# Wick rotation of the propagator in quantum mechanics

I am told that making the substitution $$t\to-i\tau$$, or a 'Wick rotation', can be used to study the propagator in imaginary time, making some problems easier. For example, this source proposes that we take the usual propagator and perform such a substitution: $$\bar{U}(x_1,0;x_2,\tau)=\int_{x_1}^{x_2}D[x]\exp\left(\frac{i}{\hbar}\int_0^t dt'\left(\frac{1}{2}m\left(\frac{dx}{dt'}\right)^2-V(x)\right)\right)|_{t\to -i\tau;dt\to-id\tau}$$ which apparently leads to: $$=\int_{x_1}^{x_2}D[x]\exp\left(\frac{1}{\hbar}\int_0^\tau d\tau'\left(-\frac{1}{2}m\left(\frac{dx}{d\tau'}\right)^2-V(x)\right)\right)$$ I do not understand how this substitution works - perhaps I am making a silly mathematical error. Taking the first line and substituting the integrand as $$dt\to -id\tau$$, I introduce a factor of $$-i$$ to the exponent, which multiplies with the existing factor of $$i$$ to yield 1. Using the chain rule, I gain a factor of $$1/(-i)^2=-1$$ on the kinetic energy term, changing its sign. And, since I am letting $$t\to-i\tau$$, I also change the upper limit on the integral for the action, giving me overall: $$=\int_{x_1}^{x_2}D[x]\exp\left(\frac{1}{\hbar}\int_0^{-i\tau} d\tau'\left(-\frac{1}{2}m\left(\frac{dx}{d\tau'}\right)^2-V(x)\right)\right)$$ This is almost the correct result, but I have a factor of $$-i$$ on the upper limit of the action integral, which I believe should be introduced by the substitution $$t\to-i\tau$$ - but the correct result doesn't have that. Is this somehow equivalent, or have I made an error? Why wouldn't the variable change affect the upper limit of the integral?

• Comment to the post (v2): Why did you change the upper limit in the last step? – Qmechanic Sep 27 '18 at 12:28
• The upper limit was t and I'm making the substitution $t\to -i\tau$. – user502382 Sep 27 '18 at 12:35
• Just a note, you should denote the upper limit as $T \mapsto iT$ under the transformation $t \mapsto -i\tau$ and remove the primes for clearer notation. – GodotMisogi Sep 28 '18 at 10:30

Let $$t_i = 0,\, t_f = T$$. The propagator is given by:
$$\bar{U}(x_1,0;x_2,T)=\int_{x_1}^{x_2}D[x]\exp\left[\frac{i}{\hbar}\int_0^T \mathrm{d}t\left(\frac{1}{2}m\left(\frac{\mathrm dx}{\mathrm dt}\right)^2-V(x)\right)\right]$$
The transformation $$t = -i\tau \implies \mathrm{d}t = -i\mathrm{d}\tau,\, \tau_i = 0, \,\tau_f = iT$$.
$$\bar{U}(x_1,0;x_2,\tau_f)=\int_{x_1}^{x_2}D[x]\exp\left[\frac{1}{\hbar}\int_0^{\tau_f=iT} \mathrm{d}\tau\left(-\frac{1}{2}m\left(\frac{\mathrm dx}{\mathrm d\tau}\right)^2-V(x)\right)\right]$$