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The vacuum to vacuum transition amplitude for a free particle with source $J$ is given by $$Z_0[J]=\int D\phi \mathrm{exp}\{-i\int [\frac{1}{2}\phi(\square +m^2-i\epsilon)\phi-\phi J]d^4x\}$$ Let $P=\square+m^2-i\epsilon$, then the integrand in the exponent is basically a quadratic form in terms of $\phi$. If we complete the square and use the inner product defined for real fields by $$(f,g)=\int d^4xfg,$$ then we will have (putting $\bar{\phi}=P^{-1}J$) \begin{equation*} \begin{split} Z_0[J]&=\int D\phi\mathrm{exp}\{-[\frac{i}{2}(\phi-\bar{\phi},P(\phi-\bar{\phi}))+\frac{i}{2}(\bar{\phi},J)-i(J,\bar{\phi})]\}\\ &=\frac{1}{\sqrt{\mathrm{det}(iP)}}\int D\phi\mathrm{exp}\{\ \frac{i}{2}(P^{-1}J,J)-i(J,P^{-1}J)\}\\ &=\frac{1}{\sqrt{\mathrm{det}(iP)}}\int D\phi\mathrm{exp}\{\ \frac{i}{2}\int JP^{-1}Jd^4x \}, \end{split} \end{equation*} which is obviously different from the correct result shown in Ryder. The difference is the integration following the determinant term. For your reference, this term should be$$\int D\phi \mathrm{exp}\{-\frac{i}{2}\int J(x)\Delta_F(x-y)J(y)d^4xd^4y\},$$ where $\Delta_F$ is the Feynman propagator. So could anyone help to debug my derivation, much appreciated.

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Your relationship (solution) $\bar{\phi}=P^{-1}J$ is an integral relationship in fact where $P^{-1}$ is a Green's function or the Feynman propagator. You have to write down the arguments properly and you will get the right result.

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  • $\begingroup$ I am aware that $P^{-1}$ is actually Feynman propagator,but I still don't see how the integration comes about. could you be more specific? thans $\endgroup$ – M. Zeng Nov 1 '14 at 10:10
  • $\begingroup$ Vladimir is correct. Try thinking of $\phi$ and $J$ as vectors with a continuous index $x$ and so $P$ and $P^{-1}$ are matrices indexed by $x$ and $y$. To make this work, the functional version of $P$ is actually $P(x,y) = (\square+m^2-i\epsilon)\delta^4(x,y)$ where $\delta$ is the Dirac delta function. $\endgroup$ – Simon Nov 1 '14 at 10:39
  • $\begingroup$ I don't see why $P$ can be put to be this form. could you provide more detail if you don't mind? $\endgroup$ – M. Zeng Nov 1 '14 at 13:48
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Actually what I got is equivalent to the form given in Ryder, we have $$P\Delta_F(x)=-\delta^4(x)$$, then $$\Delta_F(x)=-P^{-1}\delta^4(x)$$, from which we also have $$\Delta_F(x-y)=-P^{-1}\delta^4(x-y)$$, where $P^{-1}$ will not be affected since it's a differentiation operator with respect to $x$. Consequently, we can extract $P^{-1}$ from the above expression by integrate with respect to $y$. The final form then can either be put as $$\int \{\int J(x)\Delta_F(x-y)dy\}J(x)dx$$ or $$\int \int J(x)\Delta_F(x-y)J(y)dxdy$$ where the latter is exactly the same with the one in Ryder.

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