# I Are virtual particles real or just math filler

#### friend

Thinking again about the math of virtual particles. They are supposed to come in pairs, but together they don't result in anything permanent. What kind of math would do that? What kind of math leads to complete annihilation or cancellation for two particles that both start at the same point at the same time and both end at the same different point at the same time? ...
The transition amplitude for a particle to go from |x> to |x'> is
$$< x'|U(t)|x > \,\,\, = \,\,\,{\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{im{{(x' - x)}^2}/2\hbar t}}.$$
I take this as true even for a virtual particle. The antiparticle is said to travel backwards in time between the same two points. So its transition amplitude would be
$$< x|U(t)|x' > \,\,\, = \,\,\,{\left( {\frac{{ - m}}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{ - im{{(x - x')}^2}/2\hbar t}}$$
by simply replacing t with -t. (Or take the complex conjugate). Yes, you can argue that these transitions are not measurable at these specific points since the |x> basis is a continuous spectrum. Granted! But bear with me because I'm trying to prove just that. I'm just considering the transition from some generic point to another generic point for a virtual particle pair that is said to go from some point to another.

The minus sign comes out of the square-root as the complex number i. So, we get
$$< x|U(t)|x' > \,\,\, = \,\,\,i{\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{ - im{{(x - x')}^2}/2\hbar t}}$$
This is not a transition from one point by a particle and then back again by the antiparticle. The two transitions happen at the same time. And a real particle might interact with one or the other, and we can't say which. So we can consider these two virtual particles to be in superposition with each other. And then the expectation value for measuring these particles would be
$$| < x'|U(t)|x > {|^2}\,\, + \,\,\,| < x|U(t)|x' > {|^2}$$
As seen from the above, the only difference between these terms is the complex number i in the antiparticle. After squaring it the only difference would be a minus sign, and the sum would be zero. This is what we are told, that they exist as wave functions but have zero expectation value of ever being measured. So you could fill space with as many virtual particle pairs as you like, even infinitely many, and it would not be noticeable by any observer.

Did I get my math right?

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#### bhobba

Mentor
I take this as true even for a virtual particle.
Since virtual particles are not real why you assume that beats me.

Yes some your math is correct but you are atrociously mixing concepts.

This is not a transition from one point by a particle and then back again by the antiparticle. The two transitions happen at the same time. And a real particle might interact with one or the other, and we can't say which. So we can consider these two virtual particles to be in superposition with each other. And then the expectation value for measuring these particles would be
That's utter nonsense.

You need to study an actual textbook.

Thanks
Bill

#### friend

Since virtual particles are not real why you assume that beats me.
If virtual particles exist at all, then they have a wave function that guides the way they transition from place to place.

I've heard professors say that space itself is made of virtual particles. And I've come to understand this in my own way. So if a particle propagates it must be through this sea of virtual particles (=space). If virtual particles have the same kind of transition amplitudes as those found in the propagator, then this provides a method of propagation through that sea. The propagation proceeds as follows: There already exists virtual particle pairs everywhere, including near a real particle. If a virtual particle-antiparticle pair appears near a real particle, then the real particle can annihilate with the antiparticle of the virtual pair. This leaves real the virtual particle that did not annihilate with its original partner. Thus the real particle use the transition amplitude of the virtual particle to jump from one position to the next. And now that the "real-ness" has been handed off to the virtual particle that did not annihilate, it is now subject to another jump by annihilating with yet another virtual pair, and so on through the path normally described with the path integral. If there is no expectation whatsoever of ever observing a virtual particle pair, then this is as good an interpretation of what's going on in the path integral as any other. If we accepted virtual particles to begin with, then maybe that would have led to the path integral formulation much sooner. Who knows what other properties can be described with them.

That's utter nonsense.
This is a pretty safe comment. No speculation there. And you're not even saying I'm wrong.

You need to study an actual textbook.
That's always a good idea. All in all, you've not said anything, though a bit snarky on your part.

I've looked, and I've not seen any books that go into a lot of detail about the math of virtual particles. It's disturbing that just about every professor in the classroom as well as the popular stage uses virtual particles to describe what's going on. But non of them go into depth into the math. Maybe that's what you're frustrated with.

#### bhobba

Mentor
If virtual particles exist at all, then they have a wave function that guides the way they transition from place to place.
Why are you starting with a falsehood? And that is not what a wave-function is.

I've looked, and I've not seen any books that go into a lot of detail about the math of virtual particles. .
This does:
https://www.amazon.com/dp/019969933X/?tag=pfamazon01-20

IIt's disturbing that just about every professor in the classroom
A number of professors post here and they don't.

This is my last post to you in this thread. Go away and study a textbook.

Thanks
Bill

#### friend

Go away and study a textbook.
Honestly, I doubt I'm going to find what I'm looking for in a text book, though it may help. I'm looking into foundational issues, why QM is the way it is? What logic justifies QM to begin with. However, most text books give a record of the history of its development. And the math they use seems to be used only because it works. But in my opinion, that does not explain why it is the way it is. It only describes that it is that way and what the implication of that are to application.

#### bhobba

Mentor
[QUOTE="friend, post: 5400050, member: 93840"But in my opinion, that does not explain why it is the way it is.[/QUOTE]

Why dindt you say that from the start.

Be enlightened
http://www.scottaaronson.com/democritus/lec9.html

If you want to pursue it start a new thread.

Thanks
Bill

#### Orodruin

Staff Emeritus
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And I've come to understand this in my own way.
This is generally a bad thing to do without any guidance. You will come out on the other side with several misunderstandings, such as the ones you have displayed in this thread and in the rest of the quoted paragraph.

I've looked, and I've not seen any books that go into a lot of detail about the math of virtual particles.
Then you have not looked very well. Essentially any introductory text on quantum field theory will cover this and the mathematics is rather straight forward given the required previous knowledge.

It's disturbing that just about every professor in the classroom as well as the popular stage uses virtual particles to describe what's going on. But non of them go into depth into the math.
This is also wrong. You will see professionals use this kind of language in popular science and perhaps in courses which do not go very deep into the underlying quantum field theory aspects. Through popular science and survey courses you will learn about science, you will not learn science.

And the math they use seems to be used only because it works.
This is the only reason to use anything in an empirical science. People who know QFT know what they are talking about when they mention virtual particles and they know how they enter into the mathematics. Just because you cannot figure it out on your own does not mean it is not already known.

With that said, I believe it is time to close this thread. The original question has been answered several times over.