QED notes

Identities & Expansions

Operator identities, frame transformations, and series expansions from the handwritten QM Reference notes.

Updated 2026-08-06

Operator Identities

Frame Transformation

For operators AA and BB,

eABeA=n=01n![A,[A,[An,B]]].e^A B e^{-A} = \sum_{n=0}^{\infty}\frac{1}{n!} \underbrace{[A,[A,\cdots[A}_{n},B]\cdots]] .

Equivalently, define adA(B)=[A,B]\operatorname{ad}_A(B)=[A,B]:

eABeA=eadAB.e^A B e^{-A}=e^{\operatorname{ad}_A}B .

For a simple oscillator with U=eiωaaatU=e^{-i\omega_a a^\dagger a t},

UaU=eiωata.U^\dagger a U=e^{-i\omega_a t}a .

Baker-Campbell-Hausdorff

If [A,B][A,B] commutes with both AA and BB, then

eA+B=eAeBe12[A,B].e^{A+B}=e^A e^B e^{-\frac12[A,B]} .

For coherent-state displacement operators,

Dα=eαaαa=eαaeαaeα2/2,D_\alpha = e^{\alpha a^\dagger-\alpha^*a} = e^{\alpha a^\dagger}e^{-\alpha^*a}e^{-|\alpha|^2/2}, DαDβ=e(αβαβ)/2Dα+β.D_\alpha D_\beta = e^{(\alpha\beta^*-\alpha^*\beta)/2}D_{\alpha+\beta}.

Schrieffer-Wolff Transformation

Start from a time-independent Hamiltonian

H=H0+V,H=H_0+V,

where H0m=EmmH_0|m\rangle=E_m|m\rangle and VV is a small perturbation. Absorb the diagonal part of VV into H0H_0, so mVm=0\langle m|V|m\rangle=0.

Define the dressed Hamiltonian by the unitary transformation

H=eSHeS.H'=e^SHe^{-S}.

For small VV, the generator SS is also small, and

H=H+[S,H]+12[S,[S,H]]+.H'=H+[S,H]+\frac12[S,[S,H]]+\cdots .

With SS chosen to remove the first-order off-diagonal coupling [S,H0]=V[S, H_0] = -V,

H=H0+12[S,V]+O(V3),H'=H_0+\frac12[S,V]+O(V^3),

where, in the eigenbasis of H0H_0,

mSm=mVmEmEm,mm,mSm=0.\langle m|S|m'\rangle =\frac{\langle m|V|m'\rangle}{E_m-E_{m'}}, \qquad m\ne m', \qquad \langle m|S|m\rangle=0.

Interaction Frame

Start from a lab-frame Hamiltonian

H(t)=H0(t)+HI(t).H(t)=H_0(t)+H_I(t).

The static or unperturbed evolution is

U0(t)=Texp[i0tH0(t)dt].U_0(t)=T\exp\left[-\frac{i}{\hbar}\int_0^tH_0(t')\,dt'\right].

Writing U(t)=U0(t)UI(t)U(t)=U_0(t)U_I(t), the interaction-frame unitary obeys

U˙I(t)=iHI(I)(t)UI(t),\dot U_I(t)=-\frac{i}{\hbar}H_I^{(I)}(t)U_I(t),

where

HI(I)(t)=U0(t)(H(t)H0(t))U0(t)=U0(t)HI(t)U0(t).H_I^{(I)}(t) =U_0^\dagger(t)\bigl(H(t)-H_0(t)\bigr)U_0(t) =U_0^\dagger(t)H_I(t)U_0(t).

Time-Dependent Expansions

Magnus Expansion

For a time-ordered exponential,

Texp(0tH(t1)dt1)=exp[0tH(t1)dt1+120tdt10t1dt2[H(t1),H(t2)]+O(H3)].T\exp\left(\int_0^t H(t_1)\,dt_1\right) = \exp\left[ \int_0^t H(t_1)\,dt_1 +\frac12\int_0^t dt_1\int_0^{t_1}dt_2\,[H(t_1),H(t_2)] +O(H^3) \right].

For Schrödinger evolution, insert the usual factor i/-i/\hbar in the generator.

Dyson Expansion

The notes only flag Dyson expansion and effective Hamiltonians. A common starting point is

U(t)=Texp[i0tH(t1)dt1],U(t)=T\exp\left[-\frac{i}{\hbar}\int_0^t H(t_1)\,dt_1\right],

expanded as a time-ordered series before collecting terms into an effective Hamiltonian.

Jacobi-Anger Expansion

The Bessel-function expansions are

eizcosθ=n=inJn(z)einθ,e^{iz\cos\theta} =\sum_{n=-\infty}^{\infty}i^nJ_n(z)e^{in\theta}, eizsinθ=n=Jn(z)einθ.e^{iz\sin\theta} =\sum_{n=-\infty}^{\infty}J_n(z)e^{in\theta}.

Here JnJ_n is the nnth Bessel function of the first kind, with

Jn(z)=(1)nJn(z),Jn(z)=(1)nJn(z).J_{-n}(z)=(-1)^nJ_n(z),\qquad J_n(-z)=(-1)^nJ_n(z).

In trigonometric form,

eizcosθ=J0(z)+2n=1inJn(z)cos(nθ).e^{iz\cos\theta} =J_0(z)+2\sum_{n=1}^{\infty}i^nJ_n(z)\cos(n\theta).

Supplementary Proof Notes

Adjoint Action

Let

O(λ)=eλABeλA.O(\lambda)=e^{\lambda A}Be^{-\lambda A}.

Then

dOdλ=AO(λ)O(λ)A=[A,O(λ)].\frac{dO}{d\lambda}=AO(\lambda)-O(\lambda)A=[A,O(\lambda)].

The nested-commutator series and O(λ)O(\lambda) solve the same first-order differential equation with the same initial value O(0)=BO(0)=B, so uniqueness gives the frame-transformation identity.

BCH For Displacements

With

A=αa,B=αa,A=\alpha a^\dagger,\qquad B=-\alpha^*a,

the commutator [A,B]=α2[A,B]=-|\alpha|^2 is a scalar, so the BCH simplification gives the normal-ordered displacement form. The same scalar-commutator step gives the product rule for DαDβD_\alpha D_\beta.

Schrieffer-Wolff Expansion

Substitute H=H0+VH=H_0+V into the adjoint-action expansion:

H=H0+V+[S,H0]+[S,V]+12[S,[S,H0]]+12[S,[S,V]]+.H'=H_0+V+[S,H_0]+[S,V] +\frac12[S,[S,H_0]]+\frac12[S,[S,V]]+\cdots .

Since S=O(V)S=O(V), the term [S,[S,V]][S,[S,V]] is O(V3)O(V^3). Choose SS so that

V+[S,H0]=0.V+[S,H_0]=0.

Then [S,[S,H0]]=[S,V][S,[S,H_0]]=-[S,V], giving

H=H0+12[S,V]+O(V3).H'=H_0+\frac12[S,V]+O(V^3).

In the eigenbasis of H0H_0,

[S,H0]mm=(EmEm)Smm,[S,H_0]_{mm'}=(E_{m'}-E_m)S_{mm'},

so the first-order cancellation condition gives

Smm=VmmEmEm,mm.S_{mm'}=\frac{V_{mm'}}{E_m-E_{m'}},\qquad m\ne m'.

For Hermitian VV, this choice satisfies S=SS^\dagger=-S, so eSe^S is unitary. If two states are degenerate or intentionally kept in the same low-energy subspace, do not divide by the small denominator; keep that block and project only after the transformation.