# Quantum Matter Lecture 16

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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
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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