Uncertainty principle and photon

  • Thread starter spidey
  • Start date
213
0

Main Question or Discussion Point

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?
 

Answers and Replies

ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
2018 Award
35,228
4,049
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 [itex]\Delta(x)[/itex], then the photon that passed through the slit was in that position, with an uncertainty of position being [itex]\Delta(x)[/itex].

You will also notice that if the width is made smaller and smaller, your ability to predict the value of [itex]p_x[/itex] 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 ([itex]\Delta(x)[/itex]) will corresponds in the spread in this momentum, i.e.[itex]\Delta(p_x)[/itex].

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.
 
213
0
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 [itex]\Delta(x)[/itex], then the photon that passed through the slit was in that position, with an uncertainty of position being [itex]\Delta(x)[/itex].

You will also notice that if the width is made smaller and smaller, your ability to predict the value of [itex]p_x[/itex] 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 ([itex]\Delta(x)[/itex]) will corresponds in the spread in this momentum, i.e.[itex]\Delta(p_x)[/itex].

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...
 
979
1
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.
 
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
2018 Award
35,228
4,049
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.
 

Related Threads for: Uncertainty principle and photon

Replies
17
Views
910
Replies
2
Views
1K
Replies
6
Views
2K
Replies
2
Views
1K
Replies
12
Views
463
Replies
3
Views
1K
Replies
34
Views
3K
  • Last Post
3
Replies
70
Views
4K
Top