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Lecture 14
An introduction to Schrödinger
Solving the time dependent Schrödinger equation
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• Read through your notes in conjunction with lecture presentations
• Try some examples from tutorial questions or Phils Problems
• Email me or come to E47 if anything is unclear

u
2

 u  Vu  i
2m
t
2
Remember Phils Problems and your notes = everything
http://www.hep.shef.ac.uk/Phil/PHY226.htm
SUMMARY of the procedure used to solve PDEs
 2 y ( x, t ) 1  2 y ( x , t )
1. We have an equation
 2
x 2
c
t 2
with supplied boundary conditions
2. We look for a solution of the form y ( x, t )  X ( x)T (t )
3. We find that the variables ‘separate’
1 d 2 X ( x)
1 d 2T (t )
 2
N
X ( x) dx 2
c T (t ) dt 2
4. We use the boundary conditions to deduce the polarity of N.
e.g. N  k 2
5. We use the boundary conditions further to find allowed values of k and hence X(x).
X ( x)  A cos kx  B sin kx so X n ( x)  Bn sin nx
L
6. We find the corresponding solution of the equation for T(t).
 nct

T
(
t
)

E
cos



n
T (t )  C cos kct  D sin kct n
L


7. We hence write down the special solutions. Yn ( x, t )  Bn sin nx cos nct  n 
L
 L

8. By the principle of superposition, the general solution is the sum of all special

nx
 nct

solutions..
www.falstad.com/mathphysics.html
y ( x, t )  B sin
cos
 

n1
n
L
 L
n

9. The Fourier series can be used to find the particular solution at all times.
y ( x, t ) 
8d  x
ct 1 3x
3ct 1
5x
5ct 1
7x
7ct

sin
cos

sin
cos

sin
cos

sin
cos



L
L 9
L
L
25
L
L
49
L
L
 2 
Introduction to Quantum Mechanics
Something weird is going on……justification of Quantum Mechanics
If you were to shine white light through sodium gas you would see an absorption line
in the total spectrum.
Passing a current through sodium gas causes the emission of monochromatic light
Sodium street light
Spectral lines come from the fact that atoms absorb or emit
particular energies of photons.
Introduction to Quantum Mechanics
Something weird is going on……justification of Quantum Mechanics
Hydrogen gas for example can therefore only absorb or emit specific wavelengths
of light. When a photon is absorbed an electron is promoted to a higher energy
level, but only specific energy photons can do this, it follows that discrete energy
levels exist in the atom.
This provides us with a spectral fingerprint
that tells us about the inner structure of the
atom.
E=hf
Looking at the continuous spectrum and comparing it with the Hydrogen emission
spectrum we see that the five emission lines are at 1.9, 2.5, 2.8, 3.0, 3.1eV.
Introduction to Quantum Mechanics
Something weird is going on……justification of Quantum Mechanics
The energy levels are said to be quantised and when the electron moves from one
energy level to another they are said to make quantum jumps.
For example when a current is passed through sodium gas, electrons are promoted
to higher energy levels, they then drop to lower energy levels and emit discrete
energies or wavelengths of light.
The energy of the photons emitted are equal to the differences between the energy
levels so the 2.1eV photons emitted by street lights tell us that there is an energy
gap between levels of 2.1eV.
Introduction to Quantum Mechanics
Something weird is going on……justification of Quantum Mechanics
The usual position of the electron in the hydrogen atom is in the ground state E1
whereas E∞ corresponds to the situation where the electron is separated from the
proton by an infinite amount.
five emission lines are at
1.9, 2.5, 2.8, 3.0, 3.1eV
E∞ - E1 corresponds to the ionisation
energy which is 13.6eV for Hydrogen.
NB. The energy of the
emission line is determined
by ΔE (the difference in
energy states)
Once the electron has been ionised, any
remaining energy increases the kinetic
energy of the proton and electron which,
existing in the continuum region can
have any value of energy.
Although there are 5 visible
lines there are many more
in the EM spectrum
Introduction to Quantum Mechanics
Prediction of atomic structure
Lots of people tried to explain atomic structure before quantum mechanics came along.
Rutherford’s atom had a central positive nucleus with a negative electron orbiting it.
The electrical attraction held them together and
the rotational motion kept them apart. In the
model the electron can exist at any radius so
long as the rotational velocity increases to
compensate.
This model therefore does not predict discrete energy
levels or explain line spectra.
Introduction to Quantum Mechanics
Prediction of atomic structure
Bohr then developed the Newtonian and electromagnetic ideas of electron motion
and managed to construct a Hydrogen atom with the discrete energy levels required
for line emission.
The model combines classical orbiting
electron with quantized electron
momenta. The electrons therefore still
orbit the nucleus but in this model they
are restricted to specific radii.
However when an electron rotates it constantly undergoes acceleration and emits
electromagnetic radiation. Therefore we would expect the electron to constantly
emit photons, losing energy constantly and eventually spiralling into the nucleus.
This doesn’t happen!!
Introduction to Quantum Mechanics
Quantum indeterminacy
If you bombard a hydrogen atom with photons of high enough energy to promote the
electron from E1 to E3 then sometimes it will do this and other times it wont !!!
The same occurs for an electron in an excited state that can either drop down one or
more energy levels.
We can never know if an individual atom has absorbed a photon or not and the best
we can do based on statistics is to assign a probability to whether or not the process
will occur.
Introduction to Quantum Mechanics
Schrödinger equation
Both the Rutherford and Bohr models of the atom are therefore flawed.
In the 1920s a group of Physicists headed by Schrodinger developed what we now
know as the Schrodinger equation. The equation did two main things.
It predicted the energy levels of the H atom.
But it also introduced the concept that the behaviour of the electron is intrinsically
indeterminate.
According to Quantum mechanics, if a H atom has a certain amount of energy it is
impossible to say in advance of the measurement what value will be obtained for
the electrons position and momentum.
This means that if we perform identical measurements on an atom with the same
energy, we will always have different outcomes. What can be predicted are the
range of possible outcomes of the measurement and the probability of each of
these outcomes.
Introduction to Quantum Mechanics
Schrödinger equation
A good way of illustrating the uncertainty of the position of the electron is to show it in
2D as an electron cloud.
If we imagine that for a single atom we measure the
position of the electron 1000 times and place a
single dot on the paper at each location we found
the electron to be then we would end up with an
electron probability cloud.
The greater the concentration of dots, the more likely the electron would be found
at this location. The maximum concentration of dots corresponds to a radius of
5.29×10-11m which is exactly the radius predicted by the Bohr model when the
electron is in its ground state.
Introduction to Quantum Mechanics
Schrödinger equation
So far we’ve only talked about the ground state but the
Schrodinger equation can be used to predict the behaviour of
the electron in any of its energy states (eigenvalues).
The figure right shows the probability density clouds of the
ground state of the H atom and also other higher energy
states.
But is this really the best way we can represent
the likelihood that an electron will be in a certain
position at a certain time?
The use of electron probability clouds to predict the
probability associated with measuring in the quantum world is
visually very clear but nowhere near as useful as the
eigenfunctions we are now so familiar with.
Introduction to Quantum Mechanics
The electron probability cloud is analogous to the probability density function given
below, expressing the probability P of finding the particle at some particular location
between b and a.
x b
*

 ( x, t )( x, t )dx  P
xa

Remembering that
*

 ( x, t ) ( x, t )dx  1

The position of the particle is, of course, one quantity we might imagine measuring
experimentally. It is an observable quantity.
But there are many physical observables. One is the energy, which we determine
from the solutions of the TISE.
2 d 2
( x)  V ( x) ( x)  E( x)
We usually see it written like this: 
2
2m dx
OK, so I know that I have to start thinking about
operators and eigenfunctions and eigenvalues but
where do we start ????!!
What you have to realise is that because there is a probability associated with
pinning the particle down to a specific energy or position or momentum, with every
calculation we need to incorporate the probability associated with the measurement.
We do this using OPERATORS such as the one below…..
You are most familiar with using the TISE to find the specific energy levels associated
within a zero potential well where:
2
2

 d  ( x)
 E ( x) for 0  x  L
2
2m dx
2 d 2
H ( x)  E ( x) where H denotes the Hamiltonian operator H  
 V ( x)
2
2m dx
The whole point of Quantum mechanics !!!!!!!!!
Quantum mechanics does not explain how a quantum particle behaves. Instead, it
gives a recipe for determining the probability of the measurement of the value of a
physical variable (e.g. energy, position or momentum). This information enables us to
calculate the average value of the measurement of the physical variable.
Introduction to Quantum Mechanics
Operators play a crucial role in the theory of quantum mechanics, as each experimental
observable is associated with an operator. A “hat” usually denotes operators.

An equation of the form
The values of
O  ( x)    ( x)
n
n
n
is called an eigenvalue equation.
n are called the eigenvalues, and the corresponding functions n (x)
are called the eigenfunctions.
The allowed values of the observable are the eigenvalues of the operator, each
corresponding to a function (the eigenfunction) which represents the state of the system
when the observable has that value.
The TISE is such an equation. The allowed values of energy are the eigenvalues of
the Hamiltonian operator, and the corresponding wavefunctions are its
eigenfunctions.
H ( x)  E ( x)
Eigenvalue equation
n ( x) 
2
 nx 
sin 

L  L 
Eigenfunction
 2 n 2 2
En 
2m L2
Eigenvalue
Introduction to Quantum Mechanics
Let’s show how we can find the eigenvalues of energy in zero V using an operator ….
H ( x)  E ( x)
Eigenvalue equation
n ( x) 
2
 nx 
sin 

L  L 
Eigenfunction
2 d 2
H 
 V ( x)
2
2m dx
Operator
 2
2 d 2  2
nx 
nx 




H 
sin
 E
sin
2 

2m dx  L
L 
L 
 L
 2
 2 n d  2
nx 
nx 




H 
cos
 E
sin



2m L dx  L
L 
L
L


 2
 2 n 2 2  2
nx 
nx 




sin
 E
sin
2 


2m L  L
L 
L
L


 2 n 2 2
So
 En
2
2m L
as expected
 2 n 2 2
En 
2m L2
Eigenvalue
Introduction to Quantum Mechanics
The list of operators is given below:
If we measure the momentum of the momentum eigenfunction,  ( x) 

The operator is, p 
 d
 d
 ( x )  p ( x )
and so
i dx
i dx
exp( ikx)
L
 d

k
exp( ikx) 
ik exp( ikx) 
exp( ikx)  p ( x)
i L dx
i L
L
k
p
exp( ikx) 
exp( ikx) so
L
L
p  k
Here p is the eigenvalue,
and ψ(x) is the eigenfunction
Introduction to Quantum Mechanics
Infinite potential well
The particle cannot exist where the potential
is infinite, so the boundary conditions are:
V ( x)  0 for 0  x  L
 ( x)  0 for x  0
V ( x)   otherwise
 ( x)  0 for x  L
 2 d 2  ( x)

 E ( x) for 0  x  L
2
2m dx
2
d
d 2  ( x)
2mE
and write  ( x)  k 2  ( x) where



(
x
)
Re-arrange as
dx 2
dx 2
2
As always set
  emx
d 2  ( x)
 m 2  ( x)
so
2
dx
General solution is  ( x)  A sin( kx)  B cos( kx)
 ( x)  0 at x  0 so B  0
2
2
and m  k
k2 
2mE
2
so m  ik
Boundary conditions are
 ( x)  0 at x  L so A sin( kL)  0 so kL  n
Introduction to Quantum Mechanics
Infinite potential well
2mE
The energies are given by k 2  2

eigenfunctions
nx 

 L 
Thus the solutions are n ( x)  A sin 
 2 k n2  2 n 2 2
n
where E n 
as
k


n
L
2m
2m L2
eigenvalues
Notice how as a consequence of the boundary conditions on ψ(x) at x = 0 and L we
must fit an integral number of half-wavelengths into the potential well of width L
Introduction to Quantum Mechanics
Infinite potential well
nx 

 L 

For each eigenfunction n ( x)  A sin 
the probability of finding the particle in

  ( x, t )( x, t )dx  1
the well is unity. Thus,
*

 nx 
dx  1 so
This determines A: A  sin 
 L 
0
L
2
2
1
 2nx  

1

cos

 dx  1
0 2 
 L 
L
A
2
L
so
1 
L
 2nx  
A 2   x 
sin 
   1 so
2n
 L   0
2 
A2 L
1
2
and therefore A  2
L
So the eigenfunctions n ( x) 
 2 k n2  2 n 2 2
En 

2m
2m L2
2
 nx 
sin 

L  L 
and their corresponding eigenvalues
have been found using the kinetic energy operator
Introduction to Quantum Mechanics
Finite potential well
V ( x)  0 for 
Notice that E = KE + PE, thus
KE(x) = E - PE = E -V(x)
L
L
x
2
2
V ( x)  V0 otherwise
Eigenfunctions with energy eigenvalues E > V0 are unbound
Eigenfunctions with energy eigenvalues E < V0 are bound.
For bound states the wavefunction penetrates the classically forbidden region. Thus, the
particle exists in a region where its kinetic energy is negative.
To find energies of these states we’ll solve the time independent Schrödinger equation:
Introduction to Quantum Mechanics
Finite potential well
V ( x)  0 for 
L
L
x
2
2
REGION I
V ( x)  V0 otherwise
We need to solve:
 2 d 2  ( x)
L
L


E

(
x
)
in
region
I
for


x

2m dx 2
2
2
In region I solutions are
I ( x)  A cos(kx)
Using same technique as for infinite well but
for different boundary conditions
NB. The value k above is actually smaller than the corresponding
value k for the inifinite potential well.
This means that
I (x) is not 0 at the potential well boundaries.
2 2

Also since E  k n
n
2m
the energy levels are lower compared to IPW
IPW
FPW
Introduction to Quantum Mechanics
Finite potential well
V ( x)  0 for 
L
L
x
2
2
REGION II
V ( x)  V0 otherwise
We need to solve:
 2 d 2  ( x)
L


V

(
x
)

E

(
x
)
in
region
II
for
x

0
2m dx 2
2
Re-arrange to
d 2 ( x) 2m(V0  E )

 ( x)
2
2
dx

d 2  ( x)
2


 ( x)
2
dx
As always set
e
where  2 
mx
2m(V0  E )
0
2

d 2  ( x)
2

m
 ( x)
so
dx 2
2
2
and m  
The general solution is then ( x)  Cex  De x
so m  
Introduction to Quantum Mechanics
Finite potential well
REGION II
Boundary conditions
( x)  Cex  De x
 ( x)  0 as x   to satisfy max probability of 1, thus C = 0.
 ( x) and
d ( x)
must be continuous at boundary between regions I and II at x = L/2.
dx
 kL 
 L 
A
cos

D
exp




I ( L 2)  II ( L 2) so
2
2
 


(i)
k  kL 

 L 
I ( L 2) II ( L 2)

A
sin


D
exp
 


so

2
2
2
2




dx
dx
(ii)
 kL  
tan
 
Dividing eqn (ii) by eqn (i) we eliminate A and D to obtain the condition
 2  k
Introduction to Quantum Mechanics
Finite potential well summary
REGION I
I ( x)  A cos(kx)
x
REGION II II ( x)  De
 kL  
tan   
 2  k
If we wanted to we could then perform the same procedure for the region to the left
of the potential well.
We would then be able to normalise the function between ± ∞ to unity and find
another expression linking the coefficients A and D

*

 ( x, t )( x, t )dx  1

1/α is defined as the penetration depth
Introduction to Quantum Mechanics
Potential wells
Introduction to Quantum Mechanics
Quantum Tunnelling
Classically, if you have a potential barrier of height V and a particle incident on that
barrier with E < V, the particle would reflect off the barrier completely.
The same system in quantum mechanics gives a non-zero probability that the particle
will be transmitted through the barrier. This is a wave phenomenon, but in quantum
mechanics particles exhibit wave-like properties.
The wavefunction of the tunneling particle decreases exponentially in the barrier. The
tunneling probability is strongly dependent on the width of the barrier, the mass of the
particle, and the quantity (V-E). For instance, the ratio of tunneling probability for
protons to electrons is around a factor of 10-91.
Introduction to Quantum Mechanics
Quantum Tunnelling: uses
The most important applications of quantum tunnelling are in semiconductor and
superconductor physics.
Phenomena such as field emission, important to flash
memory, are explained by quantum tunnelling.
Another major application is in
scanning tunnelling microscopes which
can resolve objects that are too small
to see using conventional microscopes,
overcoming the limiting effects of
conventional microscopes (optical
aberrations, wavelength limitations) by
scanning the surface of an object with
tunnelling electrons.
Introduction to PDEs
In many physical situations we encounter quantities which depend on two or
more variables, for example the displacement of a string varies with space
and time: y(x, t). Handing such functions mathematically involves partial
differentiation and partial differential equations (PDEs).
2
1

u
2
 u 2 2
c t
Wave equation
Elastic waves, sound
waves, electromagnetic
waves, etc.
2 2
u

 u  Vu  i
2m
t
1 u
 2u  2
h t
Schrödinger’s
equation
Quantum mechanics
Diffusion
equation
Heat flow, chemical
diffusion, etc.
 u0
Laplace’s
equation
Electromagnetism,
gravitation,
hydrodynamics, heat flow.

2
 u
0
Poisson’s
equation
As (4) in regions
containing mass, charge,
sources of heat, etc.
2
SUMMARY of the procedure used to solve PDEs
 2 y ( x, t ) 1  2 y ( x , t )
1. We have an equation
 2
x 2
c
t 2
with supplied boundary conditions
2. We look for a solution of the form y ( x, t )  X ( x)T (t )
3. We find that the variables ‘separate’
1 d 2 X ( x)
1 d 2T (t )
 2
N
X ( x) dx 2
c T (t ) dt 2
4. We use the boundary conditions to deduce the polarity of N.
e.g. N  k 2
5. We use the boundary conditions further to find allowed values of k and hence X(x).
X ( x)  A cos kx  B sin kx so X n ( x)  Bn sin nx
L
6. We find the corresponding solution of the equation for T(t).
 nct

T
(
t
)

E
cos



n
T (t )  C cos kct  D sin kct n
L


7. We hence write down the special solutions. Yn ( x, t )  Bn sin nx cos nct  n 
L
 L

8. By the principle of superposition, the general solution is the sum of all special

nx
 nct

solutions..
www.falstad.com/mathphysics.html
y ( x, t )  B sin
cos
 

n1
n
L
 L
n

9. The Fourier series can be used to find the particular solution at all times.
y ( x, t ) 
8d  x
ct 1 3x
3ct 1
5x
5ct 1
7x
7ct

sin
cos

sin
cos

sin
cos

sin
cos



L
L 9
L
L
25
L
L
49
L
L
 2 
Solving the time dependent Schrödinger equation
The TDSE is a linear equation, so any superposition of solutions is
also a solution. For example, consider two different energy
eigenvalues, with energies E1 and E2. Their complete normalised
wavefunctions at t = 0 are:
1 ( x,0) 
2
 x 
sin  
L L
2 ( x,0) 
2
 2x 
sin 

L  L 
But any superposition such as ( x,0)  C11 ( x,0)  C2 2 ( x,0) also satisfies the
TDSE, and thus represents a possible state of the system.
Recall that all wavefunctions must obey the normalization condition:

  ( x, t ) dx  1
2

When we superpose, the resulting wavefunction is no longer normalised.
However it can be shown that the normalisation condition is fulfilled so long as:
C1  C2  1
2
2
Solving the time dependent Schrödinger equation
Consider the time dependent Schrödinger equation in 1 dimensional space:
 2  2  ( x, t )
 ( x, t )


V
(
x
,
t
)

(
x
,
t
)

i

2m x 2
t
Within a quantum well in a region of zero potential, V(x,t) = 0, this simplifies to:
 2  2  ( x, t )
 ( x, t )

 i
2m x 2
t
Question
Let’s solve the TDSE subject to boundary conditions
(0, t) = (L, t) = 0 (as for the infinite potential well)
For all real values of time t
and for the condition that the particle exists in a
superposition of eigenstates given below at t = 0 .
 ( x,0) 
2 1
x 1
2x 1
3x 
sin

sin

sin
L  3
L
L
L 
3
3
Solving the time dependent Schrödinger equation
1.5
total amplitude
1
n=1
Question
0.5
Let’s solve the TDSE subject to boundary conditions
(0, t) = (L, t) = 0 (as for the infinite potential well)
0
-0.5
-1
-1.5
displacement from x = 0 to x = L
and for the condition that the particle exists in a
superposition of eigenstates given below at t = 0 .
1.5
1.5
n=2
0.5
1
total amplitude
total amplitude
1
0
-0.5
0.5
0
-0.5
-1
-1
-1.5
-1.5
Superposition at t = 0
displacement from x = 0 to x = L
displacement from x = 0 to x = L
1.5
total amplitude
1
n=3
0.5
0
-0.5
 ( x,0) 
-1
-1.5
displacement from x = 0 to x = L
2 1
x 1
2x 1
3x 
sin

sin

sin
L  3
L
L
L 
3
3
Solving the time dependent Schrödinger equation
 2  2  ( x, t )
 ( x, t )
In a region of zero potential, V(x,t) = 0, so : 

i

2m x 2
t
Step 1: Separation of the Variables
Our boundary conditions are true at special values of x, for all values of time, so we
look for solutions of the form (x, t) = X(x)T(t). Substitute this into the Schrödinger
equation:
 2 d 2 X ( x)
dT (t )

T
(
t
)

i

X
(
x
)
2m dx 2
dt
Step 2: Rearrange the equation
Separating variables:
 2 1 d 2 X i dT


2
2m X dx
T dt
Solving the time dependent Schrödinger equation
Step 3: Equate to a constant
Now we have separated the variables. The above equation can only be true for all x, t
if both sides are equal to a constant. It is conventional (see PHY202!) to call the
constant E.
So we have
2
2 1 d 2 X
d
X
2mE


E
which
rearranges
to


X
2
2
2m X dx 2
dx

i dT
dT
iE
 E which rearranges to
 T
T dt
dt

(i)
(ii)
Step 4: Decide based on situation if E is positive or negative
We have ordinary differential equations for X(x) and T(t) which we can solve but the
polarity of N affects the solution …..
For X(x)
Our boundary conditions are (0, t) = (L, t) = 0, which means X(0) = X(L) = 0.
So clearly we need E > 0, so that equation (i) has the form of the harmonic
oscillator equation.
2mE
d2X
2
k 2  2 giving X ( x)  A cos kx  B sin kx
where
It is simpler to write (i) as


k
X
2
dx

Solving the time dependent Schrödinger equation
Step 5: Solve for the boundary conditions for X(x)
For X(x)
Our boundary conditions are (0, t) = (L, t) = 0, which means X(0) = X(L) = 0.
If X ( x)  A cos kx  B sin kx where k 2 
2mE
then applying boundary conditions
2
gives X(0) = 0 gives A = 0 ; we must have B ≠ 0 so X(L) = 0 requires
i.e. k n 
n
L
so X n ( x)  Bn sin
sin kL  0 ,
nx
for n = 1, 2, 3, ….
L
Step 6: Solve for the boundary conditions for T(t)
dT
iE
  T Equation (ii) has solution T  T0 e iEt  as it’s only a 1st order ODE
dt

Step 7: Write down the special solution for  (x, t)
2
nx iEnt  where
2kn
n 2  2 2
En 

n ( x, t )  X n ( x)Tn (t )  Bn sin
e
2
m
2mL2
L
(These are the energy
eigenstates of the system.)
Solving the time dependent Schrödinger equation
Step 8: Constructing the general solution for  (x, t)
We have special solutions: n ( x, t )  X n ( x)Tn (t )  Bn sin
nx iEnt 
e
L
The general solution of our equation is the sum of all special solutions:
1.5
total amplitude
0.5
0
-0.5
-1


n 1
n 1
 ( x, t )   n ( x, t )   Bn sin
1
Superposition at t = 0
nx
exp(  iE n t )
L
(In general therefore a particle will be in a
superposition of eigenstates.)
-1.5
displacement from x = 0 to x = L
Step 10: Finding the particular solution for all times
If we know the state of the system at t = 0, we can find the state at any later time.
Since we said that  ( x,0) 
Then we can say
 ( x, t ) 
2 1
x 1
2x 1
3x 
sin

sin

sin
L  3
L
L
L 
3
3
2 1
x
1
2x
1
3x

sin
exp(

iE
t

)

sin
exp(

iE
t

)

sin
exp(

iE
t

)
1
2
3

L  3
L
L
L
3
3

 2 2
where E1 
2mL2
4 2 2
9 2 2
and E3 
E2 
2
2mL
2mL2
Particular solution to the time dependent
Schrödinger equation
 ( x, t ) 
2 1
x
1
2x
1
3x

sin
exp(

iE
t

)

sin
exp(

iE
t

)

sin
exp(

iE
t

)
1
2
3

L  3
L
L
L
3
3

1st Eigenfunction
1
3
2
3rd Eigenfunction
E1 
4 2 2
E2 
2mL2

2mL2 2nd Eigenvalue
2
2
1st Eigenvalue
9 2 2
E3 
2mL2
3rd Eigenvalue