2

While reading some questions about the polarisation of a photon, I was wondering about its quantum mechanical representation as a classical vector in the electromagnetic field operator : \begin{equation}\tag{1} \hat{\vec{A}}(t, \vec{r}) = \sum_{k, \, \sigma} \big( \hat{a}_{k,\, \sigma} \, \vec{\varepsilon}_{k, \, \sigma} \, e^{i \vec{k} \cdot \vec{r} \,-\, i \omega_k \, t} + \text{h.c}\big). \end{equation} Here, the polarisation state is represented by a classical vector $\vec{\varepsilon}_{k, \, \sigma}$ in $\mathbb{R}^3$ (or $\mathbb{C}^3$ for the circular polarisation), where $\sigma = \pm 1$ are the two orthogonal circular polarisation states, such that (the wave is transverse): \begin{align}\tag{2} \vec{k} \cdot \vec{\varepsilon}_{k, \, \sigma} &= 0, \\[12pt] \vec{\varepsilon}_{k, \, \sigma}^{\, *} \cdot \vec{\varepsilon}_{k, \, \sigma'} &= \delta_{\sigma \sigma'}. \tag{3} \end{align} The wave's amplitude is an operator : $\hat{a}_{k, \, \sigma}$, with the usual bosonic commutation relations.

Is there a Heisenberg relation on the polarisation states, or is this just a non-sense question since it's a "classical property" of the wave?

Alan Whitteaker
  • 354
Cham
  • 8,015

1 Answers1

2

That depends on what polarization states you're talking about.

There's no uncertainty relation between clockwise and counterclockwise polarization, because they're orthogonal; it's just like how there's no uncertainty relation between $p_x$ and $p_y$. To be more precise, focusing on the subspace of photons with fixed momentum $k$ and suppressing $k$, the operators that test for clockwise and counterclockwise polarization are $$n_+ = a_+^\dagger a_+, \quad n_- = a_-^\dagger a_-$$ where $a_+/a_-$ annihilate clockwise and counterclockwise polarized photons. These two operators commute, so there's no uncertainty relation between them.

The same thing goes for any polarization states used in the expansion of a quantum field operator, because they are defined so that the creation and annihilation operators of different polarizations commute. For example, for a spinor field the polarizations could correspond to spin up and spin down. And indeed there's no uncertainty relation; if you know an electron is spin up, you also know it definitely isn't spin down.

Is there an Heisenberg relation on the polarisation states, or is this just a non-sense question since it's a "classical property" of the wave?

That's a misinterpretation of what uncertianty means. Uncertainty relations only exist if you use different, incompatible bases. As such, they have more to do with linear algebra than quantum mechanics. You only associate quantum mechanics with uncertainty because it yields one new, very famous uncertainty relation.

For example, there exists an uncertainty relation between clockwise polarization and any linear polarization. And there exists an uncertainty relation between horizontal/vertical polarization and diagonal polarization, which I used here. It holds just as well in classical mechanics: if you know a wave is diagonally polarized, it must be in a superposition of horizontally and vertically polarized.

knzhou
  • 107,105