The low-energy QED Hamiltonian that I would like to derive is ($c=\hbar=1$): $$ H = \frac{(p-eA)^2}{2m} = \frac{p^2}{2m} - \frac{e}{m} \vec{p}\cdot \vec{A} + \frac{e^2}{2m}A^2$$ where $A$ is the photon field operator. I can't get the third term. Here is what I did: Start with the Dirac Lagrangian and Legendre transform to get Hamiltonian: $$H = \int\!d^3x\,\psi^\dagger(x)\left[ \vec{\alpha}\cdot (-i\vec{\nabla}-e\vec{A}) - m\beta \right]\psi(x)$$ where $\alpha^i = \gamma^0 \gamma^i$ and $\beta = \gamma^0$. Now, in the Dirac, a.k.a standard representation of the Clifford algebra (as opposed to chiral representation which is better suited to high energy), we have $$H = \int\!d^3x\,\psi^\dagger(x)\left(\begin{array}{cc}-m &\vec{\sigma}\cdot (\vec{p}-e\vec{A})\\\vec{\sigma}\cdot (\vec{p}-e\vec{A}) & m\end{array}\right)\psi(x)$$ and $$\psi(x) = \sum_s\int\!\frac{d^3p}{(2\pi)^3}\,\sqrt{\frac{E+m}{2E}}\left(\begin{array}{c}\xi^s \\\frac{\vec{p}\cdot\vec{\sigma}}{E+m}\xi^s\end{array}\right)c_{ps}e^{i\vec{p}\cdot\vec{x}}\quad+\quad{\rm positrons}\;.$$ Now I want to consider the Hamiltonian only in the 1-electron Hilbert space, given by states with only one electron operator $c^\dagger_{ps}$ acting on the vacuum and any number of photon creation operators $a^\dagger_k$. If you plug this into the Hamiltonian, you get the $p^2/2m$ and the $p\cdot A$ terms with no problem. You also get a term that depends on spin $\xi^\dagger_{s'}\vec{\sigma}\xi_s$. But there is no sign of the $A^2$ term. What am I doing wrong?

Edit: The answer is obvious if you take a constant $A$. Maybe that's a hint as to how to get the answer in the general case?

  • $\begingroup$ I've seen it but never used it. Are you saying that if I do one here, I'll get the right Hamiltonian? Somehow the transformation will have to depend on $\vec{A}$. Are you sure that's legal? $\endgroup$ – Eric David Kramer May 23 at 14:08
  • $\begingroup$ Thanks! This is very helpful. I guess it's ok for A to be an operator too? $\endgroup$ – Eric David Kramer May 23 at 15:56
  • $\begingroup$ I suppose that in the case where A is an operator, the FW transformation will just correspond to a redefinition of the electron states in the rest frame, |↑⟩→𝐹(𝐴̂ )|↑⟩+G(𝐴̂ )|↓⟩, with 𝐹 and G operator-valued functions. And then we recover the ordinary low energy Hamiltonian. So the "electron" state also has to contain some photons (it has to contain photons anyway if you impose gauge invariance: aip.scitation.org/doi/10.1063/1.523291, so this would just be an extension of that). $\endgroup$ – Eric David Kramer May 25 at 10:00
  • $\begingroup$ I just meant that whatever FW transformation acts in the space of spinors can also be used to redefine spin up and spin down to get the Hamiltonian in the form we need. The momentum dependence just means that the definition of spin up/down will depend on p, i.e. we don't need to define additional action on the states. $\endgroup$ – Eric David Kramer May 26 at 13:45
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    $\begingroup$ If you put your comments in an answer, I'll vote & accept it. $\endgroup$ – Eric David Kramer May 31 at 16:27

When $A(x)$ is a background field, we can use the Foldy-Wouthuysen transformation. It's iterative, so we only get the answer to a given order in the $1/m$ expansion, but that's good enough for the goal stated at the beginning of the question. Detailed references include:

The Foldy-Wouthuysen assumes that $A$ is a background field. It constructs (in the $1/m$ expansion) a $U(x)$ that acts as $\psi(x)\to U(x)\psi(x)$. It's unitary as an "operator" on the space of $\psi(x)$s, but its action on the Hilbert space is undefined, because it involves derivatives with respect to $x$. (This is quantum field theory, so the Hilbert space is the space on which the field operators $\psi(x)$ act. The field operators are parameterized by $x$, but the Hilbert space on which they act is not.)

When $A$ is an operator, we would need to find a unitary operator $U$ on the Hilbert space such that the positive/negative-frequency parts of $\psi$ are decoupled by writing the Hamiltonian in terms of $U\psi(x)U^\dagger$, order by order in the $1/m$ expansion. Since products of field operators at a single point are undefined, we need use renormalization, which this typically changes coefficients in the resulting $1/m$ expansion. The coefficients need to be adjusted to make the predictions match before and after the transformation: if the result has $N$ terms, then we need to check $N$ predictions to fix their coefficients. This is the idea behind effective field theory: we write down all terms that are consistent with the theory's symmetries and field content, to a given order in the $1/m$ expansion, and then match a few predictions to fix the unknown coefficients. I cited some references here.

Altogether, the $1/m$ expansion given by the Foldy-Wouthuysen when $A$ is a background field produces the same terms that we need to include in the effective-field-theory approach when $A$ is a field operator, because they both involve terms consistent with the theory's symmetries and field content to the given order in $1/m$. The coefficients are generally different quantitatively, but the structure of the $1/m$ expansion is essentially the same.

  • $\begingroup$ These guys actually managed to do it using modern techniques: arxiv.org/pdf/2106.12586.pdf (Eq. 9), and the spin-dependent field redefinition is in Eq. 11. $\endgroup$ – Eric David Kramer Jun 28 at 9:34
  • $\begingroup$ @EricDavidKramer Good example! When extended to higher orders in the expansion, their approach ends up being what I described. If eq (9) doesn't already include some of the terms that are consistent with the theory's symmetries and field content, those terms will still be generated by renormalization. Since the number of terms is infinite, the only way to determine the coefficients of the first $N$ terms is to compare $N$ predictions. Their eqs (6)-(9) don't avoid this, because that's just the classical lagrangian. Quantization slips in later through the back door, starting in eq (19). $\endgroup$ – Chiral Anomaly Jun 28 at 12:36
  • $\begingroup$ @EricDavidKramer ...but since the goal of the original question was to obtain the lowest-order result, their approach is a nice way to do it. $\endgroup$ – Chiral Anomaly Jun 28 at 12:38
  • $\begingroup$ Good point about the normalization $\endgroup$ – Eric David Kramer Jun 28 at 13:41

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