What is the Correct Form of the Free Propagator in QFT?

In summary, the free propagator between two "sources" on the field is given by D(x_mu) = -i \int \frac{d^3k}{(2\pi)^3 2 \omega_k}[e^{-i(\omega_kt-k\bullet x)} \Theta(x_0) + e^{i(\omega_k t-k\bullet x)} \Theta(-x_0) for a space like separation (x_0 = 0). Zee obtains -i\int\frac{d^3k}{(2\pi)^3 2 \omega_k}e^{-i k\bullet x} assuming \Theta(0) = 1/2. However, the speaker
  • #1
Phymath
184
0

Homework Statement



from Zee QFT in a nutshell

the free propagator between two "sources" on the field is given by[tex] D(x_\mu) = -i \int \frac{d^3k}{(2\pi)^3 2 \omega_k}[e^{-i(\omega_kt-k\bullet x)} \Theta(x_0) + e^{i(\omega_k t-k\bullet x)} \Theta(-x_0) [/tex]

for a space like separation ([tex] x_0 = 0 [/tex]) Zee gets

[tex]
-i\int\frac{d^3k}{(2\pi)^3 2 \omega_k}e^{-i k\bullet x}
[/tex]

with assumption that [tex] \Theta(0) = 1/2 [/tex]

with that assumption i don't agree with Zee i get

[tex]
-i\int\frac{d^3k}{(2\pi)^3 2 \omega_k}cos(k \bullet x)
[/tex]

where am I going wrong?
 
Physics news on Phys.org
  • #2
The two expressions are equal. If you write the complex exponential as a sum of a sine and cosine, the sine term will integrate to zero because it is odd in k.
 
  • #3
In the same book, in this definition of the D(x). Why do we get a term exp^-i(ωt-kx) when X_o in positive and a term exp^i(ωt-kx) when X_o is negative?
 

1. What is the QFT free propagator?

The QFT free propagator is a mathematical expression used in quantum field theory to describe the probability amplitude of a particle traveling from one point to another in space-time. It takes into account the effects of the field and the mass of the particle.

2. How is the QFT free propagator derived?

The QFT free propagator is derived from the Feynman path integral, which is a sum over all possible paths a particle can take between two points. The path with the lowest action, or energy, contributes the most to the propagator.

3. What is the physical significance of the QFT free propagator?

The QFT free propagator is used to calculate the probability of a particle being in a specific location at a specific time. It is also used to calculate the probability of interactions between particles, such as scattering processes.

4. How does the QFT free propagator relate to the Heisenberg uncertainty principle?

The QFT free propagator is a manifestation of the Heisenberg uncertainty principle, as it takes into account the uncertainty in the position and momentum of a particle. The propagator is also used to calculate the spread of a particle's wavefunction over time.

5. What are the limitations of the QFT free propagator?

The QFT free propagator is limited in its applicability to free particles, meaning those not affected by external forces. It also does not take into account interactions with other particles, which can significantly alter the behavior of a particle. Additionally, it only applies to relativistic quantum systems.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
530
  • Advanced Physics Homework Help
Replies
8
Views
788
  • Advanced Physics Homework Help
Replies
7
Views
1K
Replies
27
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
3K
  • Advanced Physics Homework Help
Replies
1
Views
680
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
15
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
466
Back
Top