Canonical transformations of a quantized Hamiltonian?

[yucheng]

Homework Statement

I am reading Scully!

Relevant Equations

N/A

Source: Scully and Zubairy, Quantum Optics, Section 1.1.2 Quantization

Questions:
1. Why are the destruction and creation operators considered a canonical transformations?
2. If these are canonical transformations, does it suggest that we are also canonically transforming the Hamiltonian? Doesn't this mean that $$H_{\text{transformed}} = H + \frac{\partial F}{\partial t}$$ for some generating function F? But how then can we merely changing variables (i.e. substitution) from $(q,p) \to (a,a^{\dagger})$ without also considering the generating function?
3. We know that canonical transformations preserve the Poisson bracket. Why does it appear that the creation and destruction operators 'preserves the commutator', of course, without the $i\hbar$ factor?
4. Why are the creation and destruction operators defined with an additional harmonic time dependence !????? Every book that I happen to come across does not do this!

Thanks in advance!
________________________________________________
Excerpt of the text:
$$
\begin{aligned}
& {\left[q_j, p_{j^{\prime}}\right]=i \hbar \delta_{j j^{\prime}},} \\
& {\left[q_j, q_{j^{\prime}}\right]=\left[p_j, p_{j^{\prime}}\right]=0 .}
\end{aligned}
$$
It is convenient to make a canonical transformation to operators $a_j$ and $a_j^{\dagger}$ :
$$
a_j e^{-i v_j t}=\frac{1}{\sqrt{2 m_j \hbar v_j}}\left(m_j v_j q_j+i p_j\right)
$$
$1.1$ Quantization of the free electromagnetic field
5
$$
a_j^{\dagger} e^{i v_j t}=\frac{1}{\sqrt{2 m_j \hbar v_j}}\left(m_j v_j q_j-i p_j\right) \text {. }
$$
In terms of $a_j$ and $a_j^{\dagger}$, the Hamiltonian (1.1.9) becomes
$$
\mathscr{H}=\hbar \sum_j v_j\left(a_j^{\dagger} a_j+\frac{1}{2}\right) .
$$
The commutation relations between $a_j$ and $a_j^{\dagger}$ follow from those between $q_j$ and $p_j$ :
$$
\begin{aligned}
& {\left[a_j, a_{j^{\prime}}^{\dagger}\right]=\delta_{j j^{\prime}},} \\
& {\left[a_j, a_{j^{\prime}}\right]=\left[a_j^{\dagger}, a_{j^{\prime}}^{\dagger}\right]=0 .}
\end{aligned}
$$

 

[yucheng]

P.S. Greiner does have some material on canonical transformations of Hamiltonians.

 

[yucheng]

I think the harmonic dependence is because the electric field E is harmonic in time (here E is a standing wave) hence $q(t)$ is harmonic. This let's us factor out the harmonic time dependence.

P.S.S. $$E_x(x,t) = \sum_j A_j q_j(t)\sin(k_j z)$$

Why did the author not specify $q_j(t)$ instead?
So that H will take the form of $H(q,p)$!

 

[vanhees71]

I'd not call the use of annihilation and creation operators to describe harmonic oscillators as "canonical transformation". It's just a clever rewriting of the Heisenberg algebra for solving the harmonic-oscillator eigenvalue problem. It's used here, because the free electromagnetic field can be understood as a collection of independent harmonic oscillators labelled by the single-photon momentum eigenstates and helicities. The field-quantization chapter in Zubairy&Scully is not very lucid. You are better off with textbooks from the HEP community to understand the subtleties of quantization of gauge theories. The most simple approach is to completely fix the gauge on the classical level (most convenient for that is the Coulomb gauge) and then quantize canonically. This is ok for most quantum-optics purposes. In HEP it's inconvenient, because you like to deal with higher-order perturbative calculations and renormalization, which are more convenient when working in manifestly covariant gauges, i.e., the Lorenz gauge, which however is only partially gauge-fixing and thus you need to deal with a bit more complicated formalism like the Gupta-Bleuler quantization or, more elegantly, path-integral quantization.

A very clear derivation of the quantization of the electromagnetic field, using the Coulomb gauge, can be found in Landau and Lifschitz vol. 4 or Greiner (Reinhardt), Field Quantization.