All Questions
17
questions
1
vote
1
answer
294
views
Verify that a field operator creates a particle
In example 4.1 of Lancaster and Blundell's "Quantum field theory for the gifted amateur", we verify that a field operator creates a particle as follow:
Let $|\Psi\rangle=\hat{\psi}^{\dagger}(...
- 53
1
vote
1
answer
374
views
Deriving the Single Body Hamiltonian in QFT
I'm having some trouble with a few steps of the reasoning.
My first issue is why kinetic energy is diagonal in momentum representation, and why that means the Hamiltonian as a whole will be diagonal ...
- 383
3
votes
2
answers
480
views
Canonical transformation to diagonalize Bosonic Hamiltonian
The Hamiltonian of the system of bosons ($a$, $a^{\dagger}$, $b^{\dagger}$ & $b$ are Bose operators) is:
\begin{equation}
H=\epsilon_{1} a^{\dagger}a+\epsilon_{2}b^{\dagger}b+\frac{\Delta}{2}\...
- 39
1
vote
2
answers
3k
views
Quantization of complex scalar field
I'm learning Peskin's qft now and I'm a little confused about problem 2.2 .
Suppose I write the field $\phi(x)$ as:
$\phi(x) =\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{p}}} (a_{p}e^{-ipx}+b_{p}e^...
- 81
1
vote
1
answer
492
views
Explicit expressions for the creation and annihilation operators
What are the explicit expressions for the creation and annihilation operators $\hat{a_\vec p}$ and $\hat{a}^{\dagger}_\vec p$ for bosons? I can't find them anywhere, as every source seems to introduce ...
- 145
0
votes
1
answer
110
views
Dirac spinor and field quantization
Can we simplify $$ Σ_s Σ_r [b_p^s u^s(p)\mathrm e^{ipx} (b_q^r)^†(u^r)^†(q)\mathrm e^{-iqy} + (b_q^r)^†(u^r)^†(q)\mathrm e^{-iqy} b_p^s u^s(p)\mathrm e^{ipx}]\tag{1}$$ as $$Σ_sΣ_r[ \{b_p^s, (b_q^r)^†...
- 369
0
votes
0
answers
293
views
Commutation relations following from quantization of a complex scalar field
As someone who has recently started doing QFT I have some (algebraic) confusion about the following derivation.
Starting with the Lagrangian of a complex scalar field $$\mathcal{L} =\partial_\mu \psi^...
- 145
2
votes
1
answer
142
views
Expressing the Schrödinger equation in 2nd quantised language
For times sake, I will only write about the non-interacting part of the Hamiltonian,
$$H_0=\sum_{j=1}\left(-\frac{\hbar^2}{2m}\frac{\partial}{\partial x_j^2}+U(x_j)\right)$$ where $U(x_j)$ is some ...
- 173
0
votes
1
answer
237
views
Proving equivalence of first and second quantisation (Pathria's way)
I'm trying to solve problem 11.1 form Pathria R. K. & Beale P. D. - Statistical mechanics book (the hyperlink will get you straight to the page of the problem). The point (b) is to show the ...
- 315
2
votes
1
answer
194
views
Spinor quantization: contradiction between covariant anticommutator and canonical rules?
Starting from the free lagrangian
$$\mathscr L = \bar\Psi(i\displaystyle{\not}\partial - m)\Psi$$
I compute the canonical momenta
$$\Pi =\frac{\partial \mathscr L}{\partial\dot{\Psi}}=i\Psi^\dagger ...
- 523
5
votes
2
answers
1k
views
Hamiltonian with periodic potential in second quantization
I'm working with the following Hamiltonian
$$\hat{H}=\int\mathrm{d}\mathbf{x}\sum_{\sigma\in\left\lbrace\uparrow,\downarrow\right\rbrace}\hat{\psi}_\sigma^\dagger(\mathbf{x})\left[-\frac{\hbar^2\...
- 93
1
vote
0
answers
1k
views
How to Fourier transform creation/annihilation operators?
Zee's QFT in a Nutshell pages 65-66. For a complex scalar QFT
$$
\varphi(\vec{x},t) = \int\frac{d^Dk}{\sqrt{(2\pi)^D2\omega_k}}\left[a(\vec{k})\mathrm{e}^{-i(\omega_kt-\vec{k}\cdot\vec{x})} + b^\...
- 1,487
3
votes
1
answer
730
views
Commutation relations in second quantization
I know that for operators $a(\chi_1), a(\chi_2)$ of the same type (fermionic or bosonic)
$$ [a(\chi_1), a(\chi_2)]_{-\xi} = [a^\dagger (\chi_1), a^\dagger (\chi_2)]_{-\xi} = 0 \tag{1}$$
where
$$\xi ...
- 1,001
1
vote
1
answer
184
views
Product of deltas in kinetic second quantization hamiltonian
I am trying to derive the result for a kinetic hamiltonian in second quantization in term of the fields, that is: $\hat{H} = \int - \Psi^\dagger (r) \frac{\hbar^2\hat{\nabla}^2}{2m} \Psi(r)$
I start ...
- 714
1
vote
1
answer
226
views
Deriving commutation relations in second quantisation
I am trying to start from:
\begin{align*}
[\phi(x),\pi(x')] = i\hbar\delta(x-x') \\
[\phi(x),\phi(x')] = [\pi(x),\pi(x')]=0
\end{align*}
to derive:
\begin{align*}
[a(k),a(k')^\dagger]=\delta_{kk'}\\
[...
- 477