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these two integrals below are equal, but I am not understanding where the $x'$ variable comes from.

\begin{align} I_0&=e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] -\big[\varphi(x)\big]^4 +J(x)\varphi(x)\right\}}\\ &=e^{ -i\int d^4x'\big[ \frac{\delta}{\delta J(x')} \big]^4 } e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}} \end{align}

The gist of what I have been doing is to write

\begin{align} I_0&=e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}}e^{ -i\int d^4x \big[\varphi(x)\big]^4} \\ &=e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}} \left[ 1+\left( -i\int d^4x \big[ \varphi(x) \big]^4 \right) +...\right]\\ &=\left[ 1+\left( -i\int d^4x \left[\dfrac{d}{dJ} \right]^4\right)+... \right]e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}} \end{align}

The point is to pull the $\varphi^4$ term out of the integral by writing $\varphi$ as $d/dJ$ except in this case I have to use the variation $\delta$ instead and I don't see why. I used the $d/dJ$ trick in simpler examples without an integral in the exponent, and I am not seeing the connection to the variational notation $\delta$ which appears in the second of the first two equations above. Obviously, I can recombine the prefactor sum in my final equation into $\text{exp}$, but I do not see the point of the $x'$ variable. I hope it is clear that if I had $\delta/\delta J(x')$ in my last equation instead of $d/dJ$ then I would get the form of the second equation which is the correct form. Why can't I just write it as $d/dJ$ like I did before? Please give me a tip, thanks.

Qmechanic
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Jonathan Tooker
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1 Answers1

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The issue is really what it means to to compute one functional derivative $\delta/\delta f$. Once we get that part, we can raise it to the nth power and get the result.

How to compute one variation?

\begin{align} \dfrac{\delta}{\delta J(x')} K&=\dfrac{\delta}{\delta J(x')} e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] +J(x)\varphi(x)\right\}} \\ &=\dfrac{\delta}{\delta J(x')} e^{ i\int d^4x J(x)\varphi(x)}e^{ i\int d^4x \left\{ \frac{1}{2}\left[ \left( \partial\varphi(x) \right)^2-\varphi(x)^2 \right] \right\}} \\ &=A\dfrac{\delta}{\delta J(x')} e^{ i\int d^4x J(x)\varphi(x)} \\ \end{align}

Let $F=e^{ i\int d^4x J(x)\varphi(x)} $. The next step is what follows. Instead of using $\Delta$, you need to use $\varepsilon$ times a Dirac delta. This is part of the definition of the functional derivative I guess, If anyone want to say a little more about that would be nice, but I think it's just part of the definition. However, wikipedia calls the Dirac delta a "test function" so maybe it is an ansatz of some type. Anyway, the answer follows:

\begin{align} \dfrac{\delta}{\delta J(x')}F&=\lim\limits_{\varepsilon\to0}\dfrac{F[J(x)+\varepsilon\delta(x-x')]-F[J(x)]}{\varepsilon}\\ &=\lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\int d^4x [J(x)+\varepsilon\delta(x-x')]\varphi(x)}-e^{ i\int d^4x J(x)\varphi(x)} }{ \varepsilon }\\ &=\lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\int d^4x [J(x)+\varepsilon\delta(x-x')]\varphi(x)}-e^{ i\int d^4x J(x)\varphi(x)} }{ \varepsilon }\\ &=F\,\lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\int d^4x \varepsilon\delta(x-x')\varphi(x)}-1 }{ \varepsilon }\\ &=F\,\lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\varepsilon\varphi(x')}-1 }{ \varepsilon } \end{align}

use l'Hopital's rule

\begin{align} \lim\limits_{\varepsilon\to0}\dfrac{ e^{ i\varepsilon\varphi(x')}-1 }{ \varepsilon }\stackrel{*}{=}\lim\limits_{\varepsilon\to0}\dfrac{i\varphi(x') e^{ i\varepsilon\varphi(x')}}{ 1 }=i\varphi(x') \end{align}

therefore

\begin{align} \left[\dfrac{\delta}{\delta J}\right]^4F=\left[\varphi(y)\right]^4 F \end{align}

then

\begin{align} AF+\left(-i\int d^4x \left[\dfrac{d}{dJ}\right]^4 \right)AF+...&=AF+\left(-i\int d^4x' \left[\varphi(x')\right]^4\right)AF+...\\ &=e^{ -i\int d^4x' \big[\varphi(x')\big]^4}AF \\ &=e^{ -i\int d^4x \big[\varphi(x)\big]^4} AF \end{align}

Plugging in F and A, we get the expected result.

Jonathan Tooker
  • 81