All Questions
32
questions
3
votes
1
answer
287
views
Time-evolution operator in QFT
I am self studying QFT on the book "A modern introduction to quantum field theory" by Maggiore and I am reading the chapter about the Dyson series (chapter 5.3).
It states the following ...
- 613
-1
votes
0
answers
39
views
How to get $ H=\int\widetilde{dk} \ \omega a^\dagger(\mathbf{k})a(\mathbf{k})+(\mathcal{E}_0-\Omega_0)V $ in Srednicki 3.30 equation?
We have integration is
\begin{align*}
H =-\Omega_0V+\frac12\int\widetilde{dk} \ \omega\Big(a^\dagger(\mathbf{k})a(\mathbf{k})+a(\mathbf{k})a^\dagger(\mathbf{k})\Big)\tag{3.26}
\end{align*}
where
\...
- 37
-3
votes
2
answers
107
views
Multi-particle Hamiltonian for the free Klein-Gordon field
The text I am reading (Peskin and Schroeder) gives the Hamiltonian for the free Klein-Gordon field as:
$$H=\int {d^3 p\over (2\pi)^3}\; E_p\; a^{\dagger}_{\vec p}a_{\vec p}$$
This does not seem to be ...
- 4,399
3
votes
1
answer
117
views
Show that $i/2m\int d^3\vec x\hat\pi(\vec x)\partial^2_i\hat\phi(\vec x)=1/(2\pi)^3\int d^3\vec p E(\vec p)\hat a(\vec p)^\dagger\hat a(\vec p)$ [closed]
Show that the quantum field for the Hamiltonian, $$\hat H=\frac{i}{2m}\int d^3 \vec x\hat{\pi}(\vec x)\partial^2_i\hat{\phi}(\vec x)\tag{1}$$
can be written as $$\int \frac{d^3\vec p}{(2\pi)^3}E(\vec ...
- 295
1
vote
0
answers
65
views
System interacting with Fermi Gas
My question denoted by a reduced dynamic for a system interacting with a reservoir.
Before asking the question, for completeness I will write in detail the statement of the problem and notation.
...
- 33
2
votes
2
answers
214
views
Gibbs state and creation and annihilation operators
Let's consider quantum Fermi or Bose gas. Let $a(\xi)$, $a^{\dagger}(\xi)$ are standard annihilation and annihilation operators. Hamiltonian of system is denoted as
$$
\hat{H} = \int_{R^3} \frac{p^2}{...
- 33
1
vote
1
answer
84
views
A simple question on creation and annihilation operators
We know that the KG solution for a Spin-0 particle has the following Hamiltonian
$$\hat{H}=∫ d^{3}p\frac{ω_{p}}{2}(\hat{a}_{p}\hat{a}^{\dagger}_{p}+\hat{a}^{\dagger}_{p}\hat{a}_{p})\hspace{2cm}[\hat{a}...
- 477
0
votes
1
answer
149
views
$S$-matrix from interacting picture
I’ve been reading a lot about the interaction picture, and I’m trying to string the ideas behind it together.
Essentially, the goal is to calculate something like $<f(\infty)|i(-\infty)>$. We ...
user310742
0
votes
1
answer
139
views
Energy of multiple particles in quantum field theory
Consider the free scalar theory with the zeroed Hamiltonian (i.e. such that the vacuum energy is zero). What is the energy of a multi-particle state $\phi^n|\Omega\rangle$ or (perhaps the more ...
- 1,221
-1
votes
1
answer
92
views
Commutation $[H_0, \phi_0(\vec{x},t)]$ in the Heisenberg picture [closed]
Studying from Schwartz "Quantum Field Theory and the Standard Model" p. 23, I got to the part where he discusses time dependence of the field operator $\phi$ and the annihilation/creation ...
- 99
0
votes
1
answer
386
views
Why does the interaction hamiltonian not commute with itself at different times?
If you have a poincare invariant Hamiltonian $H$, then the Hamiltonian must commute with itself at different times and not explicitly depend on time. If the Hamiltonian $H$ can be written as $H$ = $H_{...
- 39
2
votes
1
answer
214
views
A question about the Weinberg QFT Vol.1 Section2.4
I'm self-studying Weinberg QFT, and I'm confused about the connection between the momentum operator and the generator of translations
In Section 2.4, Weinberg shows the Lie algebra of Poincare Group,
\...
- 23
3
votes
1
answer
443
views
Why do we "normal-order" instead of just subtracting off vacuum energy?
The Hamiltonian is arbitrary upto a constant anyway. Why don't we just subtract off the vacuum energy? The Hamiltonian was always observable only upto a constant.
Instead, we do normal-ordering, in ...
- 6,355
0
votes
2
answers
104
views
What is the role of Hermitian Hamiltonians in relativistic QFT?
In single-particle quantum mechanics, the probability of finding the particle in all space is conserved due to the hermiticity of the Hamiltonians (and remains equal to unity for all times, if ...
- 11.8k
2
votes
1
answer
250
views
How to commute a Hamiltonian (integral form) with its operators?
How could I solve the following commutation $[\hat{H} , \hat{a}] $ when there seems to be no cancellations in its expansion? In this:
$$\hat{H} = \int \frac{d^3 p}{(2\pi)^3} a^\dagger a \tag{1}$$
I ...
- 109