Transcript The Schrödinger equation

The Quantum Theory of
Atoms and Molecules
The Schrödinger equation and
how to use wavefunctions
Dr Grant Ritchie
An equation for matter waves?
De Broglie postulated that every particles has an associated wave of wavelength:
  h/ p
Wave nature of matter confirmed by electron diffraction studies etc (see earlier).
If matter has wave-like properties then there must be a mathematical function that is the
solution to a differential equation that describes electrons, atoms and molecules.
The differential equation is called the Schrödinger equation and its solution is called the
wavefunction, .
What is the form of the Schrödinger equation ?
The classical wave equation
We have seen previously that the wave equation in 1–d is:
 2
1  2
 2
2
x
v t 2
Where v is the speed of the wave. Can this be used for matter waves in free space?
Try a solution: e.g.
( x, t )  ei ( kx t )
Not correct! For a free particle we know that E=p2/2m.
An alternative….
Try a modified wave equation of the following type:
( is a constant)
Now try same solution as before: e.g.
( x, t )  ei ( kx t )
Hence, the equation for matter waves in free space is:
For ( x, t )  e
i ( kx t )
 2


2
t
x
 2  2


 i
2
2m x
t
2 2
k

then we have
 ( x, t )    ( x, t )
2m
which has the form:
(KE)  wavefunction = (Total energy)  wavefunction
The time-dependent Schrödinger equation
For a particle in a potential V (x,t) then
p2
E
 V ( x, t )
2m
and we have (KE + PE)  wavefunction = (Total energy)  wavefunction
 2  2



V
(
x
,
t
)


i

2m x 2
t
TDSE 
Points of note:
1. The TDSE is one of the postulates of quantum mechanics. Though the SE
cannot be derived, it has been shown to be consistent with all experiments.
2. SE is first order with respect to time (cf. classical wave equation).
3. SE involves the complex number i and so its solutions are essentially complex.
This is different from classical waves where complex numbers are used imply
for convenience – see later.
The Hamiltonian operator
 2 2

 2  2
LHS of TDSE can


 V ( x, t )    
 V ( x, t )   Hˆ 
2
2
be written as:
2m x
 2m x

where Ĥ is called the Hamiltonian operator which is the differential operator that
represents the total energy of the particle.
Thus
2
2
2


ˆ
p


x
ˆ ( x)

Hˆ   

V
(
x
)


V
2
 2m
2
m

x



x
where the momentum operator is
pˆ x  i
Thus shorthand for TDSE is:

ˆ
H  i
t
Solving the TDSE – Aaargh!
Suppose the potential is independent of
time i.e. V(x, t) = V(x) then TDSE is:

2  2
i

 V ( x) 
2
t
2m x
LHS involves variation of  with t while RHS involves variation of  with x. Hence
look for a separated solution:
( x, t )   ( x)T (t )
then
Now divide by  T:
 2  2
T

T 2  V ( x)T  i
2m x
t
 2 1  2
1 T


V
(
x
)

i

2m  x 2
T t
LHS depends only upon x, RHS only on t. True for all x and t so both sides must equal a
constant, E (E = separation constant).
1 T
i
E
T t
Thus we have:
 2 1  2

 V ( x)  E
2m  x 2
Time-independent Schrödinger equation
Solving the time equation:
i
1 dT
E
T dt

dT
iE
  dt
T

 T (t )  Ae iEt / 
This is exactly like a wave e-it with E = ћ. Therefore T(t) depends upon the energy E.
To find out what the energy actually is we must solve the space part of the problem....
The space equation becomes
 2  2

 V ( x)  E
2
2m x
or
Hˆ   E
This is the time independent Schrödinger equation (TISE) .
** All Prelims problems are concerned with solving TISE rather than the
TDSE!
The TISE can often be very difficult to solve – it depends upon V(x)!
Eigenvalue equations
The Schrödinger Equation is the form of an Eigenvalue Equation:
where Ĥ is the Hamiltonian operator,
Hˆ   E
2
2

d
Hˆ  Tˆ  Vˆ  
 V ( x)
2
2m dx
 is the wavefunction and is an eigenfunction of Ĥ;
E is the total energy (T + V) and an eigenvalue of Ĥ. E is just a constant!
Later in the course we will see that the eigenvalues of an operator give the possible
results that can be obtained when the corresponding physical quantity is measured.
TISE for a free-particle
For a free particle V (x) = 0 and TISE is:
and has solutions
 e
ikx
or e
Thus the full solution to the full TDSE is:
 2  2

 E
2m x 2
ikx
where
k 2 2
E
2m
( x, t )   ( x) T (t )  ei (  kx  Et /  )
Corresponds to waves travelling in either  x direction with:
(i) an angular frequency,  = E / ћ  E = ћ !

(ii) a wavevector, k = (2mE)1/2 / ћ = p / ћ  p = h /  !
WAVE-PARTICLE DUALITY!

Interpretation of (x,t)
As mentioned previously the TDSE has solutions that are inherently complex  (x,t)
cannot be a physical wave (e.g. electromagnetic waves). Therefore how can  (x,t)
relate to real physical measurements on a system?
The Born Interpretation
Probability of finding a particle in a small length dx at position x and time t is equal to
2
 ( x, t )( x, t )dx  ( x, t ) dx  P( x, t )dx
*
* is real as required for a probability distribution and is the probability per unit
length (or volume in 3d).
The Born interpretation therefore calls  the probability amplitude, * (= P(x,t) )
the probability density and * dx the probability.
Expectation values
Thus if we know (x, t) (a solution of TDSE), then knowledge of * dx allows the
average position to be calculated:
x   xi P( xi ) x
i
In the limit that x 0 then the summation becomes:

x x 


xP( x)dx 


2
x  ( x, t ) dx

The average is also know as the expectation value and are very important in quantum
mechanics as they provide us with the average values of physical properties because in
many cases precise values cannot, even in principle, be determined – see later.

Similarly
x2 



x 2 P ( x)dx 


2
x 2  ( x, t ) dx
Normalisation
Total probability of finding a particle anywhere must be 1:



P( x)dx 


2
 ( x, t ) dx  1

This requirement is known as the Normalisation condition. (This condition arises
because the SE is linear in  and therefore if  is a solution of TDSE then so is c
where c is a constant.)
Hence if original unnormalised wavefunction is (x,t), then the normalisation integral is:

N2 

2
 ( x, t ) dx

And the (re-scaled) normalised wavefunction norm = (1/N) .
Example 1: What value of N normalises the function N x (x  L) of 0  x  L?
Example 2: Find the probability that a system described by the function 21/2sin (x) where
0  x  1 is found anywhere in the interval 0  x  0.25.
Boundary conditions for 
In order for  to be a solution of the Schrödinger equation to represent a physically
observable system,  must satisfy certain constraints:
1.
2.
3.
4.
Must be a single-valued function of x and t;
Must be normalisable; This implies that the   0 as x  ;
 (x) must be a continuous function of x;
The slope of  must be continuous, specifically d (x)/dx must be
continuous (except at points where potential is infinite).
(x)
(x)
x
(x)
x
(x)
x
x
Stationary states
Earlier in the lecture we saw that even when the potential is independent of time the
wavefunction still oscillates in time:
Solution to the full TDSE is:
( x, t )   ( x) T (t )   ( x)e iEt / 
But probability distribution is static:
P( x, t )  ( x, t )   * ( x)e iEt /   ( x)e iEt /    ( x)
2
Thus a solution of the TISE is known as a Stationary State.
2
What other information can you get from ?
(and how!)
We have seen how we can use the probability distribution  to calculate the
average position of a particle. What happens if we want to calculate the average
energy or momentum because they are represented by the following differential
operators:
2 d2

Hˆ  
 V ( x)
2
2m dx
pˆ x 
Do the operators work on , or on , or on  alone?
Take TISE and multiply from
left by  and integrate:
NB  is normalised.
 
.
i x
Hˆ  n  En n
 ˆ



H

d
x


E

d
x

E

 n n  n n n n  n ndx  En
Suggest that in order to calculate the
average value of the physical quantity
associated with the QM operator we
carry out the following integration:
 ˆ

 n  ndx
Momentum and energy expectation values
The expectation value of momentum involves the representation of momentum as a
quantum mechanical operator:

px 

 * ( x, t )

 
 ( x, t )dx
i x
pˆ x 
where
 
.
i x
is the operator for the x component of momentum.
Example: Derive an expression for the average
energy of a free particle.
p2
E
2m
then
p2
E 
2m
Since V = 0 the expectation value for energy for a particle moving in one dimension is

 2 2 
  ( x, t )dx
E   * ( x, t ) 
2 
 2m x 


Our definition of the expectation value is one of the postulates of QM – see later
lectures.
Summary
TDSE:
Born interpretation:
 2  2



V
(
x
,
t
)


i

2m x 2
t
2
* ( x, t )( x, t )dx  ( x, t ) dx  P( x, t )dx

Normalisation:


P( x)dx 

TISE:

2
 ( x, t ) dx  1

 2  2

 V ( x)  E
2
2m x
or
Hˆ   E
( x, t )   ( x) T (t )   ( x)e iEt / 
Boundary conditions on : single-valued, continuous, normalisable, continuous first derivative.

Expectation value of operator Ω:

ˆ   ( x, t )dx

(
x
,
t
)



It’s never as bad as it seems….