Electron wave addition using phasor diagram

In summary, the textbook uses a phasor diagram to find the probability of detecting an electron at the screen in a double slit experiment. This involves using complex numbers and considering relative phases in superposition. However, the concept of phasors may not be familiar to some, and an explanation may be needed to understand the solution provided in the textbook.
  • #1
jjson775
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TL;DR Summary
To find the probability of detecting an electron at the screen in a double slit experiment, my textbook uses a phasor diagram as shown. Although I am familiar with the addition of waves similar to vector addition I cannot follow this step and will appreciate an explanation
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  • #2
The diagram rotates around the left hand point of the triangle at the frequency. Components 1 and 2 then add to give the value represented by the hypotenuse. The angle phi is the phase difference between the two waves at a given point on the screen.
 
  • #3
jjson775 said:
TL;DR Summary: To find the probability of detecting an electron at the screen in a double slit experiment, my textbook uses a phasor diagram as shown. Although I am familiar with the addition of waves similar to vector addition I cannot follow this step and will appreciate an explanation

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It's not your fault that you don't understand this calculation, because it does not make any sense. Which textbook are you using? In a complex vector space, as is the separable Hilbert space used in QM, there is no notion of "angle" between vectors.

Also sometimes I stumble over the word "phasor" in these forums. I've never heard this before? Can somebody explain to me, what this means?
 
  • #4
vanhees71 said:
Also sometimes I stumble over the word "phasor" in these forums. I've never heard this before? Can somebody explain to me, what this means?
What's wrong with wikipedia's explanation of Phasor?
In physics and engineering, a phasor (a portmanteau of phase vector) is a complex number representing a sinusoidal function whose amplitude (A), angular frequency (ω), and initial phase (θ) are time-invariant.
 
  • #5
I was taught phasors in high school, a handy tool when one is considering basics of AC-currents. Halliday and co. also uses it in the context of waves and their superposition.
 
  • #6
I see. So it's simply phase factors, ##\exp(\mathrm{i} \varphi)##. So what's discussed here is
$$\psi=\psi_1 + \psi_2=R_1 \exp(\mathrm{i} \varphi_1) + R_2 \exp(\mathrm{i} \varphi_2),$$
with ##R_1,R_2>0##. Then of course, with ##\phi=\varphi_2-\varphi_1|##:
$$|\psi|^2=\psi^* \psi = |R_1+R_2 \exp(\mathrm{i} \phi)|^2=[R_1+R_2 \exp(\mathrm{i} \phi)][R_1+R_2 \exp(-\mathrm{i} \phi)]=R_1^2 + R_2^2 +R_1 R_2 [\exp(\mathrm{i} \phi) + \exp(-\mathrm{i} \phi)]=R_1^2+R_2^2 + 2 R_1 R_2 \cos \phi.$$
Since ##R_1=|\psi_1|## and ##R_2=|\psi_2|## that's indeed the equation given in the OP.

The important point is that only relative phases in superposition play a role. Wave functions, which differ only in a phase factor (or "a phasor") represent the same pure quantum state of the electron, i.e., it's not the normalized vectors but the (unit) rays that represent pure states in quantum mechanics.
 
  • #7
vanhees71 said:
I see. So it's simply phase factors, ##\exp(\mathrm{i} \varphi)##. So what's discussed here is
$$\psi=\psi_1 + \psi_2=R_1 \exp(\mathrm{i} \varphi_1) + R_2 \exp(\mathrm{i} \varphi_2),$$
with ##R_1,R_2>0##. Then of course, with ##\phi=\varphi_2-\varphi_1|##:
$$|\psi|^2=\psi^* \psi = |R_1+R_2 \exp(\mathrm{i} \phi)|^2=[R_1+R_2 \exp(\mathrm{i} \phi)][R_1+R_2 \exp(-\mathrm{i} \phi)]=R_1^2 + R_2^2 +R_1 R_2 [\exp(\mathrm{i} \phi) + \exp(-\mathrm{i} \phi)]=R_1^2+R_2^2 + 2 R_1 R_2 \cos \phi.$$
Since ##R_1=|\psi_1|## and ##R_2=|\psi_2|## that's indeed the equation given in the OP.

The important point is that only relative phases in superposition play a role. Wave functions, which differ only in a phase factor (or "a phasor") represent the same pure quantum state of the electron, i.e., it's not the normalized vectors but the (unit) rays that represent pure states in quantum mechanics.
The book I am using for self study is Modern Physics by Serway/Moses/Moyer, 3rd edition. It presumes previous completion of a calculus based physics course, which I have done. I took a series of physics classes as part of an engineering curriculum 60+ years ago that did not include modern physics. I have become fascinated with modern physics and went through the modern physics section of a textbook after first reviewing classical topics like wave motion, electricity, magnetism and light. I was able to solve the homework problems for special relativity, quantum mechanics and other modern physics topics and finished that book.

I am now studying the more advanced textbook mentioned above. I suspect that the book wants you to accept the solution in the OP to the phasor diagram without showing the details, because it involves concepts not yet introduced. See attached excerpt from the book. I just can’t get to the solution from what is given. Do I have the right idea?
 

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  • #8
I don't understand the problem. I've given the derivation in detail. I never understood the hype about drawing such "phasor diagrams". We had them in high-school, and I had an agreement with my physics teacher that I could use complex numbers instead, because I was never good in drawing such diagrams ;-).
 
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  • #9
jjson775 said:
See attached excerpt from the book. I just can’t get to the solution from what is given. Do I have the right idea?
You need to uderstand how phasors are just a less useful way of writing complex numbers. All of AC rlectronics can (and should IMHO also) be taught using complex numbers (with complex impedance etc etc) This is much easier. There is a reason all the rest of science uses complex numbers. Once you see the correspondence you, too, will agree with @vanhees71 in #8 above and neverr look back!
 

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