Propagator D for a particle is basically the Green's function

[NanakiXIII]

The propagator D for a particle is basically the Green's function of the differential operator that describes that particle, e.g.

$$(\partial^2 + m^2) D(x-y) = \delta^4 (x-y).$$

This propagator is supposed to give the probability of the particle propagating from $x$ to $y$. Why does this make sense? Why would the inverse of the differential operator have something to do with probabilities?

 

[jostpuur]

I asked about the same thing some months ago, in the context of non-relativistic Schrödinger equation. I didn't get an answer yet.

Propagator and delta functions

 

[NanakiXIII]

Your discussion seems to have been a little more in-depth than what I'm looking for. I just need a basic explanation to make it plausible to me that it makes sense. Is it really such a mystery?

 

[The_Duck]

I don't have this completely clear in my mind myself, but I think it's something like this. If

$$(\partial^2 + m^2) \psi = 0$$

then $$\psi$$ describes a wave (particle) propagating freely. If we have some D that satisfies

$$(\partial^2 + m^2) D(x) = \delta (x)$$

then D describes a wave propagating freely everywhere except at x = 0. I imagine that at x = 0 the medium gets hit with a sharp hammer. If D is the retarded propagator, so that D = 0 for t < 0, then we can view this hammer blow as the source of the propagating wave described by D. If D is the advanced propagator, then we can view the hammer blow as absorbing the wave. And the Feynman propagator would be half and half.

If D is the retarded propagator, then it describes a wave (particle) propagating out from x = 0. D(x) for any x > 0 is the amplitude of the wave at x, so it's the amplitude (not the probability) for the particle to go from 0 to x.

I may have this all wrong though; someone please correct me if I do!

 

[NanakiXIII]

I hadn't considered looking at $D$ itself as a shape for the field, but that might just make sense. Thanks.

 

[NanakiXIII]

Thanks for trying to answer my question. There are a couple of things that don't make sense to me, though.

So, for any state vector $\psi$ satisfying the KG equation, we have
$$\psi(x) ~=~ \int dy \, D(x,y) \, \psi(y)$$

This doesn't look right to me. This would imply that $D(x,y) = \delta(x-y)$. Acting on this equation with the KG differential operator gives

$$(\partial^2 + m^2) \psi(x) = \int dy (\partial^2 + m^2) D(x,y) \psi(y) = \int dy \delta(x-y) \psi(y) = \psi(x).$$

But the left-most expression should yield zero if $\psi(x)$ is a solution.

From this we can see that $\psi_x$ satisfies the KG equation for all x
if $\psi_y$ does (for all y).

Well, they're the same function... or so I assume since you call both $\psi$.

$$|x\rangle ~=~ D_x^{~y} \, |y\rangle$$

(...)

Putting it all together, we get
$$\langle x | y \rangle ~=~ \langle x | D \, | y \rangle$$

I don't see this. In the latter equation $D$ seems to have become an identity operator, it doesn't turn $|y \rangle$ into $|x \rangle$ but into itself.

 

[strangerep]

Thanks for trying to answer my question. There are a couple of things that don't make sense to me, though.
[...]

Aaargh! Inappropriate notation in the first part of my post. Sorry.
I've deleted my post, since the current wording doesn't work.
I'll try to rewrite and repost later.

 

[strangerep]

The propagator D for a particle is basically the Green's function of the differential operator that describes that particle, e.g.

$$(\partial^2 + m^2) D(x-y) = \delta^4 (x-y).$$

This propagator is supposed to give the probability of the particle propagating from $x$ to $y$. Why does this make sense? Why would the inverse of the differential operator have something to do with probabilities?

OK, I decided to forget about trying to explain this in a nonstandard fancy-schmancy
way. Let's just go back to a textbook treatment...

Have you studied Peskin & Schroeder pp29-31 (subsection titled "The Klein Gordon
Propagator") ? If not, that's a reasonably direct route to get an answer (imho).

P&S give the calculation for the retarded propagator

$$D_R(x - y) ~:=~ \theta(x^0 - y^0) \; \langle 0|\, [\phi(x), \phi(y)]\, |0\rangle$$

and show that when you apply the KG operator to the RHS you get a delta function.
I.e., the retarded propagator as defined above is indeed a KG Green's function.

This example might be a little difficult to relate to simple particle propagation
from x to y since there's a commutator in there. But if you study their
calculation in eq(2.56) and then apply the same techniques to the Feynman
propagator eq(2.60), i.e.,

$$D_F(x - y) ~:=~ \theta(x^0 - y^0) \; \langle 0| \phi(x) \phi(y) |0\rangle ~+~ \theta(y^0 - x^0) \; \langle 0| \phi(y) \phi(x) |0\rangle$$

you can still show that it's also a KG Greens function, and it's more obviously
got something to do with particle propagation, -- correctly time-ordered.

I'm not sure whether this is enough for you to complete the details of the
answer yourself, so I'll add a couple more remarks...

$\phi(x)$ satisfies the KG equation, and is interpreted as the operator for
creating a particle at x. Its adjoint is correspondingly for destroying a particle at x.
A particle "propagating from x to y" means creating it at x and destroying it at y.
That's why we use expressions of the form
$$\langle 0| \phi(y) \phi(x) |0\rangle$$
to mean propagation from x to y. The extra step functions involving the
time component are just to get the physical time ordering right.

If you don't have access to P&S, or have trouble following their
calculation leading to eq(2.56), let me know and I'll try to post
a more detailed calculation.

 

[NanakiXIII]

It seems I completely misinterpreted what $\phi$ is. I though the field describes amplitudes much like a wave function does in quantum mechanics. I'm not really sure how to interpret it now.

I don't have the book you refer to, and I don't have access to a library at the moment. But perhaps if you answer my above question, I can try to get through it. Right now, I'm a little lost as to the meaning of things.

 

[vanhees71]

A single-particle wave function like in nonrelativistic quantum theory is not a good concept to describe relativistic (interacting) particles. The reason is that at relatilivistic particles tend to produce new particles or to become absorbed when colliding with other particles.

The formal reason lies in the space-time structure of Minkowski space. One can show that it is impossible to construct a Poincare covariant quantum theory of single interacting particles in terms of a single-particle wave function. Especially there are no Poincare-covariant four-currents with positive semidefinite time components which could serve as a probability measure in the sense of Born's rule.

In strangerep's posting, the $$\phi$$ symbols stand thus for field operators. In local, microcausal quantum-field theory any particle has an anti particle with the same mass but opposite charge-quantum numbers. In strangerep's answer, the fields are tacidly assumed to be self-adjoint scalar fields, which restriction means that the antiparticles are the same as the particles. As he said, the Feynman propagator (which in the vacuum-quantum field theory is identical with the time-ordered Green's function) describes indeed the causal propagation of particles.

I really recommend to read a good book about quantum-field theory. It's not so easy to explain this delicate issue of relativistic quantum physics in short forum postings. As you said that you have no access to a library for the time being, perhaps you like to have a look at my quantum-field theory notes on my homepage:

http://theorie.physik.uni-giessen.de/~hees/publ/lect.pdf

 

[NanakiXIII]

Thanks, vanhees. I am reading a book on Quantum Field Theory (Zee's QFT in a Nutshell), but I'm only in the introductory chapter. The propagator I was asking about is introduced but not elaborated upon much. I was hoping there would be a simple answer to my question, but it would seem there isn't, judging from your replies. For the mean time I suppose I'll put it on ice, until I advance a little further.

I do have some questions, though, about what you posted. You say $\phi$, i.e. the field, is an operator. But what does it act upon, then? Or am I misunderstanding, and is $\phi$ not the field itself? Like I said, I haven't advanced very far yet with my QFT, but while reading an image of what the fields are formed in my head, and if that image is incorrect I'd prefer to find out sooner rather than later.

I thought of a quantum field like a wave-function for all particles of a certain type. For example, I imagined a photon-field, which represents all photons. I imagined there being another field like that for, say, electrons, and I was expecting to learn about some kind of interaction between the two. These fields I imagined to describe probabilities of finding particles somewhere in spacetime sort of like wavefunctions do for single particles in QM.

 

[vanhees71]

Your idea is not really a misconception, but it's rather vague. Before starting with relativistic quantum field theory, you should have good knowledge about nonrelativistic quantum theory in its abstract formulation in terms of Hilbert-space vectors and operators (symbolized by Dirac's bra-ket formalism).

It should be clear that (pure) states are represented by hilbert-space vectors, $$\ket{\psi}$$ and observables by linear self-adjoint operators like $$\hat{x}$$, $$\hat{p}$$, etc. These are characterized by commutation relations like $$[\hat{x},\hat{p}]=\mathrm{i} \hbar$$. The operators act on the vectors in Hilbert space.

This is also known as "1st quantization", and this works well only for nonrelativistic particles. "Quantization" means to take the observables like position or momentum components in classical physics to self-adjoint operators in Hilbert space and somehow guess (or postulate by group-theoretical methods, using Noether's theorem) the commutation relations, which then are used to build quantum-mechanical descriptions of the behavior of particles.

As you might guess, there also exists a formalism called "2nd quantization" or field quantization. Here you take fields (classical fields like the electromagnetic fields or quantum-mechanical wave functions with their equations of motion) and "quantize" them in the same way as you quantized the point-mechanical observables in the "1st quantization".

To interpret their physical meaning, you have to construct the Hilbert space on which these field operators (which are not necessarily self-adjoint) are realized. As it turns out, what you describe with the field operators are many-particle systems, where the number of particles is not fixed, i.e., with the "2nd quantization" you can describe situations where particles are destroyed and created in collisions or decays of other particles. E.g., if an atom in an excited state gets deexcited, its electrons go over from an excited state to a lower energy state and sends out a photon. This is described by quantum electrodynamics, where the Maxwell em. field is quantized. In fact, that was historically the first field which has been quantized (Dirac 1928). So initially you have an (excited) atom and finally you have a (less excited) atom plus a photon which hasn't been there before.

As it turns out the names "1st quantization" and "2nd quantization" are not so well describing the true state of affairs. One can show that in the special case, where the "1st quantiation formalism" is applicable, i.e., in situations where the particle number is conserved (as is the case in a wide range of applications in nonrelativistic quantum theory) the same situations can also be described with help of the "2nd-quantization" formalism, and the results are identical to the results from the "1st-quantization" formalism. This in turn means that "1st and 2nd quantization" are only different ways to write down the same quantum theory in the special situation that one looks only at situations where the particle numbers are conserved. Quantum field theory is more general, because it allows an easy description of changing particle numbers in creation and annihilation processes as in the case of the excited atom emitting a photon and getting less excited in the process.

 

[NanakiXIII]

Thanks for your elaborate explanation, vanhees. I had seen the term "second quantization" before in a QM class, but I can't say I understood what it was. Your succinct explanation has made it quite a bit more clear.

I've still got a lot of questions, but I don't want to get ahead of myself too much. I still want to clear one thing up, though. In the above postings, the $\phi$ is not the field, right? And if it's not, why act on it with the KG operator? Also, the propagator has to do with amplitudes in the sense that

$$\psi(x) ~=~ \int dy \, D(x,y) \, \psi(y)$$

where these $\psi$ are operators having to do with creation of things at points in spacetimes ($x$ and $y$)? This would make some sense to me, but I got very confused because I thought strangerep was referring to a field in all these instances.

 

[ismaili]

Thanks for your elaborate explanation, vanhees. I had seen the term "second quantization" before in a QM class, but I can't say I understood what it was. Your succinct explanation has made it quite a bit more clear.

I've still got a lot of questions, but I don't want to get ahead of myself too much. I still want to clear one thing up, though. In the above postings, the $\phi$ is not the field, right? And if it's not, why act on it with the KG operator? Also, the propagator has to do with amplitudes in the sense that

$$\psi(x) ~=~ \int dy \, D(x,y) \, \psi(y)$$

where these $\psi$ are operators having to do with creation of things at points in spacetimes ($x$ and $y$)? This would make some sense to me, but I got very confused because I thought strangerep was referring to a field in all these instances.

Yes. The $$\phi(x)$$ appeared in Strangerep's post is indeed a field.
Why do you think it's not a field?
It makes sense if you regard all the $$\phi$$ appearing in Strangerep's post as field operators.

 

[Dickfore]

The propagator discussed in the OP is the free particle propagator. The short answer is this:

Suppose there's a free field with a discrete set (the index also includes the time coordinate of the field) of real components $\varphi_{i}$ (for simplicity) subject to a quadratic action with respect to the fields:

$$S[\{\varphi\}] = \frac{1}{2} \, \sum_{i, j}{\varphi_{i} K_{i \, j} \, \varphi_{j}}, \; K_{j \, i} = K_{i \, j}$$

Then, the partition function is a Gaussian path integral:

$$Z_{0}(J) = \int{\prod_{i}{d\varphi_{i}} \, e^{i \, S[\{\varphi\}] + i \, \sum_{k}{J_{k} \, \varphi_{k}}}}$$

To evaluate this integral, we "complete the square":

$$\varphi_{i} = \tilde{\varphi}_{i} - \varphi^{(0)}_{i}, \; d\varphi_{i} = d\tilde{\varphi}_{i}$$

$$\begin{array}{l} S[\{\varphi\}] + (J, \varphi) \equiv \frac{1}{2} \, \sum_{i, j} {\tilde{\varphi}_{i} \, K_{i \, j} \, \tilde{\varphi_{j}}} - \sum_{i}{\tilde{\varphi}_{i} \, (\sum_{j}{K_{i \, j} \, \varphi^{0}_{j}} - J_{i})} \\ + \frac{1}{2} \, \sum_{i, j}{\varphi^{(0)}_{i} \, K_{i \, j} \, \varphi^{(0)}_{j}} - \sum_{i}{J_{i} \, \varphi^{(0)}_{i}} \end{array}$$

We choose the shifts $\varphi^{(0)}$ so that the term linear in $\tilde{\varphi}$ vanishes:

$$\sum_{j}{K_{i \, j} \, \varphi^{(0)}_{j}} = J_{i} \Rightarrow \varphi^{(0)}_{i} = \sum_{j}{K^{-1}_{i \, j} \, J_{k}}$$

With this, we have:

$$S[\{\varphi\}] + (J, \varphi) = \frac{1}{2} \, \sum_{i, j}{\tilde{\varphi}_{i} \, K_{i \, j} \, \tilde{\varphi}_{j}} - \frac{1}{2} \, \sum_{i, j}{J_{i} \, K^{-1}_{i \, j} \, J_{j}}$$

Now, the Gaussian integral does not depend on the external sources $J$ and we can write:

$$Z_{0}(J) = Z_{0}(0) \, e^{-\frac{i}{2} \, \sum_{i, j}{J_{i} \, K^{-1}_{i \, j} \, J_{j}}$$

With the help of the partition function, we can evaluate the "propagator":

$$\Delta_{i \, j} \equiv -i \langle T \varphi_{i} \, \varphi_{j} \rangle = -i \, \left. \frac{1}{i} \frac{\partial}{\partial J_{i}} \frac{1}{i} \frac{\partial}{\partial J_{j}} Z_{0}(J)\right|_{J = 0}$$

The evaluation of the derivatives is straightforward and leads to the result:

$$\Delta_{i \, j} = K^{-1}_{i \, j}$$

We see that the (free) propagator is actually the inverse of the kernel of the quadratic part of the action. We may write the condition for the inverse matrix as:

$$\sum_{k}{K_{i \, k} \, \Delta_{k \, j}} = \delta_{i \, j}$$

Going from the discrete to the continuous variant, we may write:

$$\int{dx'' \, K(x, x'') \, \Delta(x'', x')} = \delta(x - x')$$

In the case of the Klein-Gordon equation:

$$(\partial^{2} + m^{2}) \, \varphi(x) = 0$$

the action to which this equation is the corresponding Euler-Lagrange equation of is:

$$S[\varphi(x)] = \frac{1}{2} \, \int{dx \, \partial^{\mu}\varphi(x) \, \partial_{\mu} \varphi(x) - m^{2} \, \varphi^{2}(x)} = -\frac{1}{2}\int{dx \, \varphi(x) \, (\partial^{2} + m^{2}) \, \varphi(x)}$$

So the kernel is:

$$K(x, x'') = -\delta(x - x'') \, (\partial''^{2} + m^{2})$$

and the corresponding equation that the propagator satisfies is:

$$-\{\partial^{2} + m^{2}\} \, \Delta(x, x') = \delta(x - x')$$

 

[Dickfore]

A very good free textbook (in electronic format!) for QFT is:

http://www.physics.ucsb.edu/~mark/qft.html"

Good luck studying on your own.

 

[NanakiXIII]

Yes. The $$\phi(x)$$ appeared in Strangerep's post is indeed a field.
Why do you think it's not a field?
It makes sense if you regard all the $$\phi$$ appearing in Strangerep's post as field operators.

Right, this is what I needed to know, my idea of what a field is is incorrect or at least incomplete. I thought the field operators were things that acted on the field. I will continue my reading keeping in mind that I still need to shape my image of fields.

Dickfore, thanks for supplying an explanation starting the other way around. At the moment, I do understand everything you said, but I recognize a couple of things that I saw in future chapters of my book, so in time I hope your explanation will provide me with a clear understanding. Also, thanks for the link, a secondary book to consult is always welcome.

 

[xepma]

The way field operators are usually introduced is indeed very confusing. It might be good te keep in mind that the field operator has nothing to do with the physical state of the system. After all, the operator is the same for all physical states.

One thing to keep in mind is the following (warning: abusive language going on here) : to a large extent the field operator $$\psi(x)$$can be interpreted as a "probe" that "measures" wether there is a particle at position x. Even better: it's simply the annihilation operator of a particle at position x. In the same sense $$\psi^\dag(x)$$ is the creation operator of a particle at position x. So the combination $$\psi^\dag(x)\psi(x)$$ can be interpreted as the number density operator of the system.

So the field operator is really an object that is independent of the state of the system. States come into play when you talk about quantum amplitudes, like scattering amplitudes. In that case you start talking about expectation values of (combinations of) these field operators. The whole Feynman diagram bussiness is nothing but a tool in this treatment.

 

[NanakiXIII]

Thanks, xepma, that clears up a lot of things that had me confused in this thread. If it is doable (if the answer will only raise more questions, I'll wait until I get further), could you explain to me what these states are in QFT? Are they like what I described I thought the fields were, i.e. wavefunctions of sorts for all particles of a certain type in the universe?

 

[ismaili]

Thanks, xepma, that clears up a lot of things that had me confused in this thread. If it is doable (if the answer will only raise more questions, I'll wait until I get further), could you explain to me what these states are in QFT? Are they like what I described I thought the fields were, i.e. wavefunctions of sorts for all particles of a certain type in the universe?

Field operators could operate on the states of the Hilbert space just like what we do in quantum mechanics.
For canonical quantisation(which means postulates certain (anti-)commutation relations), we treat all the fields as operators.
We constructed the Hilbert space by the Fock space, meaning, you define some Fock vacuum which is annihilated by all the annihilation operators, and we could operate creation operators on the vacuum to build up various multi-particle states. Note that in this Fock space approach, the number of particles is variate. This is what we can't do by just using first quantisation.

One more thing to clarify, for the path integral formalism, which is equivalent to the canonical formalism, all the fields are just c-numbers. This makes the calculation easy, and hence people usually do real calculation with path integral formalism.

hope this helps.

 

[NanakiXIII]

Thanks, ismaili.

 

[Anamitra]

We consider the expression:

psi(y)=integral k(y:x)psi(x)d^4x ------- (1)
[I have not used bold letters for the four dimensional variables x and y in the subsequent parts]
k(y:x) seems to indicate a probability density function related to the propagation of the psi-function from x to y.

Now let us consider our problem at hand:

(del^2+ m^2)D(y-x)= del^4(Y-x)

We construct the function:

f(y)= integral D(y-x)psi(x)del^4(x-x0) d^x ----------------- (2)
The delta function on the right hand side ensures that the psi-function is localized around the point x0 at time t=t0

D(y-x) plays the same role in equation (2) as k does in equation (1).But the important point to investigate is whether f(y) is at all a solution of the Klein-Gordon equation.Let us test the issue:

[ (del^2+ m^2)f(y)] (at y=y0)=(del^2+ m^2)[integral D(y-x)psi(x)del^4(x-x0) d^4x]

It should be noted that the differential operator on the left hand side(and hence on the right hand side) operates on "y" and not on "x".

We have,
[(del^2+ m^2)f(y)]at y=y0= integral[(del^2+ m^2)f(y) D(y-x)psi(x)del(x-x0) ]d^4x
= integral[del^4(y-x)psi(x)del^4(x-x0)]d^4x
=psi(y)del^4(y-x0)

Or,
(del^2+ m^2)f(y)= psi(y)del^4(y-x0)
For points Y not equal to x0 the right hand side =0 and we have,
[ (del^2+ m^2)f(y)] (at y=y0) =0
That is, f(y) satisfies the Klein Gordon equation for all points other than x0 in particular y=y0.
The above calculations perhaps provide some indication as to why D(x-y) could represent the probability of a particle traveling from x0 to y0.

 

[NanakiXIII]

Thanks, Anamitra. One thing, though.

f(y)= integral D(y-x)psi(x)del^4(x-x0) d^x ----------------- (2)
The delta function on the right hand side ensures that the psi-function is localized around the point x0 at time t=t0

This seems a little ad hoc. I'm not really sure why it makes sense to add that delta function, but the rest of your story hinges on this.