- #1
greypilgrim
- 547
- 38
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?
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?