# Derivative $\Lambda(\phi,\psi)=\frac{\langle\phi|A|\psi\rangle}{\langle\phi|\psi\rangle}$

I' m so sorry I found this expression on a handwritten sheet so I would like to check that it has sense and exactly what it means because I have not found a similar expression on any book.

Let the functional and the generic operator A: $$\Lambda(\phi,\psi)=\frac{\langle\phi|A|\psi\rangle}{\langle\phi|\psi\rangle}$$ $$\Lambda(\phi+\epsilon\alpha,\psi)-\Lambda(\phi,\psi)=\epsilon \left ( \frac{\langle\alpha|A|\psi\rangle}{\langle\phi|\psi\rangle} - \frac{\langle\phi|A|\psi\rangle}{\langle\phi|\psi\rangle^2} \langle\alpha |\psi\rangle \right ) + o(\epsilon^2)$$

• I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than $<, >$ does :) – Patrick Stevens May 16 '18 at 21:00
• I'm the \left\langle \right\rangle fairy, if you want to take it up a notch. – J.G. May 16 '18 at 21:09
• I' m so sorry guys – Stefano Barone May 16 '18 at 21:11
• By tradition, in physics journals (at least those that frequently have Quantum Field Theory content at a foundational level) the bra and ket vectors are written with the old fashioned $<a | b>$ brackets rather than as $\left\langle a | b \right\rangle$ which makes a bit more sense purely from a LaTeX viewpoint. But this dates back to the hand-typesetting days, where the peculiar-looking spacing would be modified before the paper hit the journal. – Mark Fischler May 16 '18 at 21:12

First let's justify the relation, which using the model of functions to represent the states, says that $$\frac{\int \left(\phi^\dagger(x) + \epsilon \alpha^\dagger(x)\right)A(\psi(x)) \,dx}{\int \left(\phi^\dagger(x) + \epsilon \alpha^\dagger(x)\right)(\psi(x)) \,dx} - \frac{\int \phi^\dagger(x) A(\psi(x)) \,dx}{\int \phi^\dagger(x) (\psi(x)) \,dx} = \epsilon \frac{\int \alpha^\dagger(x)A(\psi(x)) \,dx}{\int \phi^\dagger(x) (\psi(x)) \,dx} - \epsilon \frac{\int \phi^\dagger(x) A(\psi(x)) \,dx} {\left(\int \phi^\dagger(x) (\psi(x)) \,dx\right)^2} \int \alpha^\dagger(x) A(\psi(x))\, dx + O(\epsilon^2)$$ Using perturbation methods, expand (for small $\epsilon$) the first term on the left using the usual trick of multiplying the denominator by $$\frac {\int \left(\phi^\dagger(x) - \epsilon \alpha^\dagger(x)\right)(\psi(x)) \,dx}{\int \left(\phi^\dagger(x) - \epsilon \alpha^\dagger(x)\right)(\psi(x)) \,dx}$$ This rearranges into the $\epsilon=0$ piece being subtracted off, plus the expression on the right of your relation. However, to get this rearrangement, two properties of the functions $\phi, \alpha, \psi$ are required:
• Unless I have done something wrong, I need to assume $\alpha$ commutes with $\psi$ (or, it turns out, with $\phi$) in order to manipulate to get that nice expression.