized and horizontally polarized states to interfere, and the factors of $\frac{1}{2}$ are the statistical weights of each polarization. Mixed states are represented by operators rather than kets.

There is a subtle difference between the interpretation of statistical weights, which are also probabilities, and the probabilities obtained by overlaps like $\vert\langle\psi\vert\phi\rangle\vert^2$: both types probabilities have different origins. One is tied to incoherent averages, as referred to in the linked wiki page.

Moreover, by symmetry, "which one" of the two polarization is completely undefined if the light is completely unpolarized. In other word, the completely unpolarized light is equally well described by the mixture $$ \rho =\frac{1}{2}\vert \nearrow \rangle\langle \nearrow\vert +\frac{1}{2}\vert \nwarrow\rangle\langle \nwarrow\vert\, . $$ (unfortunately, slanted double arrows are not so easily accessible so $\nearrow$ is a standin for $\updownarrow$ rotated by $45^\circ$ to the right, and $\nwarrow$ is a standin for $\leftrightarrow$ rotated by $45^\circ$ to the right) or for that matter, using any slanting of $\updownarrow$ and $\leftrightarrow$.

If you set your polarizer along the vertical axis, you are making (in the parlance) a projective measurement and the outcome is always the pure state $\vert \updownarrow\rangle\langle \updownarrow\vert $. Since the statistical weight of this state is $1/2$, you eliminate half of the intensity. Since there is nothing special about $\vert \updownarrow\rangle\langle \updownarrow\vert $, this will be true of any orientation. This is, in the formalism of mixed states, the content of the statement immediately below your figure.

After the first polarizer, the state is pure, and pure states are those that you are familiar with. We can do away with all the density matrix stuff and think of the state as the ket $\vert \updownarrow\rangle$, or any slanted thereof where the slanting would be parallel to the axis of the polarizer. Your $V'$ state is described by a ket $$ \vert V'\rangle =\cos\theta \vert \updownarrow\rangle +\sin\theta \vert \leftrightarrow\rangle\, . \tag{2} $$

What is the difference between (1) and (2)? In (1) is it not possible to find an orientation of the polarizers that will result in no transmission of the light. If there was such an orientation, light would be polarized at $90^\circ$ w/r to this direction. In (2) on the other hand, the ket $$ \vert H'\rangle = -\sin\theta \vert \updownarrow\rangle +\cos\theta \vert \leftrightarrow\rangle $$ is orthogonal to $\vert V'\rangle$ so the probability of obtaining the outcome $H'$ as a result of having the system initially in $\vert V'\rangle$ is $ \vert \langle H'\vert V'\rangle\vert^2=0$, i.e. no light would pass through a filter aligned parallel to the $V'$ direction if light was initially described by the polarization ket $\vert H'\rangle$.

Note that, from (2), you can recover Malus' law. The probability of light initially polarized as $\vert \updownarrow\rangle$ to exit in the state $\vert V'\rangle$ is $$ \vert \langle \updownarrow\vert V'\rangle \vert^2 =\cos^2\theta $$ if $V'$ makes and angle $\theta$ with the vertical. The attenuation in intensity from Malus' law follows immediately from the (discrete) probability of a photon going though two polarizers sets at a relative angle $\theta$.

Finally, it is possible to have partially polarized states, in the sense that (1) generalizes to $$ \rho= a\vert H\rangle\langle H\vert +b\vert V\rangle\langle V\vert $$ where $a+b=1$, where $a$ is the statistical probability of having an "H" photon and $b$ the prob of finding a $V$ photon in the beam. In this case, and contrary to (1), there is an orientation where the intensity of light will be minimal but non-zero.

The best reference on density matrix in 2-level systems, with discussion that is done in terms of Stern-Gerlach magnet but otherwise immediately applicable to polarization, is the first chapter of Blum, Karl. Density matrix theory and applications. Springer Science & Business Media, 2013, available as a google books. Indeed, section 1.1.2 is on polarization (in terms of Pauli matrices).

$\endgroup$
  • $\begingroup$ This is very helpful, thanks a bunch. Just a couple of questions: a) Why is it that half of the light has one polarization and half the other, strictly? b) Are you saying it is pointless to suggest $\vert \psi \rangle = \vert V'\rangle + \vert H'\rangle$ because there cannot be a superposition of these two states once polarized? $\endgroup$ – sangstar Apr 5 '18 at 12:29
  • $\begingroup$ Finally, is the bit with partially polarized states when we fire single photons instead of a beam? My lecturer states that with a vertically oriented polarizer, it is representable by $\vert \psi\rangle = a\vert V\rangle + b\vert H\rangle$. I think I'm beginning to realize the mixed states POV is better for a beam of light in this case, and me trying to represent it using a ket vector was my problem? $\endgroup$ – sangstar Apr 5 '18 at 12:38
  • $\begingroup$ @sangstar by definition, unpolarized light is an equal mixture of any two orthogonal polarizations. Unpolarized light cannot be written at the superposition of two polarization - that the difference between mixed and pure states. The states in a superposition can interfere, but not the (orthogonal) states in a mixture. Mixed states cannot be represented by a ket. $\endgroup$ – ZeroTheHero Apr 5 '18 at 12:41
  • $\begingroup$ @sangstar Only once it's polarized can you write the polarization state of the photon as a superposition like $\vert\psi\rangle = \frac{1}{\sqrt{2}}\left(\vert V'\rangle+\vert H'\rangle\right).$ A polarized state is pure, not mixed. A partially polarized mixed state would be one with an incoherent mixture of two orthogonal polarization. The various statistical weights would not be the same, i.e. could be $7/10$ and $3/10$. $\endgroup$ – ZeroTheHero Apr 5 '18 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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