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Lecture 15
Solving the time dependent Schrödinger
equation
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
u
2

 u  Vu  i
2m
t
2
Remember Phils Problems and your notes = everything
http://www.hep.shef.ac.uk/Phil/PHY226.htm
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
Introduction to PDEs
A wave equation is an example of a partial differential equation
 2 y ( x, t ) 1  2 y ( x , t )
 2
2
x
c
t 2
You all know from last year that a possible solution to the above is
y  A cos(kx  t )
and that this can be demonstrated by substituting this into the wave equation
If you make either x or t constant, then you return to the expected SHM case
Think of a photo of waves (i.e. time is fixed)
y  A cos( kx  const )
y
x
Or a movie of the sea in which you focus on a
specific spot (i.e. x is fixed)
y  A cos( t  const )
y
t
Introduction to PDEs
2
d
x(t )
2
Unstable equilibrium


x(t )
2
dt
2
dx
d
x
mt
mt So
Step 1: Let the trial solution be
 me  mx and 2  m 2e mt  m 2 x
xe
dt
dt
Step 2: The auxiliary is then
m x  x
2
Step 3: General solution for real roots
Step 4: Boundary conditions could
then be applied to find A and B
Thing to notice is that x(t) only
tends towards x=0 in one
direction of t, increasing
exponentially in the other
2
and so roots are
m  
m   is x(t )  Ae t  Be  t
x(t )  Aet
t
Introduction to PDEs
d 2 x(t )
2
Harmonic oscillator



0 x(t )
2
dt
Step 1: Let the trial solution be
Step 2: The auxiliary is then
x
d 2x
mt So dx
mt
 me  mx and 2  m 2e mt  m 2 x
e
dt
dt
m x   x
2
Step 3: General solution for complex
where  = 0 and b =  so
2
and so roots are
m   i
m    ib is x  e t (C sin b t  D cos b t )
x  C sin  t  D cos  t
Step 4: Boundary conditions could then be
applied to find C and D
Thing to notice is that x(t) passes
through the equilibrium position
(x=0) more than once !!!!
x  C sin  t
Introduction to PDEs
Half-range Fourier sine series

f ( x)   bn sin
n 1
nx
d
where
d
12
d
4
2 d
nx
bn   f ( x) sin
dx
0
d
d
A guitarist plucks a string of
length d such that it is displaced
from the equilibrium position as
shown. This shape can then be
represented by the half range
sine (or cosine) series.
f ( x) 
x
3
d
x 1
f ( x)   
9 9
2 d4  x 
nx
2 d 1 x
nx
bn     sin
dx  d    sin
dx
0
d
d
d 4 9 9
d
3
nx
f ( x)   bn sin
d
n 1

The One-Dimensional Wave Equation
 y ( x, t ) 1  y ( x , t )
 2
2
2
x
c
t
2
2
A guitarist plucks a string of length L such that it is
displaced from the equilibrium position as shown at t = 0
and then released.
Find the solution to the wave equation to predict the
displacement of the guitar string at any later time t
Let’s go thorugh the steps to solve the PDE
for our specific case …..
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..
y ( x, t )  B sin
cos
 

n1
n
L
 L
n

www.falstad.com/mathphysics.html
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
eigenstates, 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
What does the wavefunction look like?
0.5
0
 ( x,0) 
-0.5
2 1
x 1
2x 1
3x 
sin

sin

sin
L  3
L
L
L 
3
3
-1
-1.5
displacement from x = 0 to x = L
These curves arent normalised –
figs intended just to show shape
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
displacement from x = 0 to x = L
Superposition at t = 0
displacement from x = 0 to x = L
1.5
total amplitude
1
n=3
If we measure the energy of the
state Ψ(x,t) described above we
will measure either E1 or E2 or E3
each with the probability of 1/3.
0.5
0
-0.5
-1
-1.5
displacement from x = 0 to x = L
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)
2
2m dx
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
 2 1 d 2 X i dT


2
2m X dx
T dt
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
 2 X
2
2m X dx 2
dx

i dT
dT
iE
 E which rearranges to
 T
T dt
dt

(ii)
(i)
Solving the time dependent Schrödinger equation
2
d
X
2mE
Step 4: Decide based on situation if E is positive or negative


X
2
2
dx

We have ordinary differential equations for X(x) and T(t) which we can solve but the
polarity of E 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.
It is simpler to write (i) as:
d2X
2


k
X
2
dx
2mE
where k 
giving X ( x)  A cos kx  B sin kx
2

2
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.
2mE
If X ( x)  A cos kx  B sin kx where k 
then applying boundary conditions
2

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
nx
for n = 1, 2, 3, ….
L
sin kL  0 ,
Solving the time dependent Schrödinger equation
Step 6: Solve for the boundary conditions for T(t)
A couple of slides back we decided that in order to have LHO style solutions for
X(x) we must have E > 0. So here we must also take E > 0.
dT
iE
  T Equation (ii) has solution T  T0 e iEt  as it’s only a 1st order ODE
dt

Proof of statement above
dT
iE
  dt
T

dT
iE


 T   dt
Replace the constant with T0
ln T  
T  T0 e

iEt
c

iE t

T e e
c

iE t

Solving the time dependent Schrödinger equation
Step 7: Write down the special solution for  (x, t)
2
2kn
nx iEnt  where
n 2  2 2
En 

n ( x, t )  X n ( x)Tn (t )  Bn sin
e
2m
2mL2
L
(These are the energy
eigenvalues of the system.)
Question asks for the solutions of the TDSE for real values of time
e
 i
 cos   i sin 
where
En t


Real values are therefore n ( x, t )  X n ( x)Tn (t )  Bn sin
Et
nx
cos n
L

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
Et
nx
cos n
L

The general solution of our equation is the sum of all special solutions:


n 1
n 1
 ( x, t )   n ( x, t )   Bn sin
Et
nx
cos n
L

(In general therefore a particle will be in a
superposition of eigenstates.)
Solving the time dependent Schrödinger equation
Step 10: Finding the full solution for all times


n 1
n 1
The general solution is  ( x, t )   n ( x, t )   Bn sin
Et
nx
cos n
L

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) 
2 1
x 1
2x 1
3x 
sin

sin

sin
L  3
L
L
L 
3
3


n 1
n 1
At t = 0 the general solution is  ( x,0)   n ( x,0)   Bn sin
Then we can say:
 ( x, t ) 
E3t 
2 1
x
E1t 1
2x
E2 t 1
3x
sin
cos

sin
cos

sin
cos
L  3
L

L

L
 
3
3
2 2


where E1 
2mL2
1.5
total amplitude
1
0.5
0
-0.5
-1
nx
L
Superposition at t = 0
-1.5
displacement from x = 0 to x = L
4 2 2
E2 
2mL2
9 2 2
and E3 
2mL2
Particular solution to the time dependent
Schrödinger equation
 ( x, t ) 
E3t 
2 1
x
E1t
1
2x
E2 t
1
3x
sin
cos

sin
cos

sin
cos
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
Solving the time dependent Schrödinger equation
 ( x, t ) 
E3t 
2 1
x
E1t 1
2x
E2 t 1
3x
sin
cos

sin
cos

sin
cos
L  3
L

L

L
 
3
3
1.5
2 2


where E1 
2mL2
total amplitude
1
0.5
0
-0.5
-1
Superposition at t = 0
-1.5
4 2 2
E2 
2mL2
9 2 2
and E3 
2mL2
displacement from x = 0 to x = L
In this particular example Ψ(x,t) is composed of eigenstates with different parity
(even and odd). Therefore Ψ(x,t) does not have a definite parity and P(x,t)
oscillates from side to side.
www.falstad.com/mathphysics.html
1. Energy eigenstates (namely, states with definite energy) are stationary states: they
have constant probability densities and definite energies.
2. Mixed states (namely, superpositions of energy eigenstates) do not have a definite
energy but have a probability of being in any one of the energy states when measured.
3. The probability densities of mixed states vary with time as do therefore the < x >.
Just for Quantum Mechanics course
The expectation value is interpreted as the average value of x that we would expect to
obtain from a large number of measurements.
NB. In order to have time dependence in any observable such as position, it is
necessary for the wavefunction to contain a superposition of states with different
energies.
This is because the probability density for a mixed state varies with time, whereas for
a pure state it is constant in time. Pure states are known as stationary states.
For example if we have the single eigenfunction 2 ( x, t )  B2 sin
an infinite potential well then 2* ( x, t )  B2 sin
and 2* ( x, t ) x2 ( x, t )  B22 x sin 2
2x iE2t 
e
within
L
2x iE2t 
e
L
2x
. Notice how there is no time dependence.
L
www.falstad.com/mathphysics.html
This means < x > is invariant with time
Just for Quantum Mechanics course
 ( x, t )  1 ( x, t )  2 ( x, t )
But if we add another eigenfunction for example:
The complex conjugate is written as:
Therefore:

 * ( x, t )  1 ( x, t )  2 ( x, t )
*
*

 * ( x, t ) ( x, t )  1 ( x, t )  2 ( x, t ) 1 ( x, t )  2 ( x, t ) 
*
*
Let eigenfunctions be:  ( x, t )  1 ( x, t )  2 ( x, t )  B1 sin
L
e iE1t   B2 sin
2x iE2t 
e
L
2x iE2t 
e
L
L
x iE1t 
2x iE2t  
x iE1t 
2x iE2t  

 * ( x, t ) ( x, t )   B1 sin
e
 B2 sin
e
e
 B2 sin
e
 B1 sin

L
L
L
L



So:  * ( x, t )  1* ( x, t )  2* ( x, t )  B1 sin
x
x
eiE1t   B2 sin
x
2x
x
2x iE1t  iE2t 

 * ( x, t ) ( x, t )   B12 sin 2
 B22 sin 2
 B1 sin
B2 sin
e
e
 e iE1t  eiE2t 
L
L
L
L



x
2x
x
2x i ( E1  E2 )t 


 * ( x, t ) ( x, t )   B12 sin 2
 B22 sin 2
 B1 sin
B2 sin
e
 e i ( E1  E2 )t  
L
L
L
L



A superposition of eigenstates having different energies is
required in order to have a time dependence in the probability
density and therefore in < x >.

Non zero so
long as E1≠ E2