9.6 Mechanical System
As outlined above, two-level atoms have distinct characteristics in connection with
lasers. Electromagnetic waves similarly confined within a one-dimensional cavity
exhibit related properties and above all have many features in common with a
harmonic oscillator [15].
We have already described several features and properties of the harmonic
oscillator (Chap. 2). Meanwhile, we have briefly discussed formation of electromagnetic stationary waves (Chap. 8). There are several resemblance and correspondence
between the harmonic oscillator and electromagnetic stationary waves when we
view them as mechanical systems. The point is that in a harmonic oscillator the
position and momentum are in-quadrature relationship; i.e., their phase difference is
π/2. For the electromagnetic stationary waves, electric field and magnetic field are
in-quadrature relationship as well.
In Chap. 8, we examine the conditions under which stationary waves are formed.
In a dielectric medium both sides of which are equipped with metal layers
(or mirrors), the electric field is described as
E ¼ 2E 1 ε e sin ωt sin kz:
ð8:201Þ
In (8.201), we assumed that forward and backward electromagnetic waves are
propagating in the direction of the z-axis. Here, we assume that the interfaces
(or walls) are positioned at z ¼ 0 and z ¼ L. Within a domain [0, L], the two
waves form a stationary wave. Since this expression assumed two waves, the
electromagnetic energy was doubled. To normalize the energy so that a single
wave is contained, the amplitude E 1 of (8.201) should be divides by
ffiffi ffi
2
p
. Therefore,
we think of a following description for E:
E ¼
ffiffi ffi
2
p
Ee 1 sin ωt sin kz or E x ¼
ffiffi ffi
2
p
E sin ωt sin kz,
ð9:140Þ
where we designated the polarization vector as a direction of the x-axis. At the same
time, we omitted the index from the amplitude. Thus, from the second equation of
(7.65) we have
∂H y
∂t
¼ À
1
μ
∂E x
∂z
¼ À
ffiffi ffi
2
p
Ek
μ
sin ωt cos kz:
ð9:141Þ
Note that this equation appears in the second equation of (8.131) as well. Integrating
both sides of (9.141), we get
H y ¼
ffiffi ffi
2
p
Ek
μω
cos ωt cos kz:
Using a relation ω ¼ vk, we have
9.6 Mechanical System
375
As outlined above, two-level atoms have distinct characteristics in connection with
lasers. Electromagnetic waves similarly confined within a one-dimensional cavity
exhibit related properties and above all have many features in common with a
harmonic oscillator [15].
We have already described several features and properties of the harmonic
oscillator (Chap. 2). Meanwhile, we have briefly discussed formation of electromagnetic stationary waves (Chap. 8). There are several resemblance and correspondence
between the harmonic oscillator and electromagnetic stationary waves when we
view them as mechanical systems. The point is that in a harmonic oscillator the
position and momentum are in-quadrature relationship; i.e., their phase difference is
π/2. For the electromagnetic stationary waves, electric field and magnetic field are
in-quadrature relationship as well.
In Chap. 8, we examine the conditions under which stationary waves are formed.
In a dielectric medium both sides of which are equipped with metal layers
(or mirrors), the electric field is described as
E ¼ 2E 1 ε e sin ωt sin kz:
ð8:201Þ
In (8.201), we assumed that forward and backward electromagnetic waves are
propagating in the direction of the z-axis. Here, we assume that the interfaces
(or walls) are positioned at z ¼ 0 and z ¼ L. Within a domain [0, L], the two
waves form a stationary wave. Since this expression assumed two waves, the
electromagnetic energy was doubled. To normalize the energy so that a single
wave is contained, the amplitude E 1 of (8.201) should be divides by
ffiffi ffi
2
p
. Therefore,
we think of a following description for E:
E ¼
ffiffi ffi
2
p
Ee 1 sin ωt sin kz or E x ¼
ffiffi ffi
2
p
E sin ωt sin kz,
ð9:140Þ
where we designated the polarization vector as a direction of the x-axis. At the same
time, we omitted the index from the amplitude. Thus, from the second equation of
(7.65) we have
∂H y
∂t
¼ À
1
μ
∂E x
∂z
¼ À
ffiffi ffi
2
p
Ek
μ
sin ωt cos kz:
ð9:141Þ
Note that this equation appears in the second equation of (8.131) as well. Integrating
both sides of (9.141), we get
H y ¼
ffiffi ffi
2
p
Ek
μω
cos ωt cos kz:
Using a relation ω ¼ vk, we have
9.6 Mechanical System
375