0
$\begingroup$

In a popular introductory QM textbook, I came across this experiment which measures the state of polarization of the photon (light) by passing it to some experimental setup.

Suppose the incident light is propagating in the x-direction, that is the electric field can be in anywhere in the y-z plane. The incident light is given by

$$ E=E_0(\cos\theta\, \hat e_2 + \sin \theta \, \hat e_3 )\, \sin (kx-\omega t) $$

which is a combination of the following two linearly polarized light in y and z-direction respectively

$$E=E_0\, \hat e_2\, sin(kx-\omega t)$$ $$E=E_0\, \hat e_3\, sin(kx-\omega t)$$

which can be written as $$\rvert\ e\rangle=\rvert e_2\rangle\cos\theta\;+\;\rvert e_3\rangle\sin \theta$$

and further, $\rvert e_2\rangle$ is the pure state corresponding to the possible result R1 and $\rvert e_3\rangle$ is the pure state corresponding to the possible result R2. Now the author says that if the measurement gives the result R1, the state of polarization just after the measurement will be $\rvert e_2\rangle$ whatever be the value of $\theta$. Similarly, if the result of measurement is R2, the state of polarization just after the measurement will be $\rvert e_3\rangle$. So the state of polarization just after the measurement is the same as that in the pure state corresponding to the result obtained. Thus a measurement changes the state of polarization from $\rvert e\rangle\; \text{to either }\rvert e_2\rangle\; \text{or}\;\rvert e_3\rangle$.

Now my question is how long is this "just"? If I get the result R1 after measurement, the result just after measurement should also be R1. So if I keep on measuring in the so-called "just" time limit I must get the polarization result again and again as R1 that is $\rvert e_2\rangle$ so in my mind, the controversy happens here

Suppose there are two light detectors D1 and D2 to receive light from $\rvert e_2\rangle\; and \;\rvert e_3\rangle $ respectively. Assume for the sake of arguments that when a photon enters a detector, the detector beeps. Let the source is very weak, the beep will be intermittently and not continuously. Occasionally, one of the two detectors will beep and there will be silence. Again there will be a beep. We never know which one will beep but just the probablities which is square of amplitudes. Now my question is if I here beep from detector D1, the result is R1 and just after the measurement the result is sure to be R1 so I must hear the beep again and again from D1 only and never from D2, but that is not the case, the detectors beep randomly. Why?

$\endgroup$
  • $\begingroup$ how long is this just ? $\endgroup$ – kaivalya Apr 7 at 4:41
0
$\begingroup$

Do not confuse a pure state with a base state. In your question the state $\vert e \rangle = \vert e_2 \rangle \cos \theta + \vert e_3 \rangle \sin \theta$ is a pure state expressed as a superposition of base states (base kets).

Measurements in quantum mechanics are a statistical outcome. If you measure the polarization of the light you will have a probability of $\cos^2 \theta$ to get $\vert e_2 \rangle$ and a probability of $\sin^2 \theta$ to get $\vert e_3 \rangle$. Technically a measurement process causes the collapse of the wave function into one of the base states of the superposition with the related probability.

If you measure and get a specific polarization (collapse to one of the base states), that state of polarization will be confirmed by any successive measure, provided you do not change the measurement basis (I mean from linear polarization to different axes or to circular polarization base states).

If, after the measure, a light is horizontally polarized, a successive measure along a vertical axis will show nothing.

The detectors beep randomly because they measure statistically the continuously incident light, which is in a superposition of base states.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.