Polarisation of photons exiting a birefringent crystal

[Badgerinapie]

Homework Statement

The problem is taken from A Modern Approach to Quantum Mechanics by Townsend, and is stated as follows:

2.17. Linearly polarized light of wavelength 5890Å is incident normally on a birefringent crystal that has its optic axis parallel to the face of the crystal, along the x axis. If the incident light is polarized at an angle of 45° to the x and y axes, what is the probability that the photons exiting a crystal of thickness 100.0 microns will be right-circularly polarized? The index of refraction for light of this wavelength polarized along y (perpendicular to the optic axis) is 1.66 and the index of refraction for light polarized along x (parallel to the optic axis) is 1.49.

The answer given in the back of the book is 0.12, though I have not managed to get this result.

Homework Equations

Classically, light traveling through a crystal of refractive index $n$ will pick up the phase factor
$\phi = \dfrac{n \omega z}{c} = \dfrac{2\pi n}{\lambda_0}$

The Attempt at a Solution

I couldn't see any obvious way of doing this with Bra-ket notation, which is odd since most of the preceding two chapters were spent building this notation up gradually; I'm probably missing something obvious. Instead I tried to calculate the intensity of right-polarised light the crystal as a proportion of the overall intensity. This classical result should equate with the probability of an individual photon being right-polarised.

The incoming light is angled 45° to the x & y-axes so the respective components of the electric field are given as
$\begin{align} E_x &= E_0 \cos (\dfrac{2\pi n_x \Delta z}{\lambda_0} - \omega t) \\ E_y &= E_0 \cos (\dfrac{2\pi n_y \Delta z}{\lambda_0} - \omega t) \end{align}$
where $\Delta z$ is the thickness of the crystal. This gives the overall phase difference between the two components upon exiting the crystal as
$\Delta \phi = \dfrac{2\pi \Delta z}{\lambda_0} (n_x - n_y)$

Now, my reasoning gets a touch hand-wavey here. Right-polarised light is where $\Delta \phi = \frac{\pi}{2}$, and the sine function happens to be a maximum for this phase difference, so I'd expect $\sin \Delta \phi$ to give the component of right-polarised light(?). So, the proportional intensity of right-polarised light should be given by $\sin^2 \Delta\phi$.
But, plugging in the numbers in the question gives the proportion/probability 0.58, which is way off from the value of 0.12 given as a solution. So I suspect I have misunderstood something crucial about this question.

 

[TSny]

Hello, Badgerinapie. Welcome to PF!

You mentioned bra-ket notation. I think that would be a nice way to work it out. Can you state what you would use for the ket representing x-polarization? y-polarization? polarization at 45^o to the x and y axes? Right hand polarization?

 

[Badgerinapie]

Hello, Badgerinapie. Welcome to PF!

Thank you :)

You mentioned bra-ket notation. I think that would be a nice way to work it out. Can you state what you would use for the ket representing x-polarization? y-polarization? polarization at 45^o to the x and y axes? Right hand polarization?

Sure, I'd represent x and y polarisations with $\left|x\right\rangle$ and $\left|y\right\rangle$ respectively. Thus, for linear polarisation at some angle $\phi$ to the x-axis I'd write $\left|\psi\right\rangle = \cos\phi \left|x\right\rangle + \sin\phi \left|y\right\rangle$. Setting $\phi=45°$ gives $\left|\psi\right\rangle = \dfrac{1}{\sqrt{2}} \left(\left|x\right\rangle + \left|y\right\rangle\right)$. The matrix representation of this with $\left|x\right\rangle$ and $\left|y\right\rangle$ as the basis vectors would then be $\left|\psi\right\rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix}1 \\ 1\end{pmatrix}$.

For right polarisation I'd use $\left|R\right\rangle = \frac{1}{\sqrt{2}} \left(\left|x\right\rangle + i\left|y\right\rangle\right)$. Or, in matrix representation we'd write $\left|R\right\rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix}1 \\ i \end{pmatrix}$. The corresponding Bra is then written as $\left\langle R\right| = \dfrac{1}{\sqrt{2}} \begin{pmatrix}1 & -i \end{pmatrix}$.

I take it that we have to create some modified wavefunction, $\left|\psi\prime\right\rangle$ say, to account for the physical effect of the crystal, and then do $|\left\langle R | \psi\prime\right\rangle|^2$ to get the probabilty(?).

Cheers.

 

[TSny]

Ok, that all looks good!

I take it that we have to create some modified wavefunction, $\left|\psi\prime\right\rangle$ say, to account for the physical effect of the crystal, and then do $|\left\langle R | \psi\prime\right\rangle|^2$ to get the probabilty(?).

Yes, that's exactly right. The effect of the crystal is to increase the phase angle of the $|x\rangle$ component relative to the $|y\rangle$ component by $\Delta \phi$. Since the state going into the crystal is $\left|\psi\right\rangle = \dfrac{1}{\sqrt{2}} \left(\left|x\right\rangle + \left|y\right\rangle\right)$, how would you modify this state to include a phase shift angle of $\Delta \phi$ of the $|x\rangle$ component relative to the $|y\rangle$ component?

 

[Badgerinapie]

Yes, that's exactly right. The effect of the crystal is to increase the phase angle of the $|x\rangle$ component relative to the $|y\rangle$ component by $\Delta \phi$. Since the state going into the crystal is $\left|\psi\right\rangle = \dfrac{1}{\sqrt{2}} \left(\left|x\right\rangle + \left|y\right\rangle\right)$, how would you modify this state to include a phase shift angle of $\Delta \phi$ of the $|x\rangle$ component relative to the $|y\rangle$ component?

Okay I think I've got it, thank you! I modified the wavefunction with a complex phase factor on each term as in $\left|\psi\right\rangle = \dfrac{1}{\sqrt{2}} \left(e^{i\phi_x} \left|x\right\rangle + e^{i\phi_y} \left|y\right\rangle\right) = \dfrac{e^{i\phi_x}}{\sqrt{2}} \left(\left|x\right\rangle + e^{i\Delta\phi} \left|y\right\rangle\right)$ where $\Delta\phi = \phi_y - \phi_x$. Now I could ignore the phase term $e^{\phi_x}$ here as it'll come out when I take the magnitude squared, but I'll leave it in for rigour's sake while I'm still getting to grips with the basics.

So, the inner product becomes:
$\left\langle R|\psi\right\rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix}1 & -i\end{pmatrix} \dfrac{e^{i\phi_x}}{\sqrt{2}} \begin{pmatrix}1 \\ e^{i\Delta\phi} \end{pmatrix} = \dfrac{e^{i\phi_x}}{2} \left(1 + e^{i(\Delta\phi - \frac{\pi}{2})}\right)$.
Multiplying this by its complex conjugate gives:
$\left|\left\langle R|\psi\right\rangle\right|^2 = \dfrac{e^{i\phi_x}}{2} \left(1 + e^{i(\Delta\phi - \frac{\pi}{2})}\right) \dfrac{e^{-i\phi_x}}{2} \left(1 + e^{-i(\Delta\phi - \frac{\pi}{2})}\right) = \dfrac{1}{2}\left(1 + \sin\Delta\phi\right)$.
Plugging $\Delta\phi = \dfrac{2\pi\Delta z}{\lambda_0} \left(n_y - n_x\right)$ into the above expression gives the value 0.12, as stated in the back of the book.

Thank you TSny, your help is much appreciated! :)

 

[TSny]

Yes, nice work.

It might be worthwhile to point out that you don't really need to use the matrix representations of the states. You can stick with the abstract bra and ket notation.

$\left\langle R|\psi\right\rangle = \dfrac{1}{\sqrt{2}}(\langle x| - i \langle y|) \dfrac{e^{i\phi_x}}{\sqrt{2}} \left(\left|x\right\rangle + e^{i\Delta\phi} \left|y\right\rangle\right) = \dfrac{e^{i\phi_x}}{2} \left(1 + e^{i(\Delta\phi - \frac{\pi}{2})}\right)$ using the orhonormality of the x and y kets.

(But I used the matrix representation also when I first worked it out.)