- #1

yucheng

- 232

- 57

- 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}

$$

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}

$$