Derivation of the QED Hamiltonian

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  Introduction to Atomic Physics

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  Frontiers in Atomic Physics

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  Quantum gases

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 $a^{\dagger }$.

This result is simple and can be obtained with less rigorous derivations. Here, we want to at least mention all the steps of the rigorous derivation, starting from classical Maxwell's equations.

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. The component of the (vectorial) Fourier component, which is parallel to the k vector, is called the longitudinal part, the component orthogonal to the k vector is the transverse part. By transforming back to the spatial domain, we can distinguish the longitudinal and transverse parts. It turns out that the longitudinal electric field is not a free degree of freedom, i.e. it follows instantaneously the positions of the particles through the Coulomb potential.

Introducing the vector potential reduces the number of field components to four.

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

In the Coulomb gauge, $\nabla \cdot A=0$, i.e. the longitudinal component of A vanished. Furthermore, the scalar potential is

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

Now we are down to two components, the transverse components of the vector poential A.

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 differential equation is identical to the equation for r and p for an harmonic oscillator potential.

We can go further 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

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

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 normal mode coordinates are a linear combination of $A$ and ${\dot {A}}$ from which we obtain the commutator

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

in full analogy to the raising and lowering operators for the simple harmonic oscillator.

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 instantaneous Coulomb field due to the charge configuration. The second term is the energy of 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

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

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 the energy in the radiation field. But it looks quantum. 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 polarization, and each one is described in its quantized form by the Hamiltonian $H=\hbar \omega (a_{k}^{\dagger }a_{k}+1/2)$.

The most important aspect of our derivation was the separation of the fields into longitudinal and transverse components. For example, a naive approach 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, eliminating the dependent components, is thus essential.

Also note that quantizing the electromagnetic field is only possible if we have an expression for the energy of the radiation field that also corresponds to a valid Lagrangian for the system. Finally, our approach is not relativistically covariant. There are covariant formulations of the quantized radiation field, but they are more complicated.

Quantum description of the radiation field

The inverse Fourier transform provides us with the field components 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 $H_{R}=\sum _{i}\hbar \omega _{i}(a_{i}^{\dagger }a_{i}+1/2)$.

We may write the total Hamiltonian as a sum of parts

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

where $H_{P}$ is the particle Hamiltonian including the Coulomb field

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

$H_{R}$ 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

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

For atoms, $H_{I1}\gg H_{I1}^{S},H_{I2}$, typically. We will often perform perturbation theory in $H_{I}$. Note that this is NOT a perturbation theory in charge or field strength, since the Coulomb field has been separated off and is fully included in the particle Hamiltonian.

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 near-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 shows that the simple ${\vec {d}}\cdot {\vec {E}}$ is exact (in the long wavelength approximation) and includes the $H_{I2}$ term.

The derivation 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 $D$.

Conclusion: What you should remember from this lengthy derivation is the separation into longitudinal and transverse fields, the fact that the Coulomb interaction is included in the particle Hamiltonian, and that the electric dipole approximation includes the A^2 term.

References

• API Appendix