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?

  • $\begingroup$ Comment to the post (v2): Why did you change the upper limit in the last step? $\endgroup$
    – Qmechanic
    Commented Sep 27, 2018 at 12:28
  • $\begingroup$ The upper limit was t and I'm making the substitution $t\to -i\tau$. $\endgroup$
    – user502382
    Commented Sep 27, 2018 at 12:35
  • $\begingroup$ 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. $\endgroup$ Commented Sep 28, 2018 at 10:30

1 Answer 1


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 $t = -i\tau \implies \mathrm{d}t = -i\mathrm{d}\tau,\, \tau_i = 0, \,\tau_f = iT$.


$$\bar{U}(x_1,0;x_2,\tau_f)=\int_{x_1}^{x_2}D[x]\exp\left[\frac{1}{\hbar}\int_0^{\tau_f=iT} \mathrm{d}\tau\left(-\frac{1}{2}m\left(\frac{\mathrm dx}{\mathrm d\tau}\right)^2-V(x)\right)\right]$$

which is exactly what the Wick rotation gives you, an integration over the imaginary line.

  • $\begingroup$ Why $x(t)\rightarrow x(-i\tau)$? It is claimed that it is actually $x(\tau)$ in here, p.55. $\endgroup$ Commented Jan 21, 2020 at 17:57
  • $\begingroup$ That appears to be a typo, although I've seen such notation, i.e. $x(t) \mapsto x(\tau)$, in other QFT textbooks IIRC. I can't see anything mathematically incorrect in the transformation I've written. $\endgroup$ Commented Feb 9, 2020 at 2:38

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