B Bell Test Configuration Question

[plmustard]

TL;DR

Assuming entangled photons created with a horizontal/vertical basis, does demonstrating Malus' law in a Bell Test require one of the polarizing filters to be set at either the horizontal or vertical basis?

Qutools makes quantum physics kits for educational purposes. Its quED kit is designed to help students learn about entanglement by performing Bell tests. In the manual section 5.1 it describes "the simplest test to verify entanglement of photon pairs."

My question is if the entangled photons have a horizontal/vertical basis as described (where 0 degrees corresponds to horizontal polarization), but the alpha polarizing filter is fixed to 30 degrees instead of 0 degrees as described, would the rotation of the beta filter still result in correlations between the two filter settings consistent with Malus' law such that sin^2 (beta - alpha)? In other words if alpha was at 30 degrees and beta at -30 degrees (or 330 degrees), and the system is set to a horizontal/vertical basis, would the photons be different 75% of the time because sin^2 (-30 - 30) = sin^2 (-60) = 0.75? Or would that calculation only apply if alpha (or beta) were fixed at 0 degrees because of the photon horizontal/vertical basis being at 0/90 degrees? Thank you for any explanation.

 

[DrChinese]

Although the nominal notation is HV/VH entangled, the pair respects the usual correlation function at any pair of angles. That being a function of the difference between the two angle settings. By convention, that difference is usually called "theta". A couple of comments:

a. Although the formula sin^2 (theta) closely resembles Malus' Law, it is actually derived independently. I won't include the derivation here, as it is a bit complicated.
b. The other common type of entanglement is HH/VV. Like the other type, it operates at any pair of angles. It takes the form cos^2(theta).

 

[vanhees71]

I can guess only from the result that the prepared two-photon state is the "singlet state":
$$|\Psi \rangle=\frac{1}{\sqrt{2}} [\hat{a}^{\dagger}(\vec{k}_1,\text{H}) \hat{a}^{\dagger}(\vec{k}_2,\text{V}) - \hat{a}^{\dagger}(\vec{k}_1,\text{V}) \hat{a}^{\dagger}(\vec{k}_2,\text{H})]|\Omega \rangle,$$
where $|\Omega \rangle$ is the photon-vacuum state.

The polarization state when measured in direction $\alpha$ ($\alpha$ is the angle between the orientation of the corresponding polarization filter relative to the direction of what we call H-polarized) is given by the creation operator
$$\hat{a}^{+}(\vec{k},\alpha)=\cos \alpha |\hat{a}^{\dagger}(\vec{k},\text{H} \rangle + \sin \alpha |\hat{a}^{\dagger}|\vec{k},\text{V} \rangle.$$
Thus the probability that a photon with momentum $\vec{k}_1$ is let through an $\alpha$ oriented filter and a photon with momentum $\vec{k}_2$ is let through a $\beta$-oriented filter is
$$P(\alpha,beta)=\langle \Omega|\hat{a}(\vec{k}_1,\alpha) \hat{a}(\vec{k}_2,\beta) |\Psi \rangle|^2.$$
This gives you, using the commutation relation
$$[\hat{a}(\vec{k}_1,i),\hat{a}^{\dagger}(\vec{k}_2,j)]=\delta_{\vec{k}_1,\vec{k}_2} \delta_{ij}$$
for $i,j \in \{\text{H},\text{V} \}$.
$$P(\alpha,\beta)=\frac{1}{2} (\cos \alpha \sin \beta-\sin \alpha \cos \beta)^2=\frac{1}{2} \sin^2(\alpha-\beta).$$
Let $\bar{\alpha}=\pi/2-\alpha$ and $\bar{\beta}=\pi/2-\beta$. Then
$$\hat{\alpha}^{\dagger}(\vec{k},\bar{\alpha})=-\sin \alpha \hat{a}^{\dagger}(\vec{k},\alpha)+\cos \alpha \hat{a}^{\dagger}(\vec{k},\alpha)$$
and analogously for $\bar{\beta}$
Then the other possibilities for the polarization measurements are
$$P(\bar{\alpha},\beta)=\frac{1}{2}(-\sin \alpha \sin \beta - \cos \alpha \cos \beta)^2=\frac{1}{2} \cos^2(\alpha-\beta),$$
$$P(\alpha,\bar{\beta})=\frac{1}{2}(\cos \alpha \cos \beta+\sin \alpha \sin \beta)^2=\frac{1}{2} \cos^2(\alpha-\beta),$$
and
$$P(\bar{\alpha},\bar{\beta})=\frac{1}{2}(-\sin \alpha \cos \beta + \cos \alpha \sin \beta)^2=\frac{1}{2} \sin^2(\alpha-\beta).$$

The single-photon probabilities are
$$P_1(\alpha)=P(\alpha,\beta)+P(\alpha,\bar{\beta})=\frac{1}{2}[\sin^2(\alpha-\beta)+\cos^2(\alpha-\beta)]=\frac{1}{2}$$
and
$$P_1(\bar{\alpha})=P(\bar{\alpha},\beta)+P(\bar{\alpha},\bar{\beta})=\frac{1}{2} [\cos^2(\alpha-\beta)+\sin^2(\alpha-\beta)=\frac{1}{2}.$$
The same comes of course out for the photons with momentum $\vec{k}_2$. The single photons are totally unpolarized.

Nevetheless due to the entanglement given by the prepared state $|\Psi \rangle$ you get a full-contrast interference pattern for the photon with $\vec{k}_1$ when only considering the case where the photon with $\vec{k}_2$ is $\beta$-polarized (i.e., when you fix $\beta$ and vary $\alpha$). That's the typical case for Bell states: Though the single photons are precisely unpolarized there are strong correlations between the measured polarizations in the same direction. E.g., if you set the polarizer for the $\vec{k}_1$ photon to $\alpha$ and consider only $\vec{k}_2$ photons for which the $\vec{k}_1$ photons go through the $\vec{k}_2$ photon is necessarily $\bar{\alpha}$ polarized (i.e., setting the polarizer for the $\vec{k}_2$ photon to $\beta=\bar{\alpha}$ also all the corresponding $\vec{k}_2$ photons go through, and this happens with 50% probability considering all photon pairs investigated.

 

[plmustard]

Thank you both DrChinese and vanhees71 - sorry for my delayed reply. I take it from your responses, which I admit I don't fully understand, that in essence the quED kit says that the alpha polarizing filter should be set on 0 degrees at horizontal not because it is required to realize Malus' law and/or the (separately derived) quantum predictions of sin^2 (beta - alpha), it's just that Qutools thinks that fixing alpha at 0 degrees makes the set-up simpler for the beginner like me.