The Schrödinger equation
Announcements:
• Next homework assignment is
available
• I will be giving a public talk about
physics at the Large Hadron Collider
(LHC) at 2pm tomorrow in G1B30. If
you are interested in particle physics,
you may find it interesting.
Erwin Schrödinger
(1887 – 1961)
• I have decided to cancel the class
scheduled for Friday, March 20.
We begin working with the Schrödinger equation.
Today will be free particle and Monday will be the
infinite square well (particle in a box).
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
1
Getting to Schrödinger’s wave equation
2E 1 2E
 2 2
2
x
c t
Works for light (photons), why
doesn’t it work for electrons?
We found that solutions to this equation are
y  A cos(kx) sin( t )  B sin( kx) cos(t )
or y  C sin( kx  t )  D sin( kx  t )

k

with the constraint
c
which can be written   kc
Multiplying by ħ we get   kc
which is just E  pc
But we know that E=pc only works for massless
particles so this equation can’t work for electrons.
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
2
Getting to Schrödinger’s wave equation
2E 1 2E
 2 2
2
x
c t
doesn’t work for electrons. What does?
y  A cos(kx) sin( t )  B sin( kx) cos(t )
or y  C sin( kx  t )  D sin( kx  t )
Note that each derivative of x gives us a k (momentum)
while each derivative of t gives us an  (energy).
Equal numbers of derivatives result in E  pc
p2
For massive particles we need K 
2m
So we need two derivatives of x for p2
but only one derivative of t for K.
If we add in potential energy as well
we get the Schrödinger equation…
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
3
The Schrödinger equation
The Schrodinger equation for a matter
wave in one dimension (x,t):
2

( x, t )
( x, t )


 V ( x, t )( x, t )  i
2
2m x
t
Total
Kinetic
Potential
= energy
+ energy
energy
2
This is the time dependent Schrödinger equation (TDSE)
(discussed in 7.11) and is also the most general form.
Note, we now use V instead of U for the potential energy.
This potential energy is a function of x and t. It gives the
potential energy of the particle for any x and t. It is not intrinsic
to the particles but something from the problem at hand.
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
4
Deriving time independent Schrödinger equation
2  2( x, t )
( x, t )


 V ( x, t )( x, t )  i
2
2m x
t
In most physics situations (like hydrogen atom) the potential
function V does not change in time so can write V(x,t) = V(x).
In this case, we can separate (x,t) into y(x)f(t): ( x, t )  y ( x)f (t )
Can prove that f(t) is always the same: f (t)  e iEt / 
Text uses f (t)  e i t which is exactly the same since E  
Putting ( x, t )  y ( x)f (t ) with f (t)  e iEt / , into the TDSE
gives the time independent Schrödinger equation (TISE):
2
d
y ( x)


 V ( x)y ( x)  Ey ( x)
2
2m dx
2
HW problem 8 takes you through these steps in detail.
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
5
Solving quantum mechanics problems
2 d 2y ( x)


 V ( x)y ( x)  Ey ( x)
2
2m dx
1. Determine V(x) for the problem.
2. Guess (or lookup) the functional form of the solution
3. Plug in solutions to check that solution is correct and find any
constraints on constants
4. Figure out boundary conditions and apply them to the general
solution to get more constraints 
2
5. Use the normalization constraint  y ( x) dx  1 to normalize the

spatial wave function.
6. Multiply by time dependence for full solution: (x, t)  y (x)eiEt / 
7. Use solution to obtain information like graphs of y(x) and
|y(x)|2, energy levels, probabilities on positions, etc.
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
6
Clicker question 1
Set frequency to DA
Given a potential energy function V(x), where would a particle
naturally want to be?
A. Where V(x) is highest
Particles want to go to
B. Where V(x) is lowest
position of lowest
C. Where V(x) < kinetic energy
potential energy, like a
D. Where V(x) > kinetic energy
ball going downhill.
E. Does not depend on V(x)
V(x)
x


Mathematically we know this because F  U
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
7
Free particle
2 d 2y ( x)


 V ( x)y ( x)  Ey ( x)
2
2m dx
Start with simplest
case of a free particle
(no forces acting on it).
What is V(x) for a free particle?
Constant. Smart (easy) choice is V(x)=0.
What are the boundary conditions for a free particle?
None.
2 d 2y ( x)

 Ey ( x)
So we need to find what functions satisfy 
2
2m dx
but we don’t need to apply boundary conditions.
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
8
Clicker question 2
Set frequency to DA
2 d 2y ( x)

Which of the following are possible solutions to 
 Ey ( x)
2
2m dx
A. A cos(kx)
d 2y ( x)
 kx  k 2y ( x)
Try
answer
B:

A
(

k
)(

k
)
e

kx
B. Ae
dx 2
C. A sin( kx)
2 2

 k y (x)  Ey (x)
This
gives
us
2m
D. A and C
E. None of A,B,C
This works as long as k 2   2mE
2
But k is a real measurable thing so it cannot
be complex and therefore k2 is positive.
Since the only energy we have is kinetic, it must also be
positive so the quantity 2mE/ħ2 must be positive.
So answer B does not work. Answers A and C both work.
2 2

k y (x)  Ey (x)
d 2y ( x)
2
2
  Ak cos(kx)  k y ( x) which gives 2m
For A:
dx 2
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
9
Clicker question 3
Set frequency to DA
2
Putting either y(x)=Acos(kx) or y(x)=Asin(kx)   2 d y ( x)  Ey ( x)
2
into the TISE we end up with 2k 2 y (x)  Ey (x) 2m dx
2m
2 2
This works as long as  k  E Since p=ħk this just means the
2m
total energy is the kinetic energy
Based on this, what can we say about the free particle energy?
A.
B.
C.
D.
E.
Quantized as En = E0n2
Quantized as En = E0n
Quantized as En = E0/n2
Quantized in some other way
Not quantized
The only requirement is E=p2/2m. There is no quantization.
The energy of a free particle is not quantized!
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
10
Free particle
The general (spatial part) solution for a free particle is:
y ( x)  A cos(kx)  B sin( kx)
Using some identities you will be proving in the homework,
this can also be written as
ikx
 ikx
y (x)  Ce
 De
To get the full wave function we multiply by the time dependence:
( x, t)  y ( x)e it  Cei(kx  t)  De i(kx  t)
This is the sum of two traveling waves,
each with momentum ħk and energy ħ.
The C wave is moving right and the D wave is moving left.
Lets pick the right going wave and check it one more time.
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
11
Check free particle solution in TDSE
TDSE:
2  2( x, t )
( x, t )



V
(
x
,
t
)

(
x
,
t
)

i

2m x 2
t
For a free particle, V(x,t)=0 and the solution
( x, t )  Cei(kx  t )
for a particle moving to the right is
If we plug this into the TDSE, what do we get?
 2( x, t )
i(kx  t )   Ck 2ei(kx  t )   k 2( x, t )

C
(
ik
)(
ik
)
e
x 2
( x, t )
  Ciei(kx  t )   i( x, t )
t
2 2

k ( x, t)  ( x, t)
Putting it all together gives:
2m
It all works!
http://www.colorado.edu/physics/phys2170/
Kinetic
energy
=
Physics 2170 – Spring 2009
Total
energy
12