# Do observables for polarizers at different angles commute?

Tags:
1. Aug 18, 2015

### greypilgrim

Hi.

We can write a polarised photon as $\left|\alpha\right\rangle=\cos(\alpha)\left|\updownarrow\right\rangle+\sin(\alpha)\left|\leftrightarrow\right\rangle$. Trigonometry gives us $$\left\langle\alpha | \beta\right\rangle=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)=\cos(\alpha-\beta)$$.
Now assume a polarizer in position $\alpha$ and an observable $$P(\alpha)=1\cdot\left|\alpha\right\rangle\left\langle\alpha\right|+0\cdot\left|\alpha+\frac{\pi}{2}\right\rangle\left\langle\alpha+\frac{\pi}{2}\right|=\left|\alpha\right\rangle\left\langle\alpha\right|$$ which indicates if a photon makes it through the polarizer. Now let's look at a vertically polarized photon $\left|\updownarrow\right\rangle=\left|0\right\rangle$ and two polarizer at different angles $P(\alpha)$, $P(\beta)$. Then $$\left\langle0\right|P(\alpha)P(\beta)\left|0\right\rangle=\left\langle0 | \alpha\right\rangle\left\langle\alpha | \beta\right\rangle\left\langle\beta | 0\right\rangle=\cos(\alpha)\cos(\alpha-\beta)\cos(\beta)=\left\langle0\right|P(\beta)P(\alpha)\left|0\right\rangle\enspace,$$
hence the observables commute for all angles.
I was under the impression that two polarizers should only commute if $\alpha=\beta$ or $\alpha=\beta\pm\frac{\pi}{2}$ (isn't that what quantum cryptography relies on?).
Assume a vertically polarized photon first passing a vertical polarizer and then one at $45°$. According to Malus' law, it passes the first polarizer undisturbed and the second one at $\cos^2(45°)=\frac{1}{2}$ probability. Now let the polarizers switch places. Now it passes the first polarizer at $\cos^2(45°)=\frac{1}{2}$ probability (and gets rotated $45°$) and the second one again at $\cos^2(45°)=\frac{1}{2}$ probability, so the probability that it makes its way through both polarizers is only $\cos^4(45°)=\frac{1}{4}$. So where did I make a mistake in the translation to Dirac notation?

2. Aug 18, 2015

### Orodruin

Staff Emeritus
You have not shown that the commutator is zero, all you have shown is that $\langle 0|[P(\alpha),P(\beta)]|0\rangle = 0$. There may be other non-zero matrix elements.

3. Aug 18, 2015

### greypilgrim

Yes, but in the general Heisenberg uncertainty relation we have to take the expectation value of the commutator for a certain state, which I did. Since above is true for any $\alpha$,$\beta$, the expectation value is zero for every state (I just rotated the system such that the polarization of the state is vertical).

4. Aug 18, 2015

### Orodruin

Staff Emeritus
But your original question was how it is compatible with Malus' law, to which the answer is that the observables do not commute. The operator $|\alpha\rangle \langle \beta| \cos(\alpha-\beta)$ is very different from the operator $|\beta\rangle \langle \alpha | \cos(\alpha-\beta)$ unless $\alpha = \beta + \pi n /2$. If we take your example with $\alpha = 0$ and $\beta = \pi/4$, the first one applied to the state $|0\rangle$ gives
$$\frac{1}{\sqrt 2} |0\rangle \langle \pi/4 | 0\rangle = \frac 12 |0\rangle$$
while the second one gives
$$\frac{1}{\sqrt 2} |\pi/4\rangle \langle 0 |0\rangle = \frac{1}{\sqrt 2} |\pi/4\rangle .$$
Now, the first of these states have norm 1/4 and the second norm 1/2, just as expected.

5. Aug 18, 2015

### greypilgrim

So, the general uncertainty relation actually does not give any insight here? The right side is always zero?