3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Comment to the post (v2): Why did you change the upper limit in the last step? $\endgroup$ – Qmechanic Sep 27 '18 at 12:28
  • $\begingroup$ The upper limit was t and I'm making the substitution $t\to -i\tau$. $\endgroup$ – user502382 Sep 27 '18 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. $\endgroup$ – GodotMisogi Sep 28 '18 at 10:30
1
$\begingroup$

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

Therefore,

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.