# Reciprocal space; Definition and Properties

the crystal structure
is something that you know ah in fact you have probably learnt off from high school
days and certainly maybe early college days you have heard of course the structure you
have heard a lot of descriptions of it and so on. so in in those descriptions you would
have heard of crystal directions in other words axis ah and crystal planes
so there are conventions based on which we ah designate ah the directions and planes
and so on ah and on that basis you can have lot of the crystal structures that ah we have
in the books okay. so ah so these are there ah normally these are ah in dimensions of
length we basically a certainly ha crystal axis when we unit vectors we say with respect
to, so we we define unit vectors dimensions of length okay.
so this is how it is done, we write xyz coordinate system we say you know ah and if you take
a real ah crystal structure you may have specific dimensions in the in the form of say ah of
the order of say 2 angstroms inter atomic spacing is of the order of 2 angstroms say
1.5 angstroms 1.8 angstroms, angstroms is 10 power -10 meters, so it is still in the
dimensions of length okay. so when you normally put such information
together okay, so we would call this as real space. a real space is what we are conventionally
used to ah which where we write an xyz coordinate system and within the framework of that xyz
coordinate system we define dimensions, we define unit vectors, we define directions,
we define ah the planes and spacing between planes and so on. so this is real space okay,
what we are going to do today is define something that we will call reciprocal space. you will
define something called reciprocal space and we will begin to see some of it’s properties
okay. ah reciprocal space in fact takes ah the basic
ah umm approach here is that the same crystal structure information that you have here or
ah and many of its important key features can be represented in another format known
as reciprocal space okay. so we are not actually changing anything in a in a fundamental sense,
because you are still talking of the same material, you are still talking of the same
kind of ah relationships between the planes between those materials, the atoms between
those present within the materials and so on. so we are not fundamentally changing anything.
all we are doing is we are representing the same information using a different set of
ah a different framework. if you want to call it that it is a different framework which
is called reciprocal space, here the dimensions of the vectors we will use will be 1 by length.
so dimensions used to be 1 by length or length power -1 is the dimension that we will use.
ah in sort of a trivial sense you can see ah that this can be related to our k vector
because k vector is already in the dimension of 1 by length okay.
so therefore ah it may be it may make it easier for us to relate certain things associated
with the wave vectors of electrons which are running across the crystal structure. to the
crystal structure itself okay, if you use the same ah ah same framework within which
you are describing both of them. the framework here we are using is 1 by length 1 by lambda.
so it would help, if we also define the crystal structure in 1 by length dimensions okay.
so that is maybe ah a little bit of a trivial way of saying it, but it will ah it is it
it suddenly conveys the immediate ah link between what we have just done up until now
and what we are planning to do okay. in reality actually it is much more than that ah it turns
out that ah ah many of the information you can represent in 1 by length in this in this
reciprocal space. ah actually conveys certain details of how interactions occur between
between a crystal structure and ah waves that are present much more elegantly. then ah ah
is then is done in real space okay. so in a broader sense this is, ah ah this
conveys certain information much more elegantly reciprocal space and actually is able to highlight
specific details a much better than the real space way of representing information does
and that is the real, that is the ah ah fundamental reason why actually if you get into ah a diffraction.
if you get into meti i mean ah diffraction as a means as a tools of as a tool for looking
at structure and such information characterization of materials. ah you will find a lot of heavy
usage of reciprocal space. ah the first time we encounter reciprocal space
it is not as intuitive and it may seem like you know we are unnecessarily complicating
ah ah the issue so to speak okay. so it looked like why go to all the ah troubles of creating
everything in 1 by length when real space is already there for us. it is just that if
you use it enough you find that there is lot of ah ah information that is much more elegantly
and clearly represented in in when you use this notation than when you use real space
notation. so it is from that perspective that this is
really caught on and it traces itself back to a person by name ewald, who actually ah
worked on this ah in around years around the year 1920 okay. so around that time frame
was when this ah work was put together. and so there is a there is notation which is which
is named after him also in in this relationship okay, so ah in this context.
so we will see that as ah ah i mentioned you know we have sort of indicated why you may
need to go in for ah ah1 by length dimensions but that is not a hard-and-fast that is not
the best way of indicating, but you can see the link that we have already got to pi by
lambda 4k wave vector and we want to see the interaction between the electron waves and
the crystal structure ah and therefore it would help if we or if you present at all
the information in the same framework. so ah you can think of that as a loose link
for what why we are doing what we are doing. as we progress forward we will see ah much
more ah a better understanding of how this system works out okay. so now we will look
at ah umm ah in the next two or three classes we will actually focus ah exclusively on reciprocal
space and we will build its relationship to the reveal real space because that is something
we need to understand. ah and to some degree this discussion may
seem a little disconnected from what it is that we have discussed so far ah but we will
need to build this framework so that we can connect it up link it up to our discussion
earlier and see what ah ah what benefits we can gain from the process okay. so but temporarily
for at least this class and in most of next class this will be a discussion exclusively
on reciprocal lattice and on how diffraction can ah as a phenomenon can occur in the reciprocal
lattice. after that only we will make a link back to
real lattice, a real space and also take up in the extend this idea the diffraction occurs
in reciprocal space and it has a certain way of being conveyed and see what is the consequence
of that discussion, on how the electrons are interacting with a crystal lattice. so that
that step is going to be 2 or that is at least 2 classes away before we get there but now
we will have to build the framework which will enable us to handle that discussion ok
so that is what we would do at this point. so ah when you look at real lattice real space,
we say that you know we are usually defining it by unit vectors
and typically we would use the notation a 1 a 2 and a 3 okay. so a 1 a 2 and a 3 are
unit vectors in real space. ah so we have a crystal structure in real space represented
by these unit vectors a 1 a 2 and a 3 okay. now we will define a reciprocal space in other words we will define a space, we
will define a coordinate system. so we are just going to define it up front
this definition is at this stage may look somewhat arbitrary, we will just accept it
as an arbitrary definition. the definition the way it is given will give the space a
lot of useful properties which we can use later on. ah how arbitrary or otherwise this
definition is we will see a little later but at the moment it is just a definition we will
just accept the definition and we will work with it, so we are defining it okay.
so this is a choice we are making, so we are defining it we are defining a reciprocal space
to consist of three vectors again unit vectors there b 1, b 2 and b 3 but these are not arbitrary
ah vectors they are being defined with respect to real, i mean real space vectors in a certain
way in other words there is in the in the process of this definition we are already
making a link between these vectors and these vectors okay and what is that link it is it
is written like this so b 1 is a 2 cross a 3 by v.
what this v is we will see in just a moment. it is a volume actually volume of this unit
cell ah of consisting of a 1 a 2 and a 3, ah b 2 is defined as a 3 cross a 1 by v and
b 3 is similarly is a 1 cross a 2 by v. so these are cross
product, these are vector cross products so that is what they are so ah so these are all
vector ah quantities b 1, b 2 and b 3 are vector quantities. so they have ah a magnitude
as well as a direction, a 1, a 2 a 3 are also vector quantities they all also have magnitude
and direction and this is a cross product. and so it is defined that way b 1 is defined
as a 2 cross a 3 by v, b 2 is defined as a 3 cross a 1 by v and b 3 is defined as a 1
cross a 2 by v okay. so this is the way they are defined. ah by defining it like this certain
properties ah become ah certain properties ah arrive for ah i mean end up being available
for b 1 b 2 and b 3 which becomes convenient for our utilization later on okay.
so now we will see what immediately based on this definition itself simply because we
have defined it like this what is the meaning of what is the consequence or what ah what
is the relationship between b b 1 and this ah reciprocal, so these are called reciprocal
lattice vectors okay. these are unit vectors in reciprocal space they are also called reciprocal
lattice vectors. these are real lattice vectors and this is real space. so by simply using this definition what is
it that we have created as a relationship between what is the consequence in in the
relationship between b 1 b 2 and b 3 with respect to what we have in real space. so
that is the first thing that we will examine okay. so to do that let us take an arbitrary
say a triclinic cell. a triclinic cell is one where by definition the a 1 a 2 and a
3 vectors need not have the same length. so a 1 need not be equal to a 2 need not be
equal to a 3 and the angles between them so alpha beta and gamma which are the three angles
that exist ah in the system need not be the same okay. so that is the so in that sense
it is of it is like a very general cell we have placing a really placing no restrictions
on it. ah now ah we are defining we will we will write b 3 again here fine a 1 cross a
2 by v. so now by definition ah see volume does not
have any is not a vector volume is just volume it is not a vector it is a scalar quantity.
so you have a 1 cross a 2 by definition of cross product if you have a cross product
the ah result is perpendicular to both a 1 and a 2 okay. so that is the meaning of a
cross product. so be 3 therefore by definition is perpendicular to a 1 and a 2. so on this
scale if you want to draw b 3 it will show up something like this.
it will show up somewhere in this direction it would be perpendicular to the plane being
described by a 1 and a 2 okay. so the planes a 1 and i mean the vectors a 1 and a 2 define
a plane which is which we are now treating as the horizontal plane on this board a sort
of in this representation and b 3 would now appear in this direction perpendicular to
a 1 and a 2. so that is the ah definition by way of definition
but what about its actual magnitude if you look at a 1 cross a 2, ah if you ah if you
see the by the standard definition of a 1 cross a 2 that is the area of this parallelogram
okay. so if you write this as the origin write this as oa ah let us say this is b and let
us say this is c, so the ah area of this parallelogram oacb is what is being given by a 1 cross a
2 okay so oacb area of parallel a 1 cross a 2 okay oacb the parallelogram.
so that is what this area is ah if you look at the ah volume of the unit cell this v here
is the volume of the unit cell in real space okay. so that is the volume that is that v.
so which is that volume if you take this a 3 vector and you complete this solid okay.
so that solid, so ah this unit cell that we have drawn here i mean if you draw it properly
you will get it appropriately this volume of this unit cell is this volume
v that we have written here fine. so that is the volume that we are looking at. so what
is that volume in terms of geometrical terms it is simply the area of this base times ah
the distance between these planes or the height of this unit cell okay. so that is all the
volume is the volume is the area of the base times the height of that structure. so that
is what we are looking at, so the height is simply this the whatever this is abc so this
is d let’s call this d od, so od is the height of this ah unit cell that we have drawn
and it is simply the ah ah projection of a 3 on this on this axis on this axis.
which is perpendicular to a 1 and a 2 right, so that is the height, so ah we will call
that ah so the height is od ok so the height is od so therefore volume is the area of oacb
ah into the height od okay. so it is a product of the area times the height that is the volume
okay. so therefore if you look at it that way b 3 okay or the modulus of b 3 is area
oacb by area oacb into od. so that is what the modulus of b 3 is, so these two will cancel
out, so it is simply one by od, so it is simply one by od okay.
so that is the ah magnitude of ah b 3, so b 3 is in this direction and the magnitude
of b 3 is simply 1 by od, what is 1 by od if you look at the conventional planes that
we are looking at if you look at the way we define planes this is the 1 0 0 play okay.
so if you look at 1 0 0 plane this is a d 1 1, d 1 0 0 plane it is spacing between 1
0 0 planes. if you go back to your elementary crystallography the 1 0 0 planes are defined
this way and then these ah are 0 0 1 plates d 0 0 1 this is perpendicular to ah a 3 axis
okay. so d 0 0 1 if you want to call it 0 0 1.
it is the spacing between 0 0 1 planes so spacing between 0 0 1 planes is what we have
now looked at. so od is the spacing between 0 0 1 plane, so d 0 0 1 okay, so b 3 is therefore
spacing. ah this is the standard notation for crystallography, so if you go and look
up crystallographic from your elemental crystallography this is d 0 0 1 is the spacing between 0 0
1 planes and modulus of b 3 is simply 1 by d 0 0 1 okay. so so therefore we find and
by analogy in fact i mean. in fact if you extend this argument the same
thing would hold for the other ones to b 1 is 1 by d 1 0 0, b 2 1 by d 0 1 0 and b 3
which we just did is 1 by d 0 0 1. so b 1 b 2 and b 3 are ah inversely proportional
to the spacing of those planes ah 1 0 0 , 0 1 0 and 0 0 1 planes. so that is the by way
of our definition we have created the situation okay. so it is not arbitrarily occur since
we defined it this way this is the way it has occurred okay. ah also if you look at
the way we have defined it we will also see that if you just take a product say ah b 3
cross a 2 or b 3 dot a 2 okay. if you just take this dot product between
b 3 and a 2, i have just arbitrarily pick these two vectors we find that since b 3 is
already perpendicular to a 1 and a 2 this dot product is 0 okay. so by definition it
will be there will be a cos 90 degrees which was up here therefore this is 0. similarly
b 3 dot a 3 sorry a 1 equals 0, because b 3 is perpendicular to both a 2 and a 1 simply
based on how we have defined it that is all it is okay. so therefore these two are 0,
if you look at b 3 dot a 3 okay. based on our definition if you go back to
our the picture we have drawn here b 3 has the is this distance od, i am sorry this is
1 by the distance od, so ah which we just saw here okay 1 by od b 3 is modulus of b
3 is 1 by od, so this is simply equal to 1 by od times the projection of a 3 on b 3 okay.
when you do a dot product that is basically what it is it is one vector times the projection
of the other vector on itself okay. so what is the projection of a 3 on b 3, projection
of a 3 on b 3 is od, a 3 on b 3 the projection of a 3 on b 3 is od as shown in this diagram
fine. so therefore the dot o dot a 3 that we are having in that dot product is simply
works out od, so to speak so so this projection will become od okay. so therefore this is
equal to 1. so we find the relationships between those vectors the the reciprocal lattice vectors
and the real lattice vectors based on how we have defined those vectors okay.
based on how we have defined those vectors creates the situation where b 3 dot a 2 is
0 b 3 dot a 1 is 0 and b 3 dot a 3 is 1 so more generally if you have b i dot a j, then
this is equal to zero when i not equal to j and is equal to one, when i=j okay. so
that is the notation that we have and that is the consequence of what the wave in which
we have defined these vectors okay. ah now we have already seen, ah if you have taken
a specific case actually where we are saying that if you have a particular vector.
so d b 1 in this case is a particular vector and we found that the way we have defined
it, it works out to be perpendicular to the two vectors a 2 and a 3 and it is equal to
the in magnitude it is equal to 1 by the spacing d 1 0 0 okay. in reciprocal lattice i we can
actually generalize this much more. we will generalize it as for us we can write
any vector ah ah h h k, h is the notation that is given for a reciprocal a general,
a generic reciprocal lattice vector okay. so where we it could be anything, so h is
a general reciprocal lattice vector we will give it subscripts h k l okay. so if this
is a reciprocal lattice vector and the unit vectors in the reciprocal space are b 1 b
2 and b 3 then this is simply equal to h b 1 + k b 2 + l b 3.
this is simply based on our definition and vectorial, standard vectorial notation, standard
vectorial notation b 1, b 2, b 3 are unit vectors and h k andl are the ah ah specific
distances we are travelling along those unit vector directions okay. so h h k l is a vector
in in reciprocal space and it is therefore equal to h be 1 + k b 2 + l b 3, those h k
l are the amounts that we are travelling on those respective dimensions okay so that is
what it is. we say that when we define when unit when
reciprocal space is defined the way we have just defined it, then when you take a general
vector h h k l okay, ah in the reciprocal space we can there are some relationships
it has to real space vectors and dimensions in real space just the way b 1, b 2, b 3 themselves
have relationships to the a 1, a 2 and a 3 okay. we just saw a relationship between b
1, b 2, b 3 and a 1, a 2, a 3. similarly we there is a very general relationship
between any h k l vector in reciprocal space and certain quantities in real space. what
is that relationship it is basically that h h k l is perpendicular
to the plane which has the miller indices h k l and modulus of h h k l=1 by d h k
l. please note in both these cases we are relating something in reciprocal space to
something in real space. this is not complicated because we just did
that already when we looked at a 1, a 2, a 3 and b 1, b 2, b 3 and we made relationships
between them when i said that know b 1 is 1 by d 1 0 0, b 2 is 1 by d 0 1 0 and b 3
is 1 by d 0 0 1, there b 3 is a reciprocal lattice quantity and d 0 0 1 is a real lattice
quantity. we found that you know it’s simply because of the way we have defined it, when
in our definition itself we have linked real space and reciprocal space
so they are not arbitrary quantities okay, so they are already linked by definition therefore
some quantity in real space can relate to something in reciprocal space within the framework
of the definition. so that is basically all we are doing here this is a ah ah reciprocal
lattice vector and it is found that it is perpendicular to a plane in the real lattice.
which has that the ah ah miller indices h k and l and the modulus of this reciprocal
lattice vector is 1 by d h k l. these are properties that the reciprocal or
in other words one by the spacing of h k l planes. so these are two properties that any
ah vector in the reciprocal space has as a result of the definition of the reciprocal
lattice okay. so right now i have just stated it in the in the next few minutes we will
prove these two once we prove these two we have a good understanding of what holds in
reciprocal space and how it relates to real space and later we can use those two okay. so we will now try an attempt to prove this,
so to do that let us actually draw a general plane in the real space. so we will say that
we have, so this is a 1 this is a 2 this is a 3 okay, so these are unit vectors in real
space and we will draw the h k l plane here okay. so this is h k l plane fine, so ah ah
now we will just say that ah in in reciprocal space we have the h h k l we at this moment
i am just arbitrarily denoting it denoting it here.
this is an arbitrary de notion at the mean i have arbitrarily denoted it this way i have
indicated it here in this figure this way ah simply for the sake of convenience to show
it in the same figure ah what relationship it actually has to h h k l is not forced upon
it simply because of how i have drawn it we will show that in fact it does have some appropriate
relationship. so that relationship we will just see we are actually going to prove it.
now by definition of h k l plane its intercept along a 1, a 2 and a 3 are simply ah a 1 by
h, a 2 by k and a 3 by l. so these are the intercepts okay, only because the plane h
k l happens to intercept a 1, a 2 and a 3 at those locations such that its intercepts
are at a 1 by h, a 2 by k and a 3 by l, that is the reason why we even call it the h k
l plane okay. so therefore if you see if you take this ah vector here, this vector here
is a 1 by h this is a 1, a 1 by h is this vector that is why you get this h notation
in the h k l plane. similarly this vector here will be a 2 by
k that is why you will get the k notation. this vector here from here to here is a 3
by l that is why you get that l in the h k l. so this is a 1 by h. i will also named
this locations where this plane intercepts a 1, a 2 and a 3 as a, b and c okay. so ah
oa is a 1 by h and ob, so this is origin o, oa ob and oc. so oa is a 1 by h, ob is a 2
by k fine, so this is what we have. so ah simply by vectorial notation if you write
ah oa + ab you should get ob right. so oa simply because of standard vectorial
notation oa these are vectors by the way oa + ab should equal ob, that is easy to see,
if you go from here to here and then you go from here to here it is the same as going
from here to here that is all it is oa + ab is ob. so this is what we have a 2 okay, so
now that we know these are the vectors we can replace them oa is a 1 by h and ob is
a 2 by k. so therefore a 1 by h + ab, vector ab should equal a 2 by k fine. so this is
what we have, we can rearrange this a little bit. so we wrote a 1 by h + ab=a 2 by k, rearranging
this we get ab is a 2 by k + sorry – a 1 by h. so we now have in terms of the a 1, a 2
vectors we have the vector ab given to us in terms of the unit vectors a 1 and a 1 and
a 2 which we have already defined for a real space. we already said that if you take a
vector h h k l, we are defining it based on this notation as h b 1 + k b 2 + l b 3.
so the h k l, h k and l are simply integers okay, so this is some vector in reciprocal
space so h k and l are just integers, so as long as they are just integers you can use
them in whichever space you wish they are just integers b 1, b 2 and b 3 are reciprocal
lattice vectors therefore h b 1 + k b 2 + l b 3 is now a reciprocal lattice vector.
whereas when you use the h k and l which are just integers in the real space a 2 by k and
a 1 by h. if you take this difference which is ab it
is a real lattice vector because these are just integers where which you are using in
the real space. but they are the same ah ah integers at this time ah we have the same
h k and l being used in two different places. the h k l plane is this results in this relationship
and the same h k l values, we have now used for this definition although we have got no
significance for it. yet there is the same we are enforcing the
same h k l in this ah definition for h h k l. so if you now take a dot product of h h
k l and ab, what will you get? so if you do h h k l dot ab okay. so we have ah b 1 here
h b 1 okay, we already saw that if you have a i dot ah b j then this is equal to 0, if
i is not equal to j and this is=1, if i=j and since it is a dot product you can
have at a i dot b j is the same as b j dot a i, if the order in which you do this dot
product is not important to us because either way you will get the same thing.
so if you look at this dot product here you have h b 1, h b 1 dot a 2, so these are just
the h is just an integer b 1 dot a 2 is zero because it is 1 and 2 here. so that is 0,
b 1 dot a 1 is 1 and you have a h b 1 here and a 1 by h here so h and h will cancel you
get -1 okay. so h b 1 times or dot product of h b 1 and a 1 by h will give us – 1 right.
if you take b 2 here k b 2 times a 2 by k, the k and k will cancel, you will have b 2
dot product of b 2 and a 2 which is 1 + 1 and this b 2 dot a 1 is going to be 0 because
it is a subscript 2 and that is a subscript 1 we already saw that by definition
so that is a plus 0, so that term will become 0 and then b 3, b 3 dot a 2 is going to be
0, b 3 dot a 1 is going to be zero. so this term does not contribute in any way it becomes
all 0, so this becomes 0, so this becomes 0 the product with this gives us a +1 the
dot product with this gives us a – 1, so this is equal to 0. so we have a situation
where a vector in reciprocal space h h k l dot a vector in real space is equal to 0.
so in other words we have a dot product between two vectors which is zero which simply implies
that h h k l is perpendicular to this vector ab. therefore h h k l is perpendicular okay.
so if you go back to our figure it means that this h h k l is perpendicular to this ab okay.
so this ab is there the way this h h k l is defined it is perpendicular to ab. using exactly
the same ah ah derivation that we have done instead of we started with this this location
being a 1 by h and this being a 2 by k. we can do the same thing with a 2 by k and
a 3 by l okay, in which case we will find that h h k l will become perpendicular to
bc. similarly we can also do it with a 3 by l and a 1 by h and exactly the same calculations
we will do we will find that h h k l is perpendicular to ac. so we find that h h k l is perpendicular
to ab it is perpendicular to bc and it is perpendicular to ca based on the same derivation
that we have done. you simply have to select the other two axis, you will come to the exact
same conclusion. the same math will the mathematics is work
out in exactly the same way we will find that h h k l is perpendicular. the vector h h k
l is perpendicular to the vectors ab, bc and ca. therefore and since all of these three
are forming a plane, if it is perpendicular to any two in fact even even if it’s perpendicular
to just two, it is certainly perpendicular to two and it is also perpendicular to all
three if it is perpendicular to them it is therefore perpendicular to the plane defined
by a b and c okay. therefore h h k l is perpendicular
to plane defined by abc ab, bc and ca and therefore which is basically the hh which
is basically the h k l plane okay. so that is how those vectors are even defined ab and
bc and ca were defined based on the intercept at 1 by h, 1 by k and 1 by l in those respective
axis. so therefore h h k l is perpendicular to h k l. so any, so again we are relating
something in reciprocal space a vector in reciprocal space to a plane in real space
okay. and again all these relations are coming about
simply because our original definitions related a reciprocal lattice vector to real lattice
vectors. so that relationship was already there within the framework of this relationship
we are finding other other relationships that hold. so we find that any h h k l plane which
is that therefore defined as h b 1 + k b 2 + l b 3, where b 1, b 2 and b 3 are the unit
vectors and reciprocal lattice reciprocal space. those h h k l vectors will automatically
be perpendicular to the h k l planes in the real space okay so this is already we have
seen this. so we have just we have just shown this okay.
so the other thing we would like to see is what is the value of modulus of h h k l and
how does this relate to the spacing of between the h k l planes. we will in fact see that
this is equal to 1 by d h k okay. so this is what we are just about to we are going
to look at. we will come back to this figure let us say that a unit vector
along h h k l okay so we will let us first define a unit vector along h h k l that is
simply h h k l by modulus of h k l of h h k l right that is the definition of a unit
vector along h h k l. we will just call this say m, m cap okay.
now when you say a h k l plane is defined the way we have just drawn it okay. so when
we say this we mean that the that there is a similar plane like it at the origin then
there is one at this location then there is one similarly spaced next to it similarly
space next to it and so on that is the way we define a family of a ah ah set of planes.
when you say a h k l plane it is not just a single plane i mean it is that set of planes
that are parallel there. we take the one process to the origin and
then take the intercept of it and that is how we come up with h k l, so there are planes
correspondingly apart. so therefore the spacing between the h k l planes is simply the distance
between the origin and this h k l plane right the closest distance ah or ah rather the perpendicular
the spacing between this origin and that location which would then be the closest distance between
the this plane and the origin is therefore the d h k l right.
so now ah we have basically seen that the ah at at that point the ah line drawn from
the origin which goes closest to this plane will then they will then go perpendicular
to that plane all right so that is how you will get the ah location. so therefore ah
if you look at the distance ah between the ah plane and the origin ah we basically see
that we can write that by saying h h k l or n cap
dot a 1 by h ok. if you take the ah ah dot product of a 1 by
h and you take its ah ah what shall i say the the dot product of oa with this ah vector
in the direction of this h h k l then you get the component of oa along that direction
and that would then be the spacing that you are interested in ok. so this is what you
will get, so this is simply a 1 by h ah dot product with ah you will have h b 1 + k b
2 + l b 3 divided by the modulus of h h k l okay.
so this is d h k l right, so d h k l is now defined this way d between those planes is
such that ah the if you take the the component of any of those ah intercepts along the perpendicular
ah to that plane okay. so that perpendicular is now going through the origin. so that perpendicular
vector that is there if you take the component of this vector along the direction of the
perpendicular you therefore get the distance in this direction okay.
so that vector unit vector is simply giving us the direction there but if you take the
component of this ah intercept oa along that direction or any of this ob along the direction
or oc along the direction if you take the component along the direction that would then
represent that spacing between the origin and that point which is then the closest spacing
between the origin and that point and that would therefore and at that point that vector
will be perpendicular to that plane which is what how we have defined it okay.
so if you take the component of the intercept along the perpendicular that is the closest
approach that the plane makes to the origin and that is therefore that d spacing of that
plane because the next plane is sitting at the origin. so the d h k l is the component
of this vector a 1 by h along this unit vector here. the unit vector is simply defined by
this and this is and since it is a dot product we can interchange the i does not matter which
order we do it so we will get this. so now if we carry out this dot product ah
what do we see again you have an a 1 here you have a b 1, b 2 and b 3 here and clearly
a 1 dot b 2 is 0 and a 1 dot b 3 is 0, the way we have already seen. so only a 1 dot
b 1 is going to count for anything else so okay. so a 1 dot b 1 is going to be 1 because
of the way it is defined and the h and h is are going to cancel. therefore d h k l is
equal to ah a 1, so we will just write it here, a 1 by h dot h b 1, b 1 or sorry a 1
dot h by h b 1 dot h b 1 divided by modulus of h h k l.
so this and this will cancel a 1 dot b 1 equals 1 therefore this is equal to 1 by h h k l
here. so we see ah which is the proof that we wanted to be set out to prove. so we see that h h k l is perpendicular to
and modulus of h h k l is equal to 1 by d h k l okay. so ah what we have seen is ah
we started off this class by saying that we need a notation or which is this reciprocal
lattice notation because it will help us understand the interaction between ah the wave vectors
corresponding to the electrons and the crystal structure which is which basically conveys
to us the periodic structure that is present between all the components that are present
all the atomic components that are present within the ah ah lattice.
so to speak the material that we have so in that context we found ah we indicated that
ah that basically we might need to develop something in this reciprocal lattice notation
and it is that notation that will help us capture this interaction between the wave
vectors and the periodic crystal structure. so in that context we define reciprocal lattice
to be consisting of b 1, b 2 and b 3 with specific relationships to the real lattice
vectors a 1, a 2 and a 3. on the strength of that relationship we found
that on the strength of the definition we found that already b 1, b 2 and b 3 had specific
relations to a 1, a 2 and a 3 which then translated to a general reciprocal lattice vector h h
k l having specific relationships to the plane h k l in real space the relationships are
that the vector h h k l is perpendicular to the plane h k l. reciprocal lattice vector
perpendicular to a plane in real lattice. and the modulus of the reciprocal lattice
vector is one by the spacing between those h k l planes okay. so this is the framework
of our reciprocal space we already seen some important relationships here. in the next
class we will see how diffraction which is the interaction of waves ah with the ah periodic
crystal structure how the diffraction phenomena can be represented in the reciprocal lattice
notation.