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Do observables for polarizers at different angles commute?

  1. Aug 18, 2015 #1
    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. jcsd
  3. Aug 18, 2015 #1

    Orodruin

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    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.
     
  4. jcsd
  5. Aug 18, 2015 #2
    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).
     
  6. Aug 18, 2015 #3

    Orodruin

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    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.
     
  7. Aug 18, 2015 #4
    So, the general uncertainty relation actually does not give any insight here? The right side is always zero?
     
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