Derivation of the QED Hamiltonian

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We describe here a rigorous derivation of atom-photon interactions. Our overall goal is to arrive at a Hamiltonian description for the energy of a system of atoms, photons, and atoms interacting with photons through the radiation field.

The basic interaction we will obtain is the dipole interaction Hamiltonian,

$H_{int}=-{\vec {d}}\cdot {\vec {E}}\,,$

where ${\vec {d}}$ is the atom's dipole moment, and ${\vec {E}}$ is the electric field at the position of the atom. In the end, the electric field will be quantized, and described by operators $a$ and $a^{\dagger }$.

Quantum electrodynamics

The classical Hamiltonian which describes one particle

$H={\frac {p^{2}}{2m}}+V(r)\,,$

and this is transformed into the quantum picture by enforcing the commutation relation $[r,p]=i\hbar$. We will do the same for an atom interacting with light.

Let us begin with Maxwell's equations.

${\begin{array}{rcl}\nabla \cdot E(r,t)&=&{\frac {1}{\epsilon _{0}}}\rho (r,t)\\\nabla \cdot B(r,t)&=&0\\\nabla \times E(r,t)&=&-\partial _{t}B(r,t)\\\nabla \times B(r,t)&=&{\frac {1}{c^{2}}}\partial _{t}E(r,t)+{\frac {1}{\epsilon _{0}c^{2}}}j(r,t)\,.\end{array}}$

How many of the field components are true degrees of freedom? We can understand this by taking a spatial Fourier transform. It turns out that the longitudinal electric field is not a free degree of freedom.

We obtain as the essential equations the vector potential

${\begin{array}{rcl}E(r,t)&=&\nabla U(r,t)-\partial _{t}A(r,t)\\B(r,t)&=&\nabla \times A(r,t)\,.\end{array}}$

In the Coulomb gauge, $\nabla \cdot A=0$, such that the scalar potential is

$U(r,t)={\frac {1}{4\pi \epsilon _{0}}}\int d^{3}r'{\frac {\rho (r',t)}{|r-r'|}}\,.$

Keep in mind that the equation of motion is now a second order differential equation for the vector potential. All that is needed to specify the field evolution is thus the initial values of $A_{\perp }$ and ${\dot {A}}_{\perp }$, the transverse vector potential. The only free degrees of freedom are the two components of the transverse vector potential.

We can further understand this by decomposing the Fourier representation of the field and potential in terms of its normal modes,

${\begin{array}{rcl}\alpha (k,t)&=&\lambda (k)\left[E_{\perp }(k,t)-c{\frac {\vec {k}}{k}}\times B(k,t)\right]\\&=&\lambda (k)[-{\dot {A}}_{\perp }(k,t)+i\omega A_{\perp }(k,t)]\,.\end{array}}$

Using this, we can represent the equations of motion for the field as

${\dot {\alpha }}(k,t)+i\omega \alpha (k,t)={\frac {i}{\sqrt {2\epsilon _{0}\hbar \omega }}}j_{\perp }(k,t)\,.$

Note

${\vec {A}}_{\perp }(r)=\int d^{3}k\sum _{\epsilon }A_{\omega }[{\vec {\epsilon }}\alpha _{\epsilon }(k)e^{i2r}+h.c.]\,.$

in the normal mode decomposition, where we now use ${\vec {\epsilon }}$ as the photon polarization, which carries the vector direction of the potential.

Let us now quantize the field. We identify an equivalence between $A$ and ${\dot {A}}$ with $r$ and $p$, and quantize accordingly. The commutator is

$[a_{\epsilon }(k),a_{\epsilon '}^{\dagger }(k')]=\delta _{\epsilon \epsilon '}\delta (k-k')\,.$

Energy in the radiation field

It is helpful to go back to consider for a moment what the energy in the field is. Recall that

${\begin{array}{rcl}H&=&{\frac {\epsilon _{0}}{2}}\int d^{3}r({\vec {E}}^{2}+c^{2}{\vec {B}}^{2})\\&=&\int d^{3}rE_{\parallel }^{2}+{\frac {\epsilon _{0}}{2}}\int d^{3}r(E_{\perp }^{2}+c^{2}B^{2})\end{array}}$

where we can identify

$\int d^{3}rE_{\parallel }^{2}={\frac {1}{8\pi \epsilon _{0}}}\int \int d^{3}rd^{3}d'{\frac {\rho (r)\rho (r')}{|r-r'|}}$

as the energy of the Coulomb field due to the charge configuration. This part of the energy is static, with respect to time evolution of the field. The second term is the energy is the transverse component of the field, $H_{\rm {trans}}$, the radiation energy, which we can understand by introducing again our expression for the vector potential.

${\begin{array}{rcl}H_{\rm {trans}}=\epsilon _{0}\int d^{3}k\left[{\frac {\Pi _{\epsilon }^{2}\Pi _{e}p}{\epsilon _{0}^{2}}}+k^{2}c^{2}A_{\epsilon }^{*}A_{\epsilon }\right]\,,\end{array}}$

where $\Pi _{\epsilon }(k)=\epsilon _{0}{\dot {A}}_{\epsilon }(k)$ is identified as the conjugate momentum. This looks much like a simple harmonic oscillator Hamiltonian. Now introduce normal modes, using

$\alpha \sim [-{\dot {A}}_{\perp }+i\omega E_{\perp }]\,.$

This gives

${\begin{array}{rcl}H_{\rm {trans}}=\int d^{3}k\sum _{\epsilon }{\frac {\hbar }{2}}[\alpha _{\epsilon }^{*}\alpha _{e}p+\alpha _{\epsilon }\alpha ^{*}\epsilon ]\,,\end{array}}$

a purely classical expression for energy in the radiation field. Where does $\hbar$ come from? It enters in the constant relating the normal modes with $\alpha$. It is just a unit used in the definitions at this moment, which is convenient to use because later on $\hbar$ appears in the quantum expression of the energy.

This expression for energy is really identical to that for the classical simple harmonic oscillator,

$H=()p^{2}+()q^{2}\,,$

where $p\sim {\dot {q}}$. For this classical oscillator, it is helpful to introduce a variable describing superpositions of $a=(q+ip)/{\sqrt {2}}$, giving

$H={\frac {\hbar \omega }{2}}(p^{2}+q^{2})={\frac {\hbar }{2}}(a^{*}a+aa^{*})\,.$

When quantized, this becomes $H=\hbar \omega (a^{\dagger }a+1/2)$. We conclude that the radiation field is just a bunch of oscillators, with one per $k$ vector, and each one is described in its quantized form by the Hamiltonian $H=\hbar \omega (a_{k}^{\dagger }a_{k}+1/2)$. Beware, however, that not all expressions for the energy are valid models for the radiation field; it is essential that they also correspond to a valid Lagrangian for the system. We have also not provide a relativistically covariant formulation of the radiation field.

Quantum description of the radiation field

Let us now go back to the quantum description of the radiation field. Keep in mind that the naive approach of using Cartesian coordinates, without eliminating the longitudinal field, would fail miserably, because

$[A_{i}(r),\Pi _{j}(r')]\neq i\hbar \delta _{ij}(r-r')\,.$

The treatment above, using the vector potential, is thus essential. We now have as the field definitions, in terms of the quantum operators,

${\begin{array}{rcl}A_{\perp }(r)&=&\int d^{3}k\sum _{\epsilon }A_{\omega }[\epsilon a_{\epsilon }(k)e^{ik\cdot r}+\epsilon a_{\epsilon }^{\dagger }(k)e^{-ik\cdot r}]\\E_{\perp }(r)&=&\int d^{3}k\sum _{\epsilon }iE_{\omega }[\epsilon a_{\epsilon }(k)e^{ik\cdot r}-\epsilon a_{\epsilon }^{\dagger }(k)e^{-ik\cdot r}]\\B(r)&=&\int d^{3}k\sum _{\epsilon }iB_{\omega }[(k\times \epsilon )a_{\epsilon }(k)e^{ik\cdot r}-(k\times \epsilon )a_{\epsilon }^{\dagger }(k)e^{-ik\cdot r}]\,.\end{array}}$

The particle operators are

${\begin{array}{rcl}\rho &=&\sum _{\alpha }q_{\alpha }\delta (r-r_{\alpha })\end{array}}$

Coupling of atoms and the radiation field

The total Hamiltonian for the radiation field and charges is

$H=\sum _{\alpha }{\frac {1}{2m_{\alpha }}}\left[p_{\alpha }-q_{\alpha }A_{\perp }(r_{\alpha })\right]^{2}+\sum _{\alpha }\left(-g{\frac {q}{2m_{\alpha }}}\right)S_{\alpha }\cdot B(r_{\alpha })+V_{\rm {Coul}}+H_{R}\,,$

where the second term, with $S_{\alpha }\cdot B(r_{\alpha })$ has been added by hand, and describes spin interacting with the magnetic field, which will be discussed later. It can be derived from first principles by starting with the Dirac equation, expressed in the non-relativistic limit. The important new term, compared with standard nonrelativistic quantum mechanics, is the replacement of momentum $p$ with $p-qA$. The Coulomb interaction energy is standard. The radiation field energy is $H_{R}\sum _{i}\hbar \omega _{i}(a_{i}^{\dagger }a_{i}+1/2)$. Simplifying this, we may write the total Hamiltonian as

$H=H_{P}+H_{R}+H_{I}\,,$

where $H_{P}$ is the particle Hamiltonian,

$H_{P}=\sum _{\alpha }{\frac {p_{\alpha }^{2}}{2m_{\alpha }}}+V_{\rm {Coul}}\,,$

$H_{R}$ is the Hamiltonian for the radiation field, and $H_{I}$ can be written as the sum of three parts, $H_{I}=H_{I1}+H_{I2}+H_{I1}^{S}$, and

${\begin{array}{rcl}H_{I1}&=&-\sum _{\alpha }{\frac {q_{\alpha }}{m_{\alpha }}}p_{\alpha }A_{\perp }(r_{\alpha })\\H_{I1}^{S}&=&-\sum _{\alpha }g_{\alpha }{\frac {q_{\alpha }}{2m_{\alpha }}}S_{\alpha }\cdot B(r_{\alpha })A_{\perp }(r_{\alpha })\\H_{I2}&=&\sum _{\alpha }{\frac {q_{\alpha }}{2m_{\alpha }}}A_{\perp }^{2}(r_{\alpha })\,.\end{array}}$

For atoms, $H_{I1}\gg H_{I1}^{S},H_{I2}$, typically.

The dipole approximation

Typically, for atomic physics, the wavelength of radiation is much much larger than the size of the atom, so that we may write the main interaction between atoms and the radiation field as

$H_{int}=H_{I1}={\frac {q}{m}}{\vec {A}}_{0}\langle 2|{\vec {p}}e^{ikr}|1{\rangle }\,,$

where $E=i\omega A$. This simplifies to

$H_{int}={\frac {iq{\vec {E}}_{0}}{m\omega }}\langle 2|p|1{\rangle }\,,$

in the limit that the field wavelength is much larger than the atom, so we can take $r=0$. Since $[r,H]=(i\hbar /m)p$, and

$\langle 2|[r,H]|1\rangle =(E_{1}-E_{2})\langle 2|r|1{\rangle }\,,$

then using $\hbar \omega _{12}=E_{1}-E_{2}$ gives

$\langle 2|p|1\rangle =-im\omega _{12}\langle 2|r|1{\rangle }\,.$

The interaction energy is thus

$H_{int}=q{\vec {E}}_{0}{\frac {\omega _{12}}{\omega }}\langle 2|{\vec {r}}|1{\rangle }\,,$

which in the limit of a resonant interaction, becomes

$H_{int}=q{\vec {E}}_{0}\langle 2|{\vec {r}}|1\rangle \sim -{\vec {d}}\cdot {\vec {E}}\,.$

Questions that arise in this loose derivation include: what happens for off-resonant interactions? And what happens with the other interaction term we derived above, $H_{I2}$?

A rigorous way to obtain the full solution, which is essentially the same as that sketched above, is given in API. It involves using a canonical transformation with the operator

$T=\exp \left[-{\frac {i}{\hbar }}{\vec {d}}\cdot A_{\perp }\right]\,,$

which is just a displacement operator acting on momentum, that transforms $p-qA(r)$ into $p$, in a new frame of reference. We approximate $p-qA(r)$ with $p-qA(0)$, which is valid in the dipole approximation, in which $r\ll \lambda$. The Hamiltonian in this frame of reference is

${\begin{array}{rcl}H'&=&THT^{\dagger }\\&\sim &{\frac {p^{2}}{2m}}+V_{Coul}+\sum _{j}{\frac {1}{2\epsilon _{0}L^{3}}}(\epsilon _{j}\cdot d)^{2}+\sum _{j}\hbar \omega _{j}(a_{j}^{\dagger }a+1/2)-d\cdot \sum _{j}E_{\omega _{j}}[ia_{j}\epsilon _{j}-ia_{j}^{\dagger }\epsilon _{j}]\,.\end{array}}$

In this frame, the dipole interaction energy appears explicitly. The transformed electric field is

${\begin{array}{rcl}E_{\perp }'(r)&=&TE_{\perp }T^{\dagger }\\&=&E_{\perp }(r)-{\frac {1}{\epsilon _{0}}}P_{\perp }(r)\,.\end{array}}$

The interaction Hamiltonian is

$H_{I}'=-d\cdot D'(0)/\epsilon _{0}=-d\cdot E_{\perp }(0)\,.$

Note that this formulation already takes into account the polarizability of matter, and the relation between $E$ and $D$.