Why Does the Path Integral in Quantum Mechanics Include an Imaginary Unit 'i'?

[unchained1978]

I've been studying the path integral approach to QM on my own, and trying to draw some analogies between the partition function of QM \begin{equation}Z_{QM}=\int D\varphi e^{\frac{i}{\hbar}S[\phi]}\end{equation} and that of statistical mechanics \begin{equation}Z_{SM}=\displaystyle\sum\limits_{i=0}^N g_{i}e^{-\beta E_{i}}\end{equation}. The thing is I don't understand why there is an $i$ in $Z_{QM}$. I've gone through a derivation and it comes from the Unitary operator $\hat{U}=e^{-i\hat{H}t}$, but I don't see why this is necessary. On wikipedia, the explanation is that the $i$ comes from the jacobian of the complex projective space or something like that. I'm not quite satisfied with that definition. The reason I'm investigating this is because in one of Hawking's papers he calculates the entropy of various spacetimes, and one thing I noticed is that the entropy $S=k_{B}lnZ+\beta<E>$ is only defined when $lnZ$ is real, which requires that $iS[g]$ is also real, and therefore the action must be complex. But I don't quite understand this argument from an intuitive point. Could anyone give me a good description (or link to one) of why this $i$ appears at all?

 

[azntoon]

Thanks in advance. A:The i appears in the path integral because the quantum state of a system can be written as an element of $L^2(\mathbb{R},d\psi)$ where $\psi$ is a complex valued function, and the inner product operation is defined to be $$\langle \psi_1 |\psi_2\rangle=\int_{-\infty}^{\infty}\psi_1^*\psi_2 d\psi$$ So the inner product of two quantum states is a measure of their overlap. The path integral formulation of QM is essentially a way of expressing the state of a system as a sum over all possible paths. The i appears in the path integral because the paths have to be weighted according to their overlap with each other. Since the overlap is a complex number, the weighting must also be a complex number. Thus the path integral must have an i in it in order to properly weight the paths.