I Schwinger-Dyson equations derivation

  • Thread starter Thread starter simonjech
  • Start date Start date
  • Tags Tags
    Derivation
Click For Summary
The discussion focuses on the derivation of the Schwinger-Dyson equations, specifically addressing confusion regarding the appearance of the commutator in the equations. The participant understands the emergence of delta functions from Heaviside functions but questions the absence of a minus sign in the terms leading to the commutator. They suggest that an anticommutator would be more logical in this context. Additionally, they propose expressing the Heaviside function ##\theta (-t)## in terms of ##\theta (t)## and mention that drawing a graph of ##\theta (-t)## or applying the chain rule for differentiation could clarify the issue. Ultimately, they acknowledge a realization that contributes to their understanding of the derivation.
simonjech
Gold Member
Messages
12
Reaction score
5
This is the part of Schwinger-Dyson equations derivation. I did not understand how can we obtain the commutator in the last line of the picture. I understand why the delta functions appeared from Heaviside functions but there is no minus sign in any term so how can we get the commutator? Anticommutator would make more sence for me.
Screenshot_20230305_212957_Drive.jpg
 
Physics news on Phys.org
Write down ##\theta (-t)## in terms of ##\theta (t)##. Drawing the graph of ##\theta (-t)## helps.
 
Last edited:
  • Like
Likes vanhees71, simonjech and malawi_glenn
Or just use the chain rule for differentiation
 
I think that i figured it out. The problem was that I did not realized that
Screenshot_20230306_154107_Math Editor.jpg
.
 
  • Like
Likes vanhees71, malawi_glenn, julian and 1 other person
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 39 ·
2
Replies
39
Views
1K
  • · Replies 39 ·
2
Replies
39
Views
5K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
Replies
1
Views
1K