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

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    $\begingroup$ I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than $<, >$ does :) $\endgroup$
    – Patrick Stevens
    May 16, 2018 at 21:00
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    $\begingroup$ I'm the \left\langle \right\rangle fairy, if you want to take it up a notch. $\endgroup$
    – J.G.
    May 16, 2018 at 21:09
  • $\begingroup$ I' m so sorry guys $\endgroup$
    – Stefano Barone
    May 16, 2018 at 21:11
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    $\begingroup$ 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. $\endgroup$
    – Mark Fischler
    May 16, 2018 at 21:12

1 Answer 1

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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:

  • All three must fall off sufficiently rapidly that the needed manipulations of the integrals are not rendered meaningless.

  • 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.

So when you have an operator which commutes with the test states, you relation gives an expression for the derivative of the matrix element for that operator. But I think you have not seen this relation much because it is wrong for operators in general, and it would be "trappy" to blithely point it out with the caveate that you must be careful to use it only in the commutative case.

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