QM: Changing Basis | Why Use T_{σa,σb}?

  • Thread starter Thread starter Niles
  • Start date Start date
  • Tags Tags
    Basis Qm
Click For Summary
SUMMARY

The discussion focuses on the representation of the kinetic energy operator, denoted as T(ri), in a specific \left| \sigma \right\rangle -representation. The matrix elements of T are expressed as T = ∑_{σ_a,σ_b} T_{σ_a,σ_b} |ψ_{σ_a}(r_i)⟩⟨ψ_{σ_b}(r_i)|. The confusion arises regarding the identity operator, where the author clarifies that the identity operator can be represented as \hat{1} = ∑_{σ_a,σ_b} |ψ_{σ_a}(r_i)⟩⟨ψ_{σ_b}(r_i)|, contrasting with the simpler form \hat{1} = ∑_{σ} |ψ_{σ}(r_i)⟩⟨ψ_{σ}(r_i)|. This distinction is crucial for understanding the contraction of terms in quantum mechanics.

PREREQUISITES
  • Understanding of quantum mechanics and operators
  • Familiarity with the concept of basis representations in quantum states
  • Knowledge of matrix elements in quantum mechanics
  • Proficiency in notation for bra-ket vectors, specifically |ψ⟩ and ⟨ψ|
NEXT STEPS
  • Study the derivation of the kinetic energy operator in quantum mechanics
  • Learn about the role of identity operators in quantum state representations
  • Explore the mathematical framework of matrix elements in quantum mechanics
  • Investigate the implications of different basis representations on quantum systems
USEFUL FOR

Students and professionals in quantum mechanics, physicists working with operator theory, and anyone interested in the mathematical foundations of quantum state representations.

Niles
Messages
1,834
Reaction score
0

Homework Statement


Hi

Say I have the kinetic energy operator denoted by T(ri) for the particle i. I wish to represent it in some \left| \sigma \right\rangle-representation. My book says it is given by

<br /> T = \sum\limits_{\sigma _a ,\sigma _b } {T_{\sigma _a ,\sigma _b } \left| {\psi _{\sigma _a } \left( {r_i } \right)} \right\rangle \left\langle {\psi _{\sigma _b } \left( {r_i } \right)} \right|},<br />

where the first part of the sum denotes the matrix elements of T. My question is why the author is using

<br /> \hat 1 = \sum\limits_{\sigma _a ,\sigma _b } {\left| {\psi _{\sigma _a } \left( {r_i } \right)} \right\rangle \left\langle {\psi _{\sigma _b } \left( {r_i } \right)} \right|} ,<br />

when in fact it is given by

<br /> \hat 1 = \sum\limits_\sigma {\left| {\psi _\sigma \left( {r_i } \right)} \right\rangle \left\langle {\psi _\sigma \left( {r_i } \right)} \right|} <br />

I hope you will shed some light on this.


Niles.
 
Physics news on Phys.org
He is, but part of the expression has been contracted to form the matrix elements of T. Specifically, start with

T = \hat{1} T \hat{1} = \left( \sum\limits_{\sigma_a} {\left| {\psi _{\sigma_a} \left( {r_i } \right)} \right\rangle \left\langle {\psi_{\sigma_a} \left( {r_i } \right)} \right|}\right) T \left( \sum\limits_{\sigma_b} {\left| {\psi _{\sigma_b} \left( {r_i } \right)} \right\rangle \left\langle {\psi_{\sigma_b} \left( {r_i } \right)} \right|} \right)

= \sum\limits_{\sigma_a,\sigma_b } {\left| {\psi _{\sigma_a} \left( {r_i } \right)} \right\rangle \left\langle {\psi_{\sigma_a} \left( {r_i } \right)} \right|} T {\left| {\psi _{\sigma_b} \left( {r_i } \right)} \right\rangle \left\langle {\psi_{\sigma_b} \left( {r_i } \right)} \right|} = \sum\limits_{\sigma_a,\sigma_b } {\left| {\psi _{\sigma_a} \left( {r_i } \right)} \right\rangle T_{\sigma_a\sigma_b \right\rangle \left\langle {\psi_{\sigma_b} \left( {r_i } \right)} \right|}
 
You are right, thanks!
 

Similar threads

Replies
16
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 16 ·
Replies
16
Views
4K
Replies
63
Views
9K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K