Mode Decompositions

A mode decomposition is a choice of independent one-particle waveforms, orbitals, field patterns, or internal degrees of freedom used to describe a system by mode occupations. It turns the abstract question “what are the subsystems?” into a concrete choice of modes.

Modes are not usually particles. They are basis elements or independently addressable degrees of freedom. A single particle can be in a superposition of modes, many bosons can occupy the same mode, and a fermionic mode can be either empty or occupied. Entanglement between modes is therefore entanglement relative to a chosen mode decomposition.

Helpful background. Mode occupations introduces number states; decomposition-dependent entanglement explains why changing the subsystems can change an entanglement statement. We develop bosonic modes here; fermionic modes also require parity restrictions and a consistent operator ordering.

What Counts as a Mode

Start with a one-particle Hilbert space $\mathcal h$. A discrete mode decomposition is often an orthonormal basis

$$\{\lvert\varphi_i\rangle\}_{i\in I}, \qquad \langle\varphi_i\vert\varphi_j\rangle = \delta_{ij}.$$

Each basis vector defines a mode. In a bosonic Fock space, the occupation basis is built from states

$$\lvert n_1,n_2,\ldots\rangle,$$

where $n_i$ is the number of excitations in mode $\varphi_i$. For fermions, $n_i$ can only be $0$ or $1$, and a fixed ordering of modes is needed to define signs.

For bosons, an orthogonal one-particle splitting gives a corresponding Fock-space factorization:

$$\mathcal F_s(\mathcal h_A\oplus\mathcal h_B) \cong\mathcal F_s(\mathcal h_A)\otimes\mathcal F_s(\mathcal h_B).$$

Each factor allows all occupation numbers. A fixed-total-particle-number sector is a subspace of this product, not itself generally a product of unrestricted local sectors.

The word “mode” can also refer to a normalized wavepacket, a cavity field pattern, a lattice orbital, a spin-orbital, a polarization component, a frequency bin, or a waveguide channel. The common feature is that there is an operator that creates or annihilates an excitation in that degree of freedom.

If $\lvert f\rangle$ is a normalized one-particle state, the corresponding bosonic creation operator may be written

$$a_f^\dagger = \sum_i \langle\varphi_i\vert f\rangle a_i^\dagger$$

in a discrete mode basis. The state $a_f^\dagger\lvert0\rangle$ is one excitation in the wavepacket mode $f$.

Spatial Modes

Spatial modes are one-particle wavefunctions localized in different regions, channels, traps, lattice sites, or beam paths. For example, two localized wavepackets $f_L(x)$ and $f_R(x)$ may define left and right modes if

$$\int dx\, f_L^*(x)f_R(x) \approx 0.$$

When the overlap is exactly zero, the corresponding bosonic creation operators obey

$$[a_L,a_R^\dagger]=0.$$

When the overlap is nonzero,

$$[a_{f},a_g^\dagger] = \langle f\vert g\rangle,$$

so their occupation labels cannot be treated as independent tensor factors. If $s=\langle f\vert g\rangle$ and $|s|<1$, define

$$\lvert g_\perp\rangle=\frac{\lvert g\rangle-s\lvert f\rangle}{\sqrt{1-|s|^2}}, \qquad a_g^\dagger=s a_f^\dagger+\sqrt{1-|s|^2}\,a_{g_\perp}^\dagger.$$

For example, $s=3/5$ resolves $g$ into amplitudes $3/5$ in $f$ and $4/5$ in an orthogonal mode. Keeping the overlap is valid, but it does not make the original pair independent subsystems.

Spatial modes are common in double wells, interferometers, waveguides, and lattice models. They are not the same as exact position eigenstates. A physical spatial mode is a normalizable wavepacket or orbital, while $\lvert x\rangle$ is a generalized eigenket.

Momentum Modes

Momentum labels block-diagonalize translation-invariant one-particle Hamiltonians; internal degrees of freedom may still need diagonalization within each momentum block. In a rectangular box with periodic boundary conditions, an orthonormal basis is

$$\varphi_{\mathbf k}(\mathbf x) = \frac{1}{\sqrt\Omega} e^{i\mathbf k\cdot\mathbf x},$$

where $\Omega=\prod_jL_j$ and $k_j=2\pi n_j/L_j$, with integers $n_j$. Hard-wall boundary conditions instead require standing-wave functions; the plane-wave basis displayed here is specifically periodic.

The annihilation field can then be expanded as

$$\psi(\mathbf x) = \frac{1}{\sqrt\Omega} \sum_{\mathbf k} e^{i\mathbf k\cdot\mathbf x} d_{\mathbf k}.$$

Momentum modes are natural for free particles, weakly interacting gases, phonons, photons in homogeneous media, and scattering calculations. Interactions or boundaries can mix them, so a definite momentum occupation is not always conserved.

In infinite volume, sums become integrals and Kronecker deltas become Dirac deltas. The precise powers of $2\pi$ depend on the Fourier convention, so the normalization must be stated.

Frequency Modes

Frequency modes appear when the system is an oscillator field or when a signal is decomposed spectrally. A cavity field has discrete resonant modes with frequencies $\omega_n$; a traveling pulse may be described by a continuum of frequency modes.

For a single oscillator mode,

$$H = \hbar\omega \left( a^\dagger a+\frac12 \right).$$

For finitely many independent modes,

$$H = \sum_j \hbar\omega_j \left( a_j^\dagger a_j+\frac12 \right),$$

before interactions or couplings are included. An infinite sum of vacuum energies needs a separate limiting or subtraction prescription. Frequency modes are especially useful in quantum optics, spectroscopy, and input-output descriptions of fields.

A frequency label is not enough by itself. A complete electromagnetic mode also includes spatial structure and polarization. Likewise, a wavepacket with a finite duration cannot have a perfectly sharp frequency; time-frequency mode decompositions always involve a resolution tradeoff.

Polarization Modes

Polarization is an internal mode label for fields such as light. If a fixed spatial-temporal mode supports two orthogonal polarizations, one may use creation operators

$$a_H^\dagger, \qquad a_V^\dagger$$

for horizontal and vertical polarization, or

$$a_R^\dagger, \qquad a_L^\dagger$$

for right- and left-circular polarization. The polarization basis is a mode basis inside a two-dimensional internal space.

A polarization qubit restricted to one photon lies in $\operatorname{span}\{\lvert1_H,0_V\rangle,\lvert0_H,1_V\rangle\}$. This two-dimensional subspace should not be confused with two full oscillator factors, each of which permits arbitrarily many photons.

The full mode label may combine several pieces:

$$\text{mode} = (\text{spatial profile},\text{frequency},\text{polarization}).$$

This matters because a phrase such as “the photon is horizontally polarized” is incomplete unless the spatial and spectral mode structure is also controlled or irrelevant. Entanglement can occur between polarization modes, path modes, frequency modes, or combinations of them.

Mode Basis Changes

Two complete orthonormal mode bases of the same one-particle Hilbert space are related by a unitary transformation. If

$$\lvert\chi_\alpha\rangle = \sum_i U_{i\alpha}\lvert\varphi_i\rangle,$$

then the corresponding creation operators transform as

$$b_\alpha^\dagger = \sum_i U_{i\alpha}a_i^\dagger.$$

Unitarity of $U$ preserves the canonical commutation relations for bosons:

$$[b_\alpha,b_\beta^\dagger] = \delta_{\alpha\beta}.$$

The same unitary change preserves fermionic anticommutation relations, but fermionic many-mode states still require a consistent mode ordering.

A two-mode example makes the point visible. Define

$$b_+^\dagger = \frac{a_A^\dagger+a_B^\dagger}{\sqrt2}, \qquad b_-^\dagger = \frac{a_A^\dagger-a_B^\dagger}{\sqrt2}.$$

The one-excitation state in the $+$ mode is

$$b_+^\dagger\lvert0\rangle = \frac{1}{\sqrt2} \left( \lvert1_A,0_B\rangle + \lvert0_A,1_B\rangle \right).$$

With respect to the $A,B$ mode split, this is a single excitation delocalized over two modes. With respect to the $+,-$ mode split, it is simply

$$\lvert1_+,0_-\rangle.$$

The state has not physically changed. The decomposition used to ask the entanglement question has changed. A physical beam splitter is different: it acts on the state while specified input and output ports identify the subsystem factors. Both descriptions use related linear algebra, so it is essential to say which interpretation is intended.

Mode Entanglement Caveats

Mode entanglement is precise only after the mode decomposition and accessible operations have been specified. Several caveats are standard:

• A mode basis change can change whether a state looks product or entangled.
• Superselection rules may restrict which mode superpositions are operationally accessible.
• Fermionic mode entanglement requires parity and sign conventions.
• Spatial-region decompositions are not always equivalent to particle or mode decompositions.
• Continuum mode labels require distributions, wavepackets, or finite-volume regularization.

None of these caveats makes mode entanglement meaningless. They say that the physical question must name the modes, the algebra of observables, and the allowed local operations.

Physical Interpretation

Mode decompositions are useful because experiments and Hamiltonians often single out natural modes:

• cavities select standing-wave spatial and frequency modes;
• beam splitters select input and output path modes;
• optical fibers and waveguides select transverse and polarization modes;
• lattices select site or Bloch modes;
• traps select oscillator or orbital modes;
• scattering setups select incoming and outgoing channels.

The best mode decomposition is usually the one in which preparations, measurements, or dynamics take their simplest form. It is a modeling choice constrained by the physical apparatus, not a matter of taste.

References

• S. L. Braunstein and P. van Loock, “Quantum information with continuous variables,” Reviews of Modern Physics 77, 513–577 (2005), doi:10.1103/RevModPhys.77.513.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill (1971).
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press (1997), doi:10.1017/CBO9780511813993.
• D. F. Walls and G. J. Milburn, Quantum Optics, 2nd ed., Springer (2008).

Exercises

1. Commutator under a mode change. Let $b_\alpha^\dagger=\sum_i U_{i\alpha}a_i^\dagger$, where $U$ is unitary and $[a_i,a_j^\dagger]=\delta_{ij}$. Show that $[b_\alpha,b_\beta^\dagger]=\delta_{\alpha\beta}$.

Solution

The annihilation operator is

$$b_\alpha = \sum_i U_{i\alpha}^*a_i.$$

Then

$$\begin{aligned} [b_\alpha,b_\beta^\dagger] &= \sum_{ij} U_{i\alpha}^*U_{j\beta} [a_i,a_j^\dagger] \\ &= \sum_i U_{i\alpha}^*U_{i\beta} = (U^\dagger U)_{\alpha\beta} = \delta_{\alpha\beta}. \end{aligned}$$

2. Two excitations in two bases. Express $b_+^\dagger b_-^\dagger\lvert0\rangle$ in the $A,B$ occupation basis. What happened to the term with one boson in each mode?

Solution

The two creation operators commute. Their mixed terms cancel, so

$$b_+^\dagger b_-^\dagger\lvert0\rangle =\frac{(a_A^\dagger)^2-(a_B^\dagger)^2}{2}\lvert0\rangle =\frac{\lvert2_A,0_B\rangle-\lvert0_A,2_B\rangle}{\sqrt2}.$$

The factors $\sqrt2$ come from normalized two-boson number states. There is no $\lvert1_A,1_B\rangle$ amplitude. A product occupation state in the $+,-$ factors is entangled in the $A,B$ factors.

3. Nonorthogonal wavepackets. Let $a_f^\dagger$ and $a_g^\dagger$ create bosons in normalized wavepackets $f$ and $g$. Show that $[a_f,a_g^\dagger]=\langle f\vert g\rangle$.

Solution

Expand in an orthonormal basis:

$$a_f^\dagger = \sum_i \langle\varphi_i\vert f\rangle a_i^\dagger, \qquad a_g^\dagger = \sum_j \langle\varphi_j\vert g\rangle a_j^\dagger.$$

Then

$$a_f = \sum_i \langle f\vert\varphi_i\rangle a_i.$$

Using $[a_i,a_j^\dagger]=\delta_{ij}$,

$$\begin{aligned} [a_f,a_g^\dagger] &= \sum_{ij} \langle f\vert\varphi_i\rangle \langle\varphi_j\vert g\rangle [a_i,a_j^\dagger] \\ &= \sum_i \langle f\vert\varphi_i\rangle \langle\varphi_i\vert g\rangle = \langle f\vert g\rangle. \end{aligned}$$

4. Complete optical mode labels. Why is “horizontal polarization” usually not a complete mode label for a photon?

Solution

Polarization is only one part of the mode label. A complete optical mode also needs spatial structure and spectral or temporal structure, at least to the accuracy relevant for the experiment. Two photons with horizontal polarization but different wavepackets, frequencies, paths, or transverse profiles occupy different modes.

5. Mode entanglement caveat. Explain why the state $b_+^\dagger\lvert0\rangle$ can look entangled in the $A,B$ mode basis but product in the $+,-$ mode basis.

Solution

The state is

$$b_+^\dagger\lvert0\rangle = \lvert1_+,0_-\rangle$$

in the $+,-$ decomposition, so it is a product of the two new mode factors. In the $A,B$ decomposition, the same vector is

$$\frac{1}{\sqrt2} \left( \lvert1_A,0_B\rangle + \lvert0_A,1_B\rangle \right).$$

The vector is the same; the tensor-product structure used to ask the entanglement question is different.