How to deduce the solution of normal modes of a cavity?

[Tspirit]

Homework Statement

In the book "Quantum Optics" written by Scully and Zubairy, there is an equation (1.1.5). The equation is presented directly and not explained how to be deduced. The content is as follows.

Homework Equations

The Attempt at a Solution

I know the solution should have the form of $Sin(kx)$, where $k=\frac{j\pi}{L},j=1,2,3,...$ . But I don't know how to get the coefficients $A$ and $q$.

 

[Simon Bridge]

Exploit orthogonality of the modes and your knowledge of the physics being described.

 

[TSny]

The coefficient of $\sin(k_jz)$ in (1.1.5) is written as $A_j q_j$ just for convenience in order to relate the cavity modes to mechanical SHM oscillators (i.e., point masses oscillating in SHM). The analogous point mass $m_j$ has displacement $q_j$ , where $q_j$ has units of distance. But the left side of equation 1.1.5 has units of electric field. Thus, one reason for the factor $A_j$ is to make the units come out right. $A_j$ must have units of electric field / distance. You should check that this is, in fact, the case.

The particular form chosen for $A_j$ leads to a nice expression for the energy of the cavity modes as given in 1.1.9 on page 4. See
https://www.amazon.com/dp/0521435951/?tag=pfamazon01-20
The total energy of the cavity radiation is expressed as the sum of the energies of a system of independent mechanical oscillators.

 

[Tspirit]

The coefficient of $\sin(k_jz)$ in (1.1.5) is written as $A_j q_j$ just for convenience in order to relate the cavity modes to mechanical SHM oscillators (i.e., point masses oscillating in SHM). The analogous point mass $m_j$ has displacement $q_j$ , where $q_j$ has units of distance. But the left side of equation 1.1.5 has units of electric field. Thus, one reason for the factor $A_j$ is to make the units come out right. $A_j$ must have units of electric field / distance. You should check that this is, in fact, the case.

The particular form chosen for $A_j$ leads to a nice expression for the energy of the cavity modes as given in 1.1.9 on page 4. See
https://www.amazon.com/dp/0521435951/?tag=pfamazon01-20
The total energy of the cavity radiation is expressed as the sum of the energies of a system of independent mechanical oscillators.

I maybe have understood it. For a SHM oscillator, its position is given by $$x(t)=x_{m}Sin(kt),$$ where $x_{m}$ is the amplitude of the oscillation, and $k$ is the natural oscillating frequency.For the cavity modes, it is actually a wave describing by $$x(z,t)=x(t)Sin(kz),$$ where $x(t)$ is the position at $t$ and stands for the amplitude of the wave and in some way the energy. There are many possible normal waves (or modes) with different amplitude (i.e. $x(t)$) and wave numbers (i.e. $k$) for a universal electromagnetic wave, so it should be a superposition of all of possible normal wave. It follows that $$x(z,t)=\underset{j}{\sum A_{j}}x_{j}(t)Sin(k_{j}z),$$ where $A_{j}$ is the normalization coefficient. Substituting $x(z,t)$ and $x_{j}(t)$ respectively with $E(z,t)$ and $q_{j}(t)$, we get $$E(z,t)=\underset{j}{\sum A_{j}}q_{j}(t)Sin(k_{j}z).$$ This is Eq. (1.1.5). QED
However, the coefficient $A_{j}$ is not determined in the derivation above. Maybe it is like TSny said that for convenience to leads to a nice expression for the energy of the cavity modes as given in Eq. (1.1.9): $$H=\frac{1}{2}\underset{j}{\sum}(mv_{j}^{2}q_{j}^{2}+m_{j}\dot{q}_{j}^{2}).$$