Let $\varphi(x)$ and $j(x)$ be two field configurations. Let $\Gamma[\varphi]$ be a functional of the field $\varphi$ defined by:

$$ \Gamma[\varphi] := \inf_j \ F[\varphi, j] = F[\varphi, j_\varphi] \tag{1}$$

where $F[\varphi, j]$ is a functional of both $\varphi$ and $j$, and for any fixed $\varphi$, the unique field configuration that extremizes $F$ is $j_\varphi$, which is to say,

$$ \frac{\delta}{\delta j} F[\varphi,j] \Big\rvert_{j=j_\varphi} = 0 \,. \tag{2}$$

Now suppose that the structure of $F$ allows the following identity to hold:

$$ \frac{\delta}{\delta\varphi} \Gamma[\varphi] + j_\varphi = 0 \,.\tag{3}$$

Please keep in mind that the above equality is an identity, but it can be easily promoted to the status of an equation as follows:

$$ \frac{\delta}{\delta\varphi} \Gamma[\varphi] + j = 0 \,,\tag{4}$$

whose solution is, of course, $$j=j_\varphi\tag{5}$$ for any given $\varphi$. Alternatively, however, we can fix $j$ and ask for the $$\varphi =: \varphi_j\tag{6}$$ that solves it.

We then have an equivalent identity:

$$ \frac{\delta}{\delta\varphi} \Gamma[\varphi]\Big|_{\varphi=\varphi_j} + j = 0 \,.\tag{7} $$

In such a scenario, would it be true to say that

$$ \Gamma[\varphi_j] = F[\varphi_j, j]\tag{8} $$

which allows us to write a functional of $j$ instead of $\varphi$? If yes, can you please provide a justification for the above claim?

Note that it is equivalent to claiming that $$ j_{\varphi_j} = j \,.\tag{9}$$

What does that even mean?

  • $\begingroup$ @Qmechanic Why did you change the $sup$ to $inf$? $\endgroup$ Jun 20, 2017 at 6:26
  • $\begingroup$ Yes, because the standard definition of the effective action has the opposite sign as compared to the standard Legendre transformation. $\endgroup$
    – Qmechanic
    Jun 20, 2017 at 6:36
  • $\begingroup$ That can be taken care of by the $F$ which I haven't explicitly defined here. No? $\endgroup$ Jun 20, 2017 at 6:38
  • $\begingroup$ Only if you use non-standard conventions. Eqs. (2) & (3) also partly fix conventions. $\endgroup$
    – Qmechanic
    Jun 20, 2017 at 6:40
  • $\begingroup$ I nowhere mentioned it is the effective action. I defined it as a functional of the field $\varphi.$ Of course, I had the effective action in the back of my mind. But the problem at hand doesn't require such preconceptions. Moreover, the convention is not non-standard. Check, for instance, Srednicki. $\endgroup$ Jun 20, 2017 at 6:45

1 Answer 1


We believe that OP's eqs. (1)-(3) essentially fixes the form of OP's function $F$ to be the underlying function $$F[\phi_{\rm cl},J]~=~W_c[J]-J_i \phi^i_{\rm cl} \tag{*}$$ of the QFT Legendre transformation $J_i\leftrightarrow \phi^i_{\rm cl}$ between the generating functional $W_c[J]$ for connected diagrams and the effective action $$\Gamma[\phi_{\rm cl}]~=~\inf_J F[\phi_{\rm cl},J] .$$ In this answer we will assume$^1$ that eq. (*) holds. Here $J_i$ are sources and $\phi^i_{\rm cl}$ are classical fields (hence the subscript "${\rm cl}$"). To make it look more familiar to the common physicists, we can recast the Legendre transformation in the language of classical mechanics via the dictionary $$ \begin{array}{cccc} v^i& L(v) & p_i & H(p) & h(v,p) \cr\cr\updownarrow &\updownarrow &\updownarrow &\updownarrow &\updownarrow \cr\cr J_i & W_c[J] & \phi^i_{\rm cl} & -\Gamma[\phi_{\rm cl}] & -F[J,\phi_{\rm cl}] \end{array} $$

Using the notation & definitions of my Phys.SE answers here & here, $$h(v,p)~:=~p_j v^j -L(v), \qquad g_j(v)~:=~\frac{\partial L(v)}{\partial v^j}, \qquad f~:=~g^{-1} ,$$ then OP's formally correct equations (1)-(9) transcribe as $$ H(p) ~:= \sup_v h(v,p)~=~h(f(p),p), \tag{1} $$ $$\left. \frac{\partial h(v,p)}{\partial v^i}\right|_{v^i=f^i(p)}~=~0,\tag{2} $$ $$f^i(p)~=~\frac{\partial H(p)}{\partial p_i},\tag{3} $$ $$v^i~\approx~\frac{\partial H(p)}{\partial p_i},\tag{4} $$ $$ v^i~\approx~f^i(p),\tag{5} $$ $$ p_i~\approx~g_i(v),\tag{6} $$ $$v^i~=~\frac{\partial H(p)}{\partial p_i}\circ g(v),\tag{7} $$ $$H\circ g(v)~=~h(v,g(v)),\tag{8} $$ $$ f^i\circ g(v)~=~v^i, \tag{9}$$ respectively. All eqs. (2)-(9) can formally be derived. For instance, eq. (3) is derived in Section V of my Phys.SE answer here.


$^1$ OP apparently wants to discuss a generalization of the Legendre transform. We have currently nothing to add to that interesting discussion.

  • $\begingroup$ With your translation, it has become so obscure that I cannot follow it. Please remember that for my own understanding, I need to translate your whole proof back to my original notation and see if you have really answered my question. And this I'm finding very difficult to do. If you prefer to work in a different notation, please do so but I'd appreciate very much if you can recast your whole answer in my notation, perhaps as an appendix. Hope you understand my difficulty. Thank you. $\endgroup$ Jun 19, 2017 at 13:34
  • $\begingroup$ I added definition of notation in an update. $\endgroup$
    – Qmechanic
    Jun 19, 2017 at 13:44
  • $\begingroup$ I am sorry it doesn't help. Your notational translation, if you would excuse my language, simply robs my question of its essential generality and simplicity. First, you have chosen an explicit expression for $F[\varphi, j]$ which in your case reads $h(p,v).$ Moreover, I have no such thing as $\varphi_{cl}$ in my question. Perhaps, by that, you simply mean $\varphi.$ Anyway, $j_\varphi(x)$ originally is meant to be a functional of the field $\varphi(x)$ (hence the notation). In your translation, however, it is not clear if $f(p)$ has any dependence on $v$. $\endgroup$ Jun 20, 2017 at 7:16
  • $\begingroup$ Your eq.(3) differs from mine by a minus sign (modulo translation). Eqns. (5) and (6) make no sense to me (as a logical consequence of what you have written before), unless you are merely translating the equations in my question to your notation. In eq.(7), by the use of \circ, do you perhaps mean evaluation of what comes before the \circ at the point specified after it? If yes, then this eqn. differs from mine by a minus sign. Okay, you have translated all my equations to your notation. Now what? I do not follow the proof. Did you mean to only translate or did you also provide a proof? $\endgroup$ Jun 20, 2017 at 7:32
  • 1
    $\begingroup$ Please remember that you are answering me (and others like me), not yourself. I may not be familiar with what appears most familiar to you. I hope you understand. Thank you. $\endgroup$ Jun 20, 2017 at 7:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.