The first order of business is to compute this
number, the inner product x with p. You recall
that p ket vector stands for a momentum Eigen
state of a particle moving on a line and x
ket vector stands for a position eigen state.
What we would like to find out is, given the
commutation relation, x commutator p is ih
cross times the unit operator, whatís the
overlap between a position Eigen state and
a momentum eigen state. We know already that
when two operators donít commute with each
other, you canít find a complete set of common
eigenstates for these two operators. So the
question is, given the commutational relation
which is an input information borrowed from
classical mechanics from the Poisson bracket,
what would be this quantity.
But we already saw last time that momentum
operator in the position basis 
is nothing but - ih cross d over dx of the
representative of the state psi in the position
basis. So we established this relationship
last time that followed from this commutational
relation. So 
it says in the position basis, the momentum
operator acts as if its - ih cross the derivate
operator with respect to x. we will use that
information to compute what this quantity
itself is here. Instead of psi, you replace
it with the momentum eigenstate in this equation
and thatís going to immediately tell us what
the answer is. So we have x p on p = - ih
cross d over dx x with p. but of course the
momentum operator acting on a momentum Eigen
state just produces the momentum eigen value.
So this p is a number you call it p0 if you
like. so this says - ih cross d over dx of
x with p = p times x with p because this is
a number that comes out when this operator
acts on it, it produces the eigen value which
comes right out of the matrix element and
its equal to this. So this is some function
of x labeled by p. that function, call it
f of x if you like, and satisfies an ordinary
differential equation in the x variable.
The solution implies that x with p is proportional
to e to the power. So this is just a constant
ip over h cross x. thatís true for every
p. we could have changed the value of this
p to some other eigen value. Whatever be the
eigen value, you get this equation here with
that eigen value present and this x sitting
here. This is a function of x because itís
a position space representative of a certain
ket vector. That ket vector is labeled by
this p here. So this gives us a fundamental
result which says x with p apart from some
normalization constant is essentially a phase
factor. notice that in that exponent there
is no question of putting operators or anything
like that both x and p are numbers actual
numbers they are eigen values actually. So
corresponding to the bra x, we have an eigen
value x thatís a number sitting on the right
and corresponding to ket p, you have an eigen
value p thatís on the right again. This immediately
implies that p with x is proportional to e
to the power - ipx over h cross. Now the labels
x and p are interchanged in this case. This
is the p space representative of a position
eigen state x.
We already proved remember that if you took
a state vector psi of t and you asked what
is its position space representative, this
was equal to psi of x ,t and similarly in
the momentum basis it was equal to psi tilda
of p, t.
And that is psi tilda of p , t was related
to integral dp x p p psi of t. I insert a
complete set of states there and this is equal
to integral dp, e to the ipx over h cross
and psi tilda of phi p, t. So it says that
psi of x t is the Fourier transform of psi
tilda of p and t and vice versa. And you can
fix the constants by adjusting by having a
Fourier transform normalization convention.
And of course psi tilda of p and t is integral
dx e to the power - ipx over h cross psi of
x, t. so that establish is my assertion made
earlier that the position space and momentum
space wave functions of a quantum mechanical
particle are Fourier transforms of each other.
That immediately tells us that if one of them
is square integrable, so is the other square
integrable. Because you know by the Parseval's
theorem, mod psi squared of x, t integrated
over dx is equal to the same thing for mod
psi tilde. So thatís just a consideration
of probability statement and both of them
belong to L2. This incidentally is also one
way of looking at the statement that the space
L2 is self-dual. So the Fourier transform
of a square integrable function is also square
integrable. So much for this overlap x , p,
one can generalize this to three dimensions.
So if you have in three dimensions, the momentum
eigen state p labeled by the moment of vector
p and you ask whatís its position space representative,
this is proportional to e to the power i p
dot r over h cross in three dimensions. And
similarly the other quantity is e to the power
-ip dot r. and in that case, this relation
here becomes r p operator acting on psi = - ih
cross the gradient operator acting on r, psi.
The gradient is with respect to r.
So we have this very important statement that
in L2 -infinity, infinity the representation
of with momentum operator is in fact - ih
cross the gradient operator. So thatís the
basic reason why differential operators appear
so naturally in quantum mechanics because
automatically starting with the original canonical
commutational relation between a coordinate
and its conjugate momentum, the derivative
operator appears naturally. So this is why
the Schrodinger equation for it turns out
to be a differential equation. We can now
write down the Schrodinger equation for a
particle moving in space. Letís do this in
three dimensions.
We are now talking about non-relativistic
physics. So the Hamiltonian of this particle
is p squared over 2 m + some potential which
is a function of the coordinates alone. The
Schrodinger equation is ih cross d over dt
psi of t equal to the Hamiltonian operator
acting on psi of t. thatís the abstract equation.
The question is what does it look like in
the position basis. And to do that all you
have to do is to take the position basis,
some particular labeled position Eigenstate
r and take scalar products on both sides with
respect to this. So out here I put r H psi
of t.
I want you to note particularly the fact that
this psi of t, the abstract state vector that
is not burden by labels like r p and so on
and so forth. Thatís just the state of the
system. Itís a vector in some Hilbert space.
The variables that describe a physical system
are in this case r and p and you talk about
eigen states of those variables. But this
is some abstract state. It doesnít have any
labels except it changes 
with time.
So I could pull this out and write ih cross
d over dt r with psi of t but this is the
wave function of the particle in the position
basis and I have got a name for this. So this
is equal to psi of r and t. just the wave
function and the position basis. But now this
looks like a function of 2 variables and I
must be careful to note that the time derivative
acts on the explicit time dependence here.
Therefore this should really be written as
ih cross delta over delta t. you make it completely
unambiguous that this t derivative is going
to act on that t dependence, which came from
the t dependence of the state vector. And
this is equal to on this side you have an
r and then you have a p squared, which is
an operator over 2 m acting on psi of t +
r, a v of r and then a psi of t once again.
We can now simplify both these terms. The
matrix element of p in the position basis
is - ih cross the gradient acting on r on
psi. Therefore p squared is - ih cross gradient
dot - ih cross gradient because p square is
just p dot p.
So it says ih cross delta over delta t psi
of r , t = - ih cross squared del squared
because del dot del is del squared when acting
on scalars, over 2 m psi of r, t + a certain
quantity. The psi acting on V(r) is unknown
but I know about r operator acting on r bra
vector as it just pulls out the real eigen
value r. Therefore any function is an eigen
state of a function of r as well and it pulls
out just the corresponding function of r that
v of r is. suppose thisV of r = r squared
+ e to the power - r as an operator, then
it acts on this bra vector it pulls out r
squared + e to the power - r as an eigen value.
Therefore we could write this as V of r and
now this is an ordinary function multiplying
bra r with psi which is psi over r, t.
Itís acting on a bra, so it should really
be V star of r star. But now I am talking
about Hermitian Hamiltonian. Since I would
like its eigen values to be real, I have assumed
this function V to be real. If its complex,
then we should have to interpret what the
imaginary part means. There are problems where
there is need for complex potential. When
you have a beam of particles impinging on
a quantum mechanical target, part of it would
be scattered and part of it would be absorbed.
Then one uses a complex V in order to take
into account absorption because then probability
needn't be conserved. The Hamiltonian need
not be Hermitian. So it is necessary to take
into account those situations but right now
I have in mind a particle in a real potential
here.
So this is the position space wave function
for particle and this is what you would have
seen in the hydrogen atom problem. This is
precisely what you have to seen in the hydrogen
atom problem where V of r is Coulomb potential
that an electron sees in a hydrogen atom.
Itís minus some constant over r. of course
some simplification occurs in this partial
differential equation when you have spherical
symmetry. if V of r is spherically symmetric
and depends only on the distance r without
a vector sign, then itís convenient to take
this del squared and write it out in spherical
polar coordinates and solve the problem by
separation of variables by assuming that this
wave function psi of r vector, t is a product
of functions 
of r, theta and phi. Itís a complicated equation.
Itís not a trivial equation even for a single
particle. Its first order in time and second
order in the space variables here. Now itís
a complicated partial differential equation.
you need to specify initial values, boundary
conditions and so on.
What boundary condition would you specify
for a physical particle moving in three dimensional
space? You would say that total probability
of the existence of the particle is 1. Therefore
you would normalize your psi of t by saying
that at all instance of time this probability
amplitude shall remain equal to one but that
written out in the position basis is nothing
but d 3 r psi of r, t whole squared. By d
3 r, I mean the volume element in real space.
And thatís true for all time. That condition
is called the normalization condition. It
implies therefore that this wave function
psi of r, t, the position space wave function
is a member of L2 but L2 with 3 independent
variables x, y and z or r, theta and phi.
It follows that this can only be true if as
the radial distance goes to infinity, mod
psi goes to 0. Thatís necessary but not a
sufficient condition. It should go to 0 sufficiently
fast that this number remains finite and bounded.
Now you can ask what sort of problems can
I solve with this. Well, by separation of
variables, you can solve all those problems
where this partial differential equation splits
in to equations for each of the independent
variables. The separation into time is not
very hard. But the separation of the space
variables depends on the form of V of r. itís
not very clear what would happen in general.
In the equation for a free particle, there
would be no V of r. then you canít normalize
it because the free particle moves all the
way from - infinity to infinity. And if itís
in a momentum eigen state, it goes like e
to the ip dot r whose mod is 1 and therefore
cannot be normalized. Then you would pretend
that you would put it inside a box of some
size l and let l go to infinity. This is called
box normalization. We will come back to this
when we do the particle in a box problem.
But you could also ask in what coordinate
systems does del squared separate in three
dimensions. Thatís the problem purely in
calculus and in turns out there are 11different
orthogonal coordinate systems which are Cartesian,
spherical polar, cylindrical polar coordinates,
oblates spheroidal, prolate spheroidal, parabolic
cylinder, elliptic coordinates and so on.
There is a mathematical condition on whatís
called the Stackel determinant which specifies
this. But this is not of too great interest
because unless the potential has that kind
of symmetry, you wonít choose that particular
given coordinate system. All spherically symmetric
potentials is what you choose. If its Cartesian
symmetry you choose Cartesian coordinates
and so on, we will solve a few such problems.
But this equation is what I would like you
to examine and appreciate. Well, the first
thing one could do is actually get rid of
the time dependence. That brings us to the
idea of eigen states of the Hamiltonian stationary
states. A stationary state of a quantum mechanical
system by definition is an eigen state of
the total Hamiltonian of the system. Any stationary
state is an eigen state of the Hamiltonian
of the system. In general, the Hamiltonian
operator may have very large number of eigen
states. in general, an infinite number of
eigen states. Anyone of them is called a stationary
state.
So eigen states of H itself says ih cross
d over dt psi of t must be equal to H psi
of t for any state. But if itís an eigen
state of the Hamiltonian, then H on psi reproduces
the same state. Letsí use phi for an eigen
state of the Hamiltonian for the moment.
So this is equal to phi (t) but this is equal
to Eigen value times phi (t). since I know
that the Hamiltonian classically in conservative
systems really represents the total energy
of a system, the value of the Hamiltonian
gives you the total energy. I borrow that
language and go over to quantum mechanics
and I say the eigen of the Hamiltonian. let
me call it E. as yet we donít know whether
this E is continuous or a discrete positive
or negative we donít care we do know its
real and the reason is I insist that the Hamiltonian
be made up of observables and therefore its
eigen values are guaranteed to be real for
the systems we are looking at.
So skip this line and use this fact here so
this guy is the eigen value and eigen value
and its real and phi (t) is the corresponding
eigen state. Of course there is no guarantee
that phi (t) is unique for a given E. there
maybe more than one such eigen state. And
after all when you diagonalize matrices you
know that an eigen maybe repeated in which
case you have two or more eigen vectors for
the same eigen value. If you have finite dimensional
matrices, the eigen values are all discrete.
n by n matrix can at best have n distinct
eigen values but thatís not true when you
have other more complicated operators. In
general, the eigen value spectrum could be
partly discrete, partly continuous and so
on. To start with, itís a real number.
Then it immediately follows that phi (t) is
e to the power ñ i E t over h cross, phi
of 0. In other words, the Schrodinger equation
is solved trivially for an eigen state of
the Hamiltonian. Moreover, the time dependence
of the Hamiltonian of the state vector is
just the initial state multiplied by a phase
vector. A factor whose modulus =1. Whereas,
thatís not true in general it is certainly
true when you have an eigen state of the Hamiltonian
that the time dependence of the state is utterly
trivial.
Any factor of the form e to the i theta is
called a phase factor or the modulus one complex
number. Apart from a phase factor, the initial
state persists. So if you have a quantum system
in an eigen system of its Hamiltonian, it
remains so. Apart from a phase factor which
is unimportant in some sense because it doesnít
change the normalization; phi (t), phi (t)
remains equal to phi of zero, phi of zero,
a single exponent oscillating with a frequency
which is equal to E over h cross. The reason
itís not true in general is because we can
use this now to ask if I can use this to write
down the Schrodinger equation solution in
general.
Well, for a general state psi of t, we could
do the following. we could write psi of t
= a summation over the eigen states of the
Hamiltonian and lets pretend for a minute
that the eigen states are labeled by some
quantum number n. just a set of quantum numbers.
Let me use the index n for it. We will see
later that this n could be a continuous variable.
But symbolically if itís written in this
form Cn phi n where these are eigen states
of the Hamiltonian .Then its easy to say that
I have two ways of looking at this.
One of them is to say I will choose as my
basis set phi n (0) at t = 0 in which case
the time dependence of this gets thrown into
Cn (t). The other way of doing it is to choose
my coefficients once and for all, in which
case the time dependence gets thrown into
this.
Now psi of t = summation over n Cn e to the
power ñ i En t over h cross, phi n of 0.
I will frequently use this without the 0 here
but without any time argument. By that you
must understand that I have chosen a basis
set once and for all which is time independent.
How do I notice itís a basis? There is a
theorem which tells me that if I choose an
appropriate set of mutually commuting observables
pertaining to a system, then these are all
adjoint operators. Then the common set of
Eigen state is can be made orthonormal, is
linearly independent and spans the space;
in other words it forms a basis.
In suspicious cases, I should check whether
I have covered everything. But right now take
it as an assertion that the Hamiltonian in
general, will provide such a basis set. You
could ask what happens about degeneracy. Take
again the hydrogen atom problem. The basis
set is provided by the stationary states of
the system which are labeled by 3 quantum
numbers n, l and m. the energy happens to
depend only on n but in the more general case,
it would depend on n, l and m as well.
So I have used a single symbol to take into
account all possible quantum numbers. In general,
there will be more than one quantum numbers.
In general, there would be as many quantum
numbers as there are degrees of freedom in
the system. And for a particle moving in space,
there are three degree of freedom. That is
why in potential problems in three dimensions
there would be at least three quantum numbers.
There maybe more if there are internal quantum
numbers in the system like spin, hypercharge,
and strangeness and so on and so forth for
elementary particles but there are at least
three quantum numbers. The coefficients are
the overlap at t = 0 of phi n with psi and
they are time independent. Now what is different
about this as opposed to this in the time
dependence? There is a superposition of states
here. There is a single state here and itís
multiplied by a oscillating phase factor.
Here you have time oscillating phase factors
but there are different phase factors for
different states. Notice carefully that as
the state changes, they even change the energy
eigen value to show that they are different
for different states. By the way e to the
i E t is the sin and cosine. So think of it
in terms of sins and cosines. You have got
sin, a single frequency. Thatís just a sinusoidal
wave but you add more and more frequencies.
You add sinusoidal waves with different frequencies
and then what happens is you could have interference.
You could have beats, groups formed, modulation
etc. so all the effects that you could have
when you superpose more than one frequency
would start appearing here. In fact in general,
if this spectrum goes up to infinity, then
all kinds of wild possibilities happen. These
different frequencies are in general not multiples
of each other. So when you add up different
time frequencies which are not even commensurate
with each other, you could even have an aperiodic
function. The function could become aperiodic
which goes right up to infinity.
Itís not a Fourier series because that would
happen if the En was just n. then you will
have cos nt and sin nt. But they are some
complicated functions of n. they may have
n squared n cube and all sorts of dependences.
So it can become horrendously complicated.
in fact all quantum interference phenomenon
happen because of this superposition implied
here.
x and y coordinates are orthogonal to each
other. A general vector in two dimensions
is a superposition of something along x and
something along y. they are orthogonal to
each other. Thatís the whole point so that
the coefficients will uniquely determine this
vector, psi of t. if they were not orthogonal,
thatís not true. Then you have oblique coordinates.
So I have assumed these have been orthogonalized
by the Gram Schmidt procedure and there is
a superposition and the time dependence is
exceedingly complicated. It need not even
be periodic. It could be almost periodic or
quasi periodic. It could be very complicated
in general. So it need not even have periodicity.
They are highly irregular functions. So itís
not a Fourier series at all. In principle
itís written down very easily. But in practice
this is a very intricate function. So this
summation here leads to quantum interference
effects. Letís see what happens in the position
basis in a similar situation.
Well, all we have to do is to plug this whole
thing here and write this as equal to E psi
r and t because I know ih cross of r and t
is E psi of r and t. you have to solve this
equation. Now what does ih cross delta over
delta t psi equal to E psi imply? This immediately
implies that psi of r, t is proportional to
e to the power - iEt over h cross. Only then
if you differentiate with respect to t, you
end up with this quantity here. What does
the constant of proportionality depend on?
It can depend on r. so I shouldnít call it
a constant. I should immediately say this
is equal to a function of r times this.
What notation shall I use for that function
of r? Itís the value of psi of r, t at t
= 0 because if I put t = 0, the exponent goes
to one. So without any loss of generality
I could really write this as psi of r, 0.
Letís change the state vectors as phi with
a subscript n. Again that this little n which
is labeling the eigenstates of the Hamiltonian,
it may actually be more than one label. It
could be a set of quantum numbers. So this
is what a given position space wave function
would do for a stationary state.
And of course this expansion that I have written
down here, if I now work in the position basis
and I write a r here, that goes right across
and hits this. This becomes phi n of r. and
what is cn? It was the overlap in the position
basis between r and the state of the system
at t = 0. So it says that general state at
time t in the position basis, the wave function
is just given by the superposition. The crucial
thing here is to find out what are these functions
and how do I solve them? Well that I solve
by putting it back in here. I use that fact
that H psi = ih d over dt psi, but I use the
fact that this phi is a stationary state by
saying H psi = E psi and I solve it and the
solution looks like that for each of these
eigen states
Now I put that back in this portion out here.
it says - h cross squared over 2 m del squared
phi n of r times e to the power - i Ent by
h cross + V (r) phi n of r e to the ñ iE
nt by h cross = En times phi e to the - iEnt
by h cross phi n of r. thatís what the Schrodinger
equation becomes for a stationary state. But
there is no derivative with respect to t.
I have removed it. I have in fact solved the
t dependence here. And this factor of course
cancels out. Thatís exactly what happens
when you separate variables 
and I am left with this equation here. For
each Eigen state, the position space representative
of that eigen state satisfies this partial
differential equation. Itís still a partial
differential equation but at least you have
got rid of the t. and of course if you work
in one dimension, itís an ordinary differential
equation. This is called the time independent
Schrodinger equation. Whatís the difference
between the time dependent Schrodinger equation
and the time independent Schrodinger equation
as 
a differential equation? The second one is
an Eigen value equation.
This quantity here says take this differential
operator del squared + V of r, act on phi
n of r and you are guaranteed to get a real
number times phi n of r. this is an eigen
value equation. Itís a homogeneous equation.
So is the other one because if you multiply
phi by a constant. Then of course you solve
this, subject to boundary conditions now,
there is no time here, so impose normalization
if you like and then outcome the eigen values.
Now what we are going to do is to take the
time independent Schrodinger equation or the
eigen value equation, and the statement is
if you give me a general state of the system,
no matter what it is and you tell me the initial
state, I will tell you the future state by
actually computing these eigen functions,
substituting in this equation here. This is
a constant which tells me at t = 0 and that
gives me the solution at anytime. So thatís
the strategy. You have to find the Eigen values
of the Hamiltonian and after that the general
time dependence of the state is obvious. Itís
just a superposition of these Eigen functions
with suitable phase factors. So this is the
way you would solve in principle, the Schrodinger
equation.
So we exploit the fact that the Hamiltonian
plays a very special role itís the generator
of time translations for the system. So itís
not surprising that the time dependence of
its eigen states is particularly simple and
that is used to put in the time dependence
of a general state of that. But the price
you pay is you have to solve an eigen value
equation. And now it turns out that very interesting
things happen when you have an eigen value
equation because you are familiar with the
vibrations of a string clamped at both ends.
You have a secondary wave operator, its eigen
values and eigen functions would depend not
only on the operator but also on the boundary
conditions. If you have a string which is
clamped at both ends or left one end loose
and the other end free, the normal modes change
completely. So not only is this differential
operator important but its boundary conditions
are also important in determining the spectrum.
We will see in a minute that if you put the
particle in a box, the energy levels get quantized.
Immediately as opposed to a free particle,
this could have any energy from 0 to infinity.
As soon as you put it inside a box, even if
itís free inside the box, the boundary conditions
will ensure that only a certain set of energy
is allowed. So thatís going to emerge as
a consequence here. So let me make this general
statement that any confinement of a quantum
system tends to quantize something. So you
have put a particle in a box, it quantizes
the energy levels. You put in a potential
which is a confining potential again, the
energy levels get quantized under suitable
conditions. You take angular momentum and
you know angular momentum is quantized in
units of Planck's constant, and the reason
is the corresponding dynamical variables,
the angles really donít run from - infinity
to infinity but run from 0 to 2 pi for the
Azimuthal angle and 0 to pi for the polar
angle. That is enough to produce quantization.
So when Schrodinger first solved the wave
equation and he wanted to reproduce the hydrogen
atom spectrum, he wasnít sure how he was
going to produce a discrete spectrum. Because
you put hydrogen atom you put an electron
in the Coulomb potential of proton, there
is no box. There is no confinement. The potential
extends all the way to infinity. So where
is the confinement? when he consulted David
Hilbert at Gottingen, the greatest mathematician
of the time, pointed out to him that the requirement
of square integrability of the wave function
which was needed for total probability =1
is sufficient to say the requirement of special
case of square integrability, meaning that
the wave function has to belong to a L2. That
boundary condition is enough to make the spectrum
discrete. In the hydrogen atom, that was certainly
the case. So even a condition like without
a physical box, just the fact that the whole
wave function mod squared must be finite,
when you integrate, that is sufficient to
produce a discrete spectrum.
So an important lesson, in eigen value equations
the eigen value is determined not just by
the operator for differential operators but
also by the boundary conditions. That specifies
the class of solutions. So the spectrum itself
is determined by the operator plus the boundary
conditions always. So letís take that problem
and simplify it enormously to do our one dimensional
particle in a box problem. We will complete
this tomorrow but let me set it up now.
Sometimes itís called free particle in a
one D box but I will point out to you that
is not as free as all that. So now I have
the x axis and on the x axis, from 0 up to
l on the x axis I have a box and there is
a quantum particle put inside it. The outside
of the box is not accessible to 
the particle. And I assume that the walls
of the box are perfect reflectors. No absorption
at the end. Classically this is a very simple
problem to solve. So letís do this completely
classically, figure out what happens and then
come back to the quantum problem. We will
write down the Schrodinger equation and solve
it. Because a fairly surprisingly large part
of quantum phenomenon can actually be understood
by this trivial list of problems. So let me
use this as a kind of exercise which we can
do very easily in order to explain all these
phenomena one by one.
The first thing we have to ask is what the
Hamiltonian of the particle is. Classically
the Hamiltonian of the particle is just p
squared over 2 m. just the kinetic energy
+ a potential. So letís write that potential
as a function of x alone. This potential is
zero inside the box and I donít want the
particle outside. So itís infinite outside
rigid walls. What does the phase diagram look
like?
There is x, here is p, this particle can only
exist between zero and l. so this is a forbidden
region and thatís a forbidden region and
then depending on what you do inside, what
the initial conditions are, the particle will
start moving. so classically if you start
at some point here and its moving with a velocity,
with a momentum p, letís say it goes up here
it slams against the wall momentum reverses
comes back here and keeps doing this. So itís
quite clear that this is going to just be
straight line trajectories. so if it starts
here it goes up there, it hits the wall, it
reverses momentum, comes back here, hits this
wall, jumps up, reverses and keeps following
this trajectory. And you have to specify for
me the number p which is equivalent to specifying
the total energy.
So this intercept is square root of 2 m and
this intercept is - square root of 2 m. thatís
the phase trajectory. Itís discontinuous
at the two ends because this particle instantaneously
reverses momentum. There is nothing more.
Thatís it for any p. it is just going to
back and forth in this fashion. Letís do
for Bohr-Somerfield quantization.
The old rule set integral for a periodic orbit
pdx was equal to Planck's constant times an
integer. Thatís the Bohr quantization rule.
This was postulated Niels Bohr. Of course
after the Schrˆdinger equation came in, this
got thrown out but its actually back and this
is the semi-classical result. I will come
back and explain why this is a very profound
result. Suppose we apply that to this problem.
The integral pdx over a closed period is the
area of this thing here.
That area is the area of the rectangle which
is twice square root of 2mE this length multiplied
by the length L times is equal to nh. That
must be true for n = 1, 2, 3, etc. So letís
label the E writing En and this immediately
implies that En = n squared h squared over
8 mL squared. So it immediately predicts quantization
of the energy. It says these are the only
allowed values for the energy.
What we have to see is whether the exact Schrodinger
equation which we are going to solve, imposing
correct boundary conditions, reproduces that.
And in this case this was believed to be true
only for sufficiently large n but it will
turn out here that in this case, this is in
fact true as an exact result. If you solve
the Schrodinger equation, this is exactly
what you get and we will try to understand
why this is true. So that will be our starting
point tomorrow. We will start with the time
independent Schrodinger equation in one dimension
for the simple potential problem and try to
solve this. Meanwhile think a little bit about
whether the energy Eigen states are also position
Eigen states are not for this particle. Thank
you!
