Let's solve the free particle in one dimension
This problem is a problem where I have a particle
of mass m, moving in one dimension,
we'll call it the x-dimension, and it's moving in a
constant potential. So here's my particle
The particle has mass m, moves only in one dimension
again the x-direction. And it can be anywhere from
x equals negative infinity to positive infinity
We start by writing down the Schrodinger equation
h-bar squared over two times the mass
times the second derivitive of the wave function
with respect to x, plus the potential
times the wave function, is equal to the total
energy time the wave function
Now I wrote the derivative as the normal derivative
in this case because this is a problem in one dimension
and I don't have any other variables expect for x
and so I can write just a normal derivative
I don't need to write a partial derivative
And I wrote the general form of the potential as being
a function of x, but in this particular case
we know that the potential is actually a constant in x
And so when I move the potential to the right side of the equation
I'm gonna re-write it as just a constant
So let's go ahead and do that
Moving that potential to the right side of the equation
So my first term stays the same
This is equal to the total energy, minus the potential
energy, times the wave function
And now let's multiply both sides by negative two times
the mass, divided by h-bar squared
So I have my second deriviative on the left side
of this equation, all by itself
So all I've done is re-written the Schrodinger equation
and by rearranging it and writing the potential as
as a constant. At this point it's convenient to define
a constant k where this constant is equal to
two times the mass divided by h-bar squared
times the total energy, minus the potential
and then we are going to take the square root of this whole thing
And when I do that, I can re-write my second order
differential equation, that is my rearranged Schrodinger equation
as being the second derivative of the wave function with
respect to x is equal to the negative of this
constant squared, times the wave function
So to solve this, what I need to do is just look for
functions that when I take their second derivative
I get the function back times a negative constant
So there's actually several functions that will do this
One is sine of kx. Another function is cosine of kx
Another function that works is e to the ikx
and yet another function is e to the -ikx
So let's just substitute in sine of kx to show that
it actually solves the Schrodinger equation
So if I take the second derivative of sine of kx
with respect to x, what I do first is I take the
first derivative with respect to x
When I take the derivative of sine of kx with respect to x
I get k cosine of kx
And I still have to take the derivative one more time
And when I take the derivative again
k is just a number but derivative of cosine kx with
respect to x is -k*sine of kx
So i'm gonna get a negative k coming out
which give me a k squared
and a negative, and I get my sine of kx back
So by taking the second derivative of sine of kx
i get sine of kx back time -k squared
so sine of kx does indeed solve this
second order differential equation. It's an eigen function
of the Schrodinger equation
You can do the same thing for cosine of kx
and also show that it's an eigen function of
the Schrodinger equation
So the general solution would thus be
that the wave funciton is equal to some constant
times sine of kx plus a different constant times
cosine of kx. Where A and B are constants that we would
determine by applying the boundry conditions for the
specific problem. In this particular case
We have a free particle, there are no boundy conditions
So A and B are arbitrary
I aso mentioned that e to ikx and e to the -ikx
are solutions to the Schrodinger equation
Let's show that for the case of e to the ikx
So we're gonna do the same thing that we did earlier
And that is we're just gonna plug in
this potential solution, e to the ikx, into our
differential equation and see what we get
So let's take the second derivative of e to the ikx
when I do that, I get hmmmm, I'm gonna take the
derivative twice, and each time I take the derivative
of e t the ikx, I'm gonna get a factor of ik times
e to the ikx. Since I take the derivative twice
I get the factor Ik twice, so I'm gonna square
ik, and I get back e to the ikx
Of course, ik squared is equal to -1, that is, i sqaured
is -1, times k squared. So I get -k squared
times e to the ikx
that is when I plug e to the ikx into this second order
differential equation, I get e to the ikx back times
a negative constant, so e to the ikx is also a solution
and I can write the general form of the solution as
A constant times e to the ikx plus another constant
let's say D, e to the minus ikx
And thats another general solution, and either one of
these works just fine. Sometimes, the solution
in terms of the exponentials, e to the ikx
and to the -ikx is convenient beacuse it turns out
that e to the ikx and e to the -ikx are eigen functions
of the momentum operator. And what we see is
that the first term represents the part of the wave
function that is the particle moving to the right
which momentum h-bar*k
and that the second term in theis general wave function
is the part of the wave function that is the particle
moving to the left with momentum -h-bar*k
This example illustrates a couple of key points
It's the easiest to problem in quantum mechanics
It's the easiest case to solve, I have constant potential
with no boundry conditions, I get two potential solutions
One in terms of sine and cosine, and a completely
one terms of e to the ikx and eto the -ikx