Optics of metals - basic

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1. Introduction

1.1 Maxwell's equations with damped charge

In metals, electrons can move freely, described with conductivity. The Maxwell equations in a medium must include the conductivity term
                                     j=σE.                                (1)
Hence (for E and H):
                            ∇×E+(μ/c)((∂ H)/(∂ t)) = 0                     (2)
                            ∇×H-(ϵ/c)((∂ E)/(∂ t)) =((4 π)/c) σ E             (3)
                                            ∇.E=((4 π)/ϵ) ρ                  (4)
                                             ∇.H=0                      (5)         
We would like to reduce these equations to wave-like solution. First, we observe the term with charge density:
                              ∇. (∇×H-(ϵ/c)((∂ E)/(∂ t)) )=((4 π)/c) σ ∇. E      
                                         -(ϵ/c)((∂ )/(∂ t))∇.  E=((4 π)/c) σ ∇. E      
Or, using Eq. (4):
                                          -(ϵ/c)((∂ρ )/(∂ t))=((4 π)/c) σ ρ 
                                            -ϵ((∂ρ )/(∂ t))=4 π σ ρ            (6)
Which gives:                 ρ=(Subscript[ρ, 0]) (e^(- t 4 π σ/ϵ)) =(Subscript[ρ, 0]) (e^(- t /τ -i Γ t))              (7)
where                                                   τ=Re[(ϵ/(4 π σ))]                   (8)

1.1.1 Consider the Drude model

From the Drude model:      σ= ((N (e^2) )/(m (β -i  ω)))  ; 
Hence:          ((4 π σ)/ϵ)= ((4 π N (e^2))/(ϵ  m (β -i  ω))) =((Subscript[ω, p])^2) (((β +i  ω))/(ϵ ((ω^2)+(β^2)))) =(β +i  ω) (((Subscript[ω, p])^2)/(ϵ ((ω^2)+(β^2))))  
where:           ((Subscript[ω, p])^2)≡((4 π N (e^2) )/m)  
Then:           ρ=(Subscript[ρ, 0]) (e^(- t 4 π σ/ϵ)) =(Subscript[ρ, 0]) (e^(-i ω' t  )) (e^(- t /τ  )) 
where                ω'=ω (((Subscript[ω, p])^2)/(ϵ ((ω^2)+(β^2))))        ;            (1/τ)=β (((Subscript[ω, p])^2)/(ϵ ((ω^2)+(β^2))))

We notice (e^(- t /τ  )) is an exponential term that indicates that any charge disturbance would have an exponential decay and vanish at large t. Usually, we'll see that this term is very small in metal, hence we can ignore this transient term in metals and let  ρ=0.
When we want to keep the term ρ in transient solution, that would be associated with charge wave.
Then:
                                       ∇×E+(μ/c)((∂ H)/(∂ t)) = 0                     (9)
                                      ∇×H-(ϵ/c)((∂ E)/(∂ t)) =((4 π)/c) σ E             (10)
                                               ∇.E=0                               (11)
                                               ∇.H=0                              (12)

1.2 Wave equations

We have similar derivation:
                                ∇×(∇×E+(μ/c)((∂ H)/(∂ t)) )= 0         
                                ∇×(∇×E)+(μ/c)((∂ )/(∂ t))∇× H= 0
                                ∇(∇.E)-(∇^2)E+(μ/c)((∂ )/(∂ t))(((4 π)/c) σ E+(ϵ/c)((∂ E)/(∂ t)))= 0      
                                   (∇^2)E-((4 π)/(c^2)) σμ((∂ E)/(∂ t)) -((μ ϵ)/(c^2))(((∂^2) E)/(∂ (t^2)))= 0           (13)

We see that the term ((4 π)/(c^2)) σμ((∂ E)/(∂ t)) in (13) acts like a damped term, which means that the field oscillation in metals will have exponential decaying behavior.
For harmonic solutions:
                                                E=E[r] (e^(-i ω t))

(∇^2)E+(((μ ϵ)/(c^2))(ω^2) +i ((4 π)/(c^2)) σμ ω )E= 0           (14)  
This is very similar to the form of Helmholtz equation:
                                               (∇^2)E+((OverHat[k])^2) E= 0                            (15)  
where:                                      ((OverHat[k])^2)=((μ ϵ)/(c^2))(ω^2)+i ((4 π)/(c^2)) σμ ω                 (16)  
                                                ((OverHat[k])^2)=(μ/(c^2))(ω^2) (ϵ+i ((4 π σ)/ω) )                   (17)  
                                                 ((OverHat[k])^2)=((μ (OverHat[ϵ]) )/(c^2))(ω^2)                                 (18) 
                                                  (OverHat[ϵ]) =ϵ+i ((4 π σ)/ω)                          (19)
The use of this format allows us to use all results of dielectric media, with the complex dielectric constant in place of just ϵ.
However, we must be mindful that σ is not necessarily just a real constant. The reason, we'll see that σ itself is a complex function of ω, hence we cannot assume that the real part of  (OverHat[ϵ])  and imaginary part is ((4 π σ)/ω). We'll see later that Re[(OverHat[ϵ]) ] can be negative for frequency < plasma frequency.
From the Drude model above:
                                                    σ= ((N (e^2) )/(m (β -i  ω)))
Hence:    
                                 (OverHat[ϵ]) =ϵ+i ((4 π σ)/ω) =ϵ+i ((4 π )/ω) ((N (e^2) )/(m (β -i  ω)))                         (20)              
                                (OverHat[ϵ]) =1+4 (Subscript[πχ, 0]) +i ((4 π N (e^2) )/(m ω (β -i  ω))) =1+4 (Subscript[πχ, 0]) -((((Subscript[ω, p])^2) )/(ω (ω +i β )))        
                                (OverHat[ϵ]) =1+4 (Subscript[πχ, 0]) -((((Subscript[ω, p])^2) )/(ω (ω +i β ))) =1+4 (Subscript[πχ, 0]) -((((Subscript[ω, p])^2)  (ω -i β ))/(ω ((ω^2) + (β^2) )))         
                                (OverHat[ϵ]) =1+4 π Re[(Subscript[χ, 0])]-((((Subscript[ω, p])^2) )/( (ω^2) + (β^2)))+i (4 π Im[(Subscript[χ, 0])]+((((Subscript[ω, p])^2)  β )/(ω ((ω^2) + (β^2) ))) )      
We see that the conductivity is just a contribution to the dielectric component.

1.3 Solutions

The solution is:
                                                E=(Subscript[E, 0])(e^(i ((OverHat[k]). r -ω t))) 
and we also define various complex quantity:
                    Phase velocity:           (OverHat[v])=(c/(Sqrt[μ (OverHat[ϵ]) ]))   
                    Phase index:               (OverHat[n])=(Sqrt[μ (OverHat[ϵ]) ])=(c/ω)(OverHat[k])
We also define:
                                               (OverHat[n])=n (1+i κ)  
where κ is known as the attenuation index.
Then:
                                        E=(Subscript[E, 0])(e^(- (Subscript[k, 0 ])κ n (OverHat[s]). r ))(e^(i ((Subscript[k, 0 ])n (OverHat[s]). r -ω t))) 
The attenuation is at the rate of (e^(- (Subscript[k, 0 ])κ n (OverHat[s]). r )), can have a length that is just a small fraction of a wavelength. The depth for intensity to drop to 1/e is:
                                            d=(1/(2 (Subscript[k, 0 ])κ n))=((Subscript[λ, 0])/(4 π κ n))
Since κ n for most metals is very large, this length is very short. This is also know as skin depth: the depth that an electric field can penerate metals.

1.4 Skin depth (low frequency)

Low frequency limit illustration

In the low frequency limit (e. g. RF or mm-wave or very long wavelength), we can approximate σ as a real number (like DC conductivity), and ϵ is also mostly real and ~ 1 because at that frequence, there is little intrinsic electronic absorption and also far away from those transitions, then:
                                          ((OverHat[n])^2)=(n^2)  (((1+i κ))^2) =ϵ μ+i ((4 π σ)/ω)μ   
Solving:                              (n^2) (1-(κ^2))=ϵ μ
                                           (n^2)κ =((2 π μ σ)/ω) =((μ σ)/ν)

Solve[ { (n^2) (1-(κ^2))== ϵ μ , (n^2)κ == ((2 π μ σ)/ω)}, {n, κ}]

We keep only the positive solutions:
                                       (n^2)->((μ (Sqrt[(ϵ^2) (ω^2)+16 (π^2) (σ^2)]))/(2 ω))+((ϵ μ)/2)
                                       κ->(((Sqrt[(ϵ^2) (ω^2)+16 (π^2) (σ^2)])-ϵ ω)/(4 π σ))=((16 (π^2) (σ^2))/(4 π σ ((Sqrt[(ϵ^2) (ω^2)+16 (π^2) (σ^2)])+ϵ ω)))
Or:
                                       (n^2)->(1/2)(Sqrt[(μ^2)  (ϵ^2) +16 (π^2) (((μ^2)(σ^2))/(ω^2))])+((ϵ μ)/2)
                                           =ϵ μ ((1/2)+(1/2)(Sqrt[1 +(((((4 π  σ)/(  ϵ ω))))^2)]))
We also substitute:             
                                       κ->((4 π σ )/( (Sqrt[(ϵ^2) (ω^2)+16 (π^2) (σ^2)])+ϵ ω))=((4 π σ μ)/( ω ((Sqrt[(μ^2)(ϵ^2) +((16 (π^2)(μ^2) (σ^2))/(ω^2))])+ϵ μ )))
                                                   κ=((4 π σ μ)/( 2 ω (n^2)))=((2 π σ μ)/(  ω (n^2)))  
We see that in the limit σ-> 0:   (n^2)->ϵ μ as expected, and κ ->0, this is the dielectric case.

Example of some metals

From periodic table (example: http://www.ptable.com/) 
({{Element, Conducting
electrons, Atomic W, Permeability
 μ/(Subscript[μ, 0]), Density 
(g/cm3), Conductivity
(/Ohm cm), Conductivity
(CGS) (Hz), Conduction
electron
density, Calc (not actual)
plasmon
frequency (eV), Calc β}, {Au, 1, 196.966569, ~1, 19.30, 4.51671×(10^5), 4.059416503352371`*^17, 5.900866179326097`*^22, 9.020178670383977`, 0.024232101192938213`}, {Cu, 1, 63.546, ~1, 8.94, 5.959×(10^5), 5.3556821100926957`*^17, 8.472279545935227`*^22, 10.808300125837441`, 0.026370866909875106`}, {Al, 3, 26.9815386, ~1, 2.70, 3.546×(10^5), 3.186985863800755`*^17, 1.8078786840940202`*^23, 15.788537366864045`, 0.09456448642375152`}, {Ni, 1, 58.6934, ~1, 8.908, 1.443×(10^5), 1.2969037229172277`*^17, 9.139909949895559`*^22, 11.226082141654867`, 0.11748247876287575`}, {Ag, 1, 107.8682, 1, 10.49, 6.3012×(10^5), 5.6632361322564346`*^17, 5.856431031304871`*^22, 8.986152277081711`, 0.01723881004273065`}})
Unit consideration:     σ :  1/Ω Meter ->2.99792458^2*10^9

Density={19.30, 8.94, 2.7,8.908, 10.49} ; AW={196.966569,63.546, 26.9815386, 58.6934, 107.8682} ;
Cond={4.51671×(10^5),5.959×(10^5), 3.546×(10^5), 1.443×(10^5), 6.3012×(10^5)} ;
CondE={1,1,3,1,1} ;
ElecDensity=AvogadroConstant*CondE*Density/AW *Mole  ;

CondMKS={4.51671×(10^5), 5.959×(10^5), 3.546×(10^5),1.443×(10^5),6.3012×(10^5)}*10^2 ; 
CondCGS=CondMKS* 2.99792458^2*10^9 
PlasmaFreqeV=4 (Sqrt[ π ElecDensity (aB^3)]) (Subscript[R, ∞])  /. aB -> 5.291772083535412`*^-9  /. (Subscript[R, ∞])-> 13.605691932782342`  
βeV=(1/(2 4.1356673336325144`*^-15 *CondCGS))(((PlasmaFreqeV))^2)

Since  σ  is very large for metal (at low frequency), ((4 π σ)/ω) is very large and  ((OverHat[n])^2)=(n^2)  (((1+i κ))^2) =ϵ μ+i ((4 π σ)/ω)μ    is nearly purely imaginary.

nM[ω_,σ_,ϵ_,μ_]:=Sqrt[ϵ  μ  ((1/2)+(1/2)(Sqrt[1 +(((((4 π  σ)/(ϵ ω))))^2)])) ] ;
κM[ω_,σ_,ϵ_,μ_]:=  ((2 π  σ  μ )/(  ω ((nM[ω,σ,ϵ,μ])^2)))  ;

LogLogPlot[nM[w,CondCGS , 1.,1] ,{w, 10^6,10^9}, Frame-> True, GridLines-> Automatic]

κM[10^12,CondCGS , 1.,1]

Then:             ((OverHat[n])^2)=(n^2)  (((1+i κ))^2) =ϵ μ+i ((4 π σ)/ω)μ   
                       (OverHat[n])=n  (1+i κ)  ~(1+ i )(Sqrt[((2 π σ)/ω)μ]) 
                       κ~ 1 as expected.

What this means is that k= (OverHat[n]) (Subscript[k, 0]) = ((2 π)/λ)(1+ i )(Sqrt[((2 π σ)/ω)μ]) 
The penetration depth is:          d=(1/(2 (Subscript[k, 0 ])κ n))=((Subscript[λ, 0])/(4 π  (Sqrt[((2 π σ)/ω)μ])))~((c  2 π )/(4 π  ω  (Sqrt[((2 π σ)/ω)μ])))  
                                                      =((c  )/(2   (Sqrt[2 π μ σω]))) 
 Skin depth is defined as 2 times this:              δ =((c  )/(Sqrt[2 π μ σω]))=((c  )/(2 π  (Sqrt[ μ σ f]))) 
A perfect conductor is when σ-> ∞ and skin depth -> 0. This is the case when there is no field inside the material as expected.

c=2.99792458 *10^10 ;
LogLogPlot[ c/(2 π Sqrt[CondCGS * f]),{f,10^6,10^9},Frame-> True, GridLines-> Automatic]

As we can see, the field hardly penetrate the metal as the electrons shield the field. The higher frequency, the smaller the skin depth as (Sqrt[ω])
It is also instructive to see skin depth in terms of wavelength:
                       (δ/λ) =((c  f)/(2 π  c  (Sqrt[ μ σ f]))) =(1/(2 π  ))(Sqrt[(f/(μ σ))])

LogLogPlot[  Sqrt[f/CondCGS ]/(2. π) ,{f,10^8,10^11},Frame-> True, GridLines-> Automatic]

While the E field is:   
                                                E=(Subscript[E, 0])(e^(i ((Subscript[k, 0 ])(OverHat[n]) z -ω t))) 
the magnetic field is:              ∇×E+(μ/c)((∂ H)/(∂ t)) = 0       
                                               i ω(μ/c)H = ∇×E       
Let (Subscript[E, 0]) be x, then                  i ω(μ/c)(Subscript[H, y]) = i (Subscript[k, 0 ])(OverHat[n])  (Subscript[E, 0])       
                                                (Subscript[H, y]) = (c/(μ  ω)) (Subscript[k, 0 ])(OverHat[n])  (Subscript[E, 0])   
                                               (Subscript[H, y])  =(((OverHat[n]) )/μ) (Subscript[E, 0])  ~ (1+ i )(Sqrt[((2 π σ)/(μ ω))]) (Subscript[E, 0])  ~ (e^(i π/4)) 2 (Sqrt[((π σ)/(μ ω))])  (Subscript[E, 0])    
with μ ~ 1, and  (Sqrt[((π σ)/(μ ω))]) is very large, it means that the magnetic field is very large compared with the E field and out of phase by 45 degree.  The field energy is thus principally magnetic

Δz=0.00001  ; δ=1./10000.  ; nArrow=10  ;
Animate[
Graphics3D[Join[Table[{Hue[0,1,1],
Arrow[{{ Δz l,0,0},{Δz l,(E^(- Δz  l/δ ))  Cos[2 π (Δz l- t)]  ,0}} ]},{l,0,nArrow}] ,
Table[{Hue[0.7,1,1],
Arrow[{{ Δz l,0,0},{Δz l, 0 ,2 (E^(- Δz  l/δ ))  Cos[2 π (Δz l- t)+Pi/4]}} ]},{l,0,nArrow}] ], ViewPoint-> {5,-8,3}, PlotRange-> {{0,Δz*nArrow },{-1,1},{-2,2}}, BoxRatios-> {1,1,1}],
{t,0,1,0.025}, AnimationRunning-> False, DisplayAllSteps-> True]

1.5 High frequency limit

We cannot use the equations
                                           ((OverHat[n])^2)=(n^2) (((1+i κ))^2)=ϵ μ+i ((4 π σ)/ω)μ   
                                                    (n^2) (1-(κ^2))=ϵ μ
                                           (n^2)κ =((2 π μ σ)/ω) =((μ σ)/ν) 
For high frequency, i. e. optical because we'll see that σ is complex. In fact, at high frequency limit, σ is nearly purely complex, and κ is very large. To deal with this case, we should refer to section 2.

2. Drude's model of conductivity

2.1 Conductivity in metal

Drude first created a model for electron response in a metal. The equation of motion under an electric field is:
                                                 m (Overscript[r, ..]) +m β (OverDot[r])= e E
where m β (OverDot[r]) represents the damping term. Let the solution be harmonic:
                                               - m (ω^2)(Subscript[r, 0]) -  m i ω β (Subscript[r, 0])= e (Subscript[E, 0])  
Then:                                         (Subscript[r, 0]) = -((e )/(m ω (ω+ i  β)))(Subscript[E, 0])   
And the velocity of the charge is:
                                             (OverDot[r])=- i  ω (Subscript[r, 0]) (e^(-i ω t))= ((e )/(m (β -i  ω)))(Subscript[E, 0])(e^(-i ω t))
The current is:                        j =N e (OverDot[r])= ((N (e^2) )/(m (β -i  ω)))E = σ E 
Hence, the conductivity is:
                                                         σ= ((N (e^2) )/(m (β -i  ω)))  
For such a simple model, it qualitatively explains quite well for a number of metals.

Let's consider some limit:
As ω-> 0 (DC), σ= ((N (e^2) )/(m (β -i  ω)))->((N (e^2) )/(m β ))  as the DC conductivity. 
We can approximate  β = ((N (e^2) )/(m  (Subscript[σ, DC])  )) Strictly speaking it is not, because the Drude model has limited validity, but it is useful nonetheless.
As ω>> β, then σ= ((N (e^2) )/(m (β -i  ω)))->i((N (e^2) )/(m ω)), which means that the current is Pi/2 out of phase with the field. And at ∞, it approaches to zero; this means the material is no longer conducting at high frequency. 
The behavior of the conductivity results in complicated behavior of dielectric function.

2.2 Dielectric function

(OverHat[ϵ]) =ϵ+i ((4 π σ)/ω)=ϵ+i ((4 π N (e^2) )/(m ω (β -i  ω)))      
Where ϵ is permittivity from bounded electrons, which is usually very close to 1 for many metals.
A common expression is:
                                       (OverHat[ϵ]) =1+4 (Subscript[πχ, 0]) +i ((4 π N (e^2) )/(m ω (β -i  ω))) =1+4 (Subscript[πχ, 0]) -((((Subscript[ω, p])^2) )/(ω (ω +i β )))  
where 4 (Subscript[πχ, 0]) is from the polarization of bound inner electrons, and  ((((Subscript[ω, p])^2) )/(ω (ω +i β ))) is the plasma response term, with ((Subscript[ω, p])^2)≡((4 π N (e^2) )/m)  is defined as the plasma resonance frequency. This is the natural frequency of the electron plasma in response to disturbance.
(Note:  (ℏ^2) ((Subscript[ω, p])^2)≡4 π N (( (ℏ^2) (e^2) )/m) =16 π N ((Subscript[a, B])^3) (((ℏ^2) )/(2 m ((Subscript[a, B])^2)))((  (e^2) )/(2 (Subscript[a, B])))=16 π N ((Subscript[a, B])^3) ((Subscript[R, ∞])^2) 
hence:   h (Subscript[ω, p])=4 (Sqrt[ π N ((Subscript[a, B])^3)]) (Subscript[R, ∞])  )
 β = ((N (e^2) )/(m  (Subscript[σ, DC])  ))=(1/(4 π (Subscript[σ, DC])))((4 π N (e^2) )/m) =(1/(4 π (Subscript[σ, DC])))((Subscript[ω, p])^2)  ;  ℏ β = (1/(4 π ℏ (Subscript[σ, DC])))(((ℏ (Subscript[ω, p])))^2) =(1/(2 h (Subscript[σ, DC])))(((ℏ (Subscript[ω, p])))^2)

((((Subscript[ω, p])^2) (ω -I β ) )/(ω Expand[(ω +I β )  (ω -I β ) ]))

We'll see that the Re and Im part of the dielectric function is:
                                               Re[ϵ]≡(n^2) (1-(κ^2))= 1+4 (Subscript[πχ, 0])-(((Subscript[ω, p])^2)/((ω^2) + (β^2)))  
                                               Im[ϵ]≡2 (n^2)κ =(β/ω) (((Subscript[ω, p])^2)/((ω^2) + (β^2)))     
The approximation for high frequency is:
                                              Re[ϵ]≡(n^2) (1-(κ^2)) ~ 1-(((Subscript[ω, p])^2)/(ω^2))  
                                              Im[ϵ]≡2 (n^2)κ =(β/( ω)) (((Subscript[ω, p])^2)/(ω^2))

We see that for ω< cut-off frequency that is ~ plasma frequency, Re[ϵ] is negative, which is well known for metals. In addition, Im[ϵ] is large and the wave is strongly attenuated.
For ω above the plasma frequency, Re[ϵ]  becomes positive, and Im[ϵ] is small, which means that the metals act just like a dielectric. One of very useful optoelectronic materials for making electrodes is ITO: indium tin oxide. The material has a low plasma frequency, and is transparent in the vis-NIR.

PlanckConstant*SpeedOfLight/(ElectronCharge)

1.2398418754199976`*^-6

1.23984

h c/λ

EpsMetal[λ_,Parm_List]:=Module[{ω, wp, β, χ},
{wp, β, χ}=Parm ; ω=1.23984187542/λ  ;
1+4 Pi χ -(((wp^2) )/(ω  ( ω +I β )))  ]

EpsMetal[λ_,Parm_List]:=Module[{ω, wp, β, χ},
{wp, β, χ}=Parm ; ω=1.23984187542/λ  ;
1+4 Pi χ -(((wp^2) )/(ω  ( ω +I β )))  ];
AlParm={15., 0.2,0.04}  ;
wp=15. ;
Manipulate[
GraphicsRow[{
Plot[Re[EpsMetal[λ,{wp, β, χ}] ],
{λ, 0.05,0.1}, Frame-> True, GridLines-> Automatic]
,Plot[Im[EpsMetal[λ,{wp, β, χ}] ],
{λ, 0.05,0.1}, Frame-> True, GridLines-> Automatic] }]
, {β,0.05,0.5},{χ,0.,0.1}]

3. Reflection and transmission

This section has been damaged and cannot be recovered.