Schrödinger Equation
&
OUTLINE OF SECTION 1
,
• Revision: differential equations and complex numbers
•The
time-dependent Schrödinger equation
Free particle
Particle in a potential
• Interpretation of the wave function
Probability
Normalization
•Boundary conditions on the wave function
• Derivation of the time-independent Schrödinger equation
Revision: differential equations
2 2 2 d d 3 5 d d f f x fx x
A differential equation is an equation to be satisfied by a particular function, e.g. f(x), that involves
derivatives of that function, e.g. df/dx
A linear differential equation is one in which there are no powers higher than the first of the unknown function (or its derivatives).
In an ordinary differential equation the unknown function depends on only one variable, e.g. f(x), and the derivatives are taken with respect to that variable.
In a partial differential equation the unknown function depends on more than one variable, e.g. f(x,t), and the equation contains partial derivatives with respect to the different variables, e.g. classical wave equation
Which of the following differential
equations for the unknown function f(x) is linear? 2 2 3 2
d
3
d
d
2
d
f
x f
x
f
xf
x
x
2 2 2 2 d d 3 5 d d f fx f xf
x x
2 2
2 2 2
1
f
f
x
c
t
Differential equations (cont)
ˆ
Lf
g
Any linear differential equation can be written in the form
where:
f is the unknown function
g is a known function
is a differential operator (an object that does something to the function – operates on it – and that operation involves
differentiation). Hats above symbols are frequently used to denote operators.
Examples
:
ˆ
L
A linear, homogeneous differential equation is one in which all the terms are linear in the unknown function, i.e. g = 0.
The most important type in quantum mechanics. 2 2 2 2 2 2 d d 3 5 d d d d 3 5 d d f f x f x x x f x x 2 2
2 2 2
2 2
2 2 2
1
1
0
f
f
x
c
t
f
x
c
t
is the same as
Solving differential equations with exponentials
For a homogeneous, linear, ordinary differential equation with constant coefficients, the substitution
(where A, λ are constants) always reduces the differential equation to an algebraic (ordinary) equation for λ.
Example:
For a homogeneous, linear equation, a linear combination of any two solutions is also a solution. This is the
superposition principle
1 2
1 2
ˆ
0
ˆ
0
ˆ
0
Lf
Lf
L f
f
( )
xf x
Ae
Boundary conditions
A differential equation by itself does not fully determine the unknown function f
For a differential equation of order n (in which the highest derivative is the nth) we have n arbitrary constants
Obtain these from additional information (boundary conditions), such as the value and/or derivative of the unknown function at specified points.
Example: motion of classical particle, subject to constant force
F 2
2
2
( )
2
d x
F
F
F
ma
x t
t
At B
dt
m
m
Revision: complex numbers
in classical physics
i t
e
z
A
Often convenient to use complex numbers in classical physics, especially in description of wave motion or vibration
Can take real or imaginary part as physical
solution. Complex numbers are a mathematical convenience.
Displacement Velocity
Acceleration
i
A
A e
|A|
Re(z)
φ
Im(z)
z(t=0)
i t+
e
z
A
An equation for matter waves: (1)
the time-dependent Schrödinger equation
2 2
2 2 2
1
x v t
wave velocity
v
Classical 1D wave equation e.g. waves on a string:
Can we use this to describe matter waves in free space?
( , )
x t
x
( , ) wave displacementx t
Try solution
x t
,
e
i kx tBut this isn’t correct! For free particles we know that
2
2
p
E
m
An equation for matter waves (2)
2 2
Try
t
x
2 2
2
i
2
t
m x
Seem to need an equation that involves the first derivative in time, but the second derivative
in space
( , ) is "wave function" associated with matter wavex t
,
i kx tx t
e
As before try solution
So equation for matter waves in free space is (free particle Schrödinger equation)
An equation for matter waves (3)
2
( , )
2
p
E
V x t
m
2 2 2
i
( , )
2
V x t
t
m x
For particle in a potential V(x,t)Suggests modification to Schrödinger equation:
Time-dependent Schrödinger equation
Total energy = KE + PE
Schrödinger
What about particles that are not free?
,
2 2
,
2
k
x t
x t
m
Has form (Total Energy)*(wavefunction) = (KE)*(wavefunction)
2 2 2
i
2
t m x
, i kx tx t e
Substitute into free particle equation
2
2
p
E
m
(Total Energy)*(wavefunction) = (KE+PE)*(wavefunction)
The Schrödinger equation: notes
2 2 2
i ( , )
2 V x t
t m x
•
This was a plausibility argument, not a derivation. We believe theSchrödinger equation not because of this argument, but because its predictions agree with experiment.
•
There are limits to its validity. In this form it applies only to a single, non-relativistic particle (i.e. one with non-zero rest mass and speed much less than c)•
The Schrödinger equation is a partial differential equation in x and t (like classical wave equation). Unlike the classical wave equation it is first order in time.•
The Schrödinger equation contains the complex number i. Therefore its solutions are essentially complex (unlike classical waves, where the use of complex numbers is just a mathematical convenience).•
Note the +ve sign of i in the Schrödinger equation. This came from our looking for plane waves of the formWe could equally well have looked for solutions of the form Then we would have got a –ve sign.
This is a matter of convention (now very well established).
i t
e
i t
e
The Hamiltonian operator
2 22
i
( , )
2
V x t
t
m x
ˆ
i
H
t
Can think of the RHS of the Schrödinger equation as a
differential operator that represents the energy of the particle.
Hence there is an alternative (shorthand) form for the time-dependent Schrödinger equation:
This operator is called the Hamiltonian of the particle, and usually given the symbol
H
ˆ
2 2
2
( , )
ˆ
2
d
V x t
H
m dx
Kinetic energy operator Potential energy operatorHamiltonian is a linear differential operator.
Schrödinger equation is a linear homogeneous partial differential equation
Time-dependent Schrödinger equation
Interpretation of the wave function
2 2
0
( , ) ( , ) d
b b
x
x a a
x t
x x t x
2 *
2
( , )
x t
x
Ψ
is a complex quantity, so how can it correspond
to real physical measurements on a system?
Remember photons: number of photons per unit volume is proportional to the electromagnetic energy per unit volume, hence to square of electromagnetic field strength.
Postulate (Born interpretation): probability of finding particle in a small length δx at position x and time t is equal to
Note: |Ψ(x,t)|2 is the probability per unit length. It
is real as required for a probability distribution. Total probability of finding particle
between positions a and b is
a b
|Ψ|2
x δx
Example
Suppose that at some instant of time a particle’s wavefunction at t=0 is
( ,0) 2
x
x
What is:
(a) The probability of finding the particle between x=1.0 and x=1.001?
(b) The probability per unit length of finding the
particle at x=1?
(c) The probability of finding the particle between x=0 and
DOUBLE-SLIT EXPERIMENT REVISITED
Detecting screen Incoming coherent
beam of particles (or light)
sin
d
D
θ
y
1
2
Schrödinger equation is linear: solution with both slits open is
1 2Observation is nonlinear
1 2
2 2 2 * *
1 2 2 1
Usual “particle” part Interference term gives fringes
Normalization
Total probability for particle to be somewhere should always be one
Suppose we have a solution to the Schrödinger equation that is not normalized. Then we can
•Calculate the normalization integral •Re-scale the wave function as
(This works because any solution to the SE multiplied by a constant remains a solution, because the SE is LINEAR and HOMOGENEOUS)
Normalization condition 2
( , ) d
x t
x
1
1
( , )
x t
( , )
x t
N
2
( , ) d
N
x t
x
New wavefunction is normalized to 1 A wavefunction which obeys this condition is said to be normalized
Conservation of probability
Total probability for particle to be somewhere should ALWAYS be one
2
( , ) d
x t
x
constant
If the Born interpretation of the wavefunction is correct then the normalization integral must be independent of time (and can always be chosen to be 1 by normalizing the wavefunction)
We can prove that this is true for physically relevant wavefunctions using the Schrödinger equation. This is a very important check on the consistency of the Born interpretation.
2 2 2
i
( , )
2
V x t
t
m x
( , ) d
x t
2x
constant
Boundary conditions for the wavefunction
The wavefunction must:
1. Be a continuous and single-valued
function of both x and t (in order that the probability density is uniquely defined)
Examples of unsuitable wavefunctions
Not single valued
Discontinuous
Gradient discontinuous x ( )x
( )x
x
( )x
x
2. Have a continuous first derivative
(except at points where the potential is infinite)
3. Have a finite normalization integral
Time-independent Schrödinger equation
Suppose potential is independent of time
2 2 2
i
( )
2
V x
t
m x
LHS involves only variation of Ψ with t
RHS involves only variation of Ψ with x (i.e. Hamiltonian operator does not depend on t)
Look for a separated solution
( , )
x t
( ) ( )
x T t
Substitute:
,
( )
V x t
V x
2 2
2
( ) ( )
( ) ( ) ( )
( ) ( )
2
m x
x T t
V x
x T t
i
t
x T t
2 22
( ) ( )
( )
2d
x T t
T t
x
dx
2 22
( )
2
d
dT
T
V x T i
m
dx
dt
N.B. Total not partial derivatives now
2 2 2
1
1
( )
2
d
dT
V x
i
m
dx
T dt
Divide by ψT
LHS depends only on x, RHS depends only on t.
True for all x and t so both sides must be a constant, A (A = separation constant)
2 2
2
( )
2
d
dT
T
V x T i
m
dx
dt
1
dT
i
A
T dt
2 2 21
( )
2
d
V x
A
m
dx
So we have two equations, one for the time dependence of the wavefunction
and one for the space dependence. We also have to determine the separation constant. This gives
/
( )
iEt
T t
ae
1
dT
i
A
T dt
dT
iA
T
dt
/
( )
iAtT t
ae
1 dT
i A
T dt
2 2
2
1
( ) 2
d
V x A m dx
SOLVING THE TIME EQUATION
• This only tells us that T(t) depends on the energy E.
It doesn’t tell us what the energy actually is. For that we have to solve the space part. • T(t) does not depend explicitly on the potential V(x). But there is an implicit dependence because the potential affects the possible values for the energy E.
2 2 2
d
( )
2
m x
d
V x
E
With
A = E
, the space equation becomes:This is the time-independent Schrödinger equation
ˆ
H
E
Solving the space equation = rest of course!
Time-independent Schrödinger equation
or
But probability
distribution is static
2 * / /
2 *
,
,
( )
( )
( ) ( )
( )
iEt iEt
P x t
x t
x e
x e
x
x
x
/
( , )
x t
( ) ( )
x T t
( )
x e
iEt
Solution to full TDSE is
Even though the potential is independent of time the wavefunction still oscillates in time
Notes
• In one space dimension, the time-independent Schrödinger equation is an
ordinary
differential equation (not a partial differential equation)
• The time-independent Schrödinger equation is an
Eigen value equation
for the Hamiltonian operator:
Operator × function = number × function (Compare Matrix × vector = number × vector)
• We will consistently use uppercase Ψ(
x,t
) for the full wavefunction
(TDSE), and lowercase ψ(
x
) for the spatial part of the wavefunction when
time and space have been separated (TISE)
ˆ
H
E
2 22
d
( )
2
m x
d
V x
E
SE in three dimensions
ˆ
H
E
To apply the Schrödinger equation in the real
(3D) world we keep the same basic structure:
i
H
ˆ
t
BUTWavefunction and potential energy are now functions of three spatial coordinates:
Kinetic energy now involves three
components of momentum
Interpretation of wavefunction:
, , , ,
x x y z
V x V V x y z
r r
2 2 2
2 2
2 2 2 2 2 2 2
2
2 2 2 2
2 2 2
2 2 2
x y z
x p p p
p
m m m
m x m m x y z
p
23
,
d
r
r
t
probability of finding particle in a volume element centred on r
2,
t
r
probability density at r
SE in three dimensions
2 2
ˆ
2
H
V
m
r
r
So 3D Hamiltonian is
2
2
,
i
,
,
,
2
t
t
V
t
t
t
m
r
r
r
r
2
2
2
m
V
E
r
r
r
r
Time-independent Schrödinger equation is
Time-dependent Schrödinger equation is
Puzzle
( )
( , ) e
x t
i kxt
2 2 2 2 4
0
E
p c
m c
The requirement that a plane waveplus the energy-momentum relationship for free-non-relativistic particles
led us to the free-particle Schrödinger equation.
Can you use a similar argument to suggest an equation for free relativistic
particles, with energy-momentum relationship:
2
2
p E
m
SUMMARY
SUMMARY
Time-dependent Schrödinger equation
2 22
,
2
m x
V x t
i
t
2 *
,
,
,
,
P x t dx
x t dx
x t
x t dx
Probability interpretation and normalizationTime-independent Schrödinger equation 2 2
2
( ) ( )
( )
2
d
V x
x
E
x
m dx
x t
,
( ) ( )
x T t
( )
x e
iEt/ Conditions on wavefunctionsingle-valued, continuous, normalizable, continuous first derivative
2,
,
1
dx P x t
dx
x t
Particle in a Box-I.
(a) The Schrödinger Equation
(b) The acceptable solution
(c) Normalization
(d) The properties of the solutions
The particle in a box...one dimension
For a particle moving in one dimension we have for the
Hamiltonian :
ˆ
H =
p
ˆ
x
2
2m
+ V(x) =
-
2
2m
d
2
dx
2
V(x)
the particle is described by the wavefunction
(
x) which is a
solution to the S.W.E
H
(x) = E
(x)
-
2
2m
d
2
(x)
dx
2
V(x)
(x) = E
(x)
The particle in a box...one dimension
-
2
2m
d
2
(x)
dx
2
V
(
x
)
(x) = E
(x)
X=0
V=
X=l
V=
V=
V
V
We shall consider a case in which V(x) is given by :
Thus inside" the box" between x
0 and x = l we have V = 0
Outside the "box" V =
We have :
I: -
2
2m
d
2
(
x)
dx
2
(
x) = E
(
x) X < 0
The particle in a box...one dimension
X=0
V=
X=l
V=
V=
V
V
II : -
2
2m
d
2
(x)
dx
2
V
(x) = E
(x) 0 < X < l
III :
2
2m
d
2
(x)
For region I:
-
2
2m
d
2
(x)
dx
2
(x) = E
(x) X < 0
The particle in a box...one dimension
Or :
-
2
2m
d
2
(
x)
dx
2
[E
]
(
x)
X=0
V=
X=l
V=
V=
V
V
Since E <<
(Physical Argument)
Thus
1
2
2m
d
2
(
x)
dx
2
(
x)
-
2
2m
d
2
(
x)
The particle in a box...one dimension
X=0
V=
X=l
V=
V=
V
V
We have
(x) =
1
2
2m
d
2
(x)
dx
2
Must be
finite
Also for region III where X > l
-Thus
:
(x) = 0 for x < 0
we have
(x) =
1
2
2m
d
2
(x)
dx
2
and must require
(x) = o for x > l
T
= -
2
2m
*
d
2
dx
2
dx
The particle in a box...one dimension
X=0
V=
X=l
V=
V=
V
V
Region I Region II Region III
We have
I
(x) =
III
(x) = 0
{ Probability of finding particel in III
"No particle outside box"
Thus
P
I
(
x
)
I
*
(x)
I
(x) = 0
P
III
(
x
)
III
*
(x)
III
(x) = 0
The particle in a box...one dimension
X=0
V=
X=l
V=
V=
V
V
Region I Region II Region III
Now for region II:
-
2
2m
d
2
(x)
dx
2
V
(x) = E
(x)
-
2
2m
d
2
(x)
dx
2= E
(x)
d
2
(
x)
dx
2
2mE
2
(
x)
or
d
2
(
x)
dx
2= - k
2
