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Context

On the top of the 4th page of Schulman's paper, A path integral of spin, he argued that:

Let the propagator $K$ from $(\phi_1,t_1)$ to $(\phi_2,t_2)$ is the sum of $e^{iS[\phi]/\hbar}$. It is impossible that $$K \sim e^{iS[\phi]/\hbar} - e^{iS[\psi]/\hbar} \tag{p.1561}$$ if the two path, $\phi$ and $\psi$, are homotopic, i.e. they can be deformed to each other. The reason is that "If we deform $\psi(t)$ continuously into $\phi(t)$, the contribution due to $\psi\left( e^{iS[\psi]/\hbar}\right)$ must continuously go over into that due to $\phi$. "

Question

I can't understand this argument. I agree that the contribution due to $\psi\left( e^{iS[\psi]/\hbar}\right)$ must continuously go over into that due to $\phi$. However, there could be a continuous function $f$ with $f(0)=0$ and $f(1)=\pi$ s.t. $$K \sim e^{iS[\phi]/\hbar} + \cdots + e^{if(\alpha)}e^{iS[\phi_{\alpha}]/\hbar} + \cdots + e^{if(1)}e^{iS[\psi]/\hbar} + \cdots $$ where $\phi_{\alpha = 0} = \phi$ and $\phi_{\alpha = 1} = \psi$. Thus, I think it is totally fine there is a relative phase between the contribution of $\psi$ and $\phi$ although these two paths are homotopic. Why is my argument incorrect?

Qmechanic
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Steven Chang
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1 Answers1

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One argument is that if we go through the usual (admittedly formal textbook) derivation of the path integral from the operator formalism, then we can use the same (possibly phase space, possibly coherent) state basis convention$^1$ for two sufficiently close homotopic paths, i.e. the phase factor in front of the Boltzmann factor $e^{\frac{i}{\hbar}S}$ must be the same for two sufficiently close homotopic paths.

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$^1$ There are typically phase factor ambiguities in the choice of basis vectors, cf. e.g. my Phys.SE answer here.

Qmechanic
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