Real Scalar Field Fourier Transform

Click For Summary
The Fourier transform of the quantized real scalar field \phi(x) includes both a positive and negative frequency term, represented by a(k)e^{-ik \cdot x} and a^{\dagger}(k)e^{ik \cdot x}, to ensure the field remains real. This duality arises because the second term acts as the complex conjugate of the first, which is essential for maintaining the hermitian property of the field when coefficients are treated as operators. The presence of both terms is standard in quantum field theory to account for particle and antiparticle contributions. Thus, the expression reflects the physical requirement that real fields must yield real values upon evaluation. Understanding this structure is crucial for grasping the foundations of quantum field theory.
Xenosum
Messages
18
Reaction score
2

Homework Statement



Silly question, but I can't seem to figure out why, in e.g. Peskin and Schroeder or Ryder's QFT, the Fourier transform of the (quantized) real scalar field \phi(x) is written as

\phi (x) = \int \frac{d^3k}{(2\pi)^3 2k_0} \left( a(k)e^{-ik \cdot x} + a^{\dagger}(k)e^{ik \cdot x} \right) .

It's just weird because the Fourier transform usually only has one term, and one coefficient.

Thanks for any help.

Homework Equations



N/A

The Attempt at a Solution


[/B]
N/A
 
Physics news on Phys.org
It just makes the field real. Note that the second term is the complex conjugate (or the hermitian conjugate once the coefficients have been promoted to operators) of the first term and in general, adding an expression to its complex conjugate gives something real.
 
  • Like
Likes Xenosum

Similar threads

  • · Replies 0 ·
Replies
0
Views
2K
Replies
1
Views
2K
Replies
18
Views
2K
Replies
2
Views
1K
Replies
1
Views
1K
  • · Replies 24 ·
Replies
24
Views
3K
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K