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What is imaginary (or complex) time? I was reading about Hawking's wave function of the universe and this topic came up. If imaginary mass and similar imaginary quantities do not make sense in physics, why should imaginary (or complex) time make sense?

By imaginary I mean a multiple of $i$, and by complex I mean having a real and an imaginary part, i.e., $\alpha + i\beta$, where $\alpha, \beta \in {\mathbb R}$.

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I've heard about rotating integrals over time into the complex-time domain. I don't know if complex time would have a physical interpretation or if its just a calculation tool. Or if your complex time and my complex time are even the same thing. Good question. –  Todd R Dec 14 '12 at 2:27
I sometimes think imaginary time should be space and complex time should mean a combination of space and time twisted together... Thanks to relativity. I speculate this since in the metrics of relativity, the signs of space and time components are opposite, so taking their square roots would... –  namehere Dec 14 '12 at 11:40
perhaps it may be seen as an 'analytic continuation ' trick so all the things are finite in the end $ t \to it $ by analytic continuation –  Jose Javier Garcia Sep 11 '13 at 10:22

5 Answers 5

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

$$ \psi(x) = \psi_0 \exp(\frac{i}{\hbar}S(x)) $$

Then we get

$$ -\frac{i\hbar}{2m}\frac{d^2S(x)}{dx^2}+\frac{1}{2m}\left(\frac{dS(x)}{dx}\right)^2+V(x)-E=0 $$

which is nonlinear. We can make progress with an $\hbar$ expansion

$$ S(x)=S_0(x)+\hbar S_1(x)+\frac{\hbar^2}{2}S_2(x)+... $$

After long calculation we can calculate various amplitudes and derive things like the barrier tunneling coefficient

$$ T=\exp(\frac{2}{\hbar}Im(S)) $$


$$ Im(S)=\int_a^b |p(x)|dx $$

($p(x)=\sqrt{2m(E-V)}$) and $a$ and $b$ are the $x$ values where the potential function is such that $E<V(x)$. Now Feynman offers another way to approach this, namely that the amplitude to get from x=a to x=b is just $$\langle x=b|\exp(\frac{iHt}{\hbar})|x=a \rangle = \int \mathcal{D}[x(t)] \exp(\frac{iS[x(t)]}{\hbar}) \ \ \ (1)$$ where the integral is over the space of classical paths $x(t)$ with the right endpoints. Now this, although very elegant, is extremely hard to compute: it's an integral over an infinite dimesional space after all! The imaginary time trick works as follows: You just make a change of variable $$t=-i\tau$$ then the action

$$ S(x(t))=\int{\frac{1}{2}}m \left({\frac{dx}{dt}}\right)^2-V(x) dt $$

becomes $$ S(x(\tau))=i\int{ \frac{1}{2}}m \left({\frac{dx}{d\tau}}\right)^2+V(x) d\tau$$ so the potential energy has swapped sign relative to the kinetic energy (and we picked up an overall i factor). Defining $$ S_E(x(\tau))=\int{(\frac{1}{2}}m ({\frac{dx}{d\tau}})^2+V(x)) d\tau$$, our path integral is now $$\langle x=b|\exp(\frac{-H\tau}{\hbar})|x=a \rangle = \int \mathcal{D}[x(\tau)] \exp(\frac{-S_E[x(\tau)]}{\hbar})\ \ (2)$$ Now the integral will be dominated by classical paths which extremize this action. Whereas an extremal path contributing to (1) would require imaginary energy to tunnel through the potential, which looks like a hill, for (2), the potential hill is now a valley and the corresponding extremal case is just that of a ball rolling down one side of the valley and up the other. Having done your computation in Euclidean space, you then proceed by taking whatever answer you got, and rotating back to Minkowski space.

So much for mechanics. You can do the same trick in field theory, where your path integral is now over classical field configurations. The Euclidean space extremal field configurations are called instantons. Now in your question, Hartle and Hawking were interested in what the equivalent, for the initial conditions of the universe, of "x=a" in our simple example is. Just like in the QM example, they were working in Euclidean time, and wanted their equivalent of "x=b" to be a de Sitter universe. Their guess was that, in the path integral, they should include all Euclidean metrics for spaces with no boundary. Just as our Euclidean extremal paths satisfy the equations of classical mechanics in Euclidean time, so the metrics included in the quantum cosmology path integral would satisfy the classical Euclidean signature Einstein equations.

So to summarize, Euclidean time is a clever trick for getting answers to extremely badly behaved path integral questions. Of course in the Planck epoch, in which the no-boundary path integral is being applied, maybe Euclidean time is the only time that makes any sense. I don't know - I don't think there's any consensus on this.

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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 way. This approach was named the "Euclidean path integral approach" to quantum gravity or simply "Euclidean quantum gravity". Hawking views are summarized in J. B. Hartle and S. W. Hawking, "Wave function of the Universe" Phys. Rev. D 28 (1983) 2960–2975.

This old approach does not work because many difficulties and limitations arise with it.

Notice that although imaginary times are sometimes used as a trick to simply some mathematical computations in statistical mechanics and quantum field theory, they do not have any physical meaning.

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Let'me just sketch an idea to inroduce "imaginary time" .

A photon in a black hole or in a singularity, has to disappear, it is energy should be 0.

If that photon had a previous existence, the black hole has to distruct its energy:

$E, A$, or its energy $a+a-=\left(N+\frac12\right)h\nu$

The simplest way is to consider the phase factor of the field $\exp^i(\Omega t-Kr)$

The photon should be destroy in setting :

$t->i\tau$, that squeezes the factor to zero when $\tau\rightarrow \infty$.

This could an idea for what imaginary tau) time should be helpful

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Another way to look at it is to imagine that time is a curved dimension, in sense it will be cyclic. To visualize that imagine a plane of two dimensions, then the third usual dimension will be a perpendicular line to this plane. Now consider this line to be bent in a circle, thus this dimension will go around and around, in sense it will have maximum value after which you will come back. For information, is approach called Compactification, and if you familiar with complex numbers, remember that purely imaginary exponential is cyclic.

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No, curving doesn't make it perpendicular, for obvious reasons. –  Dimensio1n0 Sep 11 '13 at 11:37

Imaginary time is a curved and cyclic dimension which can be reached by entering a large enough wormhole (probably the centre of our milkyway is large enough) with lightning speed. But there was another civilization that beat us to reach this time-traveling technology and time and time again they have destroyed the 'leading' creatures (us humans in this case) at the moment they were close to reaching the same (god-like) technology. These creatures (time-travellers) then destroy most of the earth but leave some marks for further civilazations as warning signs. For example The Pyramids (combine this with illuminati and the all-seeing eye) and Stonehenge (combine this with druids, whiches and circles).

I make this statement not out of opinion but out of many personal experiences (i.g. visions, divine signs and meetings with timetravellers) but cannot prove any of it. All I can say is that is would be wise not to travel with lightning speed to black holes because that will mean the last Messias (the time-travellers) will have to come to actually end humankind (and judge our souls of you believe in that....).

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This is science fiction as it doesn't have any support in current physics. –  WetSavannaAnimal aka Rod Vance Dec 26 '14 at 22:46
I now see what happens when someone takes too much LSD! –  David White Aug 9 at 19:54
Wow.. This is the best answer one could have ever imagined! This could easily be the "most generic" answer to any questions that have been asked or will be asked in this universe. Hats off Peter! –  pinu Aug 9 at 22:15

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