Schrödinger equation for hydrogen - Atomic Structure Video Lecture | Chemistry for GRE Paper II

so we'll look at the Schrodinger
equation we're going to work for bound
states we could work with scattering
States in spherical coordinates but it's
usually done in advanced courses the
most important thing is the calculation
of the bound States this is kind of a
neat part of quantum mechanics udas in a
sense you get accustomed in quantum
mechanics about uncertainties you cannot
predict the probability that the photo
will go through this branch of the
interferometer or this other probability
you can't be sure and all this but here
you get the energy levels and you get
the exact energy level so it's a nice
thing that in quantum mechanics energy
levels are things that you calculate
exactly of course when you try to
measure energy levels experimentally the
uncertainties arise again you get the
photon of some energy and they're just
an energy time uncertainty or something
like that that can bother you but the
end result is that these systems have
beautifully the term in fixed numbers
called energy levels and that's what
we're aiming at it's a nice thing
because it's the most physical example
it has many applications the energy
levels of hydrogen atom the more you
study the more complicated they are
because you can include fine effects
like the effects of the spins of the
particles what does it do the specs of
relativity what does it do all kinds of
things you can put in and do more and
more accurate results so it's really
unbelievable how much you can learn with
the fine spectrum of the hydrogen atom
our equation is minus H squared over 2m
ii V R squared the radial equation plus
h squared L times L plus 1 over 2m R
squared minus Z e squared over R U is
equal to e
you so this is the radial equation
remember it was like a sure dinner
equation for a variable U and the wave
function was u over our time subsea LM
or a wire LEM actually a spherical
harmonic here I could have labeled you
with E and L because certainly u depends
use U of R and u depends on the energy
that we're gonna get and it will depend
on L that is there in the differential
equation this is the effective potential
that we discussed before was the
original potential to which you add the
centrifugal barrier and what are you
supposed to solve here you're supposed
to solve for L equal zero to find some
states where I go one two three four
infinity you should find all the energy
levels of this thing so the wave
function is I of R theta and Phi will be
this U of R over R times a ylm of theta
and Phi and in fact plugging that into
the Schrodinger equation was what gave
us this radial equation so this was
called the radial equation which we
talked about we've never quite solved it
for any particular example okay as you
know if we like in this course to get
rid of units and constants so first step
is to replace R by a unit free X X and
we have the right quantity a 0 a 0 would
be the perfect thing so you if you do
that you will be able to clean up the
units
but if you do that you might not quite
be able to clean up the Z from all the
places that you would like to clean it
up so we can improve that maybe it's not
too obvious by putting a 2 over Z here Z
has no units of course so so that you
would probably do by trial and error you
would say well I want to get rid of the
Z as well and I have it in this form and
though you will see how it works out so
at this moment I have to plug our into
this equation and just clean it up and
what should happen is that everything
all the unit all the units should give
you a factor with units of energy
because at the end of the if whatever is
left is not going to have units
everything out must have units of energy
so this takes about one line to do I'll
skip some algebra on these things I'll
try to post notes soon on this so if you
leave a line you could do the
calculation for yourselves and though it
might be worth it the claim is that we
get 2 Z squared e squared over a 0
multiplying minus ii be x squared plus L
times L plus 1 over x squared minus 1
over x times u equals e you
that will be the common factor that will
come out of everything by the time you
you solve it and it has the units of
energy as you can imagine a squared over
a naught has a units of energy so might
as well move that to the other side to
get the final form of the equation which
would be minus d ii the x squared plus L
comes L plus 1 over x squared minus one
over X U is equal to minus Kappa squared
u / Kappa squared it's going to be minus
e over 2 Z squared little a squared over
a 0 ok the trick on all these things is
to not lose track of our variables and
that it's a little challenging so let me
just reinforce what's going on here
we've passed from our variables to u
variables that we will still do more
things and we have a knot in there Z in
there and then the energy is really
encapsulated by Kappa squared here
um and it's Kappa is unit free so if you
know Kappa you know the energies that's
what you should be looking for Kappa is
the thing you want to figure out knowing
Kappa is the same as knowing the
energies because this is just a constant
that gives you the scale of
energies so this is what we want to
solve so an equation doesn't look all
that complicated but it's actually not
yet that simple
unfortunately so we have to keep working
with it a bit so one reason you can see
that the an equation like that would not
be too simple is to look at it and see
what kind of recursion relation it would
give you and that's a good thing to look
at the beginning and you see okay here
I'm going to lower the number of powers
by two here I'm going to lower the
number of powers by one and here I'm not
going to lower the number of powers
there are three terms so it's not a
simple recursion relation which the next
term is determined by the previous one
so that suggests you better do some work
still with this equation to simplify the
situation and several things that you
can do one thing is to look at the
behavior of the equation near infinity
near zero and see if you you see
patterns going on one thing we're going
to do is to which is so not urgent but
it's usually done and and people do it
in different ways is to look at our X
goes to infinity and see what the
solutions may look like so as X goes to
infinity the differential equation
probably can be approximated to keep
this term to see how things vary and X
goes to infinity throw these throw this
and keep that so you would have this
second you the x squared is equal to
Kappa squared
so this suggests that you goes like e to
the plus minus Kappa X so exponential
behavior e to the plus minus Kappa X I
really of course for our solutions we
would like the minus one but we will see
what the equations do now this is just
yet another transformation that people
do which is look Kappa is dimensionless
and we had X that is unit free also so
Kappa has no units X has no units so
let's move to yet another variable you
see we started with our being
proportional to X and now we can put
factors that may help us without units
here so I'm not suggesting that this is
something that would occur to me if I'm
doing this problem but it's certainly a
possible thing to do to say okay I'm
going to define our row as Kappa X and
with X being given by this row would be
2 Kappa Z over a not R so Rho is going
to be my new coordinate so from our with
I'm sorry we've gone to X and now we've
gone to row
and a new variable over here so what
happens to the differential equation
well it's going to be a little better
but in particular the solutions may be a
little better but here it is if you look
at the differential equation the
differential equation here think of
moving the Kappa squared here below and
then you see immediately it fits
perfectly well in the first two terms so
you get minus the second zero squared
plus L times L plus 1 Rho squared and
here it doesn't fit all that well you
would have a 1 Kappa left over so 1 over
Kappa Rho u equals minus u yes it's kind
of a suggestive the Kappa has almost
disappeared from everywhere but they
better not disappear from everywhere if
it would have disappeared from
everywhere would have been improved
because the equation would have not fix
the energy you see we're hoping that the
differential equation will have initial
solutions for some values of the energy
and for the others not so the energy
better not disappear it came close to
disappearing the Kappa but it's still
here so we're still okay and now you
would say okay this is nice so if you
look at the Rho going to infinity again
it works as we wanted you get the sake
and you the Rose squared is equal to you
and that means you goes to e to the plus
minus Rho which is what inspired this
and the other part of the solution is
the solution at near are going to zero
or your X going to 0 or near Rho equal
going to 0 so near Rho going to 0 you
have these two terms and we actually did
it last time we analyzed what was going
on with this equation
last time near Rho equal to 0 and we
found out that you must behave like Rho
to the L plus 1 for Rho going to 0 so it
was from these two terms the
differential equation and for Rho going
to 0 remember the wave function must
vanish and how fast it should vanish
should vanish to this power to the L
plus 1 okay so you know lots of things
about this function so first of all that
this thing doesn't have necessarily
polynomial solutions because it behaves
exponentially moreover you know it
doesn't start with constant plus Rho
plus Rho squared it starts with Row to
the L plus 1 so it is pretty important
to see all these things before you try
to do a recursion relation because
regeneration might lead you to funny
things so so here we go what do we do
based on all this we try something
better which is we set U of Rho the
solution we write an answer is going to
be Rho to the L plus 1 that will
have the right behavior for ro going 2-0
an unknown function W of Rho times e to
the minus Rho which is the right
behavior we want now there's no
assumption whatsoever when you're
writing and sets of this form
you're just expressing your knowledge
because at the end of the day if Rho if
Omega or W here actually is undetermined
this W could have an e to the Rho that
cancels this factor and maybe off with
some funny power that cancels this this
is just a hope that we're expressing
that by writing the solution in this
form this quantity may be simple because
we know this is present in the solution
for far large Rho this is present for
small row well in between we might have
that and when you write analysis of this
form you you hope for a simple
differential equation for W so what do
we get
well we can plug that answers into the
differential equation that we already
have and see what happens
and indeed that's what we'll do I'll
skip the calculation in my notes it took
me half a page and I write big so it's
not too long so what do we get
you get an equation that doesn't look
that simple I'm sorry it just that looks
like a step back but it's not the second
W D Rho squared plus two times L plus 1
minus Rho strange V W D Rho plus one
over Kappa minus 2 2 times L plus 1 W
equals 0 ok
aesthetically looks worse certainly that
equation on the Left board looked nicer
but ah actually it's it's pretty good
because again now look at your recursion
relation how will it be if you take some
power fix power here you lose 1 power
here with the L plus 1 on this you lose
1 power here you lose nothing and here
you lose nothing so you have either one
powerless or your power so it's a
one-step recursive relation without the
gap it's not like the two steps that we
had for the harmonic oscillator for the
Legendre polynomials here is one step
recursion relation a k plus 1 determined
by a K so we
say excuse me to the equation you don't
look that good but you are very solvable
so we can proceed