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Quantum Mechanics

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 {φi(x)}\{\varphi_i(\mathbf x)\}, the nonrelativistic annihilation field is written

ψ(x)=∑iφi(x)di,ψ†(x)=∑iφi∗(x)di†.\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 di,di†d_i,d_i^\dagger stand for either bosonic mode operators ai,ai†a_i,a_i^\dagger or fermionic mode operators ci,ci†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.

Helpful background. Mode Occupations explains orthonormal modes, and Creation and Annihilation Operators introduces the expansion coefficients.

Orthonormal modes and completeness

Section titled “Orthonormal modes and completeness”

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

∫d3x φi∗(x)φj(x)=δij.\int d^3x\, \varphi_i^*(\mathbf x) \varphi_j(\mathbf x) = \delta_{ij}.

Completeness means

∑iφi(x)φi∗(y)=δ(3)(x−y),\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 basis may use trap eigenfunctions, periodic-box plane waves, spin-orbitals, or orthonormal wavepackets. Bound eigenstates alone are not complete if the one-particle Hamiltonian also has scattering states; its spectral expansion then requires the continuum contribution. Lattice-site orbitals generally describe a retained band subspace. The mode label ii is shorthand for all quantum numbers needed to specify one complete one-particle state.

Mode Annihilation Operators

Section titled “Mode Annihilation Operators”

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

di†∣0⟩=∣1i⟩,di∣0⟩=0.d_i^\dagger\lvert0\rangle = \lvert1_i\rangle, \qquad d_i\lvert0\rangle = 0.

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

df=∑i⟨f∣φi⟩di,df†=∑i⟨φi∣f⟩di†.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

df=∫d3x f∗(x)ψ(x),df†=∫d3x f(x)ψ†(x).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 ψ(x)\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

Section titled “Position-Space Field Operators”

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

di=∫d3x φi∗(x)ψ(x).d_i = \int d^3x\, \varphi_i^*(\mathbf x)\psi(\mathbf x).

The adjoint formula is

di†=∫d3x φi(x)ψ†(x).d_i^\dagger = \int d^3x\, \varphi_i(\mathbf x)\psi^\dagger(\mathbf x).

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

∫d3x φi∗(x)ψ(x)=∑j(∫d3x φi∗(x)φj(x))dj=∑jδijdj=di.\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

Section titled “Completeness and Field Algebra”

For bosonic modes,

[ai,aj†]=δij,[ai,aj]=0.[a_i,a_j^\dagger] = \delta_{ij}, \qquad [a_i,a_j]=0.

Using the field expansion,

[ψ(x),ψ†(y)]=∑ijφi(x)φj∗(y)[ai,aj†]=∑iφi(x)φi∗(y)=δ(3)(x−y).\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,

{ci,cj†}=δij,{ci,cj}=0,\{c_i,c_j^\dagger\} = \delta_{ij}, \qquad \{c_i,c_j\}=0,

and the same completeness calculation gives

{ψ(x),ψ†(y)}=δ(3)(x−y).\{\psi(\mathbf x),\psi^\dagger(\mathbf y)\} = \delta^{(3)}(\mathbf x-\mathbf y).

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

PS(x,y)=∑i∈Sφi(x)φi∗(y).P_{\mathcal S}(\mathbf x,\mathbf y) = \sum_{i\in\mathcal S} \varphi_i(\mathbf x) \varphi_i^*(\mathbf y).

For a one-dimensional ring, retaining kn=2πn/Lk_n=2\pi n/L with −K≤n≤K-K\le n\le K gives

PK(x,y)=1L∑n=−KKe2πin(x−y)/L=sin⁡[(2K+1)π(x−y)/L]Lsin⁡[π(x−y)/L].P_K(x,y)=\frac1L\sum_{n=-K}^K e^{2\pi in(x-y)/L} =\frac{\sin[(2K+1)\pi(x-y)/L]}{L\sin[\pi(x-y)/L]}.

At coincident points its limiting value is (2K+1)/L(2K+1)/L, not an infinite delta-function peak. This projected kernel approaches the periodic delta distribution as the cutoff grows; it is not a pointwise approximation to an ordinary function. Spatial projection is distinct from a bosonic occupation cutoff, which changes the ladder commutator itself.

Basis Changes

Section titled “Basis Changes”

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

∣χα⟩=∑i∣φi⟩Uiα.\lvert\chi_\alpha\rangle = \sum_i \lvert\varphi_i\rangle U_{i\alpha}.

In wavefunction form,

χα(x)=∑iφi(x)Uiα.\chi_\alpha(\mathbf x) = \sum_i \varphi_i(\mathbf x)U_{i\alpha}.

The field operator is basis-independent:

ψ(x)=∑iφi(x)di=∑αχα(x)d~α.\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

d~α=∑iUiα∗di,d~α†=∑iUiαdi†.\widetilde d_\alpha = \sum_i U_{i\alpha}^* d_i, \qquad \widetilde d_\alpha^\dagger = \sum_i U_{i\alpha} d_i^\dagger.

Because UU 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

Section titled “Momentum Modes”

For a cubic box of volume V=L3V=L^3 with periodic boundary conditions, the normalized plane-wave modes are

φk(x)=1Veik⋅x,k=2πLn,n∈Z3.\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

ψ(x)=1V∑keik⋅xdk,\psi(\mathbf x) = \frac{1}{\sqrt V} \sum_{\mathbf k} e^{i\mathbf k\cdot\mathbf x}d_{\mathbf k},

with inverse

dk=1V∫Vd3x e−ik⋅xψ(x).d_{\mathbf k} = \frac{1}{\sqrt V} \int_V d^3x\, e^{-i\mathbf k\cdot\mathbf x} \psi(\mathbf x).

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

In infinite volume, a common continuum convention is

ψ(x)=∫d3k(2π)3/2 eik⋅xd(k),\psi(\mathbf x) = \int \frac{d^3k}{(2\pi)^{3/2}}\, e^{i\mathbf k\cdot\mathbf x} d(\mathbf k),

with

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

for bosons, or the corresponding anticommutator for fermions. Since ∑k→V∫d3k/(2π)3\sum_{\mathbf k}\to V\int d^3k/(2\pi)^3, this convention relates the operators by d(k)=V/(2π)3 dkd(\mathbf k)=\sqrt{V/(2\pi)^3}\,d_{\mathbf k} in the continuum normalization limit. The discrete operator is dimensionless; d(k)d(\mathbf k) has units of length3/2^{3/2} and ψ(x)\psi(\mathbf x) has units of length−3/2^{-3/2}. Other Fourier conventions redistribute powers of 2π2\pi consistently.

Spinful Modes

Section titled “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-1/21/2 particles is

∣n,s⟩=∣ϕn⟩⊗∣s⟩,s∈{↑,↓}.\lvert n,s\rangle = \lvert\phi_n\rangle\otimes\lvert s\rangle, \qquad s\in\{\uparrow,\downarrow\}.

The field has components

ψs(x)=∑nϕn(x)dns,ψs†(x)=∑nϕn∗(x)dns†.\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,

{ψs(x),ψt†(y)}=δstδ(3)(x−y),\{\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(x)=∑sψs†(x)ψs(x),n(\mathbf x) = \sum_s \psi_s^\dagger(\mathbf x)\psi_s(\mathbf x),

and a spin-density component is written

Sa(x)=∑stψs†(x)(Sa)stψt(x).S^a(\mathbf x) = \sum_{st} \psi_s^\dagger(\mathbf x) (S^a)_{st} \psi_t(\mathbf x).

For physical spin density, use Sa=ℏσa/2S^a=\hbar\sigma^a/2 for spin one-half; using Pauli matrices without this factor instead defines a dimensionless polarization density. In a spin–orbit eigenbasis the expansion uses spinor components φi,s(x)\varphi_{i,s}(\mathbf x) rather than a factorized orbital and spin label.

The internal labels matter: 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.

From field expressions to mode matrices

Section titled “From field expressions to mode matrices”

Inserting the expansion into A^=∫d3x d3y ψ†(x)A(x,y)ψ(y)\widehat A=\int d^3x\,d^3y\,\psi^\dagger(\mathbf x)A(\mathbf x,\mathbf y)\psi(\mathbf y) gives A^=∑ij⟨φi∣A∣φj⟩di†dj\widehat A=\sum_{ij}\langle\varphi_i|A|\varphi_j\rangle d_i^\dagger d_j. The one-body operator derivation explains why this acts as the sum of the same one-particle observable over all particles. Expanding four fields similarly produces two-body matrix elements.

Exercises

Section titled “Exercises”
  1. Invert the mode expansion. Starting from ψ(x)=∑jφj(x)dj\psi(\mathbf x)=\sum_j\varphi_j(\mathbf x)d_j, prove that di=∫d3x φi∗(x)ψ(x)d_i=\int d^3x\,\varphi_i^*(\mathbf x)\psi(\mathbf x).
Solution

Insert the expansion:

∫d3x φi∗(x)ψ(x)=∑j(∫d3x φi∗(x)φj(x))dj.\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 δij\delta_{ij}, so the sum reduces to did_i.

  1. Check a basis change. If ∣χα⟩=∑i∣φi⟩Uiα\lvert\chi_\alpha\rangle=\sum_i\lvert\varphi_i\rangle U_{i\alpha} and d~α=∑iUiα∗di\widetilde d_\alpha=\sum_iU_{i\alpha}^*d_i, show that ψ(x)=∑iφi(x)di\psi(\mathbf x)=\sum_i\varphi_i(\mathbf x)d_i is also ∑αχα(x)d~α\sum_\alpha\chi_\alpha(\mathbf x)\widetilde d_\alpha.

As a finite-lattice check, take NsN_s sites at rj=jar_j=ja, j=0,…,Ns−1j=0,\ldots,N_s-1, with periodic boundary conditions. For km=2πm/(Nsa)k_m=2\pi m/(N_sa), m=0,…,Ns−1m=0,\ldots,N_s-1, define ckmσ=Ns−1/2∑je−ikmrjcjσc_{k_m\sigma}=N_s^{-1/2}\sum_j e^{-ik_mr_j}c_{j\sigma}. Show explicitly that these fermionic operators satisfy {ckmσ,ckm′σ′†}=δmm′δσσ′\{c_{k_m\sigma},c_{k_{m'}\sigma'}^\dagger\}=\delta_{mm'}\delta_{\sigma\sigma'}.

Solution

Substitute both definitions:

∑αχα(x)d~α=∑αijφi(x)UiαUjα∗dj.\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

∑αUiαUjα∗=δij.\sum_\alpha U_{i\alpha}U_{j\alpha}^* = \delta_{ij}.

Therefore

∑αχα(x)d~α=∑jφj(x)dj=ψ(x).\sum_\alpha \chi_\alpha(\mathbf x)\widetilde d_\alpha = \sum_j \varphi_j(\mathbf x)d_j = \psi(\mathbf x).

For the lattice transform, use the site anticommutator to obtain

{ckmσ,ckm′σ′†}=δσσ′Ns∑j=0Ns−1e2πi(m′−m)j/Ns=δmm′δσσ′.\begin{aligned} \{c_{k_m\sigma},c_{k_{m'}\sigma'}^\dagger\} &=\frac{\delta_{\sigma\sigma'}}{N_s} \sum_{j=0}^{N_s-1}e^{2\pi i(m'-m)j/N_s}\\ &=\delta_{mm'}\delta_{\sigma\sigma'}. \end{aligned}

The finite geometric sum is NsN_s for m=m′m=m' and zero for distinct momenta in the stated grid. The symmetric Ns−1/2N_s^{-1/2} normalization therefore preserves the canonical algebra.

  1. Spinful number density. For spin-1/21/2 fields, show that
N=∫d3x ∑sψs†(x)ψs(x)N = \int d^3x\, \sum_s \psi_s^\dagger(\mathbf x)\psi_s(\mathbf x)

equals ∑nsdns†dns\sum_{ns}d_{ns}^\dagger d_{ns} when {ϕn}\{\phi_n\} is orthonormal.

Solution

Substitute

ψs(x)=∑nϕn(x)dns.\psi_s(\mathbf x) = \sum_n \phi_n(\mathbf x)d_{ns}.

Then

N=∑s∑mn(∫d3x ϕm∗(x)ϕn(x))dms†dns=∑s∑mnδmndms†dns=∑nsdns†dns.\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.

References

Section titled “References”
  • A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
  • 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.
  • M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.
  • R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.