Undergrad Wave function when there is coupling between spin and position

Click For Summary
The discussion centers on the form of the wave function in the presence of coupling between spin and position. It questions why the general state cannot be expressed as a combination of symmetric and anti-symmetric components for the spatial wave function and spinors. Participants clarify that while Griffiths uses the notation of symmetric and anti-symmetric wave functions, it may not apply uniformly in this context. The confusion arises from the interpretation of the symbols used for the spatial wave functions associated with the spinors. Ultimately, the nature of the coupling dictates the symmetry requirements of the coefficients in the wave function.
Happiness
Messages
686
Reaction score
30
Screenshot 2019-08-09 at 12.07.34 PM.png

Screenshot 2019-08-09 at 12.07.55 PM.png


Why can't the general state, in the presence of coupling, take the form $$\psi_-(r)\chi_++\psi_+(r)\chi_-$$ where ##\psi_+(r)## and ##\psi_-(r)## are respectively the symmetric and anti-symmetric part of the wave function, and ##\chi_+## and ##\chi_-## are respectively the spinors representing spin up and spin down? In other words, why must the "coefficient" of the spin-up spinor be symmetric, in the presence of coupling?

Reference: Intro to QM, David J Griffiths, p210
 
Physics news on Phys.org
You just changed the orientation of the z axis (by swapping what up and down is). Why would it be symmetric and antisymmetric? I don't know, needs more context, I don't have Griffiths around.
 
The part of the spatial wave function associated/coupled with ##\chi_+## does not need to be symmetric, right?

On second thought, I think ##\psi_+## does not represent a symmetric wave function, but it’s just to denote the part of the spatial wave function associated with ##\chi_+##. Throughout the book, Griffiths has been using ##\psi_+## to mean a symmetric wave function, but I think it does not mean that in this particular case, hence the confusion.
 
Last edited:
That makes more sense.
 
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?

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K