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  • Scalar field expansion in terms of plane wave modes is given by $$\phi(x)=\int\frac{d^3{\vec p}}{\sqrt{(2\pi)^{3}2\omega_{\vec p}}}\left(a_{\vec p}e^{-ip\cdot x}+a_p^\dagger e^{+ip\cdot x}\right)$$ where $p\cdot x={\vec p}\cdot{\vec x}-\omega_{\vec p}t$. If we define the linear combination $$a_{\vec p}=\alpha b_{\vec p}+\beta b^\dagger_{\vec p},\\ a^\dagger_{\vec p}=\alpha^* b^\dagger_{\vec p}+\beta^* b_{\vec p},$$ the original field expansion becomes $$\phi(x)=\int\frac{d^3{\vec p}}{\sqrt{(2\pi)^{3}2\omega_{\vec p}}}\left(b_{\vec p}f_{\vec p}(x)+b^\dagger_{\vec p}f^*_{\vec p}(x)\right)$$ where $f_{\vec p}=\alpha e^{-ip\cdot x}+\beta^*e^{ip\cdot x}$. In addition, if we impose canonical commutation rules to the newly defined cretation and annihilation operators, $b^\dagger_{\vec p}$ and $b_{\vec p}$, we find $|\alpha|^2-|\beta|^2=1$. This is the standard Bogoliubov transformation.

If we are interested only in inertial observers, flat spacetime, and a Lorentz invariant notion of the vacuum state, I have the following questions.

  • Is it possible that a given inertial observer uses two different choices of creation and annihilation operators related by Bogiliubov transformation? If so, does the 'particle' interpretation become ambiguous?

  • Is it possible that two observers, moving uniformly w.r.t each other, uses two different choices of creation and annihilation operators related by Bogiliubov transformation? If so, what does it mean? Does it mean that one observer a 'particle' is different from what the other calls a 'particle'?

Qmechanic
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Solidification
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2 Answers2

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Bogoliubov transformations between inertial frames will preserve positive frequency sectors, and so agree on the meaning of vacuum.

However, the idea that observers "use" different Bogoliubov transformations is a misconception. (Although note that this can still be a useful way to think when connecting locally inertial coordinates in different asymptotically flat regions). Bogoliubov transformations merely describe the same state in a different basis. The fact that non-inertial particle detectors click in vacuum is a separate phenomenon. This is a common misconception because for uniformly accelerating Rindler observers, the two effects are the same. In general they are not. See Padmanabhan and Singh 1987 https://doi.org/10.1088/0264-9381/4/5/033.

Joe Schindler
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You can build an creation/annihilation operator pair for any solution of the field equation, correctly normalized. With a complete set of solutions and the right orthogonality condition, you get to expand the field in terms of the modes.

Which set of modes you use is arbitrary (it does not depend on the reference frame, but on the preference of the person doing the calculations).

Every inertial operator will agree on the vacuum. For particle state, they might not agree on which modes are populated. However, if an observable is written in terms of the field, everyone can expand it using his own set of modes (they will get different expansions), but will get the same expectation value.

For more details, see : Jacobson T. (2005) Introduction to Quantum Fields in Curved Spacetime and the Hawking Effect. In: Gomberoff A., Marolf D. (eds) Lectures on Quantum Gravity. Series of the Centro De Estudios Científicos. Springer, Boston, MA. https://doi.org/10.1007/0-387-24992-3_2

SolubleFish
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  • I am not interested in curved spacetime. I am asking if the first expansion is valid whether the second expansion is also valid, and whether the particle interpretations are different. For example, in the first expansion, one-particle states are $a_p|0\rangle$ and in the second case, $b_p|0\rangle$. Why is the plane wave expansion preferred in ordinary treatments of QFT. – Solidification May 03 '21 at 14:41