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In his Quantum Field Theory In a Nutshell, in page 12, (Second Ed), A Zee says that conventionally, the amplitude $\langle0|e^{-iHT}|0\rangle$ is denoted by $Z$. In the next paragraph, he considers the amplitude in eq. (6), and performs a Wick rotation to Euclidean time, $t\rightarrow -it$, and writes, $$ Z=\int Dq(t)e^{-\int_0^Tdt[\frac{1}{2}m\dot{q}^2+V(q)]} $$ which is the Euclidean Path Integral. I am confused because he denotes this quantity by $Z$. Is it the same $Z$ as $\langle0|e^{-iHT}|0\rangle$? If so, then I cannot see how. Can anybody help me?

I understand that $$ Z=\langle0|e^{-iHT}|0\rangle=\int dq_F\int dq_I\langle F|q_F\rangle\langle q_F|e^{-iHT}|q_I\rangle\langle q_I|I\rangle $$ with $|F\rangle=|I\rangle=|0\rangle$. How is this similar to the former integral?

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    $\begingroup$ Why the transition amplitude is the path integral is covered in every basic derivation of the path integral, e.g. here. $\endgroup$ – ACuriousMind Dec 17 '14 at 11:47
  • $\begingroup$ @ACuriousMind - that is clear to me. $\endgroup$ – Sayan Mandal Dec 17 '14 at 13:26
  • $\begingroup$ Then I don't understand what you are asking. $\endgroup$ – ACuriousMind Dec 17 '14 at 13:27
  • $\begingroup$ I am asking whether the two integrals I have written are the same, because the author uses the same symbol $Z$ to denote them. In the first integral, $q_I$ and $q_F$ are fixed, while they are being integrated over in the second. My whole confusion stems precisely from the author using $Z$ for either. $\endgroup$ – Sayan Mandal Dec 17 '14 at 16:09
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    $\begingroup$ I think I understand what you're asking. The concern is that the path integral is usual defined as the time evolution operator $e^{-iHT}$ sandwiched between position eigenstates. However here in $Z$ it is sandwiched between two vacuum states. It looks like these are rather different, with the latter expressible in terms of the first only if we introduce two resolutions of the identity in the form of integrals over initial and final position. It's not clear to me either why this is the same thing. Perhaps try Srednicki, page 47. $\endgroup$ – gj255 Apr 27 '15 at 12:55
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The step of evaluating your second expression for $Z$ and arriving at the path integral form, which is the first expression you mention, is a nontrivial but standard derivation in the path integral formulation of QFT. It is quite lengthy, but can be found in most books that cover the path integral formulation. The general idea it to split the time interval $T$ into $N$ intervals (with $N$ large), and to insert a full set of eigenstates at each intermediate time, such that $$ \langle q_F|e^{-iHT}|q_I\rangle= \langle q_F|\left(e^{-iHT/N}\right)^N|q_I\rangle\\ =\int dq_1\ldots dq_{N-1}\langle q_F|e^{-iHT/N}|q_{N-1}\rangle\langle q_{N-1}|e^{-iHT/N}|q_{N-2}\rangle\langle q_{N-2}|\ldots|q_1\rangle\langle q_{1}|e^{-iHT/N}|q_I\rangle $$ This can be solved in the following way. Since Hamiltonian also contains momentum operators you additionally have to insert a complete set of conjugate momentum eigenstates at each intermediate time. Then, all the operators in the Hamiltonian can be evaluated on these inserted states and you are left with C-number functions only. Since the Hamiltonian is quadratic in the momentum operator the momentum integrals are Gaussian and can be performed explicitly. The $N$ integral over intermediate field configurations enter the definition of the path integral (in essence, the path integral is just defined as such an integral over $n$ intermediate time field configurations in the limit where $N\to\infty$).

For more details, see for example the Advanced QFT lecture notes of Hugh Osborn, which can be found here. The derivation you are looking for is at the very beggining of the script.

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  • $\begingroup$ Thanks for the help. One doubt is, that the first integral is between fixed endpoints in space, $q_I$ and $q_F$, while in the second integral, we are integrating over the end points. Isn't there a contradiction? $\endgroup$ – Sayan Mandal Dec 17 '14 at 13:28
  • $\begingroup$ The second integral leaves qI and qF fixed. $\endgroup$ – Harry Wilson Dec 17 '14 at 14:04
  • $\begingroup$ There are integrals over $q_F$ and $q_I$. $\endgroup$ – Sayan Mandal Dec 17 '14 at 16:07

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