One Photon Polarization
The polarization of a classical electromagnetic wave was introduced in the quantization of a single mode of the electromagnetic field here.
The most general case of an electric field in absence of charges, can be written as:
\mathbf E(\mathbf r, t) = \sum_\lambda \mathbf e_\lambda \left( E_\lambda(t) e^{i \mathbf k_\lambda \cdot \mathbf r} + \bar E_\lambda(t) e^{-i \mathbf k_\lambda \cdot \mathbf r}\right)
This formula express a sum of independent modes, and we considered the cases of polarized monochromatic traveling waves. Each of this modes in a characterized by a \mathbf k_\lambda vector, which define the direction of propagation and the frequency \omega_\lambda = c k_\lambda. For each mode, the electric field is perpendicular to the \mathbf k_\lambda vector.
It therefore belongs to a plane that can be spanned by two perpendicular unit vectors \mathbf e_{\lambda_1} and \mathbf e_{\lambda_2}, so that along \mathbf k_\lambda form a coordinate system.
So the modes \mathbf e_{\lambda_1} and \mathbf e_{\lambda_2} have the same \mathbf k_\lambda vector, but a different polarization associate with these two modes.
In my analysis, often quantized radiation discussions have simplified polarization.
We now introduce a special case with two modes: these modes share a wavevector \mathbf k (along \mathbf z) but have orthogonal polarizations.
Without any loss of generalization we are going to orient the axes so that two modes associated with the same \mathbf k_\lambda are along \mathbf x and \mathbf y:
\begin{aligned} \mathbf k_\lambda & = k \, \mathbf k \\ \mathbf e_{\lambda_1} & = e_x \, \mathbf i \\ \mathbf e_{\lambda_2} & = e_y \, \mathbf j \end{aligned}
Since there is a single \mathbf k vector, the classical field represents a traveling electromagnetic wave with dual polarization:
\mathbf{E}(\mathbf{r}) = \left(e_x E_x + e_y E_y \right) e^{i(\mathbf k \cdot \mathbf r -\omega t)} + \left(e_x \bar E_x + e_y \bar E_y \right) e^{-i(\mathbf k \cdot \mathbf r -\omega t)}
The most general quantum state for this traveling wave, described by two orthogonally polarized modes, resides in the tensor product Hilbert space of the individual mode states.
A convenient basis for this space is the number state basis, formed by tensor products of number states |\mathbf n_x, \mathbf n_y\rangle, representing n_x photons in the x-polarized mode and n_y photons in the y-polarized mode. The general state is thus a superposition of these basis states:
| \boldsymbol \Psi \rangle = \sum_{n_x, n_y} c_{n_x, n_y} |\mathbf n_x, \mathbf n_y\rangle
Even in this simplified scenario, the state space dimension grows exponentially with the number of photons per mode. Restricting each mode to a maximum of one photon still yields a four-dimensional state space, encompassing entangled states:
| \boldsymbol \Psi \rangle = c_{0, 0} |\mathbf 0, \mathbf 0 \rangle + c_{0, 1} |\mathbf 0, \mathbf 1 \rangle + c_{1, 0} |\mathbf 1, \mathbf 0 \rangle + c_{1, 1} |\mathbf 0, \mathbf 0 \rangle
In this analysis I will focus only on the case of one-photon polarized radiation and keep only the state with one photon neglecting entangled states:
| \boldsymbol \Psi \rangle = c_{0, 1} |\mathbf 0, \mathbf 1 \rangle + c_{1, 0} |\mathbf 1, \mathbf 0 \rangle
With the previous convention:
| \boldsymbol \Psi^{(1)} \rangle = \alpha |\mathbf 0_x, \mathbf 1_y \rangle + \beta |\mathbf 1_x, \mathbf 0_y \rangle, \quad | \alpha|^2 + |\beta|^2 = 1
Ignoring the state with no photons:
| \boldsymbol \Psi^{(1)} \rangle = \alpha |\mathbf 1_y \rangle + \beta |\mathbf 1_x \rangle, \quad \alpha^2 + \beta^2 = 1
This is one photon state and it is an eigenstate of the number operator \mathbf N = \sum_\lambda \mathbf a_\lambda^\dag \mathbf a_\lambda:
\begin{aligned} \mathbf N | \boldsymbol \Psi^{(1)} \rangle & = \sum_{\lambda=x,y} \mathbf a_\lambda^\dag \mathbf a_\lambda \left(\alpha |\mathbf 1_y \rangle + \beta |\mathbf 1_x \rangle\right) \\ & = \mathbf a_x^\dag \mathbf a_x \left(\alpha |\mathbf 1_y \rangle + \beta |\mathbf 1_x \rangle\right) + \mathbf a_y^\dag \mathbf a_y \left(\alpha |\mathbf 1_y \rangle + \beta |\mathbf 1_x \rangle\right) \\ & = \alpha \mathbf a_x^\dag \mathbf a_x |\mathbf 1_y \rangle + \beta \mathbf a_x^\dag \mathbf a_x |\mathbf 1_x \rangle + \alpha \mathbf a_y^\dag \mathbf a_y |\mathbf 1_y \rangle + \beta \mathbf a_y^\dag \mathbf a_y |\mathbf 1_x \rangle \\ & = \alpha \mathbf a_x^\dag \left( \mathbf a_x |\mathbf 1_y \rangle \right) + \beta \mathbf a_x^\dag \left( \mathbf a_x |\mathbf 1_x \rangle \right) + \alpha \mathbf a_y^\dag \left( \mathbf a_y |\mathbf 1_y \rangle \right) + \beta \mathbf a_y^\dag \left( \mathbf a_y |\mathbf 1_x \rangle \right) \end{aligned}
We know that the annihilation operator acting on a vacuum state gives zero, and the state |\mathbf 1_\lambda \rangle can be seen as |\mathbf 1_\lambda \rangle = \mathbf a_\lambda^\dag |\mathbf 0_\lambda \rangle:
\mathbf a_\lambda |\mathbf n_\lambda \rangle = \sqrt{n_\lambda} |\mathbf {n_\lambda - 1} \rangle
Therefore:
\begin{aligned} & \mathbf a_x |\mathbf 1_y \rangle = \mathbf a_x |\mathbf 0_x, \mathbf 1_y \rangle = 0 \\ & \mathbf a_x |\mathbf 1_x \rangle = \mathbf a_x |\mathbf 1_x, \mathbf 0_y \rangle = \sqrt{1} |\mathbf 0_x, \mathbf 0_y \rangle = |\mathbf 0_x \rangle \\ & \mathbf a_y |\mathbf 1_y \rangle = \mathbf a_y |\mathbf 0_x, \mathbf 1_y \rangle = \sqrt{1} |\mathbf 0_x, \mathbf 0_y \rangle = |\mathbf 0_y \rangle \\ & \mathbf a_y |\mathbf 1_x \rangle = \mathbf a_y |\mathbf 1_x, \mathbf 0_y \rangle = 0 \end{aligned}
Substituting back:
\begin{aligned} \mathbf N | \boldsymbol \Psi^{(1)} \rangle & = \alpha \mathbf a_x^\dag (0) + \beta \mathbf a_x^\dag |\mathbf 0_x \rangle + \alpha \mathbf a_y^\dag |\mathbf 0_y \rangle + \beta \mathbf a_y^\dag (0) \\ & = 0 + \beta |\mathbf 1_x \rangle + \alpha |\mathbf 1_y \rangle + 0 \\ & = \alpha |\mathbf 1_y \rangle + \beta |\mathbf 1_x \rangle \\ & = |\boldsymbol \Psi^{(1)} \rangle \end{aligned}
The one-photon state space is two-dimensional, mirroring the classical polarization vector space and the linear polarization basis (\mathbf e_x, \mathbf e_y) corresponds to the quantum basis (|\mathbf 1_x\rangle, |\mathbf 1_y\rangle).
A polarization \mathbf e at an angle \theta from the x-axis corresponds to the quantum state:
|\mathbf 1_\theta\rangle = \cos \left( \theta \right)|\mathbf 1_x\rangle + \sin \left( \theta \right)|\mathbf 1_y\rangle
This state describes a single photon with polarization \mathbf e_\theta at an angle \theta relative to \mathbf x.
Polarizing beam-splitter
Polarization beam-splitters measure photon polarization by directing photons into different output channels, each with a detector.
For a single photon, only one detector clicks (we arbitrarily choose +1 for transmission, -1 for reflection).
The quantum observable for this measurement has |\mathbf 1_x\rangle and |\mathbf 1_y\rangle as eigenstates, with eigenvalues +1 and -1 respectively; it express the sum of the projections associated with each polarization:
\mathbf A = | \mathbf 1_x \rangle \langle \mathbf 1_x | - | \mathbf 1_y \rangle \langle \mathbf 1_y |
This observable, in the (|1_x\rangle, |1_y\rangle) basis, is represented by a diagonal matrix:
\mathbf A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}
Consider a photon polarized at an angle \theta from the x-axis, described by the state:
|\mathbf 1_\theta\rangle = \cos \left( \theta \right)|\mathbf 1_x\rangle + \sin \left( \theta \right)|\mathbf 1_y\rangle
To find the probabilities of measurement outcomes using an x-aligned polarizing beam-splitter, we project |\mathbf 1_\theta \rangle onto the eigenstates of the measurement operator.
The eigenstates corresponding to transmission (+1 outcome) and reflection (-1 outcome) are |\mathbf 1_x\rangle and |\mathbf 1_y\rangle, respectively.
For the transmission, we project |\mathbf 1\rangle onto |\mathbf 1_x\rangle and take the squared modulus:
\mathcal P(+1) = \left|\langle \mathbf 1_x | \mathbf 1_\theta \rangle\right|^2 = |\langle \mathbf 1_x \left| (\cos \left( \theta \right)|\mathbf 1_x\rangle + \sin \left( \theta \right)|\mathbf 1_y\rangle)\right|^2
Using the orthonormality of |\mathbf 1_x\rangle and |\mathbf 1_y\rangle (\langle \mathbf 1_x | \mathbf 1_x \rangle = 1, \langle \mathbf 1_x | \mathbf 1_y \rangle = 0):
\mathcal P (+1) = \left|\cos \left( \theta \right)\langle \mathbf 1_x | \mathbf 1_x \rangle + \sin \left( \theta \right)\langle \mathbf 1_x \right| \mathbf 1_y \rangle|^2 = \left|\cos \left( \theta \right)\cdot 1 + \sin \left( \theta \right)\cdot 0\right|^2 = \cos^2 \left( \theta \right)
For the reflection, we project |\mathbf 1_\theta\rangle onto |\mathbf 1_y\rangle and take the squared modulus:
\mathcal P(-1) = \left|\langle \mathbf 1_y | \mathbf 1_\theta \rangle\right|^2 = \left|\langle \mathbf 1_y | (\cos \left( \theta \right)|\mathbf 1_x\rangle + \sin \left( \theta \right)|\mathbf 1_y\rangle)\right|^2
Using orthonormality:
\mathcal P(-1) = \left|\cos \left( \theta \right)\langle \mathbf 1_y | \mathbf 1_x \rangle + \sin \left( \theta \right)\langle \mathbf 1_y | \mathbf 1_y \rangle\right|^2 = \left|\cos \left( \theta \right)\cdot 0 + \sin \left( \theta \right)\cdot 1\right|^2 = \sin^2 \left( \theta \right)
These probabilities, \cos^2 \left(\theta\right) for transmission and \sin^2 \left(\theta\right) for reflection, are the expected results for polarizing beam-splitter measurements.
A single polarizing beam-splitter measurement on a photon yields a binary outcome (+1 or -1), providing only one bit of information. This single bit is insufficient to determine the photon’s polarization angle \theta.
To estimate \theta, one must determine the probabilities of both outcomes, requiring repeated measurements on identically prepared photons. Therefore, knowing the polarization of a single photon is impossible; we need many copies. In general, quantum state estimation necessitates numerous measurements of non-commuting observables on identically prepared systems to reliably characterize a quantum state.
While light amplifiers might seem like a solution to create multiple copies of a photon for polarization measurement, the no-cloning theorem (here) in quantum mechanics prohibits perfect copying of arbitrary quantum states. Therefore, amplifying a single photon to obtain multiple identical copies for polarization measurement is fundamentally impossible.
If the polarization is computed at an angle \theta from the x-axis the observable becomes:
\mathbf A_\theta = | \mathbf 1_\theta \rangle \langle \mathbf 1_\theta | - | \mathbf 1_{\theta + \frac{\pi}{2}} \rangle \langle \mathbf 1_{\theta + \frac{\pi}{2}} |
We can represent it in the | \mathbf 1_x \rangle and | \mathbf 1_y \rangle basis:
\begin{aligned} & | \mathbf 1_\theta \rangle = \cos\left(\theta\right)| \mathbf 1_x \rangle + \sin\left(\theta\right) | \mathbf 1_y \rangle \\ & | \mathbf 1_\theta \rangle = -\sin\left(\theta\right)| \mathbf 1_x \rangle + \cos\left(\theta\right) | \mathbf 1_y \rangle \end{aligned}
Computing the first term:
\begin{aligned} | \mathbf 1_\theta \rangle \langle \mathbf 1_\theta | =& (\cos\left(\theta\right)| \mathbf 1_x \rangle + \sin\left(\theta\right) | \mathbf 1_y \rangle) (\cos\left(\theta\right) \langle \mathbf 1_x | + \sin\left(\theta\right) \langle \mathbf 1_y |) \\ = & \cos^2\left(\theta\right) | \mathbf 1_x \rangle \langle \mathbf 1_x | + \cos\left(\theta\right)\sin\left(\theta\right) | \mathbf 1_x \rangle \langle \mathbf 1_y | \\ & + \sin\left(\theta\right)\cos\left(\theta\right) | \mathbf 1_y \rangle \langle \mathbf 1_x | + \sin^2\left(\theta\right) | \mathbf 1_y \rangle \langle \mathbf 1_y | \end{aligned}
Computing the second term:
\begin{aligned} | \mathbf 1_{\theta + \frac{\pi}{2}} \rangle \langle \mathbf 1_{\theta + \frac{\pi}{2}} | = & (-\sin\left(\theta\right)| \mathbf 1_x \rangle + \cos\left(\theta\right) | \mathbf 1_y \rangle) (-\sin\left(\theta\right) \langle \mathbf 1_x | + \cos\left(\theta\right) \langle \mathbf 1_y |) \\ = & \sin^2\left(\theta\right) | \mathbf 1_x \rangle \langle \mathbf 1_x | - \sin\left(\theta\right)\cos\left(\theta\right) | \mathbf 1_x \rangle \langle \mathbf 1_y | \\ & - \cos\left(\theta\right)\sin\left(\theta\right) | \mathbf 1_y \rangle \langle \mathbf 1_x | + \cos^2\left(\theta\right) | \mathbf 1_y \rangle \langle \mathbf 1_y | \end{aligned}
Summing:
\begin{aligned} \mathbf A_\theta &= | \mathbf 1_\theta \rangle \langle \mathbf 1_\theta | - | \mathbf 1_{\theta + \frac{\pi}{2}} \rangle \langle \mathbf 1_{\theta + \frac{\pi}{2}} | \\ &= (\cos^2\left(\theta\right) - \sin^2\left(\theta\right)) | \mathbf 1_x \rangle \langle \mathbf 1_x | + (\cos\left(\theta\right)\sin\left(\theta\right) - (-\sin\left(\theta\right)\cos\left(\theta\right))) | \mathbf 1_x \rangle \langle \mathbf 1_y | \\ &+ (\sin\left(\theta\right)\cos\left(\theta\right) - (-\cos\left(\theta\right)\sin\left(\theta\right))) | \mathbf 1_y \rangle \langle \mathbf 1_x | + (\sin^2\left(\theta\right) - \cos^2\left(\theta\right)) | \mathbf 1_y \rangle \langle \mathbf 1_y | \\ &= \cos\left(2\theta\right) | \mathbf 1_x \rangle \langle \mathbf 1_x | + \sin\left(2\theta\right) | \mathbf 1_x \rangle \langle \mathbf 1_y | + \sin\left(2\theta\right) | \mathbf 1_y \rangle \langle \mathbf 1_x | - \cos\left(2\theta\right) | \mathbf 1_y \rangle \langle \mathbf 1_y | \end{aligned}
In matrix form:
\mathbf A_\theta = \begin{bmatrix} \cos\left(2\theta\right) & \sin\left(2\theta\right) \\ \sin\left(2\theta\right) & -\cos\left(2\theta\right) \end{bmatrix}
We can check the eigenvalues \lambda of the matrix \mathbf A_\theta which are given by the characteristic equation \left|\mathbf A_\theta - \lambda \mathbf I\right| = 0:
\begin{aligned} \begin{vmatrix} \cos\left(2\theta\right) - \lambda & \sin\left(2\theta\right) \\ \sin\left(2\theta\right) & -\cos\left(2\theta\right) - \lambda \end{vmatrix} & = (\cos\left(2\theta\right) - \lambda)(-\cos\left(2\theta\right) - \lambda) - \sin^2\left(2\theta\right)\\ & = -\cos^2\left(2\theta\right) - \lambda \cos\left(2\theta\right) + \lambda \cos\left(2\theta\right) + \lambda^2 - \sin^2\left(2\theta\right) \\ & = \lambda^2 - (\cos^2\left(2\theta\right) + \sin^2\left(2\theta\right)) \\ & = \lambda^2 - 1 = 0 \end{aligned}
which gives, as expected:
\lambda = \pm 1
For eigenvalue \lambda = 1, we solve (\mathbf A_\theta - \mathbf I) \mathbf v = \mathbf 0:
\begin{bmatrix} \cos\left(2\theta\right) - 1 & \sin\left(2\theta\right) \\ \sin\left(2\theta\right) & -\cos\left(2\theta\right) - 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
From the first row:
(\cos\left(2\theta\right) - 1) x + \sin\left(2\theta\right) y = 0
Using 1 - \cos\left(2\theta\right) = 2\sin^2\left(\theta\right) and \sin\left(2\theta\right) = 2\sin\left(\theta\right)\cos\left(\theta\right), we get:
(-2\sin^2\left(\theta\right)) x + (2\sin\left(\theta\right)\cos\left(\theta\right)) y = 0
which simplifies (assuming \sin\left(\theta\right) \neq 0) to:
-\sin\left(\theta\right) x + \cos\left(\theta\right) y = 0
Then:
y = \frac{\sin\left(\theta\right)}{\cos\left(\theta\right)} x = \tan\left(\theta\right) x
Choosing x = \cos\left(\theta\right), we get y = \sin\left(\theta\right).
The eigenvector is:
\begin{bmatrix} \cos\left(\theta\right) \\ \sin\left(\theta\right) \end{bmatrix}
which corresponds to | \mathbf 1_\theta \rangle.
For eigenvalue \lambda = -1, we solve (\mathbf A_\theta + \mathbf I) \mathbf v = \mathbf 0:
\begin{bmatrix} \cos\left(2\theta\right) + 1 & \sin\left(2\theta\right) \\ \sin\left(2\theta\right) & -\cos\left(2\theta\right) + 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
From the first row:
(\cos\left(2\theta\right) + 1) x + \sin\left(2\theta\right) y = 0
Using \cos\left(2\theta\right) + 1 = 2\cos^2\left(\theta\right) and \sin\left(2\theta\right) = 2\sin\left(\theta\right)\cos\left(\theta\right), we get:
(2\cos^2\left(\theta\right)) x + (2\sin\left(\theta\right)\cos\left(\theta\right)) y = 0
which simplifies (assuming \cos\left(\theta\right) \neq 0:
\cos\left(\theta\right) x + \sin\left(\theta\right) y = 0
Therefore:
y = - \frac{\cos\left(\theta\right)}{\sin\left(\theta\right)} x = - \cot\left(\theta\right) x
Choosing x = -\sin\left(\theta\right), we get y = \cos\left(\theta\right).
The eigenvector is:
\begin{bmatrix} -\sin\left(\theta\right) \\ \cos\left(\theta\right) \end{bmatrix}
This corresponds to | \mathbf 1_{\theta + \frac{\pi}{2}} \rangle.
As expected, the eigenvalues are +1 and -1, and the associated eigenstates are | \mathbf 1_\theta \rangle and | \mathbf 1_{\theta + \frac{\pi}{2}} \rangle.
The notion of qubit
Quantum bits, or qubits, form the core of quantum information science by enabling both data transmission and processing at the quantum level.
They can be represented by any physical system with two discrete energy levels can serve as a qubit, from photon polarization to spin-1/2 degrees of freedom:
| \boldsymbol \psi \rangle = \alpha | \boldsymbol + \rangle + \beta | \boldsymbol - \rangle
Polarized photons are essential “flying qubits”, carrying quantum information across long distances, but their brief lifetime limits their use in static memories:
|\mathbf 1_\theta\rangle = \cos \left( \theta \right)|\mathbf 1_x\rangle + \sin \left( \theta \right)|\mathbf 1_y\rangle
Spins of electrons or protons provide more stable qubits, ideal for quantum storage. Trapped atoms or ions offer additional possibilities, with targeted lasers controlling specific transitions to form effective two-level qubit systems.
Nanofabricated qubits, such as Josephson junction devices, exploit quantized current levels addressed by microwaves to create engineered quantum circuits that expand the diversity of qubit implementations.
Classical bits, representing information as 0 or 1, have been extensively used for classical computation:
x \in \{ 0, 1 \}
However, qubits, also based on two states (analogous to 0 and 1), introduce a distinction which is superposition:
| \boldsymbol \psi \rangle \in \{ | \mathbf 0 \rangle; | \mathbf 1 \rangle \}
Unlike classical bits which must be either 0 or 1, a qubit can exist in a superposition, simultaneously being in both states 0 and 1 with specific probabilities.
This superposition property, precisely within quantum mechanics, is meaningless for classical bits. This ability of qubits to exist in superpositions unlocks the vast unique capabilities of quantum information processing.
A benchmark for any physical qubit implementation is its coherence time – the duration it can maintain a superposition.
Quantum information schemes are often abstract and applicable to any qubit type, but visualizing photon polarization as a concrete example can aid in understanding these general concepts.