# Gauge covariant derivative of a creation operator

Suppose we define the (gauge) covariant derivative or as $$\tilde{\nabla}=\nabla+ie\textbf{A},$$ where the vector potential $\textbf{A}$ has a matrix structure where only the diagonal has nonzero entries, but not all entries on the diagonal are the same. In calculating the covariant derivative of a product of an annihilation and a creation operator it appears that $$\tilde{\nabla}(\psi\psi^\dagger)=(\nabla\psi+ie\textbf{A}\psi)\psi^\dagger+\psi(\nabla\psi^\dagger-ie\psi^\dagger\textbf{A}),$$ where there's a minus sign appearing in the second term. This suggests that the covariant derivative works differently when acting on a creation operator, namely: $$\tilde{\nabla}\psi^\dagger=\psi^\dagger\tilde{\nabla}^\dagger=\nabla\psi^\dagger-ie\psi^\dagger\textbf{A}.$$ Question: Why does the covariant derivative act differently on the creation operator, why is it conjugated? If it's any help, the annihilation and creation operators are defined as $$\psi=\begin{pmatrix}\psi_\uparrow\\\psi_\downarrow\\\psi_\uparrow^\dagger\\\psi_\downarrow^\dagger\end{pmatrix},\qquad\psi^\dagger=\begin{pmatrix}\psi_\uparrow^\dagger&\psi_\downarrow^\dagger&\psi_\uparrow&\psi_\downarrow\end{pmatrix}.$$

• Obviously the covariant derivative should be different for $\psi$ and $\psi^\dagger$, because covariant derivative basically tells you how the particles couple to $A$, or in another word the charge of the particles. While $\psi$ annihilates a particle, $\psi^\dagger$ can be thought as annihilating a hole. Particles and holes obviously carry opposite charges under $A$. Oct 2 '15 at 0:30
• @Winther Yes, it's a 4x4 matrix, that's true. This term appears in the Eilenberger equation, and would most likely pop up in other equations involving electron-hole pairs as well. The $\psi^\dagger\psi$ term is easier to deal with, hence it's not mentioned here. I said "it appears that" because this is what a statement in a paper boils down to. Oct 2 '15 at 12:27

Let us look at this prolem from a (relativistic) field theory perspective.

The Hamiltonian for $\psi$ must contain a term of the form $$\nabla\psi^\dagger\cdot\nabla\psi$$ due to Lorentz invariance. Assuming $\psi$ to transform in a representation of U(1). Resulting in the following simultanious transformations: \begin{aligned}\psi\rightarrow U(x)\psi \\\\\psi^\dagger \rightarrow \psi^\dagger U^\dagger(x)\end{aligned} where U(x) is a unitary (1x1) matrix describing the result of gauge transforming $\psi$. It is then easy to verify the gauge invariance of $\psi^\dagger \psi$, using $U^\dagger(x)U(x)=1$. In the case of U(1) we can write $U(x)=e^{i\xi(x)}$ and therefore $U^\dagger(x)=e^{-i\xi(x)}$, where U is the fundamental representation and $U^\dagger$ is in the anti-fundamental one. The above term in the Hamiltonian is somewhat problematic however, since $$\nabla [\psi U(x)] = \nabla \psi U(x)+\psi \nabla U(x)$$ contains the nasty inhomogeneos $\psi \nabla U(x)$ term. To solve this problem, the covariant derivative was introduced as a variant of $\nabla$ that commutes with the transformation, that is under a gauge transformation $$\tilde{\nabla} \psi \rightarrow U(x)(\tilde {\nabla}\psi)$$ we can take this as a defining property of $\tilde{\nabla}$. Notice however that the exact form of U(x) is determined by the transformation of $\psi$ and thus depends on the representation of $\psi$. In order to learn more about $\tilde{\nabla}$ we might express it in terms of $\nabla$ and `some other object' that we will call X for now. $$\tilde{\nabla}=\nabla+X$$ we would then like it to satisfy $$\tilde{\nabla}[\psi U(x)]=\tilde{\nabla}[\psi] U(x)$$ expanding both in terms of the ordinary derivative: $$(\nabla+X)[\psi U(x)]=\nabla\psi U(x) + \psi \nabla U(x) + X\psi U(x)$$ and we would like this to be equivalent to $$[\nabla\psi] U(x) + X\psi U(x)$$ without the nasty inhomogeneity. Notice how the offending term could be expressed as $$\nabla U(x)=i\nabla[\xi(x)] U(x)$$ Simmilarly: $$\nabla U^\dagger(x)=-i[\nabla\xi(x)] U^\dagger(x)$$ It is this sign difference that confuses you. To illuminate, express X in terms of A. Since the physical part of A remains unchanged under a transformation $$A \rightarrow A+\nabla\xi(x)$$ (when accompanied by a suitable transformation of V). A suitable choice of X is $$X=\pm iA$$ It is however impossible to use the same expression for X when dealing with the anti-fundamental, to get rid of both the $\nabla U$ and $\nabla U^\dagger$ term using the same transformation. In stead, we need to pick a different X, depending on the representation it is acting on \begin{aligned} XU(x)=\pm iAU(X)\\\\ XU^\dagger(x)=\mp iAU^\dagger(X)\\\\\end{aligned} More often, X is expressed in terms of the relevant generator: $$X=A\tau$$ where $\tau$ is the generator of the group, in the prepresentation it is acting on. In the case of U(1), $\tau$ can be either equal to $i$ or equal to $-i$, depending on your conventions, either can be assignt to the fundamental representation, of the anti-fundamental one.

edit fixed some typography

further edit

when acting on $\psi\psi^\dagger$ in stead, simply use the Leibnitz rule (which still holds for the covariant derivative) and when expanding, let the two different X operators act from both sides, make sure the are expanded in different generators, again resulting in the sign difference.

tldr edit

TLDR: Your definition of the covariant derivative is not the actual definition, it is a consequence thereof. Indeed, the actual covariant derivative can be found to act on $\psi$ differently from $\psi^\dagger$.

Given that the regular (non-covariant) derivative of the adjoint satisfies $$\nabla \psi^\dagger = \left(\nabla \psi \right)^\dagger$$

one likewise expects and defines $$\tilde{\nabla}\psi^\dagger = \left( \tilde{\nabla}\psi \right)^\dagger = \left( \nabla \psi +ie\bf{A}\psi \right)^\dagger = \nabla \psi^\dagger -ie\bf{A} \psi^\dagger$$

• Yes, but the question is, why does the regular derivative satisfy that? And can one really just "define" an operator like that? If it were defined with the opposite sign there would be a physically different result. Oct 2 '15 at 10:25
• @Betohaku The detailed reason why $\tilde{\nabla} \psi$ is defined how it is defined and why it still stands that $\tilde{\nabla} \psi^\dagger = \left( \tilde{\nabla} \psi \right)^\dagger$ is addressed in the nice answer below. As to why the regular derivative satisfies a similar relation under hermitian conjugation think about the way $\nabla \psi$ and $\nabla \psi^\dagger$ are defined in terms of $\psi$ and $\psi^\dagger$. It comes down to (limits of) linear combinations that transform the same way.
– udrv
Oct 2 '15 at 18:55
• @Betohaku nah there wouldn't, the sign can be absorbed into the coupling constant $e$. Oct 6 '15 at 1:36