Wave Functions for Particles in a Box Video Lecture - Civil Engineering (CE)

in the last video we introduced the wave
equation for matter waves and which is
called the Schrodinger equation in this
video we're going to use the form called
the time independence for under equation
that we mentioned in the last video to
to solve for the wave functions inside a
certain a certain a certain situation
and that situation is a problem called a
particle in a box and it's it's usually
the first the first problem someone
someone will make you do if you ever
take a quantum mechanics course so what
we have here is is a particle that's
free to move in one dimension you can
move to the left or to the right just in
this one X dimension and it can only
exist between between x equals 0 and x
equals L and the way that we say that
physically is we say that the potential
energy inside the well inside the box or
sometimes it's called
the potential well and so that's why if
I if I use the word well that's just
another name for the box that that
people use so I'm I might say well and
all I mean all I'm referring to is the
box but we say that but how we say that
there's a box here is that the potential
energy inside the box is 0 or some
constant and as is always the case with
energy we can pick we can pick some
constant we can pick we can pick our
zero-point of energy so inside inside
the well we have zero potential energy
of the particle but outside the well we
have infinite potential energy and if we
think about say a ball rolling up a hill
and and say the hill is infinitely tall
we know that you know no finite amount
of energy can ever can ever push the
ball all the way up the hill and so so
so that's our argument for for why the
the particle cannot exist
here and this isn't this isn't a very
realistic situation but it's a what it's
a it's a it's a fairly easy situation to
solve actually solve the Schrodinger
equation and see some of the properties
of these wave functions and so that's
what we're doing in this video and so
already we know what this wave function
looks like outside of this outside of
this box or 4x is less than 0 or X is
greater than L and then and let's write
that down size of X our wave function as
a function of X equals 0 for for the
cases where X is less than 0 or or when
X is greater than L and that's just
saying that there's 0 probability for
for the particle to exist outside of
this box so I'll I'll put this you know
word where our mission in this video is
to to find the wave functions and so so
this is knowledge about the wave
function I'm going to store it in this
little red red rectangle is so we we
don't forget about it because we've
accomplished part of our mission and now
let's let's move on to the other case
where or X or where the where the
particle is inside the well the wave
function inside the well so so inside
the well we say that at the potential
energy of the of the of the particles so
this term equals 0 right this equals 0
multiplied by the wave function whatever
it might be is 0 so this this term in
the equation goes away so down below I'm
going to rewrite this equation and I'll
just copy and paste it to save some time
there it is
we're down here where we have some room
so we're left with this differential
equation to solve for for what the wave
function looks like inside the box so
our first our first thing we do will be
to take both of these constants and get
them on the same side just so we have
less to worry about as far as constants
goes well now rewrite the equation I'll
divide both sides by this and we'll
rewrite it so we have two times M times
the energy M is the mass of the particle
by the way not sure if I've mentioned
that but that's what it is
divided by H bar squared the negative of
that times our wave function I've
accidentally switched to a slightly
different purple but what we can manage
I think is equal to second derivative
with respect to X of our wave function
again so if you watched the video about
solving solving the wave function of a
classical wave the classical wave
equation in the situation that we we
picked before you'll you'll recognize
this this will look very very similar to
what we had before we have some constant
times the wave function is equal to the
second derivative with respect to X of
that wave function and without doing a
general solution I'll just recall in
this case that that the possible
solutions for to this this wave equation
are the wave functions these wave
functions and so sines and cosines are
the are the are the possible solutions
of this
of this wave equation so sine of KX plus
cosine of K times X and I'll include
constants in front of these because we
don't know how much of this wave
function is composed of sines and how
much is composed so of cosines but so we
need to mean to figure these out later
and I've I've included these constants
multiplying K because we know that when
we take the derivative two times of this
this this constant will come out in
front and this constant will be related
to this other constant and if we don't
put these constants here when we take
two derivatives we'll just end up with
with with the exact same function again
with a negative sign because because
they're trig functions so that the
derivative of sine is cosine and the
derivative of cosine is negative sine so
we'll just get the function back with a
negative sign and then we'll then that
will mean that this constant has to be 0
or it has to be 1 but since we don't
necessarily know that we have to include
these extra factors in front of these in
front of our X values just to take into
account that this might be be something
different than 1 so now let's use our
boundary conditions to to try to figure
out what these constants are or find out
something about these constants so our
first boundary condition is that at X X
is less than 0 we know that the wave
function is zero right wave function is
0 up to here and then and then it's
allowed to change and do whatever it
will inside of this inside of inside of
the box but at either end it needs to be
0 and and and a lot like the string the
the the wave function for a matter wave
has to be continuous just like just like
the matter or just like the wave
function or a wave on a string or other
types of waves that we've mentioned
because if it has two values at one
point all of this stuff we're doing kind
of starts to break down so so if if this
is a good way to analyze things this has
to has one have one value and this has
to be continuous so so we have the
boundary condition that at x equals zero
the wave function has to equal zero so
if we plug that value x equals zero into
this into this wave function we get sine
of zero and the sine of zero is zero so
so we don't get any information about
this term so so we'll say that psy of
zero equals this whole term goes to zero
so I'll write a zero here plus B B times
the cosine of zero cosine of zero and we
know that since the cosine of zero is
one the only way to make this equal zero
which we know it equals zero because
this is our boundary condition that size
zero equals zero we know that B has to
equal zero right that's the only way to
to make this equation true so B equals
zero it's more information about what
our wave function looks like so I'll put
in this red box here and so so now we
know that that our wave function has to
has to look like a sine function in this
situation and now to figure out we still
have to figure out what this K looks
like what are their restrictions on this
K that that that a lot dead does this K
have to equal certain things for this
wave equation to be satisfied so let's
look at our second boundary condition
just like X has to or just like the wave
function has to equal zero at x equals
zero it also equals zero at the other
side of the box at x equals L so we have
sy of l equals zero our wave function
evaluated at l equals zero I'll write
this out here sy of L
goals are constant a times the sine sine
of K times L all that equals zero equals
zero and just like before in the in the
other video about solving the classical
wave function or the classical wave
equation we know that at the sign of the
sine function equals zero when when its
argument equals some multiple of Pi so
if K L K L must equal K almost equal
some multiple of pi n times pi over N
and give you 1 2 3 etc etc and if we
follow this just divide both sides by L
you get that K equals n PI over L so we
have this requirement on K so this is
really just like the other the other
time we solved for the wave the wave
functions of the other situation in the
classical classical wave situation so
over here I've drawn these these these
wave functions just to just to drive
home the analogy to the other the other
system that at these these wave
functions for matter waves in this
situation look just like the wave
functions of the waves on a string so we
have this one bump who are assigned the
sine function goes from sine of 0 to 2
sine of pi and it goes as positive and
it comes back to 0 when x equals L for
the case that N equals 1 in and
similarly for N equals 2 it it goes it
goes twice as far through a so goes from
sine of 0 to sy sine of pi to sine of 2
pi
so we see that at least in this
situation we have the same the same
really the same shape wave functions so
we've maybe a knowledge analogy to the
other wave functions of a classical wave
but now we can we can think about how
when we take the derivative of our wave
function will actually get K to come out
side because of the chain rule so we can
actually make a relationship between K
in E and then in the next video I'll
I'll solve for e and we'll talk about
the energies of the of this system see
you in the next video