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2

Does this mean, that the probability of detecting the particle it the SAME everywhere? No, it does not. This is quite a common mistake, stemming from the idea that the Green function $\mathcal{M}$ can be used in the role of the $\psi$ function of free particle with the Born interpretation of $|\psi|^2$ as probability density. But that is not possible, ...

2

[By statistical mechanics I mean classical statistical mechanics throughout this answer. If you are curious to think about the complications added with making the statistical side of the story quantum mechanical, that sounds like a very good exercise.] The analogy between Euclidean quantum field theories and equilibrium statistical mechanics is exact, once ...

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In my naive view, this is merely a mathematical trick that should not be taken too seriously in term of physical interpretation. After all, a "Wick rotation" applied to the Schrodinger equation yields a diffusion equation. This is helpful for some mathematical problems but the physics it describes is very very different from quantum mechanics; not even ...

0

Temperature in the classical model is mapped to imaginary time in the quantum model. By analytic continuation, one can obtain the real-time evolution. The matrix elements of the time-evolution operator of the quantum model at zero temperature will get mapped to the matrix elements of the transfer matrix of the classical model at an appropriate temperature ...

6

I think it will depend the kind of statistical mechanics. For classical statistical mechanics, there is no time, so it is really hard to imagine a nice physical picture of the propagation of something. But nevertheless we still talk of loops as propagating "particles" (we give the "momenta", for instance, which is conserved, etc.). Interestingly, ...

1

Let's say $f$ admits a taylor series $f(\bar\theta \theta) = A + B \bar\theta \theta + C\bar\theta \theta \bar\theta \theta + \dots$. Now, $\bar\theta \theta \bar\theta \theta = -\bar\theta^2 \theta^2 = 0$, etc., so our function terminates at the linear term. Furthermore, the integral of $f$ over $d \bar\theta\, d\theta$, by the rules of Berezin integration, ...

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