### SUBTITLES:

Subtitles generated by robot

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

22:51

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

30:29

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

31:03

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

33:45

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

34:18

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

41:39

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

43:57

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

48:20

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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