Quantum Matter Lecture 6

Quantum Matter Lecture 6

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00:09
welcome back where we left off last time we were talking about interacting bosons from a more microscopic standpoint I'm gonna review a couple of things we did at the end of last lecture and we'll continue on from there just to remind you where we were we wrote down a Hamiltonian for our interacting bosons we can write in first quantized notation it was sum over particles this is a single particle kinetic energy then there's a single particle potential
00:39
energy as well and then there might be an interaction term I guess the interaction term we were using was you over to sum over i J we used to Delta function interaction or I'm sorry J but in general you could use more complicated interaction we rewrote rewrote that in second to quantize form as integral dr and then we have side dagger hat operator the kinetic energy
01:10
operator 2m plus v bar yes this is a function of r also and sie hat of our tamar and then plus you over to an obese i dagger are like this and so these are
01:44
the Hamiltonians we're working with that we chose to Delta function interaction because it's reasonable and it's it's easy as well and we started by choosing a trial state which was a coherent state we wrote Alpha Phi we're going to multiply occupy the the Phi or mole which is just e to the a dagger Phi on on the vacuum state and the advantage of working with such a coherent state trial
02:15
state is it turns the sigh operator into a complex number which we also called bar which is alpha 5 our our health is just square root of the number of number of particles in the system what we then calculated was the expectation of the Hamiltonian which just with a simple integration by parts well basically it just converts all of the sy operators in
02:49
this expression for the Hamiltonian into size scalars and then we do one integration by parts to rewrite it the del square del Phi squared instead plus v of r sy squared plus nu over to the fourth and then we minimize or extra mais the energy of the system with respect to sy to get the gross pdfs key
03:22
equation GP or ginsburg Landau equation or non-linear Schrodinger equation equation which is minus h-bar squared l squared over 2m plus V of R + u sy squared applied to side of R equals zero as we mentioned before this is just a nominator Schrodinger equation it's the leading term over here just looks like
03:54
the regular Schrodinger equation with the eigen value absorbed in the V of R and then this term is the interaction between particles okay so in the disadvantage of the advantage of using the coherent state is that it turns it turns operators into into scalars and that's very convenient the disadvantage is that now we're working with an indefinite number of particles and we asked at the end of the last lecture if
04:25
we could have done things in an easier way and indeed we could have done things an easier way we could have also also we would get the same result if we write a trial state and we can do that everything now in first quantized notation the trial state sigh this is a product I equals 1 to N of Phi bar I calculate the energy of the expectation and the energy of the
04:54
Hamiltonian we get exactly the same we get exactly the same Hamiltonian where we define again the sigh to be square root of n times Phi here so we'll get the same same expression for the expectation in the Hamiltonian and at least in larger limit we get the same expression for the from the Hamiltonian we can differentiate the Hamiltonian with respect the field again and get
05:26
them down in their short equation so either approach either way so either way we are getting the best possible single orbital single orbital which we call Phi here to multiply occupy so solving the nonlinear Schrodinger equation of Gordievsky equation gives us an expression for the best possible
06:05
single orbital wave function that we can multiply occupy for this interacting bose gas now our natural question that I raised at the end of the last lecture is if we could do everything in first quantized notation here and get exactly the same result as we did when we had a coherent state up here why did we go through all the effort of working with coherent States where we have you know
06:37
indefinite number particles in the system and so forth and we have to worry about second quantization and the reason for this is because on very generically even when this simple coherent state form or this simple product state form are not particularly accurate representations of the ground state we can still choose sy hat of our expectation to be the order parameter to
07:07
represent the the wave function of the superfluid the quantity that goes to zero at at the critical temperature and represents the wave function below that below the critical temperature and the reason why these wave functions these trial state wave functions that we've used are not accurate is because the the interaction between particles inevitably is going to kick some of the particles out of the at the lowest out of this
07:42
single eigenstate so you try to multiply occupy the same eigen state many many times but the interaction between particles then kick some of the particles out of that of that orbital and so we already know that the the wave function we're using is not going now this this is where as we start today's lecture the idea of using the expectation of annihilation operator as the order parameter or the wave function of of the common set and we'll
08:13
do some examples of this as we go on now most people in the field agree that the use of the expectation of annihilation operator is an order parameter is very effective for describing superfluids there are some people who feel like this is not really a physical thing to do and one should avoid using this object as an order parameter and one person who is very vocal about this opinion is Tony
08:44
Leggett Nobel laureate he is um he won his Nobel Prize for work on superfluids actually for superfluid helium 3 we can only three is a Fermi on it but the helium-3 particles they pair up to form oppose on very much like superconductivity and becomes a superfluid at extremely low temperatures I think we'll discuss that a little bit more later in the in the Electric course so he's a real expert on super fluids and he really feels that describing your
09:14
superfluid in terms of this expectation of the annihilation operator is is just it is wrong it's deceptive and why is it that this is wrong the reason he feels this is wrong is that you can certainly have a physical system with a fixed number of particles if you have a system with a fixed number of particles the expectation of the annihilation operator is strictly zero because you know if you take a wave function on the right hand
09:46
side over here you apply the annihilation operator it changes the number of particles then if it has a fixed number and over on the right hand side it will have n minus 1 of them on the left hand side and the and the expectation of this will will then then be 0 now the way we get around this is we say well we're gonna work with some circle here in state which is a mixture of different numbers of particles but he said well you don't have to do you still have super fluidity in systems with a fixed number of particles so you're obviously missing some of the physics if
10:16
you if you try to use an order parameter that's not conserving numbers of particles fortunately there is a way around this problem which is to think in a little bit more complicated ways so I'm going to go through that because I think it's actually fairly enlightening to try to understand how you describe this order parameter of the of the condensate even if you're if you're forced to have a fixed number of particles in your system so the first thing we're gonna do is we're gonna
10:47
define the so called one body density matrix density matrix I should probably tell you a little bit about twenty like a tenant Tony Leggett is rather brilliant physicist he was an undergraduate at Oxford and he studied when he was at Oxford as an undergraduate he studied classics but then at the end of his his classics degree he convinced a tutor to take him
11:18
on as a physics student as as well he did the physics degree and I think did the physics degree in two years rather than three years he said he struggled with doing it but then went on to graduate school where he was essentially orphaned by his his supervisor who didn't pay much attention to him but he did quite well anyway by finding his own path studying studying superfluids there's a great story in his his noble art autobiography about how it when he
11:47
was a.m. he was a starting student at Oxford like everyone else he he went out for rowing and so he joined the boat and he thought he was doing really well in in the in this boat and then they discovered that he was actually a lot lighter than the Coxon so they switched him and he became the Cox and he thought that was a lot less interesting than actually rowing and anyway so this is sort of the story of physicists anyway
12:20
so he spent his time doing physics instead of studying physics instead of it's not rowing which wasn't a bad anyway so we'll start with the one body density matrix this is the technique which is espoused by by Leggett in particular so it's it's a function of two positions R and R Prime and I'm going to find it as the expectation of side dagger of our prime sigh of R so it
12:52
removes a particle from position hardware puts it back in position our our prime so this operator can have a non zero expectation value a little bit better says it not can have a non zero exponent K that's not a whole lot better we try it one more time it can draw this maasai ok so this operator has
13:21
a non zero expectation value even in a system with a fixed number of particles it removes a particle but then it puts back again but locally it doesn't concern number if R and R prime are very far apart from each other it can remove a particle from position R and then very very far away it can put a particle back in to position our prime we can write this operator in first quantized notation and very often you know we as theorists we like to work in second
13:54
quantized notation because it's very very convenient for doing a lot of complicated calculations but often you can see the physics more readily in first quantized notation and I'm not sure if this is enlightening or not lightning but let me give the xbm the expression for the density matrix in first planted notation so it's an integral d r2 through dr and the end
14:24
particles in the system of size star the mini body wave function where are the first coordinate is at put at position R Prime and then R 2 through our and then sign anybody protocol wavefunction sigh and here the first coordinate is present position R and then these are all at R to R M okay so this is what the one body density matrix is in terms of many-body wave functions that you keep are two through RN fixed
14:55
and integrate over their positions and you put sy at position R inside a star at position R prime for the first particle okay we can even do this at finite temperature if we like finite T we would write one over the partition function the system sum over all the eigenstates to the system e to the minus beta energy of the eigenstate and the same integral TR to 4dr and then we'll have capital sy star of
15:27
the nth eigen state our prime R to do our n and sy r and then R - okay so good so this is the expression for the one body density matrix and one thing to notice is that Rho 1 if I fix both coordinates at the same position R this is just regular density regular density
16:04
see if I can draw them write that a little bit nicer regular usual density Rho bar okay so if I put R in our prime at the same position this will be side at your side at both of the same position or integral over all the I mean the size squared here will be size squared which is the the face base 10 ceiling integrate over R 2 through RN and you just look measuring the density
16:36
of the first coordinate okay um good so it's sometimes useful to think about let me write again Brar one of our comma R prime is sometimes useful to think of think of our and our prime as discrete variables discrete discrete variables in other words they live on a lattice and
17:10
the lattice might even be extremely very very small spacing or something like that and you can take a limit of what this Basin gets extremely small but our prime an R so it's just it just makes a lot easier to think about discrete coordinates than it is to think about a continuum of coordinates if you do that then this this object here row of our row 1 over R prime comma R is then a matrix then this thing equals a matrix
17:43
in fact it's a relation matrix its matrix is a function of two indices this is the first discreet index this is a second discrete index and bro one of of R comma R is diagonal because diagonal of the matrix then when R is not equal to R prime that's off diagonal so
18:24
I'm terrible here because I draw ours in two ways I sometimes do this as are and sometimes I do this is our and I really apologize about that it's it's I'm not sure why this is habit but it but this is how I do it anyway so this is our and our prime so for this when our it's not equal to our prime this is off that angle okay so if we have a hermitian
18:59
matrix in any number of dimensions is diagonalizable so we have a diagnosable or each permission matrix and let's write out its its its eigenvalues so it's call its eigenvalues let's call it n sub alpha and let's order it that n 0 is the largest eigen values greater than and 1 greater than n 2 and so forth and the eigenvectors of x will call them phi
19:33
alpha and they'll be normalized the way eigenvectors are usually org on normalized that well in a in when the discrete coordinates it would be sum over I PI alpha okay so this is maybe some over are some of our positions are I should write it this way sum over positions are Phi Alpha R squared equals
20:06
1 but if we go to a continuum when we eventually want to go to a continuum it would just be integral okay so we can then write in terms of you can write your recent matrix in terms of its eigenvalues and eigenvectors in the in the usual way so we'll write what Row 1 of R comma R prime is it then going to be sum over alpha and alpha phi alpha star of our by
20:42
alpha of our so this is just the usual eigen value eigen vector decomposition of a sigh this one's R prime of a of a hermitian matrix okay now in terms of this one body density matrix the DEF definition of superfluidity is that and zero the largest eigen value
21:15
is order order and the total number of particles in the entire system and then Phi naught is the wave function of the condensate condensate with the usual translation that we write the order parameter psy as square root of M naught times final okay so this is the the new definition of the of the order parameter
21:49
so these are all a function of position this ok now I'm doing that thing with different writing the arms in different ways okay well so this is the U this is the definition of the order parameter we can and we can use this even for non interacting sorry even for interacting bosons no matter how complicated the wave function is we can always define the one body density matrix we can the one done by a density matrix is always hermitian matrix we can always
22:20
diagonalize the hermitian matrix and we can write it's the largest eigenvalue if the largest side value is order the number of particles in the system the corresponding eigenvector is then the wave function of the condensation - this is normalization so we can do an example of this a really simple example of this example is if we go back to the non-interacting VEC now throwing out the interesting stuff we're getting rid of interactions again let's
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do a non int B C at T equals zero so our wave function or many-body wave function is then e to the Alpha a dagger Phi 0 so I'm multiplying the same the same same orbital and then while okay we'll go back to the original definition Row 1 of Arkham our prime is then side dagger our
23:25
sy prime and in in the usual way aside these are these are operators these are operators these second quantized operators here in the usual way since we're using a coherent state the second quantized operators just get turned into numbers so we get psystar of our side our prime and then we can write this as as n not the total number of particles
23:57
in the system Phi naught star R prime so this is maybe not surprising that that we that we recover the same definition with the same result that sy our order parameter is square root and naught times Phi naught and and that's what we how we define the order parameter previously for for non-interacting Boza bosons but and
24:31
actually let's let's do a simple a simple case so the case is if we have a translationally invariant system so if we have translationally in vans ley tional in variant in other words v of r equals a constant then our wave function is just 1 over the square root of the volume of the system the lowest energy wave function for a b c is just a constant wave function if you have
25:03
periodic boundary conditions so maybe the translational invariant periodic BC's okay so now we have a problem here this would be is the potential up here this is the potential is a constant and there's a V is the volume of the system so sorry about that notation in which case of Row one is of R comma R prime is Jen then just n over V and which is the
25:36
density and this is independent R prime okay okay that's maybe maybe a little bit not exciting more interesting is if we consider a b ec at finite temperature at finite e okay if we consider a beat you see a finite temperature then Row 1 of R comma R prime will be well let's again write it in second quantized
26:06
notation side dagger r sy prime and it's convenient now to to write sy of R as an operator in terms of plane wave States so I want to reveal sum over k a k bi K dot R so that Rho 1 is it's real one of our come on our prime is then one over the volume of the
26:41
system this is all so it'll be sum over K and K Prime expectation of a K dagger a K Prime and then e to the minus I K dot R plus K prime R prime okay now a couple things to notice that that this expectation by translational invariance in what has to be proportional to Delta KK prime if you remove something with
27:12
momentum K prime but you better put it back with the same momentum in order to get something nonzero so the expectation in a translational invariance system the expectations can only be nonzero if when you remove some of the meant I mean put them lent them the momentum back so that means I can rewrite this as one over volume just sum over K a dagger k a k K
27:44
prime and this expectation now a dagger K a dagger a at momentum K is just for non interacting Bose gasifier temperature this is just the bose factor and bose of energy of ke and why does the chemical potential okay now so maybe let's we write that out one volume sum over K and Bose
28:16
okay - potential I guess times e to the minus I K dot R prime good now for case one is if T is greater than the than the critical temperature then if T is greater than the chemical potential then the Bo's factor is is completely smooth
28:47
so NB is smooth near K equals zero nothing special about the K equals zero State if if you're above the critical temperature and so we have Row 1 of R comma R prime is the Fourier transform transform okay I guess what this tells us here is it's it's only a function of R minus R prime actually right if we
29:19
look up here the role of our - it's a whole whole expression is Rho of R comma R prime here it's just a function of R minus R prime by translational invariance so this is going to be transporting a transform of some smooth function again how smooth function the buzz factor which is a smooth smooth function so when you have a Fourier transform of a smooth function this will be some
29:55
smooth function decaying decaying for large R minus R prime if you can read my writing better than me sometimes I write things I have no idea what it is I've written however case two is what happens if we have here's the expression again we're evaluating as small as you can see it okay here's the expression where we're
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evaluating here the bhojas factor of beta and that's inverse temperature energy minus mu and then the Fourier transform factor so for T less than TC then we know that expectation of a dagger not a naught is and not of T which is large macroscopically large the number of bosons in the zero state is a macroscopic fraction so in which case in
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this expression here in the sum over over K the K equals zero state is special it's macroscopically occupied and all the other all the other modes are not microscopically occupied so what we have then is Row one of our - R prime is and not over V and not being number of particles in the in the lowest energy
31:30
state plus some smooth decaying decaying function which looks very similar to this smooth decaying function that we calculated up here's the Fourier transform the bose factor but without the the large piece of cables at K equals zero so what we have then is let me just draw a picture of some pictures of this so let's draw some axes okay so this
32:08
axis is absolute R minus R Prime that and this axis is Rho a one of our minus R Prime this is a function of the distance between them by translational invariance and and this is zero this is zero now remember that if R equals R prime you are always just measuring the density of the system so it always starts out at just Rho 1 equals Rho the constant
32:39
density of the system at any temperature this is always true by definition the the one body density matrix when you take the two corners equal to each other you just get the density so as long as the fluid you'll just get a uniform constant density Rho bar the constant density that's alright R prime equal equals 0 now for T above TC this decays rather rapidly yes so this is for T
33:09
greater than TC 40 equals zero you get BBC at t equals zero this is constant independent of temperature that didn't draw that so well let's draw this a little bit more flat do that draw this little flatter yeah okay completely flat so this is uh for T equals zero non-interacting BC B
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EC at T equals zero and what we just calculated was that for a non int B C at T between 0 and the critical temperature we saturate to okay so if Rho 1 is n over V we saturate to and naught over V
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which is a function of temperature um a you get smooth decaying part here and then you saturate to the constant and not okay better um good so um we haven't showed this but it turns out
34:50
that when you consider an interacting VEC VEC for any T less than less than its critical temperature it looks like this middle curve looks like the middle curve the saturates to some finite value saturates to some finite value which which is the order of the order
35:24
parameter it's fine like super fluid density at or superfluid order even for our very far from from our prime and this can be true for an interacting VEC even at T equals zero even at T equals zero you it will still look like this middle curve because as we mentioned before that some of the bosons are kicked out of the actual ground state
35:56
wave function due to the interactions so it's both temperature and interactions will kick some of the particles out of the of the load of a single single orbital but nonetheless that will remain some superfluid order even at very long distance so actually and let me try to draw this better this sort of looks like it's um okay so this type of order this
36:31
order is called you might call superfluid order but it's actually called is called off-diagonal long-range order sometimes that's called oh D oh D L R o off-diagonal long-range order and it's a it's it's the way to define superfluid order in terms of the one-particle
37:12
density matrix and remember it's off diagonal because we're considering Row one that are not equal to R Prime diagonal order would be some sort of order in the diagonal part of this matrix in other words in the regular density where R equals R equals R prime ok now in in genuine superfluids like like you know for the healin forest is highly interacting and helium-4 even if
37:43
he went down to zero temperature the amount of you know the amplitude of the superfluid fraction is actually a very very small fraction of the total density so it's it's only about 1/10 as high as the total density is so that the the off diagonal Longridge orders is less than 1/10 of the of the total density where it's a non interacting VEC the off diagonal order is exactly equal to the
38:15
total density okay now the next thing we're gonna do if we have time and I think maybe we should try to have time to do this yeah let's try to have time to do this because I'm gonna go through this fairly quickly anyway is I'm going to try to discuss a microscopic picture of how particles get kicked out of the ground state in detail so how is it that that interactions kick some particles out of the ground state into so you can
38:49
hear if you'd listen careful in the background you can hear my daughter screaming because she knows what's gonna happen here we're going to start doing Boogaloo buff transforms in that mixer upset um okay hopefully she will stop screaming in in a moment or so no she's still screaming um maybe I should pause the video at this point okay I'm gonna pause the video at this point so interruption she was very very angry okay I I promised I would discuss in
39:27
some detail how particles kick get kicked out of of the ground state wave function and doo-doo-doo to interactions this calculation calculation using Bo glue both technique is something that that a number of people may have seen if you took CID Pharma swarns course earlier this this year you have to have already seen this calculation but it's good to go through it again first of all for those people who haven't seen it and also because it's important to what
39:58
we're we're studying now so we're gonna consider it exactly the same problem we've been kissing so far but we're gonna consider a translational invariance system with no trap so we set the potential equal to zero it's convenient to then switch to working with K States so instead of writing side dagger our I'll write in terms of a sum over K modes even the minus K dot R so K
40:33
our Hamiltonian is it's a single single particle part side dagger to the kinetic energy term this side is AI plus there's the you over to we assuming the Delta function interaction dagger so I sigh this and then all we have to do is substitute in this Fourier mode
41:06
expression into into our Hamiltonian to get an expression for the Hamiltonian in terms of these operators so doing that without too much fanfare the first term is exactly what you would expect it's the kinetic term is you count the number of bosons in mode K and each one and McKay gets an energy H bar squared K squared over 2 number 2 m and the interesting term is the is the
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interaction term u over 2 times the volume and then there'll be a sum or a k1 k2 k3 k4 a dagger okay 1 a dagger k2 ak3 ak4 and then there's a delta function which guarantees the total the total momentum is conserved or total wave vectors conserved so K 1 plus K 2 minus K 3 minus K 4 better equals 0 and it's so we
42:11
should think of that here we when we destroy a k4 destroy a k3 we subtract momentum K 3 and then we add a particle k2 add a particle K 1 we were adding momentum k1 and k2 and the sum of all these terms has to be 0 since we have a translationally invariant system and a translationally invariant interaction as well ok so the approach here is that we expect a 0 K equals 0 to be multiplied
42:45
occupied to be Multipla occupied but we have to note that a that the number of particles in in the zero mode that operator does not commute with the interaction term it's just easy to check that a dagger a zero does not commute with this term here and that's just another way of saying that the that this
43:23
interaction term kicks particles out of the K equals zero State now the way we're going to handle this is we're still going to assume that the cable 0 mode is highly occupied so we can you know as we've mentioned a number of times it doesn't matter if you think of it as a coherent state or a fixed number state with a large number of particles in it let's assume it's a coherent state just because it's simpler and that allows us to take a 0 as an operator and
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we write it as square root of the number of particles in the state and similarly a 0 dagger can be written as square root of number of particles in this state as well this is a usual thing that working with a coherent state you turn operators into into numbers the assumption here is that the number of particles in the state is is large so the fluctuation or the comparatively the fluctuation in the
44:29
numbers is a small number so we're going to assume that n 0 the number of particles in the k equals 0 state is is a large number so it's gonna be approximately n but it's it's a little bit less than n because some of the particles have been kicked out of the the the K equals zero state so say and - and not the total number of particles - those in the in the zero momentum state is much less than n and our
45:00
justification for this is that we're considering weakly interacting bosons so this this a whole approach is appropriately as appropriate for for weakly interacting bosons now the way one handles this this Hamiltonian when we have weakly interacting bosons and we have a highly multiply occupied a not state is we want to treat and not here as a large parameter and so the largest
45:32
terms in this Hamiltonian are those with the largest number of factors and not so in these sums the sums some of these sums sometimes K happens to be zero and whenever K happens to be zero then you pick up a factor of a square root of N naught which is a large parameter and the largest terms we have in this in this Hamiltonian are then the terms where and not occurs the most number of times so the leading term leading term
46:06
is a term where we get the most factors of square root of n not so in fact we give four factors of a not and not and that occurs when all four of these KS happened to be 0 the value of that let's call that so this will be for k1 equals k2 equals k 3 equals K 4 and that gives
46:39
you a value H naught equals u over 2 times the volume times n naught squared that's the leading term now the first sub leading term you it your next term you might think we should look for terms which have only three factors of had not instead of instead of four factors than or not but if you look back up at this at the Hamiltonian here if we tried to set three of these days
47:11
to zero then the Delta function here forces the first a the fourth K we fit trying to fit if we try to fix three of these case to 0 the Delta function then fixes a fourth one to be 0 as well so there's no such term we rewrite this way there's no term no term with exactly three three factors and so the next term
47:46
is next term you want to have two factors square root or not and that can happen again going back to our Hamiltonian in several waves well one you might expect that while these two could both be at Kate was zero but if that would if if you set k equal to zero in this first term in the kinetic term here then this K is zero and it has a value of zero so that one gets gets thrown out you don't have to worry about that but you do have to worry about this
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term if two out of these four a k's happen to be at K equals zero so either at K 1 and K 2 or 0 K 1 and K 3 K 1 and K 4 K 2 and K 3 K 2 and K 4 and so forth ok so it all possibilities there's six possibilities of assigning which two of these four A's come out to be zero so let's write down all those terms so write down all those terms you you get two factors are square root of n not so
48:52
Princeton and a lot we divide by two times the volume I'm gonna sum over the remaining K K not equal to zero uh and we'll have for a dagger k a k plus a dagger K a dagger minus k plus K minus K a K so how do we get all these terms let's back up to here there's four ways in which you can choose one of these two and one of these two
49:23
to be to be zero and one of them to be nonzero so we choose one a to be nonzero and one of these eight which is one of these two to be nonzero one of these two A's to be in the Haun's ero there's two ways to choose this two ways to choose this and that gives us four possible terms where the remaining terms these leave one a not a tear on one and one other a knotted zero and that and the momentum conservation guarantees of these two K's all right the same value sorry about doing all the scrolling here
49:55
then there's also the possibility that you choose these two K's to be 0 only these two K's nonzero or these 2 K 0 I mean these two K's non 0 in which case for energy for our momentum conservation these case have to be minus each other and the or these two K's have to mines each other so the statement so that with the different ways that you can conserve momentum it's either you take a particle you destroy a particle at K and you put
50:27
it back in K or you destroy particle at k and destroy a particle minus K or you create a particle at K and you create a particle at minus K all of these things conserve momentum and this is the total of six terms okay so the the two terms that we have in our Hamiltonian now this is the leading term this is a sub leading term we also want to use to our advantage that we have that n 0 the
50:58
number of particles in the zero state is the same as well that's a dagger 0 0 and we can write that as the total number of particles which is fixed number minus sum over K not equal to 0 of a tiger k k so we're going to I'm going to need that plugged in here as well something here something less important anyway if we if we if we do that if we replace the n zeros particularly up here
51:32
replace the N 0 with n minus a there okay a K and we add it to this the total Hamiltonian at the end of the day is then rewritten as H equals u over 2 times n plus and such as a constant plus K not equal to 0 and we have H bar squared K squared over 2m just the kinetic term which we haven't done anything with plus u times rho k EK + u
52:04
rho or - hey dagger okay hey dagger minus k plus a minus k a k okay um where Rho is total number divided by V so this is the Hamiltonian that we now want to want to solve and just so you you have the idea of what's going on here so this this term is just the kinetic term and
52:36
it's changed by an interaction so each each particle at any K has some additional energy to the interaction but then these two terms the so called anomalous terms the terms of have either two creations or two annihilations what they're actually doing is they're taking or or taking two particles out of the content set to here you take two particles out of the countenance that you put one at k + 1 - k to conserve momentum or this term destroys two
53:09
particles and puts them into the condoms it so it looks like it's not conserving particle number and the reason it looks like it's not a conserving particle number is because when we dump particles into the condensate we sort of lose track of them okay the particle now you know we have the part of the condensate is in a coherent state anyway it has an indefinite particle number so we can just dump particles you know into into the contents two at a time so that we conserve momentum or take particles out of the Communist it two at a time so as to
53:40
conserve momentum now this Hamiltonian is quadratic in these a operators which means it's solvable it's solvable by so-called Blubaugh transformation Bogoliubov transform after nikolai you're boffin was somewhat more famous as a mathematician than as a physicist but also a physicist if you're if you're familiar with stat mech you might
54:11
remember the BB gky hierarchy he's one of the one of the B's in in BB gky I think he might be the first B I'm not sure whether there's a convention as to which one is named first but one of one of the bees is born I think and the other one is Bogle liubov and I forget what order they're written in anyway the the trick of the transform is to define some new operators B K and B dagger
54:44
minus K in terms of these old a operators via a matrix since theta K since theta K gosh yeah okay times a K a dagger minus K now notice that on the right hand side both of these terms destroy K units of momentum here you
55:18
destroy K explicitly and here you create minus K so both of these terms have the same same momentum in both of these terms would have the same momentum as well but the bees are a linear combination of the term that creates a part go with minus K and the term that destroys a particle with with plus K now it's easy to check that BP the these bees satisfy
55:48
canonical commutation vinegar P dagger Q equals 0 and B P u dagger Q equals Delta P Q okay no SID when he teaches this he he has a his favorite comment that he likes to to mention very frequently is that it kind of looks like this transformation is is not unitary this matrix here is is it's
56:21
not a unitary matrix so how do we get away with that we're very used to making basis transformations and in quantum mechanics and we like our basis transformations to be to be unitary but this is not a unitary matrix so so why is it it's okay well the reason why it's okay is because we're making a transform on operators not on the basis the basis is the transformation we're actually making on the basis infox space is actually still
56:55
unitary the only thing that's required to have it be a unitary transformation is that our new I was trying to just erase this you should always have your commentators and look good right yeah so in right what was I saying the the only thing that's required in order to have a unitary transformation of our bases
57:26
infox space is that these operators still satisfy canonical commutation relations as long as that's true then we have a perfectly good unitary transformation infox space even though the operators transform into each other in a way that looks like it's nonunitary okay then we can rewrite our Hamiltonian way up here our Hamiltonian here in terms of the B operators instead of the a operators I won't go I won't belabor
57:56
that too much but if you do so you find that if you it's a good exercise to go through that if you choose tangent of 2 theta K to be u Rho over H bar squared K squared over 2m plus u row then the Hamiltonian comes out to be completely diagonal and I'll write down what it is the Hamiltonian then simplifies to a constant which were not interested in plus sum of k not
58:28
equal to 0 energy sub K B dagger K became completely diagonalized the Hamiltonian these B operators are sometimes known as local new bonds and they are the elementary excitations of the system the e the energy of the bull Galoob on excitations are as follows again this is a good exercise to work through it if you've never done the P for some amount of algebra hoops minus
59:01
zero squared and for small K this is square root of u Rho over m absolute h-bar okay plus dot dot dot or small K and it's it's rather crucial here that it's rather crucial here that this is linear in let me just draw the spectrum here so spectrum looks kind of like this it starts out linear and small K and
59:34
that eventually it turns up to be quadratic the linear at small K this is the so this is K nice okay linear is small K this is the sound mode velocity sound velocity velocity in fact you can calculate the velocity is just square root of u Rho over m it's this coefficient this coefficient here and
01:00:07
that's important that it's a sound velocity that doesn't start quadratically it's linearly in order that it satisfies the landau criterion for super fluidity so a weakly interacting both gas will SuperFlow because it has a sound mode velocity at small k that the spectrum is linear and energy with respect to momentum and that then guarantees that it is a this
01:00:42
actually is superfluid now what's in looking back at our Hamiltonian here again in in this language the ground state of this Hamiltonian is clearly so the ground state is the vacuum of Bholu bonds it's a it's a state which all the B's annihilate it if you then take the ground state and yeah so the B dagger K on the ground state will then have
01:01:14
energy K on the ground state and so forth or maybe I should say energy K okay if if sorry this is ignore that the Hamiltonian if the Hamiltonian on the ground state gives you zero on the ground state then the Hamiltonian on be dead okay on the ground state that gives you you
01:01:45
know hot plus plus ek on the ground state okay so you know it was the ground state energy many sub k is the energy of the blue bonds and you can just you can check that by by the commutation of the of the B operators now the fact that the the ground state is the vacuum of the a operators remember that the the B operators are actually some linear
01:02:16
combination of the 8k operators we can show you what the linear combination is here it is the B operators is some linear combination of VA operators and conversely we can just invert this matrix ba operators are some linear combination and the B operators again I'm scrolling here and I probably shouldn't be because it makes it hard for you to follow but sorry about that but if we have this expression here we
01:02:46
know the a operators say a K is going to be some linear combination B K + V B dagger minus K and unfortunately a K or fortunately or unfortunately on the ground state is then not equal to zero because some portion of the a operator has a B dagger creates a bug loop on
01:03:20
okay so this means that this means that the ground state the ground state is not a vacuum for the a operators for a for a K then the interpretation of that is that some K states are occupied K not equal 0
01:03:52
should say K not equal to 0 states okay not equal to 0 states are occupied in the ground state which is exactly what I was saying before that due to these interactions some of the bosons get kicked out of the of the K equals 0 state in terms of the Boogaloo bonds in the ground state of Bo glue bonds is the
01:04:22
absence of all Bo glue bonds but this ground state is some superposition of of K States being occupied along with the zero energy state now one last thing to check is the the occupancy of the this worth doing worth checking the occupants okay exercise it's a good exercise to
01:04:51
try is to see how many bosons are not in k equals 0 in the ground state and it's easy to well with a little bit more calculation you'll discover that I think I do it in the notes because it does it all so that it's proportional to u to the three-halves power for small
01:05:27
for small by u so as you increase the interaction you you get more more particles kicked out of the ground state remember that n minus n not and when it's a not is this sum K not equal to zero of a dagger k a k count some particles that are not in McKay was zero state I should also point out that that
01:06:00
this this object here was at least for the case where the interaction is a delta function this object here was was part of what we called our off-diagonal long-range order so the N 0 over V is is the off-diagonal OD all or oh and if you remember we drew this diagram before as a function of distance between RR prime
01:06:33
the density over the this axis is Rho 1 of R minus R prime the single particle density matrix on this axis it always starts at 0 it starts at n over V and then it drops down over some distance scale to and naught over V which is the off-diagonal long-range order in this case it's just the number of bosons that remain in the k10 state well that's not
01:07:08
generally a valid definition of lro the general definition is well the odl area we defined before is a expectation of the eigen value of the expectation of sigh Dekker times sigh okay so that is a very very quick summary of the book lock method for treating a weakly interacting Bose gas next week or next week next
01:07:40
lecture if whenever I do it tomorrow next day is afternoon I don't know whenever my daughter deems it appropriate to take her next nap the next lecture will be on findings method of studying superfluid helium and then I will also do an example of exam problem because we've been told as lecturers that we don't do enough example problems
01:08:10
like exams so I'm going to work through one of the problems from last year's exam okay I see that

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