I How does the thickness of a beam splitter affect the phase shift produced?

confused_man
Messages
15
Reaction score
1
I'm having trouble understanding the phase shift produced by a beam splitter. I seem to be finding conflicting information.

I'm specifically looking to understand a 50/50 beam splitter where one side has a dielectric mirror, as shown in this figure from wikipedia:
Wavesplitter1.gif


I understand the pi phase shift (due to air/dielectric reflection, where n_air < n_dielectric). Why doesn't the thickness of the beam splitter have any effect? Shouldn't there be an additional phase?

On a related note, I'm trying to see how such a beam splitter would be able to act like a Hadamard gate in quantum information, if the additional phase shift due to the thickness of the plate is included. I've seen many articles/documents say that a beam splitter can implement a Hadamard gate, but this only seems to work if the thickness of the splitter is ignored. What am I missing?

Thank you for any insight.
 

Attachments

  • Wavesplitter1.gif
    Wavesplitter1.gif
    924 bytes · Views: 1,819
Physics news on Phys.org
confused_man said:
I'm having trouble understanding the phase shift produced by a beam splitter. I seem to be finding conflicting information.

I'm specifically looking to understand a 50/50 beam splitter where one side has a dielectric mirror, as shown in this figure from wikipedia:
View attachment 223275

I understand the pi phase shift (due to air/dielectric reflection, where n_air < n_dielectric). Why doesn't the thickness of the beam splitter have any effect? Shouldn't there be an additional phase?

On a related note, I'm trying to see how such a beam splitter would be able to act like a Hadamard gate in quantum information, if the additional phase shift due to the thickness of the plate is included. I've seen many articles/documents say that a beam splitter can implement a Hadamard gate, but this only seems to work if the thickness of the splitter is ignored. What am I missing?

Thank you for any insight.

The phase shift of the reflected beam can be compensated by putting a phase-shifter in the transmtted beam.

To make a Hadamard gate the outputs from the input state |0>|1> must be recombined in a second BS to give (|0>|1>+i|1>|0>)√2
 
Last edited:
Thank you for the reply. What about the phase shift due in the transmitted beam due to the plate thickness? This is the one that I am confused about, as it seems that many sources just ignore the fact that the beam splitter has some finite width, which would introduce some additional phase shift in the transmitted and reflected blue beams. I don't see how you could compensate for this.
 
confused_man said:
Thank you for the reply. What about the phase shift due in the transmitted beam due to the plate thickness? This is the one that I am confused about, as it seems that many sources just ignore the fact that the beam splitter has some finite width, which would introduce some additional phase shift in the transmitted and reflected blue beams. I don't see how you could compensate for this.
There is another recent post dealing with this but I cannot find it. Try searching posts in the last 3 months.
 
Most texts do not discuss the phase shift introduced by the dimensions of the crystal because it does not matter. Let us consider all possible phase deviations. Let \phi_0 be th einitial phase difference between the red and the blue beam, \phi the phase shift introduced by the beam splitter width from the reflecting surface to the right and \phi&#039; the phase shift due to light going from the lower surface to the reflecting surface.
For the for beams you will find the following phases:
red one from the left to the top: \phi_0+\pi
blue one from bottom to top: \phi&#039;
red one from the left to the right: \phi_0+\phi
blue one from the bottom to the right: \phi+\phi&#039;

Now the first thing that matters is the phase difference at the output ports:
top: \phi_0+\pi-\phi&#039;
right:\phi_0+\phi-\phi-\phi&#039;=\phi_0-\phi&#039;

Now the final quantity of interest is the phase difference of these differences and as you can see, this will always be \pi regardless of how the other parameters are chosen.
 
  • Like
Likes Derek P, DrChinese and Mentz114
I read Hanbury Brown and Twiss's experiment is using one beam but split into two to test their correlation. It said the traditional correlation test were using two beams........ This confused me, sorry. All the correlation tests I learnt such as Stern-Gerlash are using one beam? (Sorry if I am wrong) I was also told traditional interferometers are concerning about amplitude but Hanbury Brown and Twiss were concerning about intensity? Isn't the square of amplitude is the intensity? Please...
I am not sure if this belongs in the biology section, but it appears more of a quantum physics question. Mike Wiest, Associate Professor of Neuroscience at Wellesley College in the US. In 2024 he published the results of an experiment on anaesthesia which purported to point to a role of quantum processes in consciousness; here is a popular exposition: https://neurosciencenews.com/quantum-process-consciousness-27624/ As my expertise in neuroscience doesn't reach up to an ant's ear...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Back
Top