Bogoliubov Theory

Bogoliubov theory describes small quantum fluctuations around a mean-field background by retaining the Hamiltonian through quadratic order and then finding canonical operators that diagonalize that quadratic form. The new operators create quasiparticles: normal modes that are coherent mixtures of particles and holes in the original basis.

Two logically distinct steps are often given the same name:

1. The Bogoliubov approximation chooses a background, expands the interacting Hamiltonian in fluctuations, and neglects cubic and higher fluctuation terms.
2. A Bogoliubov transformation is an exact linear canonical change of creation and annihilation operators for the resulting quadratic Hamiltonian.

The first step is approximate and needs a control parameter. The second is algebraically exact once the quadratic Hamiltonian has been specified. Keeping that distinction visible prevents a successful diagonalization from being mistaken for a proof that the original interacting model is weakly correlated.

For bosons, the transformation preserves commutators through an indefinite particle–hole metric. It is paraunitary rather than ordinarily unitary in Nambu space. That minus sign is responsible for hyperbolic normalization, squeezed vacua, positive- and negative-norm mode pairs, and the possibility of complex frequencies when the chosen background is dynamically unstable.

Helpful background. The Gross–Pitaevskii equation supplies the condensate background; bosonic operator algebra supplies the commutators, and normal ordering explains the constant terms.

Anomalous pair terms around a condensate

Suppose an interacting bosonic Hamiltonian is expanded around a classical field or other mean-field reference. For a condensate,

$$\hat\psi = \Phi + \delta\hat\psi.$$

For a stationary condensate at fixed mean number, expand the rotating-frame grand Hamiltonian $K=H-\mu N$. Its linear term vanishes when the background solves the Gross–Pitaevskii equation. For a general unconstrained stationary reference one can use $H$ directly. Schematically, writing the appropriate generator as $\hat H$, the expansion is:

$$\hat H = H^{[0]} + \hat H^{[1]} + \hat H^{[2]} + \hat H^{[3]} + \cdots.$$

The terms have distinct roles:

Fluctuation order	Interpretation
zero	mean-field energy
one	vanishes when the background is stationary
two	independent normal modes after canonical diagonalization
three and higher	quasiparticle interactions and corrections beyond leading order

The quadratic part generally contains both number-conserving bilinears,

$$\hat a_i^\dagger\hat a_j,$$

and anomalous pair terms,

$$\hat a_i^\dagger\hat a_j^\dagger, \qquad \hat a_j\hat a_i.$$

The anomalous terms mean that bare-particle number is not conserved by the truncated fluctuation Hamiltonian. They do not imply that the microscopic Hamiltonian violates total particle-number conservation. They arise because the background acts as a phase and particle reservoir in a symmetry-breaking description, or because condensate particles are transferred into and out of the noncondensed sector in a number-conserving formulation.

General Quadratic Bosonic Hamiltonian

For $M$ bosonic modes, write

$$\hat H_2 = E_{\mathrm c} + \sum_{i,j} A_{ij}\hat a_i^\dagger\hat a_j + \frac{1}{2} \sum_{i,j} \left( B_{ij}\hat a_i^\dagger\hat a_j^\dagger + B_{ij}^*\hat a_j\hat a_i \right).$$

Hermiticity and bosonic commutation require

$$A = A^\dagger, \qquad B = B^T.$$

The symmetry of $B$ reflects

$$\hat a_i^\dagger\hat a_j^\dagger = \hat a_j^\dagger\hat a_i^\dagger.$$

An antisymmetric part would multiply a vanishing bosonic pair operator and carries no physical information.

Nambu vector

Introduce

$$\hat\Gamma = \begin{pmatrix} \hat{\mathbf a} \\ \hat{\mathbf a}^\dagger \end{pmatrix},$$

where

$$\hat{\mathbf a} = \begin{pmatrix} \hat a_1 \\ \vdots \\ \hat a_M \end{pmatrix}.$$

Define the Hermitian bosonic Nambu matrix

$$\mathcal H_{\mathrm B} = \begin{pmatrix} A & B \\ B^* & A^T \end{pmatrix}.$$

Then

$$\hat H_2 = E_{\mathrm c} - \frac{1}{2}\operatorname{Tr}A + \frac{1}{2} \hat\Gamma^\dagger \mathcal H_{\mathrm B} \hat\Gamma.$$

The subtraction is essential. Nambu notation writes particle and hole components together, so the lower block reproduces the normal term plus a commutator constant:

$$\hat{\mathbf a} A^T \hat{\mathbf a}^\dagger = \operatorname{Tr}A + \hat{\mathbf a}^\dagger A \hat{\mathbf a}.$$

Forgetting the factor $1/2$ or the trace subtraction double counts degrees of freedom and shifts the vacuum energy incorrectly.

The Bosonic Particle–Hole Metric

The Nambu components do not obey a positive-definite canonical algebra. Instead,

$$[ \hat\Gamma_\alpha, \hat\Gamma_\beta^\dagger ] = (\Sigma_z)_{\alpha\beta},$$

with

$$\Sigma_z = \begin{pmatrix} I_M & 0 \\ 0 & -I_M \end{pmatrix}.$$

The minus sign comes from

$$[ \hat a_i^\dagger, \hat a_j ] = -\delta_{ij}.$$

It is not an optional convention. It determines which linear transformations preserve the canonical commutators and which eigenvalue problem governs the dynamics.

The Heisenberg equation is

$$i\hbar\partial_t\hat\Gamma = \Sigma_z \mathcal H_{\mathrm B} \hat\Gamma.$$

Thus the dynamical matrix is

$$\mathcal D_{\mathrm B} = \Sigma_z\mathcal H_{\mathrm B}.$$

Although $\mathcal H_{\mathrm B}$ is Hermitian, $\mathcal D_{\mathrm B}$ is generally not Hermitian in the ordinary Euclidean inner product. It satisfies

$$\mathcal D_{\mathrm B}^\dagger = \Sigma_z \mathcal D_{\mathrm B} \Sigma_z,$$

which is a particle–hole metric relation. Diagonalizing $\mathcal H_{\mathrm B}$ as though it were an ordinary one-particle Hamiltonian therefore gives the wrong normal-mode problem.

Bosonic Canonical Transformation

Let

$$\hat\Gamma = T\hat\Gamma_\beta, \qquad \hat\Gamma_\beta = \begin{pmatrix} \hat{\boldsymbol\beta} \\ \hat{\boldsymbol\beta}^\dagger \end{pmatrix}.$$

To preserve commutators, $T$ must satisfy

$$T\Sigma_zT^\dagger = \Sigma_z.$$

For an invertible square transformation, this is equivalent to

$$T^\dagger\Sigma_zT = \Sigma_z,$$

and

$$T^{-1} = \Sigma_zT^\dagger\Sigma_z.$$

Such a matrix is called paraunitary. A useful block form is

$$T = \begin{pmatrix} U & V^* \\ V & U^* \end{pmatrix},$$

corresponding to

$$\hat{\mathbf a} = U\hat{\boldsymbol\beta} + V^*\hat{\boldsymbol\beta}^\dagger.$$

The block conditions include

$$U^\dagger U - V^\dagger V = I,$$

and

$$U^TV - V^TU = 0.$$

Signs assigned to $V$ differ across references. The invariant content is the preserved commutator metric, not one preferred sign convention.

For one real pair of amplitudes, the normalization is

$$u^2-v^2 = 1.$$

It can be parameterized by

$$u = \cosh r, \qquad v = \sinh r.$$

This hyperbolic geometry is the algebraic reason bosonic Bogoliubov vacua are squeezed states.

Real-coefficient slices with $u\ge0$ of the canonical constraints. Bosons preserve $u^2-v^2=1$, giving a hyperbola and unbounded $v^2$. Fermions preserve $u^2+v^2=1$, giving a circle and $0\leq v^2\leq1$. Complex multimode transformations obey the corresponding matrix conditions.

Generalized Eigenvalue Problem

The normal modes solve

$$\mathcal D_{\mathrm B}w_\nu = E_\nu w_\nu,$$

or equivalently

$$\mathcal H_{\mathrm B}w_\nu = E_\nu\Sigma_z w_\nu.$$

For a stable positive-energy mode, choose the metric normalization

$$w_\nu^\dagger \Sigma_z w_\mu = \delta_{\nu\mu}.$$

If

$$w_\nu = \begin{pmatrix} u_\nu \\ v_\nu \end{pmatrix},$$

then

$$u_\nu^\dagger u_\nu - v_\nu^\dagger v_\nu = 1.$$

Particle–hole symmetry generates a partner

$$\widetilde w_\nu = \Sigma_x w_\nu^*,$$

where

$$\Sigma_x = \begin{pmatrix} 0 & I_M \\ I_M & 0 \end{pmatrix}.$$

The partner has eigenvalue $-E_\nu^*$ and opposite metric norm. Positive and negative branches are therefore Nambu partners, not two independent sets of physical bosonic excitations.

When a complete stable paraunitary basis exists,

$$T^\dagger \mathcal H_{\mathrm B} T = \begin{pmatrix} \mathcal E & 0 \\ 0 & \mathcal E \end{pmatrix},$$

with

$$\mathcal E = \operatorname{diag}(E_1,\ldots,E_M), \qquad E_\nu>0.$$

The Hamiltonian becomes

$$\hat H_2 = E_{\mathrm{vac}} + \sum_{\nu=1}^{M} E_\nu \hat\beta_\nu^\dagger\hat\beta_\nu,$$

where

$$E_{\mathrm{vac}} = E_{\mathrm c} - \frac{1}{2}\operatorname{Tr}A + \frac{1}{2} \sum_{\nu=1}^{M}E_\nu.$$

The vacuum shift is physical only after the same regulator and parameter matching used to define the quadratic model have been applied.

Stability Is More Than Real-Looking Algebra

Three ideas should be separated.

Energetic stability

An equilibrium background is energetically stable at quadratic order when the constrained second variation is nonnegative. For a finite bosonic system without zero modes, positive definiteness of $\mathcal H_{\mathrm B}$ is a strong sufficient condition for a conventional positive-energy diagonalization.

Dynamical stability

For a finite mode problem, bounded linear evolution requires real frequencies and a complete, diagonalizable mode basis. Real eigenvalues alone exclude exponential growth but do not exclude polynomial growth from a Jordan block. A complex eigenvalue of

$$\Sigma_z\mathcal H_{\mathrm B}$$

produces exponential growth or decay. It is not the energy of a long-lived quasiparticle.

Negative-energy and zero modes

A real spectrum can still contain a negative-energy mode relative to the chosen frame or constraint. Such a mode signals energetic instability even when isolated linear dynamics oscillates.

Continuous symmetries can produce zero modes. Their metric norm may vanish, and the dynamical matrix can require generalized eigenvectors rather than an ordinary normalized basis. Zero modes must be treated separately; dividing by their norm or counting both Nambu partners as ordinary oscillators is invalid.

In rotating systems, the relevant energy is often

$$E - \Omega L_z.$$

A mode that is positive in the laboratory frame can become negative in the rotating frame. Stability claims must therefore state the frame and constraints.

Single-Mode Squeezing Example

Consider

$$\hat H = A\hat a^\dagger\hat a + \frac{B}{2} \left( \hat a^{\dagger 2} + \hat a^2 \right),$$

where $A$ and $B$ are real. Use

$$\hat a = \cosh r\,\hat b - \sinh r\,\hat b^\dagger.$$

The anomalous terms vanish when

$$\tanh 2r = \frac{B}{A}.$$

A real $r$ and a positive-energy oscillator result for

$$A > |B|.$$

The excitation energy is

$$E = \sqrt{A^2-B^2},$$

and the diagonal Hamiltonian is

$$\hat H = E\hat b^\dagger\hat b + \frac{E-A}{2}.$$

The new vacuum is lower in energy than the original empty-$a$ vacuum when $B\ne0$. It contains

$$\langle0_b| \hat a^\dagger\hat a |0_b\rangle = \sinh^2r.$$

The full classification follows from the quadrature coefficients $A+B$ and $A-B$:

• $A>|B|$: a positive oscillator with a normalizable ground state.
• $|A|<|B|$: an imaginary frequency and exponential growth.
• $A<-|B|$: real oscillations but negative oscillator energy, unbounded below.
• $|A|=|B|\ne0$: a soft quadrature and generally a zero-frequency Jordan block; motion need not remain bounded.

For $A<-|B|$, a real squeeze exists because $|B/A|<1$, but its diagonal number coefficient is $-\sqrt{A^2-B^2}$. Thus a real transformation does not by itself establish energetic stability. The special case $A=B=0$ is the zero Hamiltonian.

Uniform Dilute Bose Gas

The canonical physical application is a homogeneous, weakly repulsive Bose gas. In a volume $V$,

$$\hat H = \sum_{\mathbf k} \epsilon_k \hat a_{\mathbf k}^\dagger \hat a_{\mathbf k} + \frac{g}{2V} \sum_{\mathbf k,\mathbf k',\mathbf q} \hat a_{\mathbf k+\mathbf q}^\dagger \hat a_{\mathbf k'-\mathbf q}^\dagger \hat a_{\mathbf k'} \hat a_{\mathbf k},$$

where

$$\epsilon_k = \frac{\hbar^2k^2}{2m}, \qquad g = \frac{4\pi\hbar^2a_s}{m}.$$

Work with

$$\hat K = \hat H - \mu\hat N.$$

For a macroscopically occupied zero-momentum mode, set

$$\hat a_{\mathbf0} \approx \sqrt{N_0}$$

in the symmetry-breaking organization and retain terms through quadratic order in nonzero-momentum operators. Stationarity fixes

$$\mu = gn_0, \qquad n_0 = \frac{N_0}{V}.$$

After separating the background constant $-g n_0^2V/2$, the quadratic fluctuation grand Hamiltonian is

$$\hat K_2 = \sum_{\mathbf k\ne\mathbf0} \Bigg[ (\epsilon_k+gn_0) \hat a_{\mathbf k}^\dagger \hat a_{\mathbf k} + \frac{gn_0}{2} \left( \hat a_{\mathbf k}^\dagger \hat a_{-\mathbf k}^\dagger + \hat a_{\mathbf k} \hat a_{-\mathbf k} \right) \Bigg].$$

Opposite momenta form independent Nambu sectors because the background is translation invariant.

Pairwise Diagonalization

Choose the convention

$$\hat a_{\mathbf k} = u_k\hat b_{\mathbf k} - v_k\hat b_{-\mathbf k}^\dagger,$$

with real, even coefficients

$$u_{-k} = u_k, \qquad v_{-k} = v_k.$$

Commutator preservation requires

$$u_k^2-v_k^2 = 1.$$

For

$$A_k = \epsilon_k+gn_0, \qquad B_k = gn_0,$$

the anomalous terms vanish when

$$\tanh 2r_k = \frac{B_k}{A_k}, \qquad u_k = \cosh r_k, \qquad v_k = \sinh r_k.$$

The positive excitation energy is

$$E_k = \sqrt{A_k^2-B_k^2} = \sqrt{ \epsilon_k (\epsilon_k+2gn_0) }.$$

The amplitudes are

$$u_k^2 = \frac{1}{2} \left( \frac{\epsilon_k+gn_0}{E_k} + 1 \right),$$ $$v_k^2 = \frac{1}{2} \left( \frac{\epsilon_k+gn_0}{E_k} - 1 \right),$$

and

$$u_kv_k = \frac{gn_0}{2E_k}.$$

The diagonal form is

$$\hat K_2 = \frac{1}{2} \sum_{\mathbf k\ne\mathbf0} \left( E_k-A_k \right) + \sum_{\mathbf k\ne\mathbf0} E_k \hat b_{\mathbf k}^\dagger \hat b_{\mathbf k}.$$

The sum over all nonzero $\mathbf k$ is correct here because the explicit factor $1/2$ in the vacuum term compensates Nambu pairing. An alternative derivation may sum only over one representative from each pair $\{\mathbf k,-\mathbf k\}$. Mixing those conventions creates a factor-of-two error.

Phonon and Particle Regimes

At long wavelength,

$$E_k \approx \hbar c k,$$

where

$$c = \sqrt{\frac{gn_0}{m}}.$$

The quasiparticle is a collective phonon. In this limit,

$$u_k \approx v_k \approx \sqrt{ \frac{mc}{2\hbar k} },$$

so neither the particle nor the hole component is a small correction.

The combinations

$$(u_k-v_k)^2 = \frac{\epsilon_k}{E_k},$$

and

$$(u_k+v_k)^2 = \frac{\epsilon_k+2gn_0}{E_k}$$

separate density-like and phase-like fluctuation weights in this convention. At small $k$, density fluctuations are suppressed while phase fluctuations are enhanced.

At high momentum within the contact theory’s range of validity,

$$E_k = \epsilon_k + gn_0 + O(\epsilon_k^{-1}),$$ $$u_k \longrightarrow 1, \qquad v_k \approx \frac{gn_0}{2\epsilon_k}.$$

The mode becomes particle-like. The crossover occurs at

$$k\xi \sim 1,$$

where

$$\xi = \frac{\hbar}{\sqrt{2mgn_0}}.$$

The dispersion and its thermodynamic consequences are developed in Weakly Interacting Bose Gas Preview.

The Quasiparticle Vacuum

Define

$$\hat b_{\mathbf k} |0_b\rangle = 0$$

for every nonzero momentum. The inverse transformation is

$$\hat b_{\mathbf k} = u_k\hat a_{\mathbf k} + v_k\hat a_{-\mathbf k}^\dagger.$$

For one representative of each pair,

$$|0_b\rangle \propto \prod_{\mathbf k>0} \exp\!\left[ -\frac{v_k}{u_k} \hat a_{\mathbf k}^\dagger \hat a_{-\mathbf k}^\dagger \right] |0_a\rangle.$$

It is a two-mode squeezed vacuum in the bare-particle basis. Its leading contractions are

$$\langle0_b| \hat a_{\mathbf k}^\dagger \hat a_{\mathbf k} |0_b\rangle = v_k^2,$$

and

$$\langle0_b| \hat a_{\mathbf k} \hat a_{-\mathbf k} |0_b\rangle = -u_kv_k.$$

Thus a quasiparticle vacuum need not be empty of particles. Conversely, one bare atom is not generally one normal-mode excitation.

At temperature $T$,

$$n_{\mathrm B}(E_k) = \frac{1}{ e^{\beta E_k}-1 },$$

and

$$\langle \hat a_{\mathbf k}^\dagger \hat a_{\mathbf k} \rangle ={} (u_k^2+v_k^2) n_{\mathrm B}(E_k) + v_k^2.$$

The first term is thermal quasiparticle occupation mapped back to particles; the second is zero-temperature quantum depletion.

For infinitely many modes, a canonical algebra transformation need not be implemented by a unitary operator on the original Fock representation. A sufficient finite-depletion condition is

$$\sum_\nu \|v_\nu\|^2 < \infty.$$

In continuum problems, volume limits, infrared behavior, and ultraviolet matching must therefore be stated before two vacua are declared to belong to the same Hilbert-space representation.

Vacuum Energy and Ultraviolet Matching

For the contact gas, the formal zero-point shift is

$$\Delta E_{\mathrm{zp}} = \frac{1}{2} \sum_{\mathbf k\ne\mathbf0} \left( E_k - \epsilon_k - gn_0 \right).$$

At large $k$,

$$E_k - \epsilon_k - gn_0 \sim - \frac{(gn_0)^2}{2\epsilon_k}.$$

The three-dimensional momentum integral is ultraviolet divergent. This is not a failure of the low-energy spectrum. It signals that a bare contact coefficient cannot be used simultaneously as a physical scattering amplitude without matching.

After expressing the coupling in terms of the physical scattering length, the leading regulated energy density may be organized as

$$\frac{E_0}{V} = \frac{gn^2}{2} + \frac{1}{2} \int\frac{d^3k}{(2\pi)^3} \Bigg[ E_k - \epsilon_k - gn + \frac{(gn)^2}{2\epsilon_k} \Bigg].$$

The counterterm cancels the displayed high-momentum asymptote. Evaluating the finite integral produces the Lee–Huang–Yang correction, whose physical interpretation is developed in the weak-gas treatment. Normal ordering alone does not perform this coupling renormalization.

Inhomogeneous Bogoliubov–de Gennes Equations

Let $\Phi(\mathbf r)$ solve the stationary Gross–Pitaevskii equation,

$$\left[ -\frac{\hbar^2}{2m}\nabla^2 + V_{\mathrm{ext}} + g|\Phi|^2 \right]\Phi = \mu\Phi.$$

Perturb the time-dependent field as

$$\Psi(\mathbf r,t) = e^{-i\mu t/\hbar} \left[ \Phi(\mathbf r) + \delta\psi(\mathbf r,t) \right].$$

Use the mode ansatz

$$\delta\psi = u_\nu(\mathbf r) e^{-i\omega_\nu t} + v_\nu^*(\mathbf r) e^{i\omega_\nu^* t}.$$

Here $v_\nu$ is a signed amplitude in the plus-sign convention of the general Nambu vector above. For the real uniform repulsive gas, this component is negative and equals the negative of the positive $v_k$ used in $a_k=u_kb_k-v_kb^\dagger_{-k}$. Linearization gives

$$\begin{pmatrix} \mathcal L & g\Phi^2 \\ -g\Phi^{*2} & -\mathcal L^* \end{pmatrix} \begin{pmatrix} u_\nu \\ v_\nu \end{pmatrix} = \hbar\omega_\nu \begin{pmatrix} u_\nu \\ v_\nu \end{pmatrix},$$

where

$$\mathcal L = -\frac{\hbar^2}{2m}\nabla^2 + V_{\mathrm{ext}} - \mu + 2g|\Phi|^2.$$

The factor $2g|\Phi|^2$ in $\mathcal L$ and the anomalous coupling $g\Phi^2$ come from different derivatives of the cubic mean field. Replacing both by one Hartree shift destroys the gapless phase mode.

The bosonic norm is

$$\int d^3r\, \left( |u_\nu|^2 - |v_\nu|^2 \right) = 1$$

for a positive-norm mode. Distinct stable modes satisfy the corresponding metric orthogonality relation.

If $(u_\nu,v_\nu)^T$ has frequency $\omega_\nu$, then

$$\begin{pmatrix} v_\nu^* \\ u_\nu^* \end{pmatrix}$$

is the particle–hole partner with frequency $-\omega_\nu^*$.

Phase zero mode

Global phase invariance produces

$$\begin{pmatrix} u_0 \\ v_0 \end{pmatrix} \propto \begin{pmatrix} \Phi \\ -\Phi^* \end{pmatrix}.$$

Substitution gives

$$\omega_0 = 0.$$

Its metric norm vanishes:

$$\int d^3r\, \left( |\Phi|^2 - |\Phi|^2 \right) = 0.$$

This vector cannot be normalized as an ordinary quasiparticle. A complete treatment pairs it with a generalized number or phase direction, or projects the condensate mode out in a number-conserving construction.

Number-Conserving Formulations

The symmetry-breaking derivation assigns a classical amplitude to the condensate and allows the fluctuation Hamiltonian to exchange particles with that background. It is efficient, but an exact fixed-$N$ state obeys

$$\langle\hat\psi\rangle = 0.$$

Number-conserving approaches instead separate the condensate orbital from the orthogonal fluctuation field. Schematically,

$$\hat\psi = \hat a_0\phi_0 + \delta\hat\psi_\perp,$$

with

$$\int d^3r\, \phi_0^* \delta\hat\psi_\perp = 0.$$

Composite fluctuation operators transfer one particle between the condensate and orthogonal sectors while preserving total $N$. To leading order in small depletion, they reproduce the same physical positive-frequency spectrum as the broken-symmetry method.

The formulations organize the same leading physics differently:

• broken symmetry makes the phase and anomalous averages explicit;
• number conservation keeps every state in one number sector;
• the condensate phase zero mode is handled differently;
• finite-$N$ corrections and projector terms become more visible in the number-conserving form.

Neither formalism licenses neglecting depletion. Both require the noncondensed population and higher fluctuation terms to remain controlled.

Fermionic Bogoliubov Preview

Fermionic pairing Hamiltonians also mix creation and annihilation operators, but anticommutation changes the canonical geometry. Consider one paired block of two distinct complete modes, denoted $\mathbf k$ and $-\mathbf k$ for compactness. A spin-singlet application also includes their opposite spin labels. With spinless modes, the full pairing function is odd under interchange and self-inverse momenta cannot form such a pair. The block is

$$\hat H_{\mathbf k} ={} \xi_k \left( \hat c_{\mathbf k}^\dagger\hat c_{\mathbf k} + \hat c_{-\mathbf k}^\dagger\hat c_{-\mathbf k} \right) - \left( \Delta \hat c_{\mathbf k}^\dagger \hat c_{-\mathbf k}^\dagger + \Delta^* \hat c_{-\mathbf k} \hat c_{\mathbf k} \right).$$

In the Nambu basis

$$\begin{pmatrix} \hat c_{\mathbf k} \\ \hat c_{-\mathbf k}^\dagger \end{pmatrix},$$

the corresponding Bogoliubov–de Gennes matrix can be written

$$\mathcal H_{\mathrm F}(\mathbf k) = \begin{pmatrix} \xi_k & -\Delta \\ -\Delta^* & -\xi_k \end{pmatrix}.$$

This matrix is Hermitian. Its eigenvalues are

$$\pm E_k, \qquad E_k = \sqrt{\xi_k^2+|\Delta|^2}.$$

A fermionic quasiparticle has the form

$$\hat\gamma_{\mathbf k} = u_k\hat c_{\mathbf k} - v_k\hat c_{-\mathbf k}^\dagger.$$

Anticommutator preservation requires

$$|u_k|^2 + |v_k|^2 = 1.$$

For real positive $\Delta$,

$$|u_k|^2 = \frac{1}{2} \left( 1+\frac{\xi_k}{E_k} \right),$$

and

$$|v_k|^2 = \frac{1}{2} \left( 1-\frac{\xi_k}{E_k} \right).$$

The transformation is a compact rotation rather than a hyperbolic squeeze. Fermionic occupation of one mode is bounded by Pauli exclusion, while bosonic $v^2$ can grow without bound as a mode softens.

The negative fermionic BdG branch is also a Nambu redundancy, not an independent negative-energy particle. The full BCS construction must additionally determine $\Delta$ self-consistently and account for constants, filling, symmetry, and the interaction cutoff. Those tasks are derived in BCS Mean-Field Theory.

Bosons and Fermions Compared

Feature	Bosons	Fermions
canonical algebra	commutator	anticommutator
pair matrix	symmetric	antisymmetric in complete labels
scalar normalization	$\lvert u\rvert^2-\lvert v\rvert^2=1$	$\lvert u\rvert^2+\lvert v\rvert^2=1$
Nambu transformation	paraunitary	unitary with particle–hole constraint
coefficient geometry	hyperbolic	circular
bare-mode occupation in vacuum	unbounded in principle	between zero and one
mode equation	generally non-Hermitian dynamical matrix	Hermitian BdG matrix in standard form
instability signal	complex frequency possible	self-consistent saddle may fail, but Hermitian block has real eigenvalues
conserved quantity in number-breaking form	quasiparticle description need not conserve bare $N$	fermion parity remains conserved

The shared phrase “Bogoliubov transformation” refers to preserving the appropriate canonical algebra. It does not mean that bosonic and fermionic Nambu problems have the same metric or stability theory.

What the Approximation Omits

Quasiparticle interactions

Cubic and quartic fluctuation terms produce scattering among quasiparticles, energy shifts, and finite lifetimes. Beliaev decay at low temperature and Landau damping at finite temperature are beyond a strictly quadratic Hamiltonian.

Self-consistent backreaction

If depletion is not negligible, the condensate equation and fluctuation spectrum must be solved with consistent normal and anomalous densities. Naive self-consistent closures can violate conservation laws or open an unphysical gap. “Self-consistent” is not by itself a guarantee of a conserving and gapless approximation.

Critical and low-dimensional fluctuations

Near a phase transition or in sufficiently low dimension, infrared fluctuations can invalidate expansion around one rigid order parameter. A formally gapless quadratic mode may make fluctuation integrals divergent, signaling the need for finite-size, algebraic-order, renormalization-group, or nonperturbative treatments.

Strong depletion and fragmentation

When no single mode dominates the one-body density matrix, a one-condensate Bogoliubov expansion is the wrong starting point. Strongly correlated lattice phases, Tonks–Girardeau gases, and fragmented condensates require different reference spaces.

Time-dependent backgrounds

For a time-dependent mean field, instantaneous eigenmodes need not evolve independently. The basis itself changes, positive- and negative-frequency components can mix, and nonadiabatic quasiparticle production can occur. Solving a static eigenproblem at each time is not automatically an exact time-evolution method.

Exercises

Phonon mixing

Show that $u_k$ and $v_k$ become equal at leading order as $k\to0$, while $u_k-v_k$ vanishes. Interpret the result.

Solution

For small $k$,

$$E_k \approx \hbar ck, \qquad A_k \approx gn_0 = mc^2.$$

Thus

$$u_k^2 \approx v_k^2 \approx \frac{A_k}{2E_k} = \frac{mc}{2\hbar k}.$$

Therefore

$$u_k \approx v_k \approx \sqrt{ \frac{mc}{2\hbar k} }.$$

But

$$(u_k-v_k)^2 = \frac{\epsilon_k}{E_k} \propto k \longrightarrow 0.$$

The particle and hole pieces are individually large but nearly cancel in the density-like combination. The mode is a collective phase-dominated phonon, not a slightly dressed single atom.

Condensate phase zero mode

For the inhomogeneous Bogoliubov–de Gennes operator, verify that

$$\begin{pmatrix} \Phi \\ -\Phi^* \end{pmatrix}$$

is a zero mode whenever $\Phi$ satisfies the stationary Gross–Pitaevskii equation. Compute its metric norm.

Solution

The stationary equation implies

$$\mathcal L\Phi = g|\Phi|^2\Phi.$$

The upper BdG component is

$$\mathcal L\Phi + g\Phi^2(-\Phi^*) = g|\Phi|^2\Phi - g|\Phi|^2\Phi = 0.$$

The lower component is the complex conjugate relation with the corresponding signs and also vanishes. Therefore $\omega=0$.

Its metric norm is

$$\int d^3r\, \left( |\Phi|^2 - |-\Phi^*|^2 \right) = 0.$$

It is a symmetry direction, not an ordinary positive-norm oscillator mode.

Fermionic pairing block

Diagonalize

$$\begin{pmatrix} \xi & -\Delta \\ -\Delta^* & -\xi \end{pmatrix}$$

and show why the fermionic normalization uses a plus sign.

Solution

The characteristic equation is

$$E^2 = \xi^2+|\Delta|^2,$$

so the eigenvalues are

$$\pm\sqrt{\xi^2+|\Delta|^2}.$$

For

$$\hat\gamma_1=u\hat c_1-v\hat c_2^\dagger, \qquad \hat\gamma_2=u\hat c_2+v\hat c_1^\dagger,$$

with two distinct modes and real $u,v$ after choosing real positive $\Delta$, the same-mode anticommutator is

$$\{ \hat\gamma_1, \hat\gamma_1^\dagger \} = |u|^2+|v|^2,$$

because

$$\{ \hat c^\dagger, \hat c \} = 1$$

rather than $-1$. Canonical normalization therefore requires

$$|u|^2+|v|^2 = 1.$$

The partner signs also give $\{\gamma_1,\gamma_2\}=uv-vu=0$, as required. Mixing one fermion mode with its own adjoint would instead give $\{uc-vc^\dagger,uc-vc^\dagger\}=-2uv$, so the normalization condition alone would not suffice. Solving the positive-energy eigenvector gives

$$|u|^2 = \frac12 \left( 1+\frac{\xi}{E} \right), \qquad |v|^2 = \frac12 \left( 1-\frac{\xi}{E} \right).$$

Real frequencies with unbounded motion

Take one bosonic mode with $A=B>0$. Show that its dynamical matrix has real eigenvalues but is not diagonalizable, and find the time evolution.

Solution

Here

$$D=A\begin{pmatrix}1&1\\-1&-1\end{pmatrix}, \qquad D^2=0,\qquad D\ne0.$$

Both eigenvalues are zero, but there is only one independent eigenvector. Hence

$$e^{-iDt/\hbar}=I-\frac{iDt}{\hbar}.$$

Generic perturbations grow linearly in time. Real frequencies alone are not enough for bounded dynamics at a defective zero mode.

References

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