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Suppose, I made a measurement of the position of an electron using a photon. How the minimum uncertainty in the measurement of the position be one wave length? Also how do we use waves to measure the position of an electron? any help will be highly appreciated.

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The Heisenberg uncertainty principle, HUP, gives a general answer to your

How the minimum uncertainty in the measurement of the position be one wave length?

$$\Delta x\Delta p > \frac{\hbar}{2}$$$$\Delta E\Delta t>\frac{\hbar}{2}$$

For uncertainty in space , the momentum of a photon is

$p=h/λ$

and one wavelength is the probable region where the photon is, when hitting an electon at rest hit. These are two quantum mechanical point particles.

BUT

Suppose, I made a measurement of the position of an electron using a photon.

You have to decide what the detectors are , which will detect a photon and an electron. These detectors themselves will introduce larger uncertainties. In addition quantum mechanics is a probabilistic theory, and one has to take distributions of many photons on many electrons to establish a measurement.

gamma electron scatter

Fig. 4.3 . An invisible gamma ray photon (top) produces an electron and a positron, short for positive electron, seen by curved tracks in a bubble chamber. Both the electron and the positron are bent into circular tracks by the instrument’s magnetic field, moving in opposite direction because of their opposite electrical charge and spiraling into a smaller circular motion as they lose energy. In this upper pair, some of the photon’s energy is taken up in displacing an atomic electron, which shoots off towards bottom left. In the lower example, all of a gamma ray’s energy goes into the production of the electron-positron pair. As a result, these particles are more energetic than the upper pair, and their tracks do not curve so tightly in the chamber’s magnetic field. (Schematic of a Lawrence Berkeley Laboratory bubble chamber image, reproduced by Frank Close, Michael Marten and Christine Sutton in The Particle Explosion, New York: Oxford University Press 1987.)

Using a bubble chamber detector, for incoming gamma hitting an electron off a hydrogen atom, one can measure the point in the picture and get (x,y,z) coordinates, but the uncertainties will be due to measurement errors and not the intrinsic quantum mechanical wavelengths, which are much smaller.

So the subject is not simple, and you have to study a course in particle physics to get a good grounding in it.

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  • $\begingroup$ Thanks for your response. I get that since one wavelength is where a photon can be found, when it hits an electron the probability of finding the electron should be in that one wave length. Therefore, smaller the wavelength of the striking photon we use to hit the electron, the more precise its position be found. But in the bubble chamber detector, you said that the uncertainities will not be due to intrinsic quantum mechanical wavelengths. Are you neglecting the possible smaller uncertainity, because of the fact that the wavelength of gamma rays is extremely small? Can you explain that. $\endgroup$ – zal916 Jan 13 at 15:21
  • $\begingroup$ No, but we cannot measure it in the bubble chamber because the measurement errors are of order of microns and if you substitute the momenta the HUP is always fulfilled. maybe with interference phenomena one could calculate the intrinsic uncertainty. ( not though with gamma rays because they are hard to manipulate, with lower energy photons) $\endgroup$ – anna v Jan 13 at 15:45

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