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I'm studying quantum field theory on my own following mainly Peskin and Schroeder. And when going through the stuff concerning spontaneous symmetry breaking and effective potential I found out that I do not fully understand the concept of path integration, especially in the context of SSB. Then I started looking through other literature devoted to the subject and discovered a deeper lack of understanding encountering new questions that I could not answer or find the answer to in the literature. They are all related so I will lay them out in this single question in the order in which I stumbled upon them.

  1. I will use the Euclidean version of path integral. It is quite easy to show that the transition amplitude from an eigenstate of the field $\Phi(x)$ with the eigenvalue $\Phi_{b}(\vec{x})$ at time $-T$ to an eigenstate with the eigenvalue $\Phi_{a}(\vec{x})$ at time $T$ equals

$\langle\Phi_{a};T|\Phi_{b};-T\rangle=\int\limits^{ \Phi(T,\vec{x})=\Phi_{a}(\vec{x})}_{ \Phi(-T,\vec{x})=\Phi_{b}(\vec{x})} D\Phi e^{-S_{E}}$.

Then one inserts on the left two resolutions of identity --- one at time $T$ and the other at time $-T$ and because $T\to\infty$ only the ground state survives and one obtains
$\int\limits^{ \Phi(T,\vec{x})=\Phi_{a}(\vec{x})}_{ \Phi(-T,\vec{x})=\Phi_{b}(\vec{x})} D\Phi e^{-S_{E}}=\langle\Phi_{a};T|\Omega;T\rangle\langle\Omega;T|\Omega;-T\rangle\langle \Omega; -T|\Phi_{b};-T\rangle$.

So one can see that the boundary conditions in the functional integral factorise and can be put into the normalisation constant whereas when performing integration in the path integral we can choose whatever boundary conditions we want. Then I thought that when it comes to spontaneous symmetry breaking this derivation is modified so that the resolution of identity should include a sum (or integral) of all the degenerate vacua of the theory, that is something like this $\int\limits^{ \Phi(T,\vec{x})=\Phi_{a}(\vec{x})}_{ \Phi(-T,\vec{x})=\Phi_{b}(\vec{x})} D\Phi e^{-S_{E}}=\int d\theta d\theta'\langle\Phi_{a};T|\Omega,\theta;T\rangle\langle\Omega,\theta;T|\Omega,\theta';-T\rangle\langle \Omega,\theta'; -T|\Phi_{b};-T\rangle$,

where $\theta$ labels the vacua of the theory with SSB. If this formula is correct then the boundary conditions clearly cannot be put into the renormalisation constant and must be treated with greater care. But the point is that I could not find anything of this kind in any treatment of SSB. If anybody could indicate the flaw in my argument or probably direct me to the literature in which this point is carefully treated I would highly appreciate it (seriously, I'm driven to despair).

  1. After that I began to think that I could take the boundary conditions into account by carefully calculating the vacuum wave functional at both infinities following the derivation by Weinberg (The Quantum Theory of Fields. Volume 1., Section 9.2, p 387, Eq. 9.2.9 ). I made a few corrections to Weinberg's calculation due to the field strength renormalisation and non-zero vacuum expectation value of the field and obtained the answer

$\langle\Phi_{a};T|\Omega;T\rangle=Nexp\left(-\frac{1}{2}\int d^3xd^3yE(\vec{x}-\vec{y})\frac{1}{Z}(\Phi_{a}(\vec{x})-\Phi_{0})(\Phi_{a}(\vec{y})-\Phi_{0})\right)$,

where $\Phi_0=\langle\Omega|\Phi(x)|\Omega\rangle$, $Z$ is the field strength renormalization constant, and $E(\vec{x}-\vec{y})$ is the same as Weinberg's. But then I thought that this could not be right because this expression clearly does not depend on $T$ and I can take $T=0$, in which case I have found the exact vacuum wave functional --- the time-independent eigenvector of the full Hamiltonian corresponding to the lowest eigenvalue. And it took me so little effort! This also means that there's something wrong with Weinberg's derivation as well. Clearly if you have found the wave function of a normalisable stationary state at temporal infinity you have found it for any other point in time due to its trivial dependence on time. If anybody could indicate the flaw in this reasoning it would be highly appreciated too.

  1. While looking at the literature I have found a few times the author noting that in case of SSB boundary conditions for the field should be treated carefully. It was only one or two times that this fact was mentioned. In particular Sydney Coleman says that it is useful to put your system in a box first because for example in scalar field theory with SSB the expectation value of the scalar field in the center of the box depends on the boundary conditions on the walls, no matter how large the walls; this is one of the easiest ways to see that the theory has many vacua. (Aspects of Symmetry, Chapter 7 The uses of instantons, p. 292). But I have found no traces of calculations of this kind in the literature. I would die to see this computation.
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  • $\begingroup$ No, I do hope you survive to read another day :) One suggestion, gather up your questions at the end, as however valid your points are, they are lost in the text. Best of luck with it. $\endgroup$ – user154420 Jul 5 '17 at 16:45
  • $\begingroup$ Concerning the boundary conditions - if you prescribe the value of the fields on the boundaries then the "preferred direction" of the vacuum vector is chosen. $\endgroup$ – Blazej Nov 17 '17 at 22:42
  • $\begingroup$ Concerning the Weinberg's argument - I think he didn't mean at all that the gaussian wave functional he proposes is the vacuum vector of the interacting theory. Instead he just wants to pick some wave functional at $+ \infty$ and $- \infty$ in such a way that it at least resembles the full vacuum vector, so that their overlap is nonzero (which is probably hopeless in the full theory, but if you are fine with thinking about your theory as defined in a box and with finite UV cutoff should be fine). Then evolution in time is hoped to project this state onto its component parallel to full vacuum. $\endgroup$ – Blazej Nov 17 '17 at 22:49
  • $\begingroup$ If you work in Euclidean signature then this process of projecting onto the ground state is quite direct - you have exponential factors $e^{- E \tau}$ supressing states excited by energy difference $E$. Here $\tau$ is Euclidean time. But even in Minkowski signature there is also a mechanism for projection - many cancellations. All the other states are in the continuum so their contibutions cancel due to oscillatory factors $e^{-iEt}$ for $t \to \infty$. Vacuum vector is the only exact eigenstate of energy so its contribution doesn't cancel. $\endgroup$ – Blazej Nov 17 '17 at 22:53
  • $\begingroup$ To see rigorous justification of this mechanism for ordinary integrals, look up Riemann-Lebesgue lemma. It goes without saying that for functional integrals rigorous proof of this is most likely impossible now, due to our poor understanding of the underlying mathemathics. $\endgroup$ – Blazej Nov 17 '17 at 22:54

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