4
$\begingroup$

Wick Rotation helps to solve the problem of the convergence of the path integral, by changing the integral contour in the complex plane. But my question is:

  • In the scalar field path integral, the field function is real-valued. If we change the parameter from $t$ to $-i\tau$ as $\phi(t,x)$ to $\phi(-i\tau,x)$, is it possible to change the original real-valued function to complex-valued function? But the convergence requires that integrand of path integral: \begin{equation} W_{\mathrm{E}}[J]=\mathcal{N}_{\mathrm{E}} \int \mathcal{D} \phi \exp \left[\frac{\mathrm{i}}{\hbar}(-\mathrm{i}) \int \mathrm{d}^{4} x_{\mathrm{E}}\left(\mathcal{L}\left(\phi, \mathrm{i} \frac{\partial \phi}{\partial x_{4}}\right)+J \phi\right)\right] \end{equation} as totally real valued.
  • Where the Lagrangian turns out to be: \begin{equation} \begin{aligned} \mathcal{L} &=\frac{\hbar^{2}}{2}\left(\partial_{0} \phi \partial_{0} \phi-\nabla \phi \cdot \nabla \phi\right)-\frac{1}{2} m^{2} \phi^{2}-V(\phi) \\ &=-\left[\frac{\hbar^{2}}{2}\left(\partial_{4} \phi \partial_{4} \phi+\nabla \phi \cdot \nabla \phi\right)+\frac{1}{2} m^{2} \phi^{2}+V(\phi)\right] \\ &=-\left[\frac{\hbar^{2}}{2} \partial_{\mathrm{E} \mu} \phi \partial_{\mathrm{E} \mu} \phi+\frac{1}{2} m^{2} \phi^{2}+V(\phi)\right] \end{aligned} \end{equation}
  • Greiner's Textbook "Field Quantization" states: \begin{equation} W_{\mathrm{E}}[J]=\mathcal{N}_{\mathrm{E}} \int \mathcal{D} \phi \exp \left[-\frac{1}{\hbar} \int \mathrm{d}^{4} x_{\mathrm{E}}\left(\frac{\hbar^{2}}{2} \partial_{\mathrm{E} \mu} \phi \partial_{\mathrm{E} \mu} \phi+\frac{1}{2} m^{2} \phi^{2}+V(\phi)-J \phi\right)\right] \end{equation} The integrand is real-valued,which require the funtion $\phi$ must be real-valued function.

I read many posts about the trick, but since the contour-integral give the equivalence of the real-axis integral and pure-imaginary axis integral, how it could be possible to change the convergence of integral without requiring the scalar-field function remains to be real-valued, or just analytic continuation?

Some people suggest there exist some deep relation between Euclidean Field Theory and Minkowskian Field Theory involving the Axiomatic description of QFT. I suspect that Wick Rotation is just a mathematical-equivalence trick, I don't know how wick rotation could be rigorously built or just an mapping relation?

$\endgroup$
0

2 Answers 2

4
$\begingroup$

Wick rotation doesn't act on the arguments of the fields. It only acts on the coefficients in the action. In particular, it acts on the $dt$ in the integral over spacetime and on the $dt$ in the denominator of the derivative $\dot\phi$. These are "coefficients."

This is more clear in lattice QFT, where both space and time are discretized. The integral over time becomes $dt\sum_n$ where $dt$ is the step-size in the time direction, and the index $n$ specifies which time-slice we're in. The action is $$ S[\phi]\sim dt\sum_n \left(\frac{\big(\phi(n+1)-\phi(n)\big)^2}{2\,dt^2}+V\big(\phi(n)\big)\right). $$ Wick rotation affects the factors of $dt$ in the coefficients, specifically the factor $dt^{-1}$ that multiplies the discretized $\dot\phi^2$ term and the factor $dt^{+1}$ that multiples the remaining terms.

We could write $\phi(t+dt)$ and $\phi(t)$ instead of $\phi(n+1)$ and $\phi(n)$, but that doesn't change the idea: the $dt$ in the argument of $\phi$ is still just an index, so it's not affected by Wick rotation. The real-valued field remains real-valued.

$\endgroup$
3
  • 1
    $\begingroup$ Thanks, but if we only change the coefficients, such as $dt$. How can we make sure the wick-rotated theory is equivalent to the original theory? Or in another words, some physical observeables are independent of whether Euclidean or Minkowskian? $\endgroup$
    – Hawk Kou
    May 4, 2020 at 14:43
  • 1
    $\begingroup$ @HawkKou The Euclidean and Minkowskian versions can be obtained from each other, but I wouldn't say they're equivalent. Wick rotation modifies how observables at different times are related to each other. In particular, the Euclidean version isn't even unitary, so it's really not even a "quantum theory" in the usual sense of the term, even though it's a useful trick. At any given time, both have the same observables, but the relationship between observables at different times is different. That relationship is determined by the action, and Wick rotation changes the action. $\endgroup$ May 4, 2020 at 18:35
  • $\begingroup$ Thanks a lot! I understand a bit. $\endgroup$
    – Hawk Kou
    May 5, 2020 at 4:04
4
$\begingroup$

Chiral Anomaly's answer is exactly right: The real scalar field $\phi_M(t_M,\vec{x})=\phi_E(t_E,\vec{x})$ remains invariant and hence manifestly real under a Wick rotation $t_E=it_M$ from Minkowski ($M$) to Euclidean ($E$) time.

In particular, the field $\phi$ is not analytically continued to a holomorphic function in the complex time plane (minus possible branch cuts and singularities). If we actually tried to analytically continue $\phi$, we could not guarantee that $\phi$ would become real on the imaginary time axis for all virtual $\phi$ configurations.

Moreover, analytical continuation of $\phi$ makes no sense whatsoever for discretized spacetimes, as pointed out in Chiral Anomaly's answer. This is an important point because physicists usually assume that continuum physics can be approximated by discrete physics.

The take-home message is that analytical continuation only takes place in spacetime (or in the Fourier-transformed spacetime, i.e. the momentum space); not in the target space of scalar fields.

For Wick rotation of non-scalar fields, see e.g. this Phys.SE post.

$\endgroup$
0

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.