This question is referred to those who are familiar with the concept of weak measurement.

In short:

How can the polarization of a photon be coupled to the position of a pointer state? What is the pointer state? The position of some particle? How to realise that in a lab?

In detail:

Let's say we want to weakly measure the polarization of a photon. We write the observable $A=|H\rangle\langle H| - |V\rangle\langle V|$ and the initial state shall be $|\psi\rangle = \alpha |H\rangle + \beta |V\rangle$. Now we weakly couple it to a pointer by sending it through a birefringent element, i.e. depending on the polarization the photon needs more time to pass the birefringent element.


The Hamiltonian whichs represents this coupling is given by $H=\lambda A \otimes P$ where $\lambda$ is the coupling strenght and $P$ the momentum operator of the pointer. Now $\exp{(-\text i H t)}$ will act as follows:

$$|\psi\rangle \otimes |g\rangle = (\alpha |H\rangle + \beta |V\rangle) \otimes |g\rangle \;\longrightarrow\; ( \alpha |H\rangle \otimes|g_+\rangle + \beta |V\rangle\otimes|g_-\rangle)$$

where the first degree of freedom is clearly the polarization of the photon and the second is the pointer state (a gaussian say). Both $g_+$ and $g_-$ refer to small shifts due to the birefringent element.

Then there is this well-known fact that a post-selection which leads to an imaginary weak value $A_w$ causes a shift in the momentum-space of the pointer proportional to $\text{Im} A_w$.

The weak value is defined as $$A_w = \frac{\langle \phi|A|\psi\rangle }{\langle \phi|\psi\rangle},$$ where $|\phi\rangle$ is the post-selected polarization state of the photon.

(reference: https://arxiv.org/abs/0706.4207 or https://arxiv.org/abs/0911.5139)

Mathematically this is easy to understand, but I have a very conceptual problem with all that: How do I have to understand the pointer state? Is it a position of a particle? If yes, how can its position be coupled (in a physically understandable way) to a photon? And how can a post-selection affect its momentum?


2 Answers 2


I find that quantum lingo always tries to be so general that it sometimes makes the simple things seem complicated.

In the case of a the experiment that you're referencing, the "pointer state" is the position of the photon when it hits the detector. Even more plainly, since we're dealing with a laser (ensemble of photons), the pointer is the position expectation value of the ensemble, which is the center of the Gaussian laser beam.

When we say that we're coupling the pointer (the position) to the system observable (the polarization state), all we really mean is that we're putting the photon through some physical process that affects one aspect of the photon (its position) depending on some other aspect of the photon (its polarization state).

In this case, the birefringent crystal does exactly that because it has two different indices of refraction for two different polarization states. When you shine that laser beam in with the initial state $|\psi\rangle = \alpha|H\rangle + \beta|V\rangle$, then the crystal will measure the photon to be in the $|H\rangle$ state with a probability of $\alpha^2$ and in the $|V\rangle$ state with a probability of $\beta^2$. This means that $\alpha^2$ of the total photons will be refracted at one angle, and the rest will be refracted at some other angle, meaning that the final position of each photon is coupled to its polarization state as measured by the crystal.

You get the same effect in a Stern-Gerlach device where atoms in different states of excitement are deflected differently by a magnetic field. In that case, the deflection is a momentum boost rather than a spatial translation, but the idea of coupling two aspects of one particle is the same.


The pointer state is indeed the position of a particle. And surprisingly it is the position of the photon itself. The realization in the lab is already expressed in your detailed question, namely, with a birefringent crystal:

Essentially the observable of the system and the pointer state are two different degrees of freedom of the same quantum system. Actually, in most applications, it is the case that the observable and the pointer correspond to two different degrees of freedom of the same system and rarely to two different systems. In your example of the photon polarization measured with the birefringent crystal, the displacement of the two different polarizations in the crystal serves as the displacement of the pointer state, i.e. the position of the beams. So due to the crystal the polarization is coupled with the displacement, i.e. the pointer.


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