# Why is the Quantum Electrodynamic Hamiltonian complex?

For reference, this is from Shankar's QM book, with the Hamiltonian $$H=\frac{1}{2m}\left(\textbf{P}\cdot\textbf{P}-\frac{q}{c}\textbf{P}\cdot\textbf{A}-\frac{q}{c}\textbf{A}\cdot\textbf{P}+\frac{q^2}{c^2}\textbf{A}\cdot\textbf{A}\right)$$ and the exact quote is "If the problem involves a magnetic field, the Hamiltonian is no longer real in the coordinate basis".

Is the Hamiltonian no longer real in the coordinate basis because $$\textbf{A}$$ and $$\textbf{P}$$ do not commute, and so the conjugate will be different?

• Why would P and A commute? – Ján Lalinský Feb 2 at 20:55
• thanks for pointing that out. It's a typo which has been fixed. A and P do not commute. – Redcrazyguy Feb 2 at 21:03
• What makes you think it is "no longer real"? It must be self-adjoint and bounded from below. – DanielC Feb 2 at 22:54
• It says in the book that if the "If the problem involves a magnetic field, the Hamiltonian is no longer real in the coordinate basis". If you have the book it's on the bottom of p.177. – Redcrazyguy Feb 3 at 0:34
• @Redcrazyguy Have you considered how $\mathbf{P}$ and $\mathbf{A}$ are represented in the coordinate basis? – Chiral Anomaly Feb 3 at 1:00

The Hamiltonian is Hermitian in that it equals its own adjoint: $$H = H^\dagger$$. In a matrix representation, the adjoint is the conjugate transpose, not simply the conjugate. Thus, a Hermitian operator can be represented by a matrix that is neither real nor symmetric, as long as conjugating and transposing together return the original matrix.
A simple example of a nonreal Hermitian operator is the momentum operator in the position basis, $$\mathbf{P} = -i \hbar \boldsymbol{\nabla}$$. Applying this to a real function generally returns a nonreal function; likewise for the Hamiltonian in the question. However, being Hermitian, these operators always have real eigenvalues.