research38 min studyLesson 826 in Physics

Bogoliubov transformations and particle definitions

Relate positive-frequency mode choices while preserving commutators, and distinguish particle number from a state-independent local observable.

On this page

What you’ll understand

  • Normalize scalar modes using the conserved Klein–Gordon product.
  • Compute occupation from a Bogoliubov beta coefficient.
  • Explain why a changed particle basis does not automatically produce a thermal state.

In a stationary spacetime, a suitable time translation supplies a natural positive-frequency convention for a free field. In a time-dependent geometry, or when comparing different physically motivated time evolutions, there may be no unique global choice. Bogoliubov transformations describe how one complete mode basis mixes positive and negative frequencies of another. This is the precise language needed for particle creation and horizon thermality, but it also exposes their limits: particle number depends on a mode choice, while the quantum state and its local correlations need independent specification.

The inner product distinguishes modes from their conjugates

For a free real Klein–Gordon field on a suitable globally hyperbolic background, the conserved Klein–Gordon product is evaluated on a Cauchy surface with future-directed surface element. With signature (−,+,+,+), choose positive-norm modes u and negative-norm conjugates u star. A complete decomposition of the solution space into these two sectors permits an expansion of the field in annihilation and creation operators. The choice of positive-frequency subspace defines the associated vacuum.

A second normalized basis v can contain both u and its conjugate. Its coefficients alpha and beta are restricted by the Klein–Gordon products so that canonical commutators are preserved. For a finite collection of modes these are symplectic normalization conditions. In an infinite field theory, convergence and unitary implementability add further requirements; an unrestricted formal infinite transformation need not act as a unitary operator on one common Fock space.

(u,v)=i∫ΣdΣa(u∗∇av−v∇au∗),vi=∑j(αijuj+βijuj∗),αα†−ββ†=I(u,v)=i\int_\Sigma d\Sigma^a\left(u^*\nabla_av-v\nabla_au^*\right),\qquad v_i=\sum_j(\alpha_{ij}u_j+\beta_{ij}u_j^*),\qquad \alpha\alpha^\dagger-\beta\beta^\dagger=I
The surface element is future directed and the scalar product is conserved for solutions with suitable boundary behavior. The ui are orthonormal positive-norm modes. A complete multimode transformation also obeys αβᵀ−βαᵀ=0; the displayed condition alone is not its only constraint.↔ Scroll wide equations horizontally.

Particle number follows from the same transformation

Taking the Klein–Gordon product of v with the field gives an annihilation operator containing the old annihilation operator and a creation part with a minus beta conjugate. The old vacuum is therefore not generally annihilated by the new operator. Its mean new particle number is the squared norm of beta. This result follows from the old vacuum commutators and does not require a classical trajectory for an emitted particle.

For one bosonic mode, the normalization says that alpha squared in magnitude exceeds beta squared by one. A squeezing parametrization automatically satisfies it, but a generic squeezed pure state is not a thermal density matrix. Thermal behavior needs additional structure, such as a particular relation across frequencies and a restriction to an accessible subsystem. A nonzero mean number alone does not specify number fluctuations, correlations, entropy, or a temperature.

aiout=∑j(αij∗ajin−βij∗ajin†),⟨0in∣Niout∣0in⟩=∑j∣βij∣2,∣α∣2−∣β∣2=1a_i^{\mathrm{out}}=\sum_j\left(\alpha_{ij}^*a_j^{\mathrm{in}}-\beta_{ij}^*a_j^{\mathrm{in}\dagger}\right),\qquad \langle0_{\mathrm{in}}|N_i^{\mathrm{out}}|0_{\mathrm{in}}\rangle=\sum_j|\beta_{ij}|^2,\qquad |\alpha|^2-|\beta|^2=1
These formulas use the explicitly stated expansion vi=Σ(αijuj+βijuj*). Other definitions of mode coefficients can move conjugates or minus signs. N is dimensionless occupation, and the final relation is the one-mode specialization.↔ Scroll wide equations horizontally.

Specify the physical choice behind a particle statement

Two inertial observers related by an ordinary Lorentz transformation in Minkowski spacetime agree on its vacuum; a coordinate relabeling alone does not create particles. The interesting comparisons involve distinct positive-frequency structures, for example early and late stationary regions separated by a changing geometry, or a boost-generated time evolution in an accelerated wedge. An operational detector additionally depends on its trajectory, coupling, switching, and the state's correlation function.

For a black-hole calculation, the early incoming state, late outgoing modes, and treatment of modes crossing the horizon must be stated. Regularity at a horizon and absence of incoming flux select different physical properties from a static vacuum condition at infinity. Renormalized stress tensors can be defined without assigning a universal local particle number, and their value is not determined by beta for one chosen mode alone. Maintaining these distinctions prevents a basis calculation from becoming an unsupported claim that every observer sees the same thermal bath.

A normalized single-mode transformation

A one-mode comparison uses real alpha=5/4 and beta=3/4 in the stated convention. Check the canonical normalization, compute the out occupation in the in vacuum, and find the mean oscillator energy in units of ħω for that out mode, including its zero-point term.

  1. The commutator condition is ∣α∣2−∣β∣2=25/16−9/16=1|\alpha|^2-|\beta|^2=25/16-9/16=1. The proposed transformed annihilation operator therefore has the required unit commutator.

  2. The in-vacuum mean out occupation is ∣β∣2=9/16=0.5625|\beta|^2=9/16=0.5625. A nonzero value refers to this pair of mode choices, not to a state-independent count.

  3. For the out oscillator, ⟨H⟩/(ℏω)=⟨N⟩+1/2=9/16+8/16=17/16\langle H\rangle/(\hbar\omega)=\langle N\rangle+1/2=9/16+8/16=17/16. This energy expectation does not by itself make the squeezed state thermal.

The result

The normalization is valid, the mean occupation is 9/16, and the oscillator energy including the zero-point term is 17ħω/16.

Explore the model

Interactive lab

Interactive lab · function

Occupation is quadratic in the mixing magnitude

For each fresh comparison, Reset and set BOTH Coefficient a=1 and Position x. Compare x=0.50 and x=1.00, predicting y=0.25 and y=1.00. Interpret x as a real beta magnitude and y as one-mode occupation, with alpha determined separately by normalization. This y=ax² graph has no quantum field, squeezing dynamics, thermal distribution, or horizon modes.

Occupation is quadratic in the mixing magnitudeThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA squared mixing coefficient gives a mean occupation, while normalization and state correlations require additional information.

Check your understanding

Does a nonzero Bogoliubov beta coefficient alone establish a thermal state?

Try it yourself

Compare two proposed one-mode operators b=Aa+Ba†: proposal I has A=√2 and B=1, while proposal II has A=1 and B=1. Check [b,b†] in each case. For the valid proposal, compute the b-mode occupation in the a vacuum. Does that mean number determine its entropy?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The commutator is |A|²−|B|². The sign chosen for B differs from the earlier beta convention only by a phase and does not change the squared occupation.

Compare with the solution

Proposal I gives [b,b†]=2−1=1[b,b^\dagger]=2-1=1 and is canonically normalized; its vacuum occupation is ∣B∣2=1|B|^2=1. Proposal II gives 1−1=01-1=0 and cannot serve as a normalized bosonic annihilation operator. For the valid finite one-mode squeezing transformation, the same underlying vacuum is a pure squeezed state in the b description, so its von Neumann entropy remains zero. A thermal state with mean number one would instead be mixed. The number expectation alone therefore does not specify the entropy.

Take it with you

  • Positive-frequency choices and Klein–Gordon normalization define particle operators.
  • The beta norm gives occupation while canonical constraints preserve commutators.
  • Thermality, detector response, stress energy, and particle number are distinct statements.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.