Cooper Pairs & BCS Theory Explained

  • Context: Graduate 
  • Thread starter Thread starter Niles
  • Start date Start date
Click For Summary
SUMMARY

Cooper pairs are formed by a single electron state and its time-reversed counterpart, specifically represented as the states | k ↑ ⟩ and | -k ↓ ⟩. The time-reversal operator, denoted as T, acts on the state | k ↑ ⟩ to yield | -k ↓ ⟩. Understanding this relationship is crucial for grasping the fundamentals of BCS theory in superconductivity.

PREREQUISITES
  • Understanding of quantum mechanics principles
  • Familiarity with the concept of time-reversal symmetry
  • Knowledge of Cooper pairs in superconductivity
  • Basic grasp of quantum state notation
NEXT STEPS
  • Research the mathematical formulation of the time-reversal operator in quantum mechanics
  • Study the BCS theory and its implications in superconductivity
  • Explore the role of Cooper pairs in the formation of superconducting states
  • Examine experimental evidence supporting BCS theory and Cooper pair formation
USEFUL FOR

Physicists, students of quantum mechanics, and researchers interested in superconductivity and BCS theory will benefit from this discussion.

Niles
Messages
1,834
Reaction score
0
Hi

In my book it says that a Cooper pair is formed by a single electron state and its time-reversed counterpart. The way I have understood it, a Cooper pair is formed by the states

[tex] \left| {k \uparrow } \right\rangle \quad \text{and} \quad \left| { - k \downarrow } \right\rangle.[/tex]

Where does the time-reversal come into play?Niles.
 
Last edited:
Physics news on Phys.org
Try to act with the time-reversal operator on the state [tex]\left| {k \uparrow } \right\rangle \quad[/tex] and see what you get. (You first need to look up how this operator works.)

Edit: Okay I might as well reveal that you should get [tex] T\left| {k \uparrow } \right\rangle \quad = \quad \left| { - k \downarrow } \right\rangle.[/tex]
 
element4 said:
Try to act with the time-reversal operator on the state [tex]\left| {k \uparrow } \right\rangle \quad[/tex] and see what you get. (You first need to look up how this operator works.)

Edit: Okay I might as well reveal that you should get [tex] T\left| {k \uparrow } \right\rangle \quad = \quad \left| { - k \downarrow } \right\rangle.[/tex]

Ahh, I see. I'll take a closer look at the operator.

Thanks.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 20 ·
Replies
20
Views
8K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 21 ·
Replies
21
Views
7K