Many Body bogoliubov transformation

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SUMMARY

The discussion centers on the Bogoliubov transformation in quantum mechanics, specifically the expression for the occupation number \( n_k \) in the ground state. The transformation is defined by the equations \( b_k = \cosh(\theta)a_k - \sinh(\theta)a_{-k} \) and \( b_k^{\dagger} = \cosh(\theta)a_k^{\dagger} - \sinh(\theta)a_{-k}^{\dagger} \). The key conclusion is that the vacuum state \( |0\rangle \) is annihilated by the operators \( a_k \), but not by \( b_k \), indicating that \( n_k \) does not equal zero unless \( \theta = 0 \).

PREREQUISITES
  • Understanding of quantum mechanics and the concept of vacuum states.
  • Familiarity with Bogoliubov transformations and their applications.
  • Knowledge of single-particle states and occupation numbers in quantum systems.
  • Basic proficiency in operator notation and quantum field theory.
NEXT STEPS
  • Study the derivation and implications of Bogoliubov transformations in quantum field theory.
  • Explore the role of the vacuum state in many-body quantum systems.
  • Learn about the significance of the parameter \( \theta \) in the context of superfluidity and Bose-Einstein condensates.
  • Investigate the mathematical properties of annihilation and creation operators in quantum mechanics.
USEFUL FOR

Students and researchers in quantum mechanics, particularly those focusing on many-body physics, quantum field theory, and superfluidity. This discussion is beneficial for anyone seeking to deepen their understanding of Bogoliubov transformations and their applications in theoretical physics.

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Homework Statement


The occupation of each single-particle state with wave vector k =/= 0 in the ground state is given by nk = <0|bkbk|0>
where b and b† are bogoliubov transformaition. Find an expression for nk.

bk = cosh(θ)ak - sinh(θ) a-k
bk = cosh(θ)ak - sinh(θ)a-k

Homework Equations

The Attempt at a Solution


I don't fully understand the notaition with zeros. I believe nk would equal to 0, however question asks for the expression for nk. thnaks in advance
 
Physics news on Phys.org
No, it is not. The state ##|0>## is the vacuum of the annihilation operators ##a_k##. That is,
$$a_k|0>=0$$ for all ##k##, but
$$b_k|0>\neq 0$$ (unless ##\theta =0##).
 

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