Mode Expansions

A mode expansion expresses position-space field operators in a basis of one-particle modes. It is the second-quantized version of expanding a wavefunction in an orthonormal basis, but the coefficients are creation and annihilation operators rather than complex amplitudes.

For a complete orthonormal one-particle basis $\{\varphi_i(\mathbf x)\}$, the nonrelativistic annihilation field is written

$$\psi(\mathbf x) = \sum_i \varphi_i(\mathbf x)d_i, \qquad \psi^\dagger(\mathbf x) = \sum_i \varphi_i^*(\mathbf x)d_i^\dagger.$$

Here $d_i,d_i^\dagger$ stand for either bosonic mode operators $a_i,a_i^\dagger$ or fermionic mode operators $c_i,c_i^\dagger$, with the appropriate algebra. The same formulas work for both statistics; the difference is in the commutators or anticommutators of the mode operators.

The expansion is not a new postulate independent of mode occupations. It is the coordinate representation of the statement that particles or excitations occupy one-particle modes.

Single-Particle Basis

Let $\mathcal h$ be the one-particle Hilbert space. In the coordinate representation, an orthonormal basis is a set of wavefunctions $\varphi_i(\mathbf x)=\langle\mathbf x\vert\varphi_i\rangle$ satisfying

$$\int d^3x\, \varphi_i^*(\mathbf x) \varphi_j(\mathbf x) = \delta_{ij}.$$

Completeness means

$$\sum_i \varphi_i(\mathbf x) \varphi_i^*(\mathbf y) = \delta^{(3)}(\mathbf x-\mathbf y),$$

understood in the distributional sense. If the basis is finite or truncated, the same sum gives the kernel of the projection onto the retained subspace, not the full delta function.

The one-particle basis may consist of bound-state orbitals, lattice-site orbitals, trap eigenfunctions, plane waves in a box, spin-orbitals, wavepackets, or normal modes. The mode label $i$ is shorthand for all quantum numbers needed to specify one complete one-particle state.

Mode Annihilation Operators

For each one-particle mode $\varphi_i$, there is an annihilation operator $d_i$ and a creation operator $d_i^\dagger$. Acting on the vacuum,

$$d_i^\dagger\lvert0\rangle = \lvert1_i\rangle, \qquad d_i\lvert0\rangle = 0.$$

For a normalized wavepacket $f(\mathbf x)$, the corresponding mode operator is a linear combination of basis-mode operators:

$$d_f = \sum_i \langle f\vert\varphi_i\rangle d_i, \qquad d_f^\dagger = \sum_i \langle\varphi_i\vert f\rangle d_i^\dagger.$$

In position-space notation this becomes the smeared field operator

$$d_f = \int d^3x\, f^*(\mathbf x)\psi(\mathbf x), \qquad d_f^\dagger = \int d^3x\, f(\mathbf x)\psi^\dagger(\mathbf x).$$

This smearing is conceptually important. The object $\psi(\mathbf x)$ is an operator-valued distribution, so sharply localized expressions should be interpreted through integrals against wavepackets, test functions, or basis modes. The operator meaning of these fields is developed in Field Operators.

Position-Space Field Operators

The field expansion is fixed by requiring the field to annihilate the mode that has wavefunction $\varphi_i(\mathbf x)$ when projected onto that mode:

$$d_i = \int d^3x\, \varphi_i^*(\mathbf x)\psi(\mathbf x).$$

The adjoint formula is

$$d_i^\dagger = \int d^3x\, \varphi_i(\mathbf x)\psi^\dagger(\mathbf x).$$

Substituting the expansion of $\psi(\mathbf x)$ into the first formula gives

$$\begin{aligned} \int d^3x\, \varphi_i^*(\mathbf x)\psi(\mathbf x) &= \sum_j \left( \int d^3x\, \varphi_i^*(\mathbf x)\varphi_j(\mathbf x) \right)d_j \\ &= \sum_j \delta_{ij}d_j = d_i. \end{aligned}$$

Thus the expansion coefficients are operators obtained by projecting the field onto the chosen one-particle basis, exactly as ordinary wavefunction coefficients are obtained by inner products.

Completeness and Field Algebra

For bosonic modes,

$$[a_i,a_j^\dagger] = \delta_{ij}, \qquad [a_i,a_j]=0.$$

Using the field expansion,

$$\begin{aligned} [\psi(\mathbf x),\psi^\dagger(\mathbf y)] &= \sum_{ij} \varphi_i(\mathbf x) \varphi_j^*(\mathbf y) [a_i,a_j^\dagger] \\ &= \sum_i \varphi_i(\mathbf x) \varphi_i^*(\mathbf y) \\ &= \delta^{(3)}(\mathbf x-\mathbf y). \end{aligned}$$

For fermionic modes,

$$\{c_i,c_j^\dagger\} = \delta_{ij}, \qquad \{c_i,c_j\}=0,$$

and the same completeness calculation gives

$$\{\psi(\mathbf x),\psi^\dagger(\mathbf y)\} = \delta^{(3)}(\mathbf x-\mathbf y).$$

If the mode sum is restricted to a subspace $\mathcal S$, the right-hand side is instead the projected kernel

$$P_{\mathcal S}(\mathbf x,\mathbf y) = \sum_{i\in\mathcal S} \varphi_i(\mathbf x) \varphi_i^*(\mathbf y).$$

This distinction matters in finite basis calculations, lattice models, numerical diagonalization, and effective low-energy descriptions.

Basis Changes

Suppose two orthonormal one-particle bases are related by a unitary matrix $U$:

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

In wavefunction form,

$$\chi_\alpha(\mathbf x) = \sum_i \varphi_i(\mathbf x)U_{i\alpha}.$$

The field operator is basis-independent:

$$\psi(\mathbf x) = \sum_i \varphi_i(\mathbf x)d_i = \sum_\alpha \chi_\alpha(\mathbf x)\widetilde d_\alpha.$$

The mode operators therefore transform as

$$\widetilde d_\alpha = \sum_i U_{i\alpha}^* d_i, \qquad \widetilde d_\alpha^\dagger = \sum_i U_{i\alpha} d_i^\dagger.$$

Because $U$ is unitary, the transformed operators obey the same bosonic commutation relations or fermionic anticommutation relations. The occupation labels change, but the field operator and the Fock-space state do not change.

This is an ordinary one-particle basis rotation. It should not be confused with a Bogoliubov transformation, which mixes creation and annihilation operators and may change which state is called the vacuum.

Momentum Modes

For a cubic box of volume $V=L^3$ with periodic boundary conditions, the normalized plane-wave modes are

$$\varphi_{\mathbf k}(\mathbf x) = \frac{1}{\sqrt V} e^{i\mathbf k\cdot\mathbf x}, \qquad \mathbf k = \frac{2\pi}{L}\mathbf n, \qquad \mathbf n\in\mathbb Z^3.$$

The field expansion is

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

with inverse

$$d_{\mathbf k} = \frac{1}{\sqrt V} \int_V d^3x\, e^{-i\mathbf k\cdot\mathbf x} \psi(\mathbf x).$$

For bosons, $[d_{\mathbf k},d_{\mathbf q}^\dagger]=\delta_{\mathbf k\mathbf q}$; for fermions, $\{d_{\mathbf k},d_{\mathbf q}^\dagger\}=\delta_{\mathbf k\mathbf q}$.

In infinite volume, a common continuum convention is

$$\psi(\mathbf x) = \int \frac{d^3k}{(2\pi)^{3/2}}\, e^{i\mathbf k\cdot\mathbf x} d(\mathbf k),$$

with

$$[d(\mathbf k),d^\dagger(\mathbf q)] = \delta^{(3)}(\mathbf k-\mathbf q)$$

for bosons, or the corresponding anticommutator for fermions. Other Fourier conventions move powers of $2\pi$ between the expansion and the delta function; the convention must be used consistently.

Spinful Modes

When the particle has spin or another discrete internal label, the complete one-particle mode includes that label. A convenient basis for spin-$1/2$ particles is

$$\lvert n,s\rangle = \lvert\phi_n\rangle\otimes\lvert s\rangle, \qquad s\in\{\uparrow,\downarrow\}.$$

The field has components

$$\psi_s(\mathbf x) = \sum_n \phi_n(\mathbf x)d_{ns}, \qquad \psi_s^\dagger(\mathbf x) = \sum_n \phi_n^*(\mathbf x)d_{ns}^\dagger.$$

For spinful fermions,

$$\{\psi_s(\mathbf x),\psi_t^\dagger(\mathbf y)\} = \delta_{st}\delta^{(3)}(\mathbf x-\mathbf y),$$

and for spinful bosons the same formula holds with a commutator instead of an anticommutator.

The local number density is

$$n(\mathbf x) = \sum_s \psi_s^\dagger(\mathbf x)\psi_s(\mathbf x),$$

and a spin-density component is written

$$S^a(\mathbf x) = \sum_{st} \psi_s^\dagger(\mathbf x) (S^a)_{st} \psi_t(\mathbf x).$$

Spin is not an optional decoration here. For electrons, the Pauli principle applies to complete spin-orbitals, so two electrons with opposite spin can occupy the same spatial orbital because they occupy different one-particle modes.

Why Mode Expansions Matter

Mode expansions are the bridge between basis-index notation and field notation. They explain why the same many-particle operator can be written in either form.

A one-body operator with position-space kernel $A(\mathbf x,\mathbf y)$ may be written as

$$\widehat A = \int d^3x\,d^3y\, \psi^\dagger(\mathbf x) A(\mathbf x,\mathbf y) \psi(\mathbf y),$$

or, after expanding in modes,

$$\widehat A = \sum_{ij} A_{ij}d_i^\dagger d_j.$$

The matrix elements are

$$A_{ij} = \int d^3x\,d^3y\, \varphi_i^*(\mathbf x) A(\mathbf x,\mathbf y) \varphi_j(\mathbf y).$$

Similarly, the field form of a two-body interaction becomes a sum over four mode indices once every field operator is expanded. This is why mode expansions are indispensable in atomic physics, condensed matter, quantum chemistry, quantum optics, and the nonrelativistic approach to QFT notation.

Common Mistakes

• Treating $\psi(\mathbf x)$ as an ordinary wavefunction rather than an operator-valued distribution.
• Forgetting the complex conjugate in the inverse formula for $d_i$.
• Using a truncated mode sum and still expecting a full Dirac delta function.
• Mixing finite-volume Kronecker-delta normalization with infinite-volume Dirac-delta normalization.
• Omitting spin or other internal labels from the complete mode index.
• Thinking an ordinary basis rotation changes the physical state or vacuum.
• Confusing mode expansions with a full relativistic field theory; the formulas here are nonrelativistic unless additional structure is supplied.

Cross-Links

• Mode Occupations
• Creation and Annihilation Operators
• Bosonic Commutation Relations
• Fermionic Anticommutation Relations
• Number Operators
• Field Operators
• Field Operators in Many-Body Models
• One-Body Operators
• Two-Body Operators
• Many-Particle Hamiltonians
• Second Quantization: Bridge to QFT
• Fock Space Examples
• Continuous-Variable Systems
• Mode Decompositions
• Formula Sheet
• Coordinate Representation
• Fourier Transform

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.
• J. W. Negele and H. Orland, Quantum Many-Particle Systems, Addison-Wesley, 1988.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.

Exercises

1. Invert the mode expansion. Starting from $\psi(\mathbf x)=\sum_j\varphi_j(\mathbf x)d_j$, prove that $d_i=\int d^3x\,\varphi_i^*(\mathbf x)\psi(\mathbf x)$.

Solution

Insert the expansion:

$$\int d^3x\, \varphi_i^*(\mathbf x)\psi(\mathbf x) = \sum_j \left( \int d^3x\, \varphi_i^*(\mathbf x)\varphi_j(\mathbf x) \right)d_j.$$

Orthonormality gives the inner integral as $\delta_{ij}$, so the sum reduces to $d_i$.

2. Derive the field algebra. For bosonic modes, use $[a_i,a_j^\dagger]=\delta_{ij}$ and completeness to derive $[\psi(\mathbf x),\psi^\dagger(\mathbf y)]=\delta^{(3)}(\mathbf x-\mathbf y)$.

Solution

Use the expansions

$$\psi(\mathbf x)=\sum_i\varphi_i(\mathbf x)a_i, \qquad \psi^\dagger(\mathbf y) = \sum_j\varphi_j^*(\mathbf y)a_j^\dagger.$$

Then

$$\begin{aligned} [\psi(\mathbf x),\psi^\dagger(\mathbf y)] &= \sum_{ij} \varphi_i(\mathbf x) \varphi_j^*(\mathbf y) [a_i,a_j^\dagger] \\ &= \sum_i \varphi_i(\mathbf x) \varphi_i^*(\mathbf y) \\ &= \delta^{(3)}(\mathbf x-\mathbf y). \end{aligned}$$

The fermionic derivation is identical after replacing the commutator by the anticommutator.

3. Check a basis change. If $\lvert\chi_\alpha\rangle=\sum_i\lvert\varphi_i\rangle U_{i\alpha}$ and $\widetilde d_\alpha=\sum_iU_{i\alpha}^*d_i$, show that $\psi(\mathbf x)=\sum_i\varphi_i(\mathbf x)d_i$ is also $\sum_\alpha\chi_\alpha(\mathbf x)\widetilde d_\alpha$.

Solution

Substitute both definitions:

$$\sum_\alpha \chi_\alpha(\mathbf x)\widetilde d_\alpha = \sum_{\alpha i j} \varphi_i(\mathbf x) U_{i\alpha} U_{j\alpha}^* d_j.$$

Unitarity gives

$$\sum_\alpha U_{i\alpha}U_{j\alpha}^* = \delta_{ij}.$$

Therefore

$$\sum_\alpha \chi_\alpha(\mathbf x)\widetilde d_\alpha = \sum_j \varphi_j(\mathbf x)d_j = \psi(\mathbf x).$$

4. Plane-wave inversion. In a periodic box, prove that

$$d_{\mathbf k} = \frac{1}{\sqrt V} \int_V d^3x\, e^{-i\mathbf k\cdot\mathbf x} \psi(\mathbf x)$$

inverts

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

Solution

Insert the plane-wave expansion into the proposed inverse:

$$\frac{1}{V} \sum_{\mathbf q} \int_V d^3x\, e^{-i\mathbf k\cdot\mathbf x} e^{i\mathbf q\cdot\mathbf x} d_{\mathbf q}.$$

Periodic-box orthogonality gives

$$\frac{1}{V} \int_V d^3x\, e^{i(\mathbf q-\mathbf k)\cdot\mathbf x} = \delta_{\mathbf k\mathbf q}.$$

The sum therefore reduces to $d_{\mathbf k}$.

5. Spinful number density. For spin-$1/2$ fields, show that

$$N = \int d^3x\, \sum_s \psi_s^\dagger(\mathbf x)\psi_s(\mathbf x)$$

equals $\sum_{ns}d_{ns}^\dagger d_{ns}$ when $\{\phi_n\}$ is orthonormal.

Solution

Substitute

$$\psi_s(\mathbf x) = \sum_n \phi_n(\mathbf x)d_{ns}.$$

Then

$$\begin{aligned} N &= \sum_s \sum_{mn} \left( \int d^3x\, \phi_m^*(\mathbf x)\phi_n(\mathbf x) \right) d_{ms}^\dagger d_{ns} \\ &= \sum_s \sum_{mn} \delta_{mn} d_{ms}^\dagger d_{ns} \\ &= \sum_{ns} d_{ns}^\dagger d_{ns}. \end{aligned}$$

The spin sum is needed because each spin component is a distinct set of modes.