Uncertainty principle and photon

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The discussion centers on the measurement of a particle's position and momentum in relation to the uncertainty principle. It clarifies that while shining a photon is a common method for measurement, it is not the only technique available; for instance, single slit diffraction can also determine a photon's position without additional light. As the slit width decreases, the uncertainty in momentum increases, illustrating the intrinsic nature of the uncertainty principle. The conversation also highlights that improvements in measurement techniques do not eliminate the uncertainty principle, as it remains a fundamental aspect of quantum mechanics. Ultimately, the uncertainty principle is inherent to the nature of quantum particles, regardless of measurement methods.
spidey
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i have always read in almost all sites that we have to shine at least a photon to measure the particle's position and momentum and hence comes the uncertainty principle...why we are using this shining photon technique always...is this the only way of measuring particle's position and momentum...is there any other method other than shining photon method to measure particle's position and momentum so that we can measure position and momentum with great accuracy...am i missing anything?
 
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spidey said:
i have always read in almost all sites that we have to shine at least a photon to measure the particle's position and momentum and hence comes the uncertainty principle...why we are using this shining photon technique always...is this the only way of measuring particle's position and momentum...is there any other method other than shining photon method to measure particle's position and momentum so that we can measure position and momentum with great accuracy...am i missing anything?

This is not actually correct. For example, in the single slit diffraction, one narrow down the position of a photon passing through the slit using just the slit width. So if the slit has a width of \Delta(x), then the photon that passed through the slit was in that position, with an uncertainty of position being \Delta(x).

You will also notice that if the width is made smaller and smaller, your ability to predict the value of p_x after it passes the slit becomes less and less accurate. The photon can acquire a larger range of momentum values as you make the slit smaller. Thus, the spread in momentum becomes larger as more and more photons passes through the slit. The uncertainty in position (\Delta(x)) will corresponds in the spread in this momentum, i.e.\Delta(p_x).

In this case, you'll notice that we did not use any light to shine on the particle that we want to measure (this works for any quantum particle such as photons, electrons, neutrons, protons, etc.). In other words, it has nothing to do with instrumentation accuracy. It is intrinsic.

Zz.
 
ZapperZ said:
This is not actually correct. For example, in the single slit diffraction, one narrow down the position of a photon passing through the slit using just the slit width. So if the slit has a width of \Delta(x), then the photon that passed through the slit was in that position, with an uncertainty of position being \Delta(x).

You will also notice that if the width is made smaller and smaller, your ability to predict the value of p_x after it passes the slit becomes less and less accurate. The photon can acquire a larger range of momentum values as you make the slit smaller. Thus, the spread in momentum becomes larger as more and more photons passes through the slit. The uncertainty in position (\Delta(x)) will corresponds in the spread in this momentum, i.e.\Delta(p_x).

In this case, you'll notice that we did not use any light to shine on the particle that we want to measure (this works for any quantum particle such as photons, electrons, neutrons, protons, etc.). In other words, it has nothing to do with instrumentation accuracy. It is intrinsic.

Zz.

thank you for clearing my doubt...
 
ZapperZ said:
In other words, it has nothing to do with instrumentation accuracy. It is intrinsic.

I always thought of it as something fundamental about physical laws (about conjugate pairs in mechanics), such that measuring devices *always* causes a back-reaction at least as large as uncertainty principle says. I guess I mean that I think it's both.
 
genneth said:
I always thought of it as something fundamental about physical laws (about conjugate pairs in mechanics), such that measuring devices *always* causes a back-reaction at least as large as uncertainty principle says. I guess I mean that I think it's both.

Maybe it does. However, we should also pay attention to the fact that the uncertainty in a single measurement can be improved with better technique and better technology. I can measure the position that an electron hit a CCD much better than using simply a charge-sensitive plate. That improves the accuracy of a position measurement. Yet, it does nothing to my knowledge of its non-commuting observable within the HUP.

Thus, improving the measurement uncertainty isn't tied to the HUP. Simply having better instruments does not make the HUP go away, or make the non-commuting observable better known.

Zz.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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