# What is the physical meaning of Bogoliubov transformation?

• I
hi guys
i recently was reading about hawking radiation and how he overcome the lack of a theory for quantum gravity by using a mathematical trick ( to see the effect of gravity on quantum fields ) and this trick was the Bogoliubov transformation .... , i just want some one to briefly explain what is the physical meaning of it and how its being applied ( i am not very familiar with creation and annihilation operators in quantum mechanics ... i tried to search online but it was somehow complicated to understand ) , could some one explain it to me with a simpler analogy ?

## Answers and Replies

atyy
Science Advisor
The use of the Bogoliubov transform in Hawking radiation is not really related to overcoming the lack of a theory of quantum gravity. Hawking was using well accepted principles, but whose consequences were not well understood.

A simple analogy for why the Bogoliubov transformation is useful is given by the Fourier transform or the Laplace transform. When a linear differential equation is Fourier or Laplace transformed, we can change the problem to an equivalent problem in simple algebra, which can be easily solved.

https://www.khanacademy.org/math/di...lace-transform-tutorial/v/laplace-transform-1

A. Neumaier
Science Advisor
A simple analogy for why the Bogoliubov transformation is useful is given by the Fourier transform or the Laplace transform. When a linear differential equation is Fourier or Laplace transformed, we can change the problem to an equivalent problem in simple algebra, which can be easily solved.
But Bogoliubov transformations change the nature of the problem in the thermodynamic limit, and produce in many important cases (e.g., superconductivity; see, e.g., https://en.wikipedia.org/wiki/Cooper_pair) phase transitions.

A good introduction to creation and annihilation operators for statistical mechanics is in the book by Linda Reichl.

Demystifier
Science Advisor
Gold Member
A good introduction to creation and annihilation operators for statistical mechanics is in the book by Linda Reichl.
The author as written in the book is L.E. Reichl. I didn't know that Reichl is a woman. EDIT: I've just realized that I had the old 2nd edition, and that now there is a significantly different 4th edition.

Last edited:
Demystifier
Science Advisor
Gold Member
i am not very familiar with creation and annihilation operators in quantum mechanics
Without that, I'm afraid, you cannot understand particle creation described by Bogoliubov transformation.

A. Neumaier
Science Advisor
• Demystifier
DrDu
Science Advisor
Maybe it is interesting to note that the extreme cases of a Bogoliubov transformation for a fermion correspond to the identity on one hand side and the interchange of a creation with an anihilation operator and vice versa. That is, we can see the absence of a particle as the presence of another particle, namely the "hole". In a metal at zero temperature, e.g. all the states below the Fermi energy are filled and the states above are empty. If we define redefine the concept of a particle for the states below the Fermi energy, i.e. we speak of the absence of holes instead of the presence of electrons, then there are neither particles (electrons) above the Fermi levels nor particles (holes) below the Fermi level. As there are no particles present after this redefinition in a metal, we can treat the ground state of the metal as a vacuum. The Bogoliubov transformation generalizes this redefinition to other mean occupation numbers apart from the extremes 0 and 1.

• Mentz114