# Commutation relation in Euclidean time

Does the usual commutation relationship $$[q,p]=i\hbar\tag{56}$$ change to $$[q,p]=\hbar\tag{55}$$ when making a Wick rotation to Euclidean time? and if so, what is the physical reason to understanding why it must change?

I have seen this article (https://physicstravelguide.com/_media/quantum_theory/path-integral.pdf) that seems to suggest this is the case. The logic in going from Eq. 55 to Eq. 56 is not clear from what is written though. My best guess is that the author must have done something similar to the following:

Starting with Eq. 55 in the article $$[q,p]=\delta(\tau-\tau'),$$ using the definition of imaginary time ($$\tau=it/\hbar$$) and the scaling property of the delta function one gets $$[q,p]= \delta\left(\frac{i}{\hbar}(t-t')\right)\stackrel{?}{=}i\hbar\delta(t-t'),$$ where I have inserted the delta functions in myself to try and rationalize the result of the article. I'm 99% sure these manipulations are not valid, they are just my guess.

Furthermore, if this result is true it seems very strange for several reasons. The main one being that in Euclidean time the commutator resembles the classical Poisson bracket. It is very counter intuitive (at least to me) that a simple change of variables could just make a system classical. I would appreciate any light that could be shed on this topic.

1. For what it's worth, the Wick rotation of the momentum operator $$\hat{p}_M=i\hat{p}_E$$ in the quantum CCRs (55) and (56) can be understood via a Wick rotation of the classical quantities in the standard manner: $$\begin{array}{lcrcl} \text{Action}:&\qquad&-S_E&=& iS_M, \cr \text{Time}:&\qquad& t_E&=& it_M, \cr \text{Lagrangian}:&\qquad& L_E&=& -L_M, \cr \text{Velocity}:&\qquad& v_M&=& iv_E, \cr \text{Momentum}:&\qquad& p_M&=& ip_E. \end{array}$$ See also e.g. this related Phys.SE post.

2. In contrast, it is mathematically delicate/intricate to try to Wick rotate the Dirac delta distribution $$\delta(t\!-\!t^{\prime})$$ directly.

References:

1. Yen Chin Ong, Note: Where is the Commutation Relation Hiding in the Path Integral Formulation? 2018 (PDF)

There is a formula valid for the composition of the delta function with another function $$f(t)$$, that reads

$$\delta(f(t))=\sum_{i} \frac{\delta\left(t-a_{i}\right)}{\left|f^{\prime}\left(a_{i}\right)\right|},$$

where $$a_i$$ are all the zeros of your function $$f(t)$$.

In your case, you have $$f(t) = i (t-t')/\hbar$$, so the only zero is $$t = t'$$ and we have $$f'(t) = i/\hbar$$. In particular, $$f'(0) = i/\hbar$$. Plugging these you can justify your computation there:

$$[q,p]= \delta\left(\frac{i}{\hbar}(t-t')\right){=}i\hbar\delta(t-t'),$$

so there is a formal justification for it in the end.

• But $|i|=1$, right?
– WAH
Aug 17, 2021 at 6:45
• I am not sure if this formula is valid when the domain of integration is complex, at least without some care needed to correctly interpret the symbols. Aug 17, 2021 at 7:58

If one constructs a path integral for the euclidean time evolution operator $$\exp\{-\tau H\}$$, for $$H= p^2+V(x)$$ by inserting sets of $$|x\rangle \langle x|$$ and $$|p\rangle \langle p|$$'s, the $$\hat x$$ and $$\hat p$$ operators are the usual ones: $$\hat p\to -i\hbar \partial_x$$, $$\hat x\to x$$, and so the commutator is unchanged.