1
vote
2answers
40 views

Hamiltonian split into Mass term and Decay Width

I have encountered the following procedure several times now, and none of the sources ever explain the physical reason behind it: The Hamiltonian $H$ is split into $M$ and $\Gamma$. WHY? Where ...
5
votes
0answers
60 views

Hamiltonian Operator for nonrenormalizable Effective Field Theories?

Assuming we have a Effective Field Theory, for example a Real Scalar Field Theory, defined through a Lagrangian density of the form $\mathcal{L}_{eff} = \frac{1}{2}\partial_\mu\phi \partial^\mu\phi - ...
3
votes
1answer
79 views

$\gamma^5$ factor in Quantum Field Theory

I have a problem with interpretation of $\gamma^5$ factor in the interaction Hamiltonian. I know that $\frac{1\pm\gamma^5}{2}$ is a helicity projection and it requires helicity conservation in ...
1
vote
3answers
224 views

Bogoliubov transformation with a slight twist

Given a Hamiltonian of the form $H=\sum_k \begin{pmatrix}a_k^\dagger & b_k^\dagger \end{pmatrix} \begin{pmatrix}\omega_0 & \Omega f_k \\ \Omega f_k^* & \omega_0\end{pmatrix} ...
4
votes
0answers
157 views

Topological Quantum Field Theories

I've asked this on Math.SE, but with no avail. So, I decided to ask it here. I was wondering about the following after reading the Wikipedia article on TQFTs. It is said that TQFTs have vanishing ...
6
votes
0answers
340 views

Exact diagonalization by Bogoliubov transformation

I am developing a model of multiple gaps in a square lattice. I simplified the associated Hamiltonian to make it quadratic. In this approximation it is given by, $$ H = \begin{pmatrix} \xi_\mathbf{k} ...
1
vote
2answers
221 views

Lagrangian and hamiltonian of interaction

How to prove that lagrangian of interaction is equal to hamiltonian of interaction with minus sign? For example, I can't prove it for special case - quantum electrodynamics.
4
votes
1answer
450 views

Why does Quantum Field Theory use Lagrangians rather than Hamiltonains? [duplicate]

Why does Quantum Field Theory use usually Lagrangians rather than Hamiltonains? I heard many reasons, but I'm not sure which is true. Some say it's just a matter of beauty, so Lagrangians are more ...
2
votes
0answers
147 views

Adiabatic quantum evolution of single photon or biphoton system

The prerequisite for adiabatic quantum evolution of single photon or biphoton system is as follows. We have to prepare a single photon or biphoton quantum system which has a ground and a higher level ...
6
votes
1answer
374 views

Second quantization

In second quantization we use Hamiltonian in form: $$H=\int d^3x [ \psi^{\dagger}(x) h \psi(x)],$$ where $h$ is Hamiltonian density. The field operators have following form: $$\psi = \sum\limits _{i} ...
2
votes
4answers
558 views

Why the Hamiltonian and the Lagrangian are used interchangeably in QFT perturbation calculations

Whenever one needs to calculate correlation functions in QFT using perturbations one encounters the following expression: $\langle 0| some\ operators \times \exp(iS_{(t)}) |0\rangle$ where, ...
1
vote
3answers
477 views

Propagators and Probabilities in the Heisenberg Picture

I'm trying to understand why $$\Bigl|\langle0|\phi(x)\phi(y)|0\rangle\Bigr|^2$$ is the probability for a particle created at $y$ to propagate to $x$ where $\phi$ is the Klein-Gordon field. What's ...
4
votes
1answer
408 views

Does the vacuum energy problem of quantum field theory only occur in the Hamiltonian approach, or also in the path integral approach and in AQFT?

In a standard QFT class, you're being indoctrinated that there is the "infinite vacuum energy density problem". (This is sometimes paraphrased as the "cosmological constant problem", which is in my ...
8
votes
2answers
566 views

Regularisation of infinite-dimensional determinants

Can a regularisation of the determinant be used to find the eigenvalues of the Hamiltonian in the normal infinite dimensional setting of QM? Edit: I failed to make myself clear. In finite ...