# Some Confusion with Path Integrals

1. Jun 29, 2008

### robousy

Hey folks,

I'm reading Zee 'Introduction to QFT' and have a quick question on some terminology.

On page 10 he describes how:

$$\langle q_f|e^{-iHT}|q_i \rangle$$ "is the amplitude for a particle to go from some initial state to some final state". He then derives the path integral:

$$\langle q_f|e^{-iHT}|q_i \rangle=Z(J)$$

Where

$$Z(J)=\mathfont{C}e^{(i/2)\int\int d^4x d^4yJ(x)D(x-y)J(y)}$$

where J(x) is some source and D(x-y) is the propogator. He then says on p23 "Physically D(x-y) describes the amplitude for a disturbance in the field to propagate from y to x".

So, am I correct in saying that Z(J) governs the amplitude of a certain state, eg from one spin state to another, and D(x-y) governs the amplitude of propagation in space from one place to another.

2. Jun 30, 2008

### Hans de Vries

This is basically correct.
Some remarks do come to mind here:

1) The term $\langle q_f|e^{-iHT}|q_i \rangle$ assumes that H is time independent

If this is not the case then it has to be written as $\langle q_f|e^{-i\int Hdt}|q_i \rangle$

2) The source J(x) is generally of a different type as the sink J(y)

For instance in QED the photon field J(y) has a source J(x) which is the transition current
(interference term) between initial and final electron field. While in case of the electron
field
J(y) the source J(x) is the interaction term between the electromagnetic field A and
the initial electron field.

So rather saying that it represents the amplitude to go from one state to another state
one should say that it's the amplitude of a source field J(x) resulting in a sink field J(y)

Regards, Hans

3. Jun 30, 2008

### Icosahedron

The Z(J) here is a functional of the field J, it gives the amplitude of the field $$\phi$$going from ground state to ground state ( or vacuum state).

D(x-y) is the propagation amplitude of one disturbance going from some x to some y. In the path integral Z(J), we integrate over all x and y, over all disturbances (see p.22 (18)). Thus you can say we have a boiling sea of quantum fluctuations.

Like in path integrals for particles where all paths of the particle contribute, here all field configurations, all disturbances of the field contribute. In the end, some or all cancel each other (depending on J(x)), and we get a non-zero Z(J), the amplitude from ground state to ground state.

Last edited: Jun 30, 2008
4. Jun 30, 2008

### Icosahedron

Wait!
Last paragraph is not quite right.
You have Z=$$\langle {0}|{0}\rangle_{J}$$, which is one. So despite sources and sinks everywhere, and disturbances propagating everywhere, you start in the vacuum state and you end in the vacuum state with probability of one. Your vacuum is a boiling sea of quantum fluctuation, as they say.

5. Jun 30, 2008

### robousy

Thank you for the detailed response Hans. Ok, so I started getting confused over why a photon should have a beginning and an end location in space(source/sink), but I think things are starting to get clearer. These are virtual photons that we are modelling, correct? They are necessary to mediate the forces between charged particles in QED. Thus, the source and sink locations correspond to the locations of physical particles.

6. Jul 1, 2008

### Icosahedron

No. If you had some better manners and had showed at least some appreciation for my posts, I would explain some more.

7. Jul 1, 2008

### Hans de Vries

Hi, Richard. I prefer to look at the real thing, QED, There it goes like this:

1) Plane wave approximation

All calculations are plane wave interactions. The scattering zone is considered to be
large enough to use a plane wave approximation. The particles all occupy the entire
scattering zone.

2) The photon source

2) The source of a 'virtual' photon is the transition current of an electron which changes
momentum. This is a moving interference pattern. The speed and direction of this moving
interference pattern is:

$$\vartheta = \frac{1}{2}\left(~\vartheta_i + \vartheta_f~\right)$$

Where a $\vartheta$ is rapidity and i and f correspond with the initial and final states of the electron.

The interference pattern is an alternating charge and spin density. These produce a
radiation field, exactly like you would expect from Maxwell's laws. (The EM 4-potential
is given by the Lienard Wiechert potentials)

The transversal component of the radiation field is caused by the alternating spin density,
see the attached image. The alternating charge density produces a longitudinal component.
There is a 90 degrees phase difference between the longitudinal and transverse components.

This interference current in QED is given by:

$$\bar{\psi}_f\gamma^\mu\psi_i ~+~ \bar{\psi}_i\gamma^\mu\psi_f$$

In QED one uses only the first term, which is OK since both terms contain the same information.
You get all the above if you work out the interference current given above. Unfortunately
the textbooks don't do this for you.

The photon propagator $1/q^2$ just implements the Lienard Wiechert potentials: The source
(the charge/current) density is propagated on the light cone as the four-potential with a 1/r
attenuation.

3) The electron field source

The source of the electron field is the EM interaction term of the Dirac equation:

$$\mbox{Source} ~=~ -ie\gamma^\mu A_\mu\psi ~=~ \left(~\gamma^\mu\partial_\mu + im~\right)\psi$$

Therefor in momentum space:

$$\psi = \frac{\mbox{source}}{\gamma^\mu p_\mu - m}$$

This explains the electron propagator in QED.

I hope this helps. Regards, Hans

PS: Attached Image. The alternating spin density (at the left) is equivalent to the
transverse currents (at the right) via Stokes law. See for instance the Gordon
Decomposition as treated in Sakurai's advanced QM (section 3.5) or in Hans Ohanian's
"What is Spin" equation 22.

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8. Jul 1, 2008

### robousy

Sorry Iso. I didn't mean to be rude. I had meant to bring up something in your post too but was distracted and ended up forgetting. Thank you for your response. From your post it suggests to me that $$\langle 0|e^{-iHT}|0\rangle$$ is an expression that one could use to calculate the vacuum energy?

Hans, I'll be replying to you shortly too, but am teaching a class in about fifteen minutes so thank you also for your reply.

9. Jul 1, 2008

### Zacku

Hi,

Excuse me but you are reading "introduction to QFT" or "quantum field theory in a nutschell" ?
Because I read the last one when I wanted to begin QFT and abviously I didn't understand anything because the book is lacking pages of calculus (i don't know if this the right word) between two following equations so that this is quite difficult to begin this difficult topic with this book only.

P.S. : sorry for being out of topic

10. Jul 1, 2008

### robousy

In a nutshell actually.

11. Jul 1, 2008

### Icosahedron

12. Jul 2, 2008

### robousy

Hey Hans,

Thanks for the in depth answer! There are so many unfamiliar ideas in here that I'm a little unsure where to start, I'll give it a go though.

You talk of the electron transition current (due to virtual photons) as a moving interference pattern. So the virtual photons are clearly interfering with each other. Ok, I think I see that.

The interference pattern is an alternating charge and spin density. I've honestly never heard of this before. I've just checked your reference to Sakurai and realized there may be a lot I need to go back and (re?)learn before proceeding to get a finer appreciation.

Thanks again for the great response!

Rich

13. Jul 2, 2008

### robousy

lol, easily done, especially when you've made the effort to share knowledge! I must admit I can be quite delayed when it comes to responding sometimes.

Thanks for the great link! I plan to watch them when I get home tonight! A truly useful resource. thank-you!

Richard

14. Jul 2, 2008

### Hans de Vries

Hi, Rich.

The electron interference between the initial and final state gives rise
to an interference pattern, usually called the transition current, which
then gives rise to a radiation field: the virtual photon.

So, the vector current associated with a Dirac electron is:

$$\bar{\psi}\gamma^\mu\psi$$

The superposition between the initial and final state is $\psi_i+\psi_f$

Plugging this in the expression for the vector current gives:

$$\bar{\psi}_i\gamma^\mu\psi_i ~~+~~ \bar{\psi}_f\gamma^\mu\psi_f$$

plus the interference terms:

$$\bar{\psi}_f\gamma^\mu\psi_i ~~+~~ \bar{\psi}_i\gamma^\mu\psi_f$$

These last two terms are complementary, (hold the same information)
QED uses the first term as the source field of the (virtual) photon.
This source field is an electric charge/current density with components
coming from both the charge and the spin of the electron. The Gordon
decomposition tells you what comes from the charge and what comes from
the spin.

For the basic ideas read section 1.10 of chapter 1 of my book here:
http://physics-quest.org/Book_Chapter_EM_basic.pdf
You also might want to read sections 1.8 and 1.9.

The detailed behavior and meaning of the electron transition current
$\bar{\psi}_f\gamma^\mu\psi_i$ will be handled in coming chapter 18 of my book

Regards, Hans.

15. Jul 2, 2008

### Fredrik

Staff Emeritus
You're writing a QFT book...that's pretty cool, but if you don't mind me asking...why? There seems to be so many of them already. The title is interesting though. Is it intended for people who have already studied at least one of the standard books? What is it that you feel that those people (people like me I guess) don't understand?

16. Jul 2, 2008

### Hans de Vries

The keywords are in the title:

Understanding Relativistic Quantum Field Theory.

"Understanding" because I was dissatisfied with only getting a recipe of how to do
some complex mathematical symbol handling which then at the end produces the right
answers. You want to now why and how the physics work rather than becoming
proficient in a certain type of mathematical juggling act.

Once the mathematics is deciphered, the much simpler visual physical picture emerges,
so my book contains more math as usual (to decipher) but at the end it should all lead
to a deeper understanding of the underlaying physics

"Relativistic" because many QFT textbooks are a halfbaked mixture of relativistic and
non-relativistic physics plus some folklore which doesn't even comply with non-relativistic
physics.

Regards, Hans

17. Jul 2, 2008

### robousy

Hey Hans,

Thanks again for the detailed answer. Thats really nice insight actually. I took a look at your book and it looks like you have put a lot of time into it. Its really nice to hear that some physicists are really making an effort to help the rest of us really understand whats going on.

I think thats one thing I found frustrating for a lot of my time at grad school and that is that we spend most of our time learning mathematical techniques without really focusing on the physics! I'm going to print out your book and see if I can't acquire some of your insight. Thanks once again!

Richard

18. Jul 3, 2008

### Hans de Vries

Thanks, Rich.

Any remarks you have are welcome. It's a work in progress.

Regards, Hans