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I'm reading "Introduction to Quantum Fields in Classical Backgrounds" - V. F. Muckhanov, S. Winitzki. I don't understand what kind of "freedom" makes Bogolyubov transformation possible. In fact it is assumed a real scalar field operator in an FLRW Universe can be written in equivalent ways $$ \hat{\chi} (\eta, r) = \frac{1}{\sqrt{2}} \int \frac{\text{d}^3 k}{ ( 2\pi )^{\frac{3}{2}}} \left( \hat{a}_k^- v_{\mathrm{k}}^\ast (\eta) e^{\mathrm{i} k\cdot r} + \hat{a}_k^+ v_{\mathrm{k}} (\eta) e^{- \mathrm{i} k\cdot r} \right) = \frac{1}{\sqrt{2}} \int \frac{\text{d}^3 k}{ ( 2\pi )^{\frac{3}{2}}} \left( \hat{b}_k^- w_{\mathrm{k}}^\ast (\eta) e^{\mathrm{i} k\cdot r} + \hat{b}_k^+ w_{\mathrm{k}} (\eta) e^{- \mathrm{i} k\cdot r} \right) $$ Since mode functions solve the same differential equation we can transform between them $$ v_{\mathrm{k}}(\eta) = \alpha_{\mathrm{k}}^\ast w_{\mathrm{k}}(\eta) + \beta_{\mathrm{k}}^\ast w_{\mathrm{k}}^\ast(\eta) $$ The mode functions sets $(v_{\mathrm{k}},v_{\mathrm{k}}^\ast),(w_{\mathrm{k}},w_{\mathrm{k}}^\ast)$ have two constraints:

  • First constraint derives from Euler-Lagrange equation, that gives the uncoupled equation $$ \ddot{v}_{\mathrm{k}} + \omega_{\mathrm{k}} v_{\mathrm{k}} = 0,\, \ddot{w}_{\mathrm{k}} + \omega_{\mathrm{k}} w_{\mathrm{k}} = 0 $$ The mode frequency $\omega_{\mathrm{k}}$ is function of conformal time $\eta$ too. This constraint gives no conditions on the Bogolyubov equation

  • Second constraint is given by the assumption of the canonical equal-time commutation relation between the field and its conjugate, plus the canonical commutation relations between ladder operators, that give $$ \dot{v}_{\mathrm{k}} v_{\mathrm{k}}^\ast - v_{\mathrm{k}} \dot{v}_{\mathrm{k}}^\ast = 2 \mathrm{i},\,\dot{w}_{\mathrm{k}} w_{\mathrm{k}}^\ast - w_{\mathrm{k}} \dot{w}_{\mathrm{k}}^\ast = 2 \mathrm{i} $$ This constraint gives a condition on the Bogolyubov transformation and is $ |\alpha_{\mathrm{k}}|^2 - |\beta_{\mathrm{k}}|^2 = 1 $

My questions are

  • Is it possible to Bogolyubov transform even in flat spacetime, or mode functions are in that case univocally determined?
  • Is there a way to think this transformation as a change of reference? In this case how I can imagine to change a reference to make a vacuum becoming a squeezed state?
  • Can you clarify what is the freedom that make Bogolygov transformation possible and doesn't determine univocally mode functions?

Hope the questions are not too trivial, I am very confused.

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2 Answers 2

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This answer is inspired by the approach taken in Wald's Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics.

When you canonically quantize a theory, you are typically trying to find a Hilbert space which provides a representation of the canonical commutation relations. There are, in fact, many Hilbert spaces that work like that, and it is very common for physicists to consider at least two when doing QM: the ones associated with the Heisenberg and Schrödinger pictures. They are different Hilbert spaces from a mathematical point of view, but they are isometric (there is an unitary transformation connecting the two, at least in finite dimension) so they lead to the same physics.

The fact that more than one Hilbert space can describe the same quantum system is pretty much the freedom that allows you to do Bogolyubov transformations.

When quantizing the field theory, you need to specify what is its Hilbert space. We use a Fock space built out of one-particle Hilbert spaces which are built out of the solutions to the classical equations of motion. In other words, you pick the space of solutions to the classical equations of motion. This is a vector space. Consider the complexification of this vector space. This gives you the vector space you'll need to construct a Hilbert space, but you still need to get an inner product. The inner product comes from the structure of the equations of motion (I won't detail it in here to avoid getting too technical) and won't be positive definite on the whole vector space of complex solutions you picked. Hence, we'll need to arbitrarily choose a subspace of the classical solutions to use as a Hilbert space. Once that arbitrary choice is made, you can use this Hilbert space $\mathcal{H}$ as a one-particle Hilbert space to build the Fock space $\mathcal{F}$, which is the Hilbert space of the quantum field theory.

Notice that there is an arbitrary step: when we choose the subspace that will become the one-particle Hilbert space. This choice is equivalent to choosing the vacuum of the theory, or alternatively, to choosing the ladder operators. Hence, it is that arbitrary choice that will lead to the freedom of Bogolyubov transformations. If you want to read a bit more, I discussed a bit more of the process in this answer, where I focus a bit more in spacetimes with time translation symmetry, in which there is a "preferred" way of choosing this one-particle Hilbert space.

Let's now discuss your specific questions.

Is it possible to Bogolyubov transform even in flat spacetime, or mode functions are in that case univocally determined?

All globally hyperbolic spacetimes (these are the spacetimes we usually consider in QFT in Curved Spacetime, of which Minkowski spacetime is an example) admit infinitely many choices of Hilbert space and hence infinitely many different Bogolyubov transformation. However, Minkowski spacetime has a time translation symmetry, which allows you to pick a "preferred" Fock space. This is precisely the Fock space of usual Quantum Field Theory. It is worth pointing out that in the $x > |t|$ region of Minkowski spacetime you have two different notions of time translation symmetry (the usual one and the one associated with boosts in the $x$ direction), which allows you to choose between two preferred notions. The second one corresponds to what is observed by accelerated observers and studying it leads you to the Unruh effect. Deep down, the idea can be understood as just a Bogolyubov transformation.

Is there a way to think this transformation as a change of reference? In this case how I can imagine to change a reference to make a vacuum becoming a squeezed state?

Not really. Different inertial observers in Minkowski spacetime, for example, have the same creation and annihilation operators. The issue in here is not related to reference frames at all, although in a few peculiar cases (e.g. the Unruh effect) it can be understood in that way. The real source of the freedom has to do with the arbitrary choice of one-particle Hilbert space. In some specific situations, this choice can be oriented according to some reference frame.

Can you clarify what is the freedom that make Bogolygov transformation possible and doesn't determine univocally mode functions?

The choice of one-particle Hilbert space. Notice that the mode functions are elements of the complexified vector space we considered and that choosing the one-particle Hilbert space means choosing which mode functions are in $\mathcal{H}$.

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  • $\begingroup$ Your answer is very good, although at a much higher level than mine at the moment, but because the subject is more complex than I expected. I also read the answer you suggested; I can't really follow it, but stimulated my curiosity and when I'm finished with this book I'll surely go for Wald's and come back here. Also, you write very clearly. Thank you very much. $\endgroup$
    – Rob Tan
    Commented Sep 4, 2022 at 8:23
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    $\begingroup$ @RobTan You're welcome, and thank you for the compliment =D. Feel free to ask for clarification if you want to $\endgroup$ Commented Sep 4, 2022 at 20:46
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I'm not sure how much this high-level overview will help on its own, but maybe the material I've linked to will be a good starting point.

transform even in flat spacetime

If you describe Minkowski spacetime in the Rindler coordinates, you get an Unruh effect due to such transformations. See e.g. Sec. III.1 here.

Is there a way to think this transformation as a change of reference?

For instance, I suppose you could think of the above example as analogous to fictitious forces when an otherwise inertial observer rotates. More generally, see this answer's final paragraph.

change a reference to make a vacuum becoming a squeezed state

Have you computed how $D,\,S$ here transform under Bogolyubov? See also this discussion of squeezed vacua (it continues into the next section).

freedom that make Bogolyubov transformation possible and doesn't determine univocally mode functions

The canonical (anti)commutation relations are equivalent to ladder operator (anti)commutators invariant under Bogolyubov transformations, which also transform other operators (e.g. the Hamiltonian) and states, and thereby change e.g. the ground state's wavefunction. Here is a simple example.

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  • $\begingroup$ Your answer is very good and detailed, but for my lack of knowledge I struggle to understand it. You talk about Unruh effect, that for the little I know is related to accelerations, and fictitious forces, but I don't see the link with the transformation. I don't know how squeezing operator transform, but I don't even know how to compute it, and what is the meaning of the result. $\endgroup$
    – Rob Tan
    Commented Sep 2, 2022 at 18:11

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