9. Light-matter interactions - metals

Complex dielectric

function and complex

refractive index

The Drude model

Optical properties of

metals- skin depth- reflectivity

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Electric field in a medium depends on n, α

/2

0( , ) ( 0)e exp[ ( )]zE z t E z j nkz t

α ω−= = −

Absorption causesattenuation of the field with increasing propagation distance z

Refractive index changes the k-vector and therefore the wavelength

0

0

medcn

c

εε

= =

…and the dielectric of the medium, εmed, is related to n:

Frequency doesn’t change!

Often denoted with the

Greek letter kappa, κ, a

dimensionless quantity!

Complex refractive index

( ) ( ) 02

0, 0jnk zz j t

E z t E z e e eα ω− −= = ⋅ ⋅ ⋅

Define the complex

refractive index: 0

0

0

2

4

2

jn n

k

n j

cn j

α

αλπ

αω

≡ +

= +

= +

%

( )0

02

0 0

jj n k z

k j tE z e e

αω

+

− = = ⋅ ⋅

k0 = k-vector in vacuum

Note the tilde!

Complex index: this is just a convenient way to group n

and α in a single quantity.

So n n jκ≡ +%

Note: κ ≥ 0 always!

Complex dielectric function

If the refractive index is complex, what about ε?

( )

2

0

2

0

medn

n j

ε ε

ε κ

=

= +%

( )2 2

0 2R I

j n j nε ε ε κ κ+ = − +

Note: this sign convention is not universal - be careful!

( )

( )0

0

1

2

1

2

= +

= −

R

R

n ε εε

κ ε εε

This can also be expressed as equations for n,κ :

Example: complex refractive Index of water

increasing frequency

radio frequencies: n ~ 9x-ray: n ~ 1 visible: n ~ 1.3

Complex dielectric function vs complex index

It is strictly correct that:

( )( )

Re

Im

•

•%

%

n

n

is related only to the phase of the wave

is related only to the attenuation of the wave

( )( )

Re

Im

•

•

εε

is related only to the phase of the wave

is related only to the attenuation of the wave

It is NOT strictly correct that:

But, if κ << n (as is often the case), then

( ) 2

0 Re• ≈ nε ε which is related only to

the phase of the wave

Note: it is NOT the

case for metals…

Eapplied(t)

nucleus electron

Does the forced oscillator model work for everything?

We started with the assumption that the electrons

are bound to nuclei.

That’s a good description

of insulating materials.

But not such a good

description of metals,

where at least some of

each atom’s electrons

are free to move

through the lattice.

Paul Drude(1863 - 1906)

� Lattice of (immobile) atomic cores (+ charges)

� Electrons are free to roam throughout the lattice

� Every once in a while, each electron collides

with something. This collision randomizes the

electron’s velocity.

The Drude Model of metallic conductivity (~1900)

The model has one single free parameter:

the average time between collisions, τ

Typical value in a metal: τ ~ 10-14 seconds

Drude theory of metals

Drude conductivity of metals

Between each collision, the electron accelerates, due to F = ma = eE.

Suppose there is an applied electric field, E.

average velocity vave = acceleration × time

e

eE

m= τ

( )( ) ( )ave

ave

number of electrons charge/electron N v t AI e

time t

∆= = ⋅

∆

The average current flowing through an area A in a time ∆t:

area A

N electrons/m3

distance vave ∆t

This is Ohm’s Law!

2

ave

e

Ne AI E

m

τ=

Drude conductivity of metals

Define: average current density J = average current

area

Plug in vave:

2

ave applied

e

NeJ E

m

τ=

Define: σ0 = Drude conductivity 2

e

Ne

m

τ=

ave 0 appliedJ E= σ

Units of σ0: ( )( )2 3 2

2 1

e

Ne meter C sec 1m sec C m volt

m kg ohm meter

−− − τ = = = ⋅

Note: 1 mho = 1 Siemens = (1 ohm)-1

Electric field in a metal

What if the electric field is oscillating? E = E0 e j(kz − ωt)

2 2 2

022 2 2

0

1∂ ∂ ∂− =∂ ∂ ∂

E E P

z t tcµ

Recall the inhomogeneous wave equation:

In this case, we do not have a polarization P, but

instead a flow of unbound electrons: a current!

How do we handle this situation? What do we use instead of P(t)?

P(t) = N·e·xe(t)

Recall that P(t) is proportional to the electron’s position:

Polarization in a metal

Replace dP/dt with the current density J(t) = σ0E(t):

2 2

0 0 022 2

0

1∂ ∂ ∂ ∂− = =∂ ∂ ∂ ∂

E E J E

z t t tcµ µ σ

P(t) = N·e·xe(t)

But the current is proportional to the average velocity of the

electrons:J(t) = N·e·ve(t) = dP/dt

So, as a guess:

(We will revisit this ‘guess’ in our next lecture…)

Wave equation for an E-field in a metal2 2

0 022 2

0

1∂ ∂ ∂− =∂ ∂ ∂

E E E

z t tcµ σ

This is almost the same situation we had when we first

encountered the polarization (lecture #7). So solve it in

a similar way.

Plug this into the above equation, and take the necessary t and z derivatives. The resulting equation is:

( ) ( )2 2

0 002 2

0 0

1∂

= − + ∂

E zj E z

z c

σωε ω

Assume that E has a time dependence of e−jωt, and that the z dependence is a slowly varying function multiplied by ejkz:

( ) ( ) ( )0,

−= j kz tE z t E z e

ω

( ) ( )2 2

0 002 2

0 0

1∂

= − + ∂

E zj E z

z c

σωε ω

Optical properties of metals

Thus, the solution is the same as the wave in empty

space if we let the speed of light be modified:0

0

0

1

=+

c

jσε ω

modified speed: c

This looks just like . So we interpret the denominator as the (complex) refractive index.

0=%

cc

n

0

0 0

1= = +%n j

σεε ε ω

complex index:

This is the same as the usual wave equation in empty

space, except for the factor in the parentheses.

But we have seen that a

complex index is just a

convenient way of grouping n and α together.

what is this value?0

0 0

1= = +%n j

σεε ε ω

Optical properties of metals

From this we can find the values of n and α:

( ) ( )0

4Re Imn n n= =

% %

παλ

which together tell us how waves propagate inside a metal:

( )0

020

0,

jj n k z

kz j tE z t E e e

+

= − = ⋅ ⋅α

ω

For real metals, σ0/ε0ω is a big number

Example: Consider copper at a frequency of 100 MHz:

100

0

10≈σε ω

In that case, the complex refractive index is given by:

( )40 0 0 0

0 0 0 0

11

2

+= + ≈ = =

%

jj

n j j eπσ σ σ σ

ε ω ε ω ε ω ε ω

in which case, the real and imaginary parts are equal:

0

02= =n

σκε ω

47.32 10= × for Cu at 100 MHz.

Skin depth

If the conductivity is high (i.e., σ0/ε0ω >>1) then

from κ we derive the absorption coefficient:

0

2

0 0 0

22= ≈c c

σ ωωκαε

As we have seen, this is a very large number for metals.

“Skin depth” or “penetration depth”:

depth of propagation of light into a metallic surface = 1/α

For metals, this depth is much less than the wavelength.

Example: copper at ν = 100 MHz

skin depth is 3.2 µm, about λ0/900,000

Vacuum (or air)

dielectric medium

metallic medium

Medium

Attenuation of waves entering a medium

Real refractive index of metals

If the conductivity is high (i.e., σ0/ε0ω >>1) then the

real refractive index is also large:

0

02= =n

σκε ω

47.32 10= × for Cu at 100 MHz.

For Cu at 100 MHz, this is: 0.9999455=R

metals make very good mirrors

We will soon learn that the reflectance of an object

depends on its refractive index:2

1

1

− = +

nR

n

Failures of Drude theory

There are several notable ways in which Drude

theory fails. These failures could not be

explained until quantum mechanics came along.

So, a tri-valent atom (like titanium Ti+3) should be a much

better conductor than a monovalent atom (like sodium, Na+1).

Here’s one: why is gold gold-colored and silver silver-colored?

Ti 5

0

Na 5

0

0.23 10 mho/cm

2.11 10 mho/cm

σ = ×

σ = ×But it’s not:

Here’s another: Drude theory predicts

2

0

e

Ne

m

τσ =

That is, the DC conductivity should increase if the

number density of electrons increases.