Quantum Field Theory – Measure of Functional Integral in Path Integral Formulation

functional-derivativesfunctional-determinantsnormalizationpath-integralquantum-field-theory

I have a question regarding the prefactor $\sqrt{\left(\frac{m}{2\pi i \hbar \Delta t}\right)}$ in $$\left<x'|e^{-iHt}|x\right> = \int D[x] \exp(\frac{i}{\hbar}\int dt' L(x, \dot{x})),$$ where $$D[x]=\lim_{N\rightarrow\infty}\sqrt{\left(\frac{m}{2\pi i \hbar \Delta t}\right)}^N \Pi dx_n .$$

From http://users.physik.fu-berlin.de/~kleinert/b6/psfiles/Chapter-13-functint.pdf, page 929, I found that for $$\int \mathcal{D}[j]\exp{\frac{1}{2}\int dxdx'j(x)A(x,x')j(x')}$$ to be $1/\sqrt{\det A}$, $\mathcal{D}[j]$ should be $\lim_{N\rightarrow\infty}\Pi_n\frac{1}{\sqrt{2\pi/\epsilon^D}}$.

Applying the above result to the path integral formulation, it seems that $\mathcal{D}[x]$ should be such that $$\mathcal{D}[x]\propto\lim_{N\rightarrow\infty}\Pi_n\frac{1}{\sqrt{2\pi/(\Delta t)}}\propto\lim_{N\rightarrow\infty}\left(\frac{\Delta t}{2\pi}\right)^\frac{N}{2}.$$ To me, the mathematical flow seems okay, so I feel the formula is correct.

Therefore, it appears that $\mathcal{D}[x]$ does not yield the proper Gaussian functional integration value but needs some sort of normalization. Could somebody please let me know what I am thinking wrong?

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Edit)
The problem I am now stuck by is the stationary phase approximation of path integral. The book I am now referring to is the Altland and Simons textbook (Condensed Matter Field Theory), which asserts that $$\left<q_f|e^{-iHt}|q_i\right> \approx \det(\frac{i}{2\pi\hbar}\frac{\partial^2 S[q]}{\partial q_i \partial q_f})^{1/2}e^{\frac{i}{\hbar}S},\tag{3.28}$$ and that this formula is obtainable using $$\int Dx e^{-F[x]}\approx e^{-F[x_{st}]}\det({\frac{A}{2\pi}})^{-1/2}\tag{3.25}$$ under the stationary phase approximation. The $Dx$ is $$Dx=\lim_{N\rightarrow\infty}\sqrt{\left(\frac{m}{2\pi i \hbar \Delta t}\right)}^N \Pi dx_n $$ and $A$ is the second functional derivative $A(t, t')=\frac{\delta^2 F[x]}{\delta x(t) \delta x(t')}$.

I am struggling in deriving (3.25) since I do not get how $\Delta t$ in $\sqrt{\left(\frac{m}{2\pi i \hbar \Delta t}\right)}$ gets canceled when following the same procedure (discretizing first, and then taking the continuum limit) as in http://users.physik.fu-berlin.de/~kleinert/b6/psfiles/Chapter-13-functint.pdf.

Best Answer

  • $\sqrt{\frac{m}{2\pi ih \Delta t}}$ is the famous Feynman fudge factor. It can be derived from the Gaussian momentum integrations in the corresponding Hamiltonian phase space path integral, see e.g. Section V of my Phys.SE answer here.

  • Altland and Simons eq. (3.25) is essentially the method of steepest descent, while eq. (3.28) is derived in my Phys.SE answer here.

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