2

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?

Qmechanic
  • 220,844
Eric David Kramer
  • 1,737

1 Answers1

1

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.

Chiral Anomaly
  • 55,710
  • 5
  • 99
  • 166