# I Unpolarised Light.

1. Nov 5, 2016

### Dadface

If I'm correct,in classical physics we can define completely unpolarised light as light where the vibrations of the electric vector occur equally in all directions simultaneously. But can such a definition apply to single photons and if so are the different vibrations assumed to be in superposition?

I'm thinking that a source that produces polarised light can be analysed as being due to photons each one of which is polarised randomly with respect to the other photons. Is that correct and if so does each photon maintain its state of polarisation until measured?

Thank you. I have searched this but haven't found the answers.

2. Nov 5, 2016

### jfizzix

Classically, the electric field has a well defined polarization at each infinitesimal point in space and time. To describe unpolarized light classically, it is where the time averaged electric field (over even a small time interval) has zero average amplitude (since different directions cancel out) but nonzero average intensity.

Quantum mechanically, the polarization state of a single photon may be a superposition of different polarization states, in which case it has a single well defined polarization once you find the right basis to measure it. However, the quantum state of a single photon could also be in a mixture of different polarization states. In this case, it is partially polarized. If it is an equal mixture of all polarization states, it is unpolarized.

Photons will maintain their state of polarization until something interacts with the photon; in the same way that optical materials can change the polarization of light, they can change the polarization of a single photon. With the right experimental conditions, you can make it so that nothing disturbs the polarization of a photon before it's measured.

3. Nov 5, 2016

### A. Neumaier

To describe the state of unpolarized light requires in classical optics the use of stochastic Maxwell equations, and in quantum optics a 2 x 2 density matrix (1/2 times the identity) in helicity space. In contrast, a rank 1 matrix (corresponding to a pure state) gives completely polarized light.

4. Nov 5, 2016

### Dadface

That's great jizzfix. Thank you very much.
Thank you also A. Neumaier. I couldn't follow your reply but that's due to my own shortcomings . I really do need to brush up on my maths.

5. Nov 5, 2016

### Strilanc

The polarization of unpolarized light is in the maximally mixed state. If you plotted it on the Bloch sphere, it'd be right in the center. Its density matrix is $\begin{bmatrix} 0.5 & 0 \\ 0 & 0.5\end{bmatrix}$.

There are two main ways to create a state like that: a) uncertainty/forgetfulness about the actual state and b) entanglement.

First uncertainty/

Suppose I prepare a qubit then give it to you. My preparation procedure is as follows: I flip a coin. If the coin lands heads, then I prepare the qubit into the $|0\rangle$ state. If the coin lands tails, then I prepare the qubit into the $|1\rangle$ state. The density matrix for $|0\rangle$ is $|0\rangle \langle 0| = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$. The density matrix for $|1\rangle$ is $|1\rangle \langle 1| = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$. I sent you one density matrix with 50% probability, and the other with 50% probability, so the overall state is $0.5 \cdot |0\rangle \langle 0| + 0.5 |1\rangle \langle 1| = \begin{bmatrix} 0.5 & 0 \\ 0 & 0.5 \end{bmatrix}$. The maximally mixed state.

There many other probabilistic preparation procedures that prepare a maximally mixed qubit state. Preparing 50% $\frac{1}{\sqrt 2} |0\rangle + \frac{1}{\sqrt 2} |1\rangle$ mixed with 50% $\frac{1}{\sqrt 2} |0\rangle - \frac{1}{\sqrt 2} |1\rangle$. Picking uniformly at random from a continuous ring around the Bloch sphere. Any combination of states where the probability-weighted sum of their Bloch-sphere coordinates lands you right on the middle.

An even easier method to produce a point at the center of the Bloch sphere is to just measure along an axis perpendicular to the current state, but lose track of the measurement result. Like this:

A good practical example of this kind of measurement is a stray photon bouncing off a qubit in your quantum computer, then heading off into space going one way if the qubit was $|0\rangle$ and some other way if the qubit was $|1\rangle$. A measurement totally out of your control. But just being forgetful is enough/

Density matrices mix together what you know about the state with what the state actually is. That's why just being forgetful, and not being able to recover from having forgotten, can pull your description of the state towards the center of the Bloch sphere (which is the "no useful information at all about the state" point).

Now entanglement.

Take the density matrix for an entangled state like $|00\rangle + |11\rangle$ and compute the density matrix for when one of the qubits isn't available (i.e. do the partial trace w.r.t. either of the qubits). The remaining density matrix will be the maximally mixed state.

This also allows us to work with quite a simple circuit, but you have to keep in mind that anyone using the second qubit does have a way to get useful information about our state. You can confuse yourself if you start thinking of the first qubit as literally being in the maximally mixed state, but haven't actually gotten rid of the second qubit. (A similar issue happens in the probabilistic preparation procedure, if I write down the coin flips used to generate the state then later use them to amaze you by predicting measurements better than someone who only knew the preparation procedure could. Entanglement is just easier to get confused about than that; i.e. see every delayed choice experiment ever)

In the end, unpolarized light could be produced in many, many ways. All that really matters is that when you add up all those ways, tally up all the details you know, you get a point in the middle of the Bloch sphere.

6. Nov 6, 2016

### Dadface

Thanks Strilanc. In my case the maths is beyond me. If I ever studied that particular branch of maths at all it was several decades ago and its all gone and forgotten.
Perhaps I need to get a good book covering the relevant maths.

7. Nov 6, 2016

### A. Neumaier

Note that the final beam behaves the same independent of how it was prepared. Someone not knowing how the beam was prepared and experimenting only with that beam has no experimental way of telling two different preparation procedures apart if the density matrix is the same. Therefore only the density matrix contains up to date physical information about the beam (or the particles in it), the detailed preparation procedure is only history. This means that the extra information is only accessible by taking the environment embodying this history into account.
It is simple matrix calculus that you need to review.

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