### SUBTITLES:

Subtitles generated by robot

00:09

okay welcome back at the end of last
lecture we were we just derived the so
called NC density response function this
is the response of arbitrary physical
system to a potential perturbation at
something wave vector K about potential
couples to the density and then we make
a measurement later of of the density
and see how the system responds hence
density density now um before going on

00:41

we're good what we're going to do in
this lecture we're going to actually
study this this expression and find out
what its properties are for various
different systems of fermions during the
simple case of non interacting fermions
and the more complicated case of
interacting fermions but before doing so
I want to just review how it is we we
got here since it's a fairly complicated
formula so the way we did this is we
started with some H Hamiltonian which is
some H naught which is which are in

01:13

general Hamiltonian it could have
interactions in it or anything but we
separate out from it the perturbation
that we're trying to measure the
response to so there's gonna be some
Delta H and T which which is arbitrary
with the only constraint that it is weak
and that at time minus infinity it
should go to zero so we chose to work
with Delta H which is the following form

01:45

integral dr a hat of of r times phi of r
comma t where we chose this a hat to be
minus e times the density but really and
that meaning that this this phi was
viewed as the external potential but we
could have just as well worked with a
response a coupling of the
to an external vector potential a couple
of current and external vector potential

02:19

or or anything that we want we can
couple a couple any operator to some
external perturbation in the same type
of calculation goes through so it will
be a little bit more general here and
call this generally a this operator
generally a and then after sometime we
measure some operator B of our we want
to measure that as a function of time
and we chose to measure again the
density operator

02:49

so we're perturbing the system with a
potential which couples to the density
perturbed with a contention a couples to
the density and then at some later time
some later time here this is different
over here we some later time we again
measure the density at some position and
some time okay
now more generally than what we what we
calculated we could we could generally

03:21

write the the response of coupling to
any operator a measuring some operator
be having coupled to any operator a and
we would we would write again having
forty making a Fourier transform first
which is what we did last time we would
write B as a function of Q and Omega or
the change in B as a function of Q and
Omega is some response function Chi sub
ba of Q and Omega times the perturbation

03:56

Phi external Q and Omega and in our case
we're interested in both B and a being
the density I guess up to this factor
minus e but more generally following
exactly the same derivation we could
have derived that Chi ba of Q and Omega
is I guess we call the cade and we call
it K naught Q K wave vectors is

04:25

okay okay okay okay okay is then going
to be one over H bar volume of the
system sum over all excited states to
the system and then you have some matrix
elements ground state B hat of K excited
state excited state a hat minus K ground
state and then divided by a denominator
Omega plus I epsilon - yeah - e ground

05:00

state divided by H bar and then subtract
from this the same thing with the orders
reversed be ok we have K ground state
divided by Omega plus I epsilon minus
the ground state - VM divided by H bar
so this is what we want we derived in

05:31

the special case that both a and B are
the density operator and that gives you
the density density response which is
what we're going to actually be
interested in so this is this formula up
here so let's actually copy this formula
up here because that's what we're
actually working with but this kupo
formula approach is is far more general
okay but it's down here again okay now
this is formula that telling us the the

06:02

response of the system to a pervasion is
completely general the hard part is
figuring out what the energies of the
system are with the energy eigenstates
of the system are if the system is a
complicated system what the energy
eigenstates are and what the eigen
energies are so you need to figure out
what these and what these matrix
elements are with the energies and
that's that's typically quite hard so
we're going to simplify our life to do
it doing a simple example first that's
what we always do so the simple example

06:32

which is already non-trivial is
non-interacting electrons
directing we're not interacting fermions
so the Hamiltonian and H naught is just
going to be the usual PI squared over 2m
and the ground state of this Hamiltonian
ground state is just the Fermi sea
filled Fermi sea filled Fermi sea gosh
okay you can read that Brown says the

07:06

fill Fermi sea all right so the hard
part we have to ask about is what is
this matrix element up here going to
well we know we know the eigenstates are
of non interacting fermions we just fill
up appropriate momentum states and we
know what their energies are and so
forth but what's not so easy as figuring
out what these with these matrix
elements are so let's think about that
for a second so first of all we're gonna
need the n hat operator and we've

07:39

calculated that before in second
quantized notation we can write it this
way sum over Q and Sigma of sum over Q
and Sigma C dagger K plus Q comma Sigma
C Q comma Sigma and that is just the
Fourier transform of the density
operator well okay it's a sum over Sigma
Fourier transform of wave vector k scy

08:07

dagger Sigma bar side dagger okay so we
when we played with the second
quantization for bosons we derived
exactly this this expression for the
density operator wave vector K so anyway
yeah we vector K so what is this this
operator do when it is applied to the
filth firm you see so let me draw a
filled firm you see here it's supposed
to be a circle okay oh it's pretty
circle and then we start with a filled

08:43

Fermi sea and if we apply this density
operator at wave vector K
Phil Fermi see that takes a firm yawn
from some arbitrary state Q out of the
pharmacy and puts it up here at wave
vector K plus Q and in order for this to
be nonzero applied to the to the ground
state we must have Q less than K F and K

09:16

plus Q greater than KF otherwise
applying the density operator to
otherwise the term in the density
operator gives you zero because if you
try to transfer a Q to some other place
within the firm you see well you can't
do that because sorry filled or if you
try to destroy a part of a state that's
empty above the Fermi sea that also
gives you zero so the the excited States
will be interested in then these are the

09:47

ones you can get to from the ground
state by applying the density operator
are of the form C dagger K plus Q Sigma
comma times C Q Sigma on the ground
state right that uh those are the only
states that you can leave many body
states you can get two by one
application of the density operator on
the firm you see and that's the kind of
matrix element we're going to need

10:18

ground state one application of the of
the density operator to get us to an
excited state okay so the energies of
these particular excited states east of
m- e ground state is gone well this is
obvious what this is gonna be h bar
squared K plus Q squared over 2m minus H
bar squared Q squared over 2m well we
can write this as epsilon naught K plus

10:50

Q minus epsilon not Q okay so that now
gives us all the information we need in
order to we know what these energy
differences are here we know what these
matrix elements are
we know which excited States em we can
we can get to so putting that all
together and also I think we use the
fact that Chi of K Omega has to equal
Chi of minus K Omega by rotational
symmetry of the system I think we use

11:23

that as well we come to the oops come
back we come to the final result which
requires a couple lines of algebra you
just still have to write in the notes
but it's not too hard it's a good
exercise to go through some over a Q and
Sigma and F it's an F here supposed to
mean be the Fermi function it's
something that gives you but we're
really considering these things it's
your temperature so it's really a step
function and F here is a step function
it gives you one if Q is less than K

11:55

affan and zero otherwise
- and F absolute Q plus K less than KF
divided by denominator H bar plus I
epsilon minus epsilon not Q plus K minus
epsilon not Q so these Epsilon's down
here are their energies H bar squared K

12:25

squared over 2m epsilon plus T this is a
positive infinitesimal that we need to
regularize the in time when we wrote
things in terms of the response in time
the positive infinitesimally regularize
things at time equals plus and minus
infinity so this quantity here is
sometimes known as Chi not of K Omega
and it is the Linde hard response

12:58

function and hard I think I spell that
right response function which is just
the response but non-interacting
electrons to a or non attracting
fermions to
finite frequency and finite wave vector
potential being applied to it this was
calculated by by Linda harden in the
1950s for the first time which is fairly
late this in this index here upstairs

13:29

zero indicates that the interactions
Lippe electrons are not interacting the
fermions are not interacting so that
means non int that's why you calling
zero so this is an important important
equation so see if I can Circle it no
work it's nice and we're going to work
with that equation for much of this
lecture
oh here's what I wanted to do so this is
a Fineman diagram you know exactly what

14:01

I'm saying I may have just been clear
that page sorry about this
okay this page I'll be right exactly the
same thing in a moment so for people who
like finding diagrams it's it's useful
sometimes to we derive or rewrite this
this expression in in terms of Fineman
diagrams and if you don't know Fineman
diagrams or we have greens functions

14:31

techniques that's fine it's not required
at all for really complicated
calculations it's often useful to work
in diagrammatic stu keep track of your
perturbation series but for us it's
easier to do things explicitly but one
reason it's useful to present the
Fineman diagnosis because when you read
the literature you will often see these
finding diagnosis it's nice to know
roughly what they correspond to so for
the Fineman diagrams so Fineman oops

15:01

black funnyman diagrams actually a lot
of the diagrammatic work and using
diagram notation for for doing
calculations and not in quantum
mechanics but in statistical mechanics
was actually worked out
by maria gap admirer around the 1940s
and 1950s as well and they got imported
into the quantum-mechanical world later
I think that was the order of how things

15:37

were developed anyway so we need the
source the source is the coupling of the
potential to the density so the source
is going to be the density operator and
of K and we know at n of K is that's
well it'll have some over Q but they
looked terms will look like C dagger of
KP well notice k plus Q before C dagger
QK plus Q C Q and we'll draw it like

16:10

this so this is something coming in at K
or being measured K and we have a Q
coming in which we destroy and a K plus
Q coming out okay so you destroy a Q and
your coupling to the den Cu you destroy
a king couple to some external potential
the density it corresponds to destroying
the Q and creating a K plus Q then we
have these straight lines which are
propagators and they correspond to

16:43

greens functions and G of this is
labeled but with a K and an Omega this
will be G of K comma Omega that's
supposed to be a big Omega a couple make
out like that is one over
isn't that a watch company Omega I don't
think my symbol Omega plus I epsilon
sign you K minus EF okay if you're not

17:17

familiar with these greens functions
techniques don't worry about it but we
can write the Lindt hard response
function then as
a diagram which is a single bubble this
like this so as Omega and K coming in
omega NK coming out so so the system is
perturbed Omega NK it creates a particle
hole pair the particle the hole is at Q
and Omega the hole is that K plus Q

17:48

Omega plus Omega sorry the particle is
this guy's the particle going forward
this is the hole going backwards and
then you make a measurement of density
at Omega NK later and if you use the
diagram technique this will then just be
the sum over Q integral D capital Omega
of G Q Omega G of K plus Q Omega plus

18:16

Omega and up two factors of 2 pi and E
and things that I dropped in this rather
simplified version this gives you
exactly the same Linde hard response
that I did I do we derive the second go
okay all right so let's let's take a
look at our Lin hard response from which
I'm going to copy this I'm gonna want it
again so copy and put it down here maybe

18:54

just turn your page there we go
put it here oh come on there there okay
great great
okay so Chi not that's Linda hard
response so the Quinton let's try to try
to understand what this Linde hard
response means well first thing to look
at is when is the response imaginary
when is imaginary of chi-chi not not
equal to zero when does have a complex

19:25

part well why is why is it important to
find out when there's an imaginary part
well if you remember when we defined
conductivity a real conductivity it
means that energy is dissipated whereas
an imaginary
conductivity is means that you have a
reactive response that outer phase
response to the finite frequency
perturbation which doesn't dissipate
energy hasn't it dissipates no power
with whatsoever well it's the same thing

19:56

for this response function except in
this case it's the imaginary product
which allows energy can be absorbed and
the real part is the reactive part which
doesn't allow energy being absorbed so
so this is energy absorbed G absorbed
and let's find out when you have a
nonzero result
well that will recall that the imaginary
part of 1 over some X plus I epsilon is

20:27

minus I PI Delta function of X so that
means that you're going to have a
imaginary part when this denominator
it's 0 so imaginary part in Chi naught
is not equal to 0 when h-bar Omega
equals you know K plus Q - you know what
of Q this okay well what is that

21:06

actually so we should be a little bit
more careful here with this in order for
the numerator to me not to be nonzero
you have to have with also when also Q
less than K F and Q plus K greater than
K F or vice-versa if they're both less
than KF then you get 1 minus 1 and that
cancels and after both greater than then

21:37

KF then you get zero minus zero and that
cancels also so you have to have either
one one has to be greater than KF and
one has to be less than caft
in order to have a nonzero result so
what is this what is this telling us
this is telling us that let's draw our
firm you see again if I can do this okay
here's our firm you see pretty good
circle and we have some particle at Q
here and it gets excited to K plus Q

22:10

outside the Fermi surface here and the
energy that excites this is H bar Omega
okay
so this is what allows you to absorb
energy when when the two states
separated by wave vector K have energy H
bar Omega and your perturbing the system
at frequency Omega then that that energy
is is absorbed okay so it's a rather

22:39

natural and physical result and let's
actually draw where in the kale make a
plane this is actually occurs so I like
to draw with resultó so this is K and
this is omega on this angle
there's Komatsu off again so this is K
on this axis and this is Omega
on this axis and let's put two K F here

23:10

and what we have is we will have a slice
coming up like this
and the slope of this line is vfq or the
FK yes and then it curves upwards this
goes like this and in this region here
the imaginary part of the response is is
nonzero and outside of that region so
maybe I'll even write that p.m. with KY

23:45

not equal to zero in this region and
outside of that region the imaginary
part of the response it's nonzero
meaning that you cannot create any
excitation with that frequency and that
wave vector now let's do an example so
you can try to understand why it is that
you have regions in the space space
where it's not possible to create
excitation of a given frequency and wave
vector let's look at the case of Omega
equals zero and K greater than 2 KF

24:17

somewhere out here ok so here we're
looking for a low-energy Omega very
close to zero transition which excites a
particle greater than 2 KF you know is
that but if we look over here at a
picture of our Fermi sea oops Fermi see
here if we take any point under the
Fermi surface any point you want on the
Fermi surface and you excited a distance

24:49

2 KF it will is guaranteed to be outside
the Fermi surface after you after you do
that because the radius of the radius of
the Fermi surface is in the situs circum
diameter is 2 K F and so if I take this
point below the Fermi surface and I
excited by 2 K F it's guaranteed to end
up outside the Fermi surface and in
particular if I want to make an
excitation which is very low energy the
only way to make something that's very
low energy is I have to get something

25:21

that starts just below the Fermi surface
an excited just above the Fermi surface
so maybe from here to to there
okay from from this point just below the
Fermi surface at this point just above
the Fermi surface the total energy
difference between those is going to be
very very small so that's what i'm doing
here i'm making a zero frequency
excitation or a low frequency excitation
of exciting something from just below
the Fermi surface to just above the
Fermi surface but you can't do that if

25:50

you if you need to make K this arrow
here if you need to make K 2 K larger
than 2 1/2 because anywhere you take
this guy you know if it's if you have to
make move it by an arrow by a distance
greater than 2 calf then you're gonna
end up in an energy very far from that
from the Fermi surface and energy is
going to be nonzero so you're gonna have
to to make an excitation you have to be
up here at finite frequency okay
it's also worth looking at why is it bit

26:22

above some finite line here with
velocity of the Fermi velocity you if
you're at a frequency higher than this
you can't make any any excitation so
let's let's look at look at small K and
for small K we have we want to make
h-bar Omega minus epsilon k plus Q minus
epsilon Q we want this to be zero but
for K small we can't why can't so let's

26:54

write write this out h-bar Omega - I
guess H bar squared over 2m this is K
plus Q squared minus Q squared these are
vectors whereas K squared equals zero
and we want we want K to be very small
so we can expand so H bar Omega that K
squared term we can throw away because K
is small so for small K I'm small okay
throw away K squared term that Q squared

27:26

terms cancel each other you have a Q
squared here and a Q squared here those
cancel and what's left is H bar over m K
q like this which and want that to be
zero K naught Q at the right this h-bar
Omega minus h-bar over m I can write
this as K Q cosine theta and I want this

27:56

to equal zero but K has to be the
original K here if the frequency is
going to be small I know the original K
is going to have to be close to has to
be close to KF this has to be close to
sorry the original Q has to be close to
KF so Q has to be close to camp Q near
KF so I can replace this with h-bar

28:30

Omega minus VF K cosine theta has to
equal zero and the highest I can be f
the f K so I have that right we have K
cosine theta has to be equal to zero and
that means that Omega cannot be greater
cosine K F depending on the relative
angle between Q and K cosine theta can
go from plus 1 to minus 1 but Omega has
to be H bar Omega has to be less than or

29:04

equal to VF times K okay so that if I'm
back up to the picture up here that
explains why we have this edge here
above which at frequencies above the FK
you don't get any excitation either okay
so it's it's worth now looking at a
couple of take our favorite equation
here so this is our favorite equation
it's our favorite for today we're thick

29:37

also maybe we'll choose another for your
equation tomorrow this can be
okay it's worth looking at some limits
of this this expression really just to
convince ourselves that we're not going
crazy that we have a correct generally
correct formula that behaves the way we
wanted to
so the first limit I'm gonna look at is

30:06

Omega going to zero and small K and
small K okay so let's write this so Chi
okay small Omega equals zero we can
write this as okay this minus e over V
there's our sum over Q and Sigma and
then all right this is NF of EQ minus NF

30:38

e Q plus K and then the only thing that
we have left in the denominator is EQ
minus e q plus K okay
so this is so I threw away the Omega
piece and I'm assuming I can get rid of
the epsilon because I'm not it will put
it back if we need it and we so you'll
notice that I mean the energy difference

31:10

because Q is small because K is small
the energy difference in the denominator
is very small and so we have a function
of something which is a small difference
divided by small difference that's
that's the definition of a derivative so
this is a function of something minus
function of something plus something
divided by the the difference here okay
so this NF here is really a step
function step function so we get the

31:41

derivative of a step function so I'll
just rewrite that as e over over vine
the system sum over Q and Sigma of a
delta function
epsilon Q minus epsilon F is having
taken that limit and we can rewrite this
in instead of a q integral in terms of
an energy integral so the energy

32:11

integral the epsilon density of states
per unit volume and dead to Rivet the
delta function e minus e happens so I
switched here from some over Q to an
integral over over energy by using a
definite density of states and I end up
with e times density of states at the
Fermi surface and that's exactly the
static compressibility which we
compressibility of non-interacting
electrons ability which we calculated

32:43

previously you'll recall that we had
this expression DN is e density of
states the Fermi surface times the
applied potential calculate yesterday ok
so that's a nice nice limit we got that
got that right the second limit is is
perhaps more interested more interest
interesting will take Omega much greater
than EF a very very high frequency
response and it's actually worthwhile to
do this calculation in three different

33:14

ways
so each one teaches us something so the
first way is to plug it into our
favorite equation let's see if it will
so let me write that equation paste Oh
beautiful ok so we're gonna just take
that limit and plug it into this this
equation here so we want to expand we
want to treat Omega as being a large
parameter so doing that and we'll expand
the denominator out I naught is minus e

33:46

over our volume sum over Q Omega we
still have that NF q minus and 1/2 EK +
q is K divided by h-bar Omega that's the
leading term and then the correction
terms will look like
the first correction term 1 minus e q +
k- e q / H bar Omega I and then plus dr
i guess that's enough that's that's up

34:18

that's all that's actually exact still
and so the meeting term so this thing in
the denominator here is is is considered
small because Omega is large so the
leading term is just the one within
brackets but if you think about this for
a second the if you if you treated this
brackets as 1 then this sum vanishes
because both terms are gonna give me the

34:49

same same number and this will give me
the total number particles this will
also give me the total number of
particles they have a minus sign on the
whole thing divided by h-bar Omega but
they'll end up canceling if they did if
this is treated one so leading term
vanishes term vanishes and the sub
leading term we need to deal we need to
simplify things a little bit let's
simplify things a little bit by writing
e Q plus K and when is e K as K dot del

35:24

Q of epsilon Q and and similarly I guess
NF e Q plus K minus NF e Q will be K dot
del Q and F absolutely Q and then
multiplying out this this term to first
order is term to for first order we end

35:56

up getting Chi not is e over V this is 1
over H bar Omega sitting out front here
and then we have a sum over Q and Sigma
we get the upstairs here is I'm going to
rewrite now
as k dot gradient Q and F I'm just like
you and then the second term is going to

36:28

be this expanded which is now just gonna
be this term and then divided by h-bar
Omega so I get okay
the gradient Q epsilon Q divided by
h-bar Omega so well okay so the other a
this H bar Omega
oops contains four Omega comes out to
here so it's now H bar Omega squared and
then this integral this this the sum I
can turn into an integral and then once
it's an integral I can integrate it by

37:01

parts so that it so that I move the
derivative from the NF on to the epsilon
Q and this gives me equals minus e
Volume one over H bar Omega squared
squared sum over Q Sigma n F of epsilon
Q and then times K dot gradient Q
squared absolutely absolutely Q this K
dot gradient of Q squared is actually on

37:32

epsilon Q since epsilon and Q here is
quadratic and Q this term here is it's
just gonna give me a scalar and the
scalar it gives me is actually H bar
squared K squared over over 2m and then
that means that can just do this sum I
can put brackets around this and do this
sum and get the total number of
particles and at the end of the day what
I get is minus K squared and energy over
m Omega squared okay so this is my first

38:04

attempt at calculating the response at
finite at high frequency K squared
density divided by times e over m Omega
squared so it turns out that this result
is actually extremely general and we can
argue that this must be the same result
must be the same
this is unusual but it's true in the
high frequency limit even for
interacting electrons even for

38:37

interacting fermions interacting
fermions and in order to see this okay
so we call this to do we call this to a
no he didn't okay okay we're gonna we're
going to do the calculation again in a
different way okay so we're gonna go
back several steps in the derivation of
the Kubo formula Chi of K and Omega we
had it in this in this sort of in this
form of an integral over over time dt e

39:13

to the I Omega plus I epsilon T and then
the correlator and hat of K comma T and
commutator and hat minus K 0 expectation
okay so this was you know on a way to
deriving the density response function
this was one of our train stops along
the way the N hat of K comma T we well

39:44

that's just these are in the interaction
representation but we can write that out
as e to the IH T over H bar and hat okay
the minus i h t over over h bar this is
what we called it h naught but that H
naught did not necessarily mean not
interacting so whatever its H is if

40:16

include it includes the interaction
doesn't include the Probation it just
concludes the interaction and if we're
going to be interested the key here is
if we're going to be interested in high
frequency that's equivalent to thinking
about small time so high Omega is small
T so expanding this expression for small
T I then get

40:43

and and I K comity is an hat of K plus I
T over H bar and the leading term is H
commuted with n hat okay okay and then
they'll be higher order terms so
plugging this in to our commutator term
here we then get KY of K Omega is
integral 0 to infinity D T here the I

41:19

Omega plus I epsilon t there's an i T
over H bar and we have a commutator H
naught and a hat of K and then that
commutator is commuted with n hat minus
K okay and this is a double commutator
and we've seen double commutator z' of a
Hamiltonian with density before and even
if we recall you know when we wrote down

41:51

before we did this for for bosons and
it's exactly the same for fermions that
H naught is you know a P squared over
over 2m the kinetic term there might be
a vor
and there might be some over I less than
J of you RI minus RJ and the point here
is that these operators n hat they
hidden in them is position operators are
and so they commute with the V and the
you and any interaction terms any
interaction with the potential and the

42:23

interaction with other particles the
only part that doesn't commute is the
kinetic energy term so we can calculate
the kinetic energy term this double
commutator is actually just a scalar
commutator and you did this before is H
bar squared K squared density divided by
M and then all we have to do is so this
whole double commutator expectation
value up here is is just a number so

42:53

this double commutator is actually just
just
number and we can plug it back in up in
our favorite favorite formula up here
the kubo formula do the integral now a
trivial integral and we end up with the
final result
H bar that Chi the response at frequency
K and frequency Omega and wave vector K
is just there's a minus e um probably

43:24

somewhere in this formula up here I
missed it there was probably a e over
volume yeah sorry
we up here I missed a I lost track of
the idea over H bar and the volume so
let me put this back in oh don't do that
so I mean with this over that was over
here
I missed equals ie over H bar volume is
supposed to be there so we get minus e

43:56

density K squared over Omega squared
which is exactly the same result we had
when we calculate explicitly from the
Linhart calculation and this this result
is sometimes known as the F Summerall
and it dictates that the frequency at
high enough and if you go up to a high
enough frequency then the the response
function always takes takes the same the
same rule the same the same form

44:29

independent of the details of the system
independently interaction between
particles interaction between
independent of any potential that might
be applied to the system as well and and
so the intuition behind this is that if
you if you perturb a system very quickly
in the first instant after the
perturbation the system doesn't remember
that it's interacting with anything each
particle thinks it's free and it
responds immediately to to the

45:01

perturbation as if it were a free
particle and only a little bit later
does it remember oh ma
with a particle of it's over there as
well so that's sort of a cartoon picture
okay we're now going to derive exactly
the same result yet a third way from
judah theory
so remember drew the theory is just the
classical behavior of particles in a
charged particles in a electro electric

45:31

or electromagnetic field here we're
going to use to our advantage that omega
is very very large it's much greater
than one over any any scattering time so
we can throw out the scattering time we
don't have to worry about scattering
time at all this is just like we did
with a superconductor we threw out the
scattering time altogether and we're
going to calculate the response of of
the system under these under these
circumstances we're going to see that
gives us the same result so we're going
to need a couple couple pieces the first
is we're going to need current
conservation Delta n dot plus the

46:04

divergence of the current equals zero so
this is the particle number density and
this is the particle current density
will move into Fourier space so we have
I Omega Delta n plus I K dotted with
j-jane the perpendicular in the powell
of direction J in the direction parallel
to K is zero so maybe I should write as
dot I'll just write it as oh my gosh no
no stop stop stop okay

46:37

fine point here I k j parallel so
magnitude K and then J in the direction
parallel to K is probably a nice way to
write it that allows us to write the J
and J in the direction parallel to to
the wave vector K is just Omega over
over K times Delta n and then we're
going to use the Drude equations of
motion as well which is M well Newton's
equation MV dot is minus B times the

47:09

electric field will also use that J is
the average density times the velocity
you might say well shouldn't we write
this is
and is n bar plus Delta n because and
can can change the function of position
and the answer is I don't need to do
that or I shouldn't do that because
these already first-order and small
right until 2n is first-order and small
if I tried to expand this to n bar the
average average density plus first-order

47:42

and small I'd be keeping track of terms
which are second order and small and I
only need to keep track of things that
are first-order and small so average
density times velocity is good enough
and that allows me to plug in to
Newton's equation over here to give me m
j dot Cousineau jello or something is
minus e times the average density times
the electric field which is nice to

48:11

write as e n bar times the gradient of a
potential and then we can take J dot and
write it as I Omega minus I Omega J and
that will and then over here on this
side we can write this as minus I K e
n bar Phi so I turned the gradient into
minus I K and this then I'm going to
plug in this expression for J over here

48:49

so this J is then Omega over K Delta N
and in the end putting me all these
together
I'll get Delta n is minus K squared and
bar e over m Omega squared times 5
exactly the same result okay so you can
get exactly the same result just from do
to theory
okay so we've studied now interacting
non-interacting electrons we studied
completely we know from the Linhart
equation we know everything we need to

49:22

know about non interacting fermions in
their response to an external frequency
and wave vector independent perturbation
but now we want to return to interacting
electrons so what how about how to treat
interacting electrons interacting
fermions okay so this can get quite
complicated but there's a fairly

49:56

straightforward method which which we
can handle here without too much
additional trouble so we'll start by
writing the response of non interacting
Q Omega Phi external of Q and Omega and
then once the system responds to the
external potential we are going to allow
the possibility that there will be an
induced potential at R and T which is

50:31

just given by integral TR prime V of R
minus R prime Delta n of R prime of T so
in other words that our scheme here is
going to allow the system to respond as
if is not interacting but then once it's
respond and it builds up some sort of
Delta N and then a an additional
potential is induced by the change in N

51:03

and we're gonna allow the system to
respond to that as well so I guess if we
if we go to Fourier space Fourier this
can be written as Phi induced of Q and
Omega is just V of Q V twiddle of Q
Delta n Q n Omega because this up here
is a convolution so and V here is the

51:31

Fourier transform Fourier transform of V
of R
and now the idea is to let the system
respond let system respond with kind not
as if it's not interacting particles but
we let it respond to chi not to both
external and induced five okay so this

52:09

is the this is the this is the trick
we're going to use we're going to we
have we calculated the non interacting
response then the nut and then the non
direct response builds up some with some
potential and the potential sorry pin
builds up some density the density gives
an induced potential and we're gonna let
this system respond with this kind on to
both the external and the induced
potential okay so the Delta n is going
to be kind not by external plus Phi

52:41

induced and Phi induced we already
decided was it's just the twiddle times
Delta n and then we can solve this
equation Delta N equals Chi Chi not if I
stir all must be twiddle Delta n to give
us
well can't move Delta n to the Delta n
both delta ends to the other side one
minus Chi not be twiddle equals KY not

53:15

by external this and so Delta n is then
a cannot over 1 minus Chi naught V
twiddle times Phi external and this
quantity here is our total response
function the response to the external
potential and it's known as Chi R PA
it's kind of
or one minus KY not v-twin law an RPA is

53:48

known as random phase approximation and
do not worry procs do not worry about
why they call it meant in phase
approximation is totally a historical
accident of the the people who derived
for the first time Bowman Pines in 1952
they you know there was some calculation
complicated calculation whereas if you
randomized over certain phases you would
get this result no one thinks in those
terms anymore it's just the name stuck

54:20

it's equivalent to self-consistent
Hartree consistent in this case time
dependent are tree so remember the idea
of our tree approximation is that you
know you recalculate the system and you
treat the potential due to all the other
particles and and you sort of iterate
that and this is exactly what we're
doing here each you know the potential
is built up and you recalculate the

54:50

response to to the potential that is
build up due to the density of the other
particles okay
so another way to write this is to write
this as Chi RPA let's expand out this
series we can write it as Chi not plus
KY naught V twiddle KY naught plus Chi
non v-twin l cannot see twiddle KY
naught plus dot and find the diagram
language we would write this as a bubble

55:22

this plus a bubble and then the bubble
is connected to another bubble by by a
Levy's in between and then we have three
bubbles so we get a chain of bubbles
there's a bubble chain and so forth and
the interpretation of this bubble chain
is is it's fairly physical you can say
well from the first term you're looking
at the direct response of the system in
the second term you're looking at the

55:55

system
bonding then it builds up some density
and the system responds to that in the
third term the new the system responds
builds up some density system responds
again built up some density Sutton
responds again and basically summing up
this series out to out to infinity okay
good okay so let's try to apply this to
some interesting cases or a particularly

56:25

interesting case so at high frequency
let's consider the case of high
frequency where we already calculated KY
naught is minus e and Bar K squared over
M Omega squared then we use the RP a
result at high frequency Chi not over 1
minus V tweedle Chi not RP a here what
we derive from this is that this thing

56:55

diverges diverges when well when V
twiddle Chi not equals equals equals 1
so let's figure out when that is so V
twiddle is just e over epsilon naught K
squared and KY naught is e and Bar K
squared over m Omega squared when that
thing equals 1 that condition is the
same as Omega equals the square root of

57:26

e squared n bar density over epsilon
naught m and this is known as the plasma
oscillation what does it mean that the
response function is is is diverging
what what it means when you have a
diversion response function it means
that there's an excitation mode there
why if you drive something resonantly
with its excitation frequency frequency
you can drive it infinitesimally weakly

58:00

and get a large response so when you
have a divergent response function it
means you know you have
a crisp sharp excitation mode that
you're that you're exciting so good if
you go to a frequency much much higher
than the then the plasma oscillation
frequency you'll recover this F some
form again but at lower frequencies you

58:33

discover that there it is a an
excitation mode at this plasma
oscillation frequency now there's a
rather simple picture of what plasma
oscillation actually is and it goes kind
of like this you start with a chunk of
your metal with which we think of as
having a background positive charge and

59:04

mobile negative charge and then you
displace the charge imagine displace
charge by some distance X in the X
direction and then what does it look
like well it looks like the following
you have here your chunk of metal again
but the so X is supposed to be small and
draws really big so this distance here

59:35

is X and we've built up the charges on
so if we if we moved the negative
charges let's see if we move the
negative charges in in I guess the
negative charges are built if we move
the negative charges in this direction
then we have minus charges built up over
at this end where where you push the
charges to and you have positive charges
over here where you remove the negative

01:00:08

charges and you're just left with the
background charges so the the surface
charge density the surface is the
average density
times X times the charge on the on the
electron okay so this is what we've done
by moving the charge is we've actually
built up a surface charge or a capacitor
and will recall from Gauss's law Delta e
equals Rho over epsilon naught that the

01:00:38

electric field is surface charge density
divided by PI epsilon naught and per
pinpointing perpendicular to the to the
surface the force on the charges is n
minus e surface charge density divided
by epsilon naught which is minus n bar X
because that surface charge density is n
bar x times e so that's too big no this
becomes a chi squared divided by epsilon
naught and we set the the force in to be
MX double dot I'm just using Newton's

01:01:11

law and examining now this equation we
realize that this is a harmonic
oscillator motion with frequency n bar V
squared over epsilon naught M and it's
important here to note that this mass
here is the bear mass of the electron
bear mass on the electron mass it's
not-me normalized by any interaction

01:01:42

between the electrons because because
the what you're really doing is you're
displacing the center of mass of all the
electrons together and if you know the
interaction doesn't care about that the
interaction energy is going to stay
exactly the same as if you move all of
the electrons together there you know
the interaction energy whatever it is
before he moved it it's the same after
you move it okay so so that's one result
a nice result out of RP a one can

01:02:15

calculate the full RP a response
function Chi R PA and I'll sketch out
what it looks like here's K and here's
Omega
and we have the same roughly the same
picture here this is 2 K f ck f and you
have excite excitation modes here

01:02:45

associated with particle hole
excitations but then up here there's an
additional sharp mode that Omega plasma
plasma frequency so in addition to this
particle whole continuum there's one
sharp mode up here at the at the plasma
frequency which then enters the part of
the whole continuum up here down here
again we have frequency being VF times Q

01:03:18

although maybe when we do a more
detailed derivation of the of the
response function maybe we'll discover
that VF should go to some VF star
associated with a mastery normalization
which we'll get to in the next next
lecture okay so the RPA approximation is
sort of the first resort so the simplest
approximation you can you can come up
with which gives you a full finite

01:03:49

frequency in finite finite frequency
pilot wave vector response of a
fermionic system and awful lot of
physics it's buried in it and it's
fairly actually it's fairly simple to to
implement all one has to do is calculate
the the non-interacting response and
then divided by one minus V twiddle Chi
not the static version of the RP a
response is actually exactly the Thomas

01:04:22

Fermi screening calculation that we did
last in last lecture I guess that's
covered in that maybe that's a good
exercise to to convince yourself of that
this is really a dynamical version of of
Thomas Fermi screening okay we'll stop
there and pick up again
next time

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