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,

/20( , ) ( 0)e exp[ ( )]zE z t E z j nkz t

Absorption causesattenuation of the field with increasing propagation distance z

Refractive index changes the k-vector and therefore the wavelength

0

0

medcnc

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

Frequency doesn’t change!

Often denoted with the Greek letter kappa, , a dimensionless quantity!

Complex refractive index

020, 0 jnk zz j tE z t E z e e e

Define the complex refractive index: 0

0

0

2

4

2

jn nk

n j

cn j

0

020 0

jj n k zk j tE z e e

k0 = k-vector in vacuum

Note the tilde!

Complex index: this is just a convenient way to group nand in a single quantity.

So n n j

Note: 0 always!

Complex dielectric functionIf the refractive index is complex, what about ?

20

20

med n

n j

2 20 2R Ij n j n

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

0

0

12

12

R

R

n

This can also be expressed as equations for n,:

n

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

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 << n (as is often the case), then

20 Re n which is related only to

the phase of the waveNote: 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,

Typical value in a metal: ~ 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 timee

eE

m

aveave

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

avee

Ne AI Em

Drude conductivity of metals

Define: average current density J = average current

area

Plug in vave:

2

ave appliede

NeJ Em

Define: 0 = Drude conductivity 2

e

Nem

ave 0 appliedJ E

Units of 2 3 2

2 1

e

Ne meter C sec 1m sec C m voltm kg ohm meter

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

Electric field in a metalWhat if the electric field is oscillating? E = E0 e j(kz t)

2 2 2

022 2 20

1

E E Pz t tc

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) = 0E(t):

2 2

0 0 022 20

1

E E J Ez t t tc

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 20

1

E E Ez t tc

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

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

0, j kz tE z t E z e

2 2

0 002 2

0 0

1

E zj E z

z c

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

j

modified speed: c

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

0

ccn

0

0 0

1 n j

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 together.

what is this value?0

0 0

1 n j

Optical properties of metals

From this we can find the values of n and :

0

4Re Imn n n

which together tell us how waves propagate inside a metal:

0

0200,

jj n k zkz j tE z t E e e

For real metals, is a big number

Example: Consider copper at a frequency of 100 MHz:

100

0

10

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

40 0 0 0

0 0 0 0

11

2

j jn j j e

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

0

02 n

47.32 10 for Cu at 100 MHz.

Skin depthIf the conductivity is high (i.e., >>1) then from we derive the absorption coefficient:

02

0 0 0

22 c c

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/

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

Example: copper at = 100 MHzskin depth is 3.2 m, about 0/900,000

Vacuum (or air)

dielectric medium

metallic medium

Medium

Attenuation of waves entering a medium

Real refractive index of metalsIf the conductivity is high (i.e., >>1) then the real refractive index is also large:

0

02 n

47.32 10 for Cu at 100 MHz.

For Cu at 100 MHz, this is: 0.9999455R

metals make very good mirrors

We will soon learn that the reflectance of an object depends on its refractive index:

211

nRn

Failures of Drude theoryThere 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 50

Na 50

0.23 10 mho/cm

2.11 10 mho/cm

But it’s not:

Here’s another: Drude theory predicts2

0e

Nem

That is, the DC conductivity should increase if the number density of electrons increases.