# Where does the 'i' come from in QFT path integral?

1. Dec 14, 2014

So I've been thinking about the axioms of quantum field theory. In particular the expression for the particle amplitudes:

G(x1,x2,...,xn) = ∫Φ(x1)Φ(x2)...Φ(xn)ei S[Φ]/ħ D[Φ] / ∫Φ(xn)ei S[Φ]/ħ D[Φ]

But I've been struggling to explain the existence of the 'i'. It seems like this is a postulate that can't be derived from anything else. It makes sense that the exponential is there since we are multiplying an infinite set of amplitudes.

Also, in classical physics the action is only defined up to a multiplicative factor - the Euler-Lagrange equations don't care about a multiplicative factor.

So if we replaced S[Φ] with 100 S[Φ] or 10000 S[Φ] or i S[Φ] we should get the same theory classically. But we should get different quantum field theories right? So how do we determine that there should be this 'i' here? It is often described as a 'phase' but this is not where phase enters into quantum mechanics. Phase comes from, for example, an electromagnetic waves emitted from oscillating source. So this is not really to do with phase.

Is there some kind of experiment which determines that this factor must be i/ħ or can it indeed be something else such as 1000? Or is it a factor like the electron mass or charge that actually does not have a fixed value but depends on your cut off?

One argument is that it comes from Unitarity and compares it to the unitary operator exp(iH) but if the action is, for example S[Φ]=∂Φ∂Φ, which doesn't come from a Hamiltonian then the argument is not valid. As far as I can tell Unitartity is only relevant to fermion wave functions (like Shrodingers equation) because it preserves particles number (and so probability).

For example if you replaced this 'i' with -1, what theoretical result would not then agree with experiment?

It seems if you replace the 'i' with a number α, then you add an extra factor of 1/α to all propagators and α to every interaction. How can this 'i' be derived from experiment?

2. Dec 14, 2014

### Avodyne

That the correct factor is $i/\hbar$ is determined from experiment.

Yes, you will not get unitary time evolution without the $i$.
This is wrong. That is the action (actually, the lagrangian density) for a free massless field, and there is a corresponding hamiltonian that you can look up in any QFT book.

3. Dec 14, 2014

### atyy

In fact most calculations do use imaginary time, and so in effect do replace the "i" with -1, so that quantum mechanics becomes statistical mechanics or Euclidean field theory. The i comes about from the usual hand-wavy derivation of the path integral. The path integral for bosons is a sum over classical paths, but quantum mechanics is about operators, Hamiltonians and unitary evolution. Under what circumstances is statistical mechanics secretly about quantum mechanics? In QFT, one answer is given by the Osterwalder-Schrader conditions, the most famous of which is reflection positivity which is needed for unitarity.

http://www.einstein-online.info/spotlights/path_integrals
http://arxiv.org/abs/hep-th/9403084
http://arxiv.org/abs/1005.3751
http://www.itp.uni-hannover.de/saalburg/Lectures/wiese.pdf

Last edited: Dec 14, 2014
4. Dec 14, 2014

I'm still not entirely convinced by the arguments. Yes, you need the i for exp(iH) to get a unitary operator. But exp(iS) is not an operator. Its often called a "phase factor".

If S was just an action for free fields, the 'i' doesn't add anything, just an overall phase to the amplitudes which doesn't alter the probabilities. All the probabilities are still conserved so it is "Unitary" in that sense.

It only seems to matter with interaction terms. But at that point, it becomes QFT and can't be derived directly from Hamiltonian quantum mechanics.

I would like to see a specific calculation using Feynman diagrams that shows that you get the wrong answer if it is anything except 'i'. What is the simplest way to show that it leads to something wrong?

For example, can it be shown that if you don't have the 'i' there then electrons would be attracted to electrons instead of repelled?

Last edited: Dec 14, 2014
5. Dec 14, 2014

### bhobba

Check out:
http://arxiv.org/abs/1204.0653

Basically the reason is nearby paths do not cancel without using complex numbers so you do not get the principle of least action - instead you get what is called a Wiener process:
http://en.wikipedia.org/wiki/Wiener_process

Its a very interesting mathematical curiosity that may be telling us something quite important that QM is basically a Wiener process in imaginary time. It certainly its very useful mathematically as shown by a Wick rotation.

Thanks
Bill

6. Dec 14, 2014

Hmm... I'll check that out. I'll get to the bottom of this!

It makes me uneasy that there is this complex number at the heart of QFT. For example, most complex numbers in quantum mechanics are not essential, such as we could write a complex spinor as a spinor with twice the components and so on. But there is this one little 'i' that won't go away! We could write it in a two component formalism as [cos(S/ħ),sin(S/ħ)] but this isn't any more satisfying. Or we could do a Wick rotation but then that just transfers the 'i' to the time variable.

It's strange that although the action, which tells you everything you need to know, is real valued, yet to get the probabilities you have to go through this process of introducing 'i'.

It makes you think if an 'i' why not a 'j' and a 'k'? Why not have a quaternion valued QFT which would contain 3 separate actions? e.g. exp(iS1+jS2+kS3) if such a thing even makes sense.

Last edited: Dec 14, 2014
7. Dec 15, 2014

### Demystifier

The path integral is derived from the unitary Hamiltonian evolution. As you can see from the derivation
http://en.wikipedia.org/wiki/Relati...ath_integral_formulation_of_quantum_mechanics
"i" in "iS" originates from "i" in "iH".

From Feynman diagrams you can calculate the S-matrix, which must be unitary. Without "i", the S-matrix would not be unitary. In particular, unitarity implies the optical theorem
http://en.wikipedia.org/wiki/Optical_theorem
and without "i" the optical theorem would not be satisfied.

Last edited: Dec 15, 2014
8. Dec 15, 2014

### dextercioby

The normal wisdom is reversed: From the S matrix you calculate (via a perturbative expansion) the Feynman diagrams.

9. Dec 15, 2014

My argument is like this. Consider the two point propagator G(x,y) = ∫Φ(x)Φ(y) exp(i∂Φ∂Φ) DΦ
Now absorb the i's into the fields making a substitution Φ' = √(i).

G(x,y) = α ∫Φ'(x)Φ'(y) exp(∂Φ'∂Φ') DΦ'

Surely this gives the same G(x,y) apart from a phase α. (But maybe that phase is undetermined from DΦ' = (i)DΦ)? Perhaps this substitution is forbidden for path integrals?

So by replacing i with -1, for example the propagators get a phase but are still proportional to 1/(x^2-t^2). Assuming an oscillating source at x=0 with energy (equiv. wavelength) E:

∫exp(iEt)/(x^2-(t-t')^2)dt' = exp(iE(t-r))/r θ(E) + exp(iE(t+r))/r θ(-E)

So this is where the phase of waves comes from (nothing to do with the i in the path integral). All the equations of optics are still valid. By replacing i with -1 we still get the same equations for free waves and when we calculate the probabilities the extra phase disappears.

So, like I said the only difference is when replacing i in exp(iS) with -1, say, occurs when we have interacting terms. For an action containing only free fields the 'i' can be absorbed into an overall phase for the propagators. And so everything is still "Unitary" for free fields.

Thus I need convincing that when we have interacting terms the 'i' becomes essential.

As for the Hamiltonian argument, consider that the Shrodinger equation has an 'i' i∂/∂t Ψ = ∇Ψ has an 'i' yet the relativistic Klein-Gordon equation does not. Most derivations of the path integral over fields tend to rely on the classical approximation hence why are we to assume the relativistic S involves the 'i'? If you have a link to deriving the exp(iS) from a relativistic Lagrangian I'd like to see it. Also does not the S-matrix depend explicitly on time? So we would expect the 'i' but S is not dependent on time as it is an integral over all time.

Or the reverse. Is it possible to derive the formula ψ(t2)=exp(i∫H(t))Ψ(t) from exp(iS)? i.e. how do we recover H from S? That would be a convincing argument.

From Wikipedia (that reliable source!) "The price of a path integral representation is that the unitarity of a theory is no longer self-evident, but it can be proven by changing variables to some canonical representation." Do you know where I can read about this? It seems to be the answer.

Last edited: Dec 15, 2014
10. Dec 15, 2014

OK... let me see if I've got this right:

For the Klein-Gordon field the Hamiltonian is:

H = ∫dx 3 (π(x)2 + ∇2Φ + m2Φ)

So to have a Unitary operator we must have U = exp(iH)

And then going from the Hamiltonian to the action we do a Fourier transform exp(iS) = ∫ exp(i∫H(t)dt) exp( i π ∂/∂t Φ) D[π]

Hmm... I guess that 'i' really is there to stay!

And I guess to get the Hamiltonian from the action is something like:

H[Φ(t),π(t)] =π δS/δ(∂tΦ) ??

Last edited: Dec 15, 2014
11. Dec 15, 2014

OK. But now I have a new confusion.

If the Klein Gordon equation is a real field. How can you have a Unitary operator on a real field?

12. Dec 15, 2014

### Jazzdude

The classical KG field equation is for a real field. After field quantisation you get a field of operators in the Heisenberg picture. Those evolve unitarily.

Cheers,

Jazz

13. Dec 15, 2014

I still need to delve deeper. These are the things that keep me up at night! If I don't fully understand things. But perhaps no-one really does.

Lets say we have defined our theory only in terms of waves and positions. Then we find of our wave functions have wavelike behaviour. So we define a wavelength or energy as the Fourier transform of the wave-functions. Thereby introducing 'i'. We find we can replace d/dt in our equations with iE. Now we postulate that energy 'E' is conserved over time. Hence all our fundamental theory is now based on iE.

So it seems that we have this 'i' because we are thinking in terms of a Fourier transform of the wave functions. It strikes me that there must be an equivalent way of writing the equations that avoids this Fourier transform in the first place? Perhaps a form that avoids complex numbers?

Didn't we just introduce i to nicely represent sinusoidal waves of light? Yet, the conservation of momentum, etc. is fundamentlly in the Fourier transformed picture. This is hurting my head.

In the position picture does momentum conservation become Lorenz invariance? And so energy conservation becomes...???

Last edited: Dec 15, 2014
14. Dec 15, 2014

I've had a Damascus moment! What I realised is that the 'i' is not an accidental hangover from writing spinors as complex numbers. But the reverse! Spinors are accidentally able to be written in complex numbers due to them being linear in the derivatives. Therefore I now realise the complex (or equiv. two component) formalism of QFT is the very essence of the theory! The fact that the Dirac or Shrodinger equation for electrons can be written in complex form is an accidental consequence of this complex structure of the path integral. So now that I've accepted the complex structure I am now ready to get rid of it! What I mean is that I only want to think of QFT without ever using the term "complex amplitude". Thus if I wanted to explain QFT to an alien who wasn't prepared to accept the concept of √(-1) I would explain it like this:

The "probability density" to find particles at x1,x2,... on the boundary of a causal surface is:

P(x1,x2...) = (∫F[Φ]cos(S[Φ])D[Φ])2 + (∫F[Φ]sin(S[Φ])D[Φ])2

where F[Φ]=Φ(x1)Φ(x2)...
This makes me somewhat happy. Since it is an equation which avoids the use of complex numbers. True it is not as compact and it says exactly the same thing. Plus it has a nice kind of Pythagorean quality to it.

You might think, what kind of Alien would not know about 'i' but be able to understand functional integrals.

Anyway, it is a fun game trying to rewrite QFT without complex numbers. Another one I came up with is the propagator:

1/(k.k+iε) can also be replaced with k.k/( (k.k)22) to avoid the

I'm not quite sure how you would rewrite the commutation rules [x,p]=iħ or the equation i∂/∂tΦ=HΦ or if you even need these. It can probably be done though.

Last edited: Dec 15, 2014
15. Dec 16, 2014

### vanhees71

You should derive the path integral from the operator formulation for one simple model (e.g., simple $\phi^4$ theory) to see, where the factors come from. Further note that quantum field theory means that the fields get operators. The most simple way to achieve this is to use what's called "canonical quantization". For an introduction see

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

There you also find the path-integral formalism for non-relativsitic "first-quantized" quantum mechanics. I think it also helps to first consider this more simple case, before tackling the pretty involved QFT formalism.

16. Dec 16, 2014

Yes, the canonical formalism helped me see how the iS is needed for unitarity. The Klein Gordon equation can be thought of as independent fields Φ and ∂0Φ for the purpose of quantisation.

It's a pity that exp(iS) doesn't converge properly without doing some magic like a Wick rotation. Are there other ways to make it converge? Such as adding a small factor exp(iS-εS2) or integrating the fields over a finite size Universe?

17. Dec 16, 2014

### atyy

But doesn't the Wick rotation answer your original question affirmatively - the i in the path integral is in some sense not needed. We can recover the i in the canonical formalism without the i in the path integral.

18. Dec 16, 2014

### vanhees71

In principle you are right, but Euclidean field theory is not the full story. In principle you can analytically continue back to real times, leading you to the retarded Green's functions, and this is in principle sufficient to reconstruct also the time-ordered Green's functions in the vacuum (and the full Keldysh Green's functions in thermal equilibrium). However, this is not an easy task in practice. E.g., in QCD the rigorous way to gain information is through lattice-gauge calculations. For the case of finite-temperature QCD that's a Euclidean path integral with the Matsubara boundary conditions in the imaginary-time coordinate. This in principles gives the Matsubara correlation functions, but it's very tough to analytically continue those to the corresponding real-time quantities, which are sometimes of high interest in heavy-ion physics. E.g., any transport coefficients in the Kubo formalism need the time-like limit of a two-point function in momentum space, i.e., you need a reliable analytical continuation of the Matsubara two-point function to a real-time retarded Green's function in the neighborhood of a quite singular point in the complex energy plane.

For the usual purposes in the vacuum case and to derive Feynman rules, all you need is an infinitesimal Wick rotation, not the full 90 degrees. This is closely related to the "adiabatic-switching procedure" a la Gell-Mann and Low in the operator formalism. It's very important to understand adiabatic switching and the meaning of the infinitesimal Wick rotation in the real-time path-integral formalism. The trouble is, that all this is (to my knowledge) not really rigorously provable from a mathematical point of view. Usually physicists have quite "robust" hand-waving treatments of all these subtle issues. The, in my opinion, best explanation of this (and the LSZ-reduction formalisms) is in Bailin and Love, Gauge Theories (for the path-integral approach) and in Srednicky (for the operator formalism).

19. Dec 16, 2014

Well, true you could wick rotate but then you would get propagators in G(x,y,z,it).

It would be nicer to have a limit in which you stuck with Minkowski space-time. Well, I think it would be nicer anyway. :)

I'm thinking about if you formalized all of QFT in a computer, the computer might have trouble with all the Wick rotating or summing of sine waves that don't converge. I'm sure there is a nice mathematically precise way of defining the limit. Although for practical purposes the Wick rotation would be quicker.

20. Dec 16, 2014

### atyy

The Osterwalder-Schrader axioms are one way of making sure that the Euclidean field theory is equivalent to QFT in Minkowski spacetime. They have been used to construct rigourous relativistic field theories in 2 and 3 spacetime dimensions. The Osterwalder-Schrader axioms are listed in http://www.rivasseau.com/resources/book.pdf (p17-18).

The rigourous construction of a relativistic field theory in 4 spacetime dimensions is an open problem. For example, no one has constructed a rigourous QCD. Also, as vanhees71 points out above, even if it were possible to do so in principle by Euclidean field theory, the imaginary time calculation is not necessarily easier in practice, especially for non-equilibrium processes.

Last edited: Dec 16, 2014