21

The easiest way to see imaginary time used is in elementary quantum mechanics in one dimension. (This is the explanation cribbed from wikipedia). Suppose we're looking at a tunneling-through-a-barrier problem. We start with the Schrodinger equation: $$ -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x) = E\psi(x) $$ Make the ansatz $$ \psi(x) ...


20

as you wrote, the spacetime invariant can be expressed as: $$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$ and from that we normally get: $$ds^2=-c^2dt^2+dx^2+dy^2+dz^2$$ This is not because of some arbitrary imaginary time unit, this is because the metric ($g_{\mu\nu}$) is a diagonal matrix with the coefficients of each term of the $ds^2$ equation: $$g_{\mu\nu}=\left(\...


18

1st comment: It's worth thinking for a second about where Wick rotation comes from. You can do this in the context of the quantum mechanics of a free particle. In QFT, all of the details are more complicated, but the basic idea is the same. In free particle, QM, we get the path integral by inserting sums over intermediate states at various times. The ...


17

The path integral, mathematically speaking, does not exist as an integral: It is not associated with any positive or complex measure. Conversely, the Euclidean path integral does exist. The Wick rotation is a way to "construct" the Feynman integral as a limit case of the well-defined Euclidean one. If, instead, you are interested in an axiomatic approach ...


17

The significance of the metric: $$ d\tau^2 = dt^2 - dx^2 $$ is that $d\tau^2$ is an invarient i.e. every observer in every frame, even accelerated frames, will agree on the value of $d\tau^2$. In contrast $dt$ and $dx$ are coordinate dependant and different observers will disagree about the relative values of $dt$ and $dx$. So while it is certainly true ...


14

but why isn't SR taught with an imaginary time coordinate as standard? From "Gravitation", page 51, via Google books. I'll type up a paraphrase later.


13

I think it will depend the kind of statistical mechanics. For classical statistical mechanics, there is no time, so it is really hard to imagine a nice physical picture of the propagation of something. But nevertheless we still talk of loops as propagating "particles" (we give the "momenta", for instance, which is conserved, etc.). Interestingly, ...


13

1) Both: it is apparently a heat equation in imaginary time and it is a wave equation because its solutions are waves. 2) Nonstationary Schrodinger equation (let us assume free particle) $$ i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^2\nabla^2}{2m}\psi $$ is essentially complex: it can never be satisfied by a real function, only by a complex one. ...


12

If you define $x^0=ict$, then I assume one takes $x_\mu=x^\mu$ so that the metric is actually $\eta_{\mu\nu}=\text{diag}(1,1,1,1)=\delta_{\mu\nu}$, i.e. you're dealing with a Euclidean metric. Then $$ds^2=\delta_{\mu\nu}dx^\mu dx^\nu$$ gives the usual outcome : $$ds^2=-c^2dt^2+d\vec{x}^2$$ The usual conventions are as follows: Option one: One defines $x^\...


12

Proposition. Let there be given two complex numbers $a,b\in \mathbb{C}$ such that ${\rm Re}(a)\geq 0$. In the case ${\rm Re}(a)=0$, we demand furthermore that ${\rm Im}(a)\neq 0$ and ${\rm Re}(b)=0$. The Gaussian integral is well-defined and is given by $$ \underbrace{\int_{\mathbb{R}}\!dx~ e^{-\frac{a}{2}x^2+bx}}_{=:~ I_{\mathbb{R}}(a,b)} ~=~\lim_{\begin{...


11

Well, the singularity does not concern the differentiable structure: Even around the tip of a cone (including the tip) you can define a smooth differentiable structure (obviously this smooth structure cannot be induced by the natural one in $R^3$ when the cone is viewed as embedded in $R^3$). Here the singularity is metrical however! Consider a $2D$ smooth ...


11

By defintion, Minkowski space $\mathbb{R}^{p,q}$ must have signature $(p,q)=(1,d-1)$, with metric, $$ds^2 = -dt^2 +dx_1^2 + dx^2_2 + \dots$$ The signature $(+,+,\dots)$ corresponds to Euclidean space, which is obtained by a Wick rotation, $$t\to -i\tau$$ to imaginary time $\tau$, and the metric is modified, in the case of Wick rotated Minkowski space, to ...


11

OBJECTIVE: how to Wick rotate a path integral using Cauchy's theorem. I like the approach of using Cauchy's theorem and I must admit I haven't seen the finite-time problem approached from this viewpoint before, so it seemed like a fun thing to think about (on this rainy Sunday afternoon!). When I started thinking about this I was more optimistic than I am ...


10

In special relativity, the metric on spacetime is $$\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2 - \mathrm{d}t^2$$ (or with inverted signs). If you now formally transform $t \mapsto \mathrm{i}t$, this becomes the familiar Euclidean metric on $\mathbb{R}^4$ $$\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2 + \mathrm{d}t^2$$ ...


10

The in- and out-states are free states, and the S-matrix definition of Mandl and Shaw is perfectly valid (with an appropriate notion of Texp). It is the one used in rigorous mathematical physics; see the treatise by Reed and Simon. It is also the one from which the LSZ formula is derived. It is the only way to define the S-matrix rigorously. The $+i\...


9

I will add to twistor59 answer. Hawking liked the concept of imaginary time $\tau=\mathrm{i}t$ because it transforms a Lorentzian metric $$ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2$$ into a four dimensional like Euclidean metric $$ds^2 = +c^2 d\tau^2 + dx^2 + dy^2 + dz^2$$ Hawking and others believed that a quantum gravity theory could be developed in this ...


9

First of all, one need to prove the following key integral formula: \begin{align*} \boxed{I = \int_{-\infty}^{+\infty} d x e^{ i a x^2} = \sqrt{\dfrac{i \pi}{a}} \qquad (a>0) } \end{align*} Usually one can pick up an analytic function $f(z)=e^{ i a z^2}$ and then performs the complex integral along the closed contour showed above. $$J = \oint d z e^{ i ...


8

The Schrödinger equation is a wave equation, not a diffusion equation. While the equations look similar, the $i$ in Schrodinger equation differentiates them; that allows non-decaying oscillatory solutions, which diffusion equations do not allow. That said there are certainly relations between the two. The Schrödinger equation is analogous to the Fokker-...


8

There is an integration formula (see "Table of integrals, series and products" 7ed, p337 section3.324 1st integral) $$\int_0^\infty d\beta \exp\left[-\frac{A}{4\beta}-B\beta\right]=\sqrt{\frac{A}{B}}K_1\left(\sqrt{AB}\right)\qquad [\mathrm{Re}A\ge0, \mathrm{Re}B>0].$$ If $\mathrm{Re}A\ge0, \mathrm{Re}B>0$ is violated, the integral will be divergence. ...


8

First of all, note that one cannot associate a temperature to a single quantum state (cf "vacuum state of the theory is defined as having zero energy and zero temperature"), and having a zero energy vacuum state is just a convention (as it is cut-off dependent, and thus renormalized). Furthermore, the OP is confused. Standard (i.e. zero-temperature) QFTs ...


8

[The following is a half-remembered comment that my doctoral advisor told me some years ago, so I may have garbled it. I welcome corrections in the comments; feel free to tell me I'm full of it as well.] One way to think about a Wick rotation is that the "Euclidean" and "Lorentzian" manifolds (both of which are four-dimensional real manifolds, with a ...


7

The role of holomorphic functions (and their generalizations in the form of holomorphic sections of vector bundles) in physics is invaluable. Please see for example the following review by B.C. Hall, discussing holomorphic methods in mathematical physics, especially in quantum mechanics. It should be emphasized that these theories cover important parts of ...


7

Carefully following Feynman's procedure one actually finds: $$\langle x''|e^{-i\frac{(z''-z')}{\hbar}H}| x' \rangle $$ $$=\langle x'', z''|x', z'\rangle =\lim_{N\to \infty\: \epsilon \to 0} \left[\frac{m}{2\pi i \hbar \epsilon} \right]^{N/2}\int_{-\infty}^{+\infty}\cdots \int_{-\infty}^{+\infty} \left(\prod_{i=1}^{N-1} dx_i \right) \exp\left\{\frac{i\...


7

I) Bosonic part: When we Wick-rotate, it is more natural to use sign convention $$\tag{1} \eta_{\mu\nu}~=~{\rm diag}(-1,+1,+1,+1)$$ for the Minkowski (M) metric, and $$\tag{2} \delta_{\mu\nu}~=~{\rm diag}(+1,+1,+1,+1)$$ for the Euclidean (E) metric. Here we will use Greek indices $\mu,\nu=0,1,2,3$, to denote spacetime indices, and Roman indices $j,k=1,...


7

The key insight is that the integral you wrote down is a Lorentz-invariant tensor, so whatever it evaluates to must also be Lorentz-invariant. To illustrate this, let's be a bit more general and consider any integral of the form \begin{align} I^{\mu\nu}[f] = \int d^4k\, f(k^2)k^\mu k^\nu \end{align} for any admissible function $f$. Then notice that for ...


7

Matt Visser's How to Wick rotate generic curved spacetime is a great reference on this subject, which basically summarizes a lot of folklore on the subject. Addendum (Summary of Paper). This turns out to be an important problem in quantum gravity and QFT in curved spacetime for the obvious reason ("How do we know the usual tricks still work in curved ...


7

That propagator is nothing but the analytic continuation of the Green function of heat equation from real positive to imaginary values of $t_b-t_a$. The cut in the complex plane to make single valued the square root has therefore to be put along the negative real axis, or however in the semiplane $x<0$. With this cut the square root is well defined. All ...


7

In some sense yes. The temperature is defined as an imaginary time in Matsubara Green's functions or some path integrals. Thus, a negative inverse imaginary temperature can be considered as a time. Here is a quotation from Alexander Altland, Ben Simons "Condensed Matter Field Theory": "Thus, real time dynamics and quantum statistical mechanics can be ...


7

This is so that there is no "conical singularity" at the horizon (not the origin). Let me elaborate. In general, how do you know if a spacetime has a curvature singularity? Well if any metric components diverge or vanish at any point, then you know you might be in trouble. However you still need to determine if you have a curvature singularity (i.e. a true ...


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