All Questions
16
questions
0
votes
1
answer
80
views
Confusion on the signs in the complex scalar field [closed]
I saw there are different ways we can write down the complex scalar field. For example, in most textbooks I can find, this is defined as
$$\phi(x) =\int \dfrac{d^3p}{(2\pi)^3}\dfrac{1}{\sqrt{2E_p}}\...
- 1,783
0
votes
0
answers
64
views
Confusion related to creation and annihilation operators
I'm studying QFT from Peskin and from the book by Ashok Das, and there seems to be a disagreement between the creation and annihilation operators, for the scalar Klein Gordon field theory.
In Das, we ...
- 1,791
0
votes
1
answer
179
views
Weinberg Chapter 10: Sign convention for momentum operator
Weinberg says that translational invariance produces a conserved momentum, i.e., $P^\mu$, such that (Eq. 10.1.1):
\begin{align*}
[P_\mu, O(x)] = +i\hbar \frac{\partial}{\partial x^\mu} O(x). \tag{...
- 433
5
votes
2
answers
311
views
Field Theory Field Operator Fourier Transform
I'm a bit confused as to whether or not the Fourier transform of the momentum and coordinate field functions need to have different signs. The notation in the book is
\begin{equation}
\begin{cases}
\...
- 383
1
vote
1
answer
235
views
Two-particle operators in QFT and the factor 1/2
I am learning about QFT through the book Quantum Field Theory for the Gifted Amateur and I am having trouble understanding the factor 1/2 in the definition of two particle field operators. In the book ...
- 203
2
votes
3
answers
1k
views
Field operators and their Fourier transform
In "QFT for the gifted amateur" page 37, the author defines the field operators:
$$ \hat{\psi}^\dagger(x)=\frac1 {\sqrt V} \sum_p{\hat a ^\dagger _p e^{-ipx}} \tag{4.7}$$
$$ \hat{\psi}(x)=\...
- 229
0
votes
2
answers
69
views
Doubt concerning the definition of $p$ and $-p$ in quantum field theory
We can define the field in term of the ladder operators as:
$$
\phi(\vec{x}) \propto \int d^3p \left( a_{\vec{p}}e^{i\vec{p}\cdot\vec{x}} + a^\dagger_{\vec{p}}e^{-i\vec{p}\cdot\vec{x}} \right)
$$
...
- 297
5
votes
2
answers
572
views
Quantization of Klein-Gordon field (what is creation operator there and what annihilation)
Recently in my class we studied quantization of fields and I'm brooding over
an argument/ motivation on the construction
of the quantization of the Klein-Gordon field. Recall the "classical"
...
- 1,395
0
votes
2
answers
265
views
What happened to the phase of creation and annihilation operators?
It's the whole day that I try to figure out at what point exactly the phase factor appearing in both creation and annihilation operators, went away maybe absorbed inside occupation number states. Here ...
- 882
1
vote
1
answer
256
views
Sign of four-dimensional translation operator
Since I started quantum field theory I had very big issues with signs, especially when I had to pass from a tridimensional euclidean space to a flat four-dimensionale spacetime, with Minkowski metric. ...
- 882
0
votes
1
answer
140
views
Are the Lowering and Raising Operators of QM the same as those of QFT?
We know that the lowering and raising operators in quantum mechanics are defined as
\begin{array}{l}
a =\frac{1}{\sqrt{2}}(X+i P) \\
a^{\dagger} =\frac{1}{\sqrt{2}}(X-i P),
\end{array}
respectively.
I ...
- 279
2
votes
1
answer
611
views
Why is the annihilation operator associated with the positive frequency solutions?
I am looking for a qualitative explanation, if possible, of the following passage from Peskin & Schroeder, bottom of page 26, concerning the quantised real Klein-Gordon field:
A positive ...
- 316
1
vote
1
answer
581
views
$(\psi_L^\dagger \psi_R)^\dagger \neq (\psi_R^\dagger \psi_L)^\dagger$ ? What is the transpose for spinors?
The dirac mass term in terms of Weyl spinors is
$$\psi_L^\dagger \psi_R + \psi_R^\dagger \psi_L.$$
My understanding is that both terms are necessary to form a hermitian term. Naively, if you take ...
- 6,936
3
votes
1
answer
220
views
Is $:A: \; \;= A - \left<0\right|A\left|0\right>$ a correct definition of normal ordering?
My course notes say that normal ordering is defined as
$$:A: \;\; = A - \left< 0\right| A \left| 0\right>.\tag{1}$$
This works for $A = aa^\dagger$ and all already normal ordered expressions.
...
- 1,386
3
votes
1
answer
202
views
Discrepancy in the form of the free Klein-Gordon field
When solving for the Klein-Gordon field $\phi$, most texts and online resources that I look at say that:
$$\phi(x) = \int \frac{ d^{3} p }{ ( 2 \pi )^{3} } \frac{1}{\sqrt{ 2 E_{\mathbf{p}} }} \left[ ...
- 3,719