Download Y (x,t

The Schrodinger Equation
For particles, forces change momentum (p=mv)
and each force has an associated potential energy.
Newton’s 2nd law related potential to the
change in momentum, via,
dp
d 2x
dV
F
 ma  m 2 and for 1 - D, F  
dt
dt
dx
dV
d 2 x i.e. given V(x), solve
thus, 
m 2
dx
dt for x(t) or v(x,t).
For matter waves, p=h/l. Thus, we need a
differential equation to calculate how forces
(described by a potential energy) affect l or p.
We will solve this problem for a wave,
which has an amplitude (y ) for each
value of position (x) and time (t).
Mathematical Forms of Waves.
Y( x,t)=Yo cos(kx-wt) , k=2p/l and w2pn
We can re-write this formula in terms of e, as
Y(x,t)Yoei(kxw t) + Yoei(kxwt)
2
Recalling eix = cosx + isinx, the simplest form
of a wave is Y(x,t)Yoei(kxw t)
Formulation of the Schrodinger Equation
Remember that one cannot derive it. This is a plausibility argument.
To relate p and V we use,
Total Energy (E) = Kinetic (K) + Potential (V)
[ Hamiltonian E = T + V]
p2
h
h
E  K +V 
+ V , where p   k , ( 
)
2m
l
2p
Now, if
Y  Y0 e
i ( kx wt )
Y

 ik Y
, then, if t  const ,
x
 2Y  
and 2
(ik Y )  ik . ik Y   k 2 Y
x
x
2

p Y 
2
2

Y
p (k ) , so
2
2m
2m  x
2
2
 Y  -i w Y , and if E  hn  w , then,
t
+ i   Y  w Y  E Y
t
Thus, 2
p + Y  Y
(
V)
i
can be written as
2m
t
2
2
  + Y  Y
(i
2 V)
2m  x
t
This is the Schrödinger equation.
Using the Schrodinger Equation.-A recipe
(i) specify V=V(x,t) (ie details of force)
(ii) specify initial conditions (ie. at t=0) and
(iii) specify boundary conditions (x=fixed values)
Time Independent Schrodinger Equation
In many cases, (eg. Coulomb potential in H
atom), V=V(x), ie. not time dependent, then
Y( x, t )  y ( x)ei w t
where y(x) satisifies
2 d 2
[
+ V ( x)]y  Ey or, Hy  Ey
2
2m dx
Properties of the T.I.S.E.
•Applies when V=V(x), ie. V is independent of
time.Eg. an e- wave acted on by a fixed nucleus,
but NOT an atom in an oscillating magnetic field.
•Applies only in the non-relativistic limit
(ie. assumes K=(1/2)mv2=p2/2m).
•Assumes p=h/l and E=hn.
•Is linear in y, ie no terms like y2, y3 etc. Laws
of superposition are valid. If ya and yb are
two possible solutions, then ycaya+cbyb
is a solution for all constants ca and cb.
Wavefunctions
We have 2 forms of Schrodinger equation,
(a) general, time-dependent solve for Y (x,t),
when V(x,t) is given, and
(b) TISE, for V=V(x) only. Solve for y(x) when
V(x) is given. Solution is a standing wave,
ie. Y (x,t)=y(x) e-iw t
Wavefunctions are solutions to the
Schrodinger wave equation. The w.function,
Y (x,t) describes physical state of the
particle, such as its momentum, energy etc.
and also where the particle is (in terms of
probability ).
Using the Born Interpretation
As in the earlier electron and photon diffraction
examples, the wave intensity is interpreted as
giving the probability of finding the particle in a
particular state. Recalling that the
Intensity =|Amplitude|2,
we interpret |Y (x,t)|2dx =probability the particle
will be found between x and x+dx at time t.
Solving the Schrodinger equation specifies
Y (x,t) completely, except for a constant,
ie. if Y is a solution, so is A xY .
From the Born interpretation the normalisation
condition of the wavefunction is that

 | Y ( x, t ) |

2
dx  1
i.e. The probability
that particle is
somewhere = 1
Note that probability is a real number.
AlthoughY is complex, |Y 2| is real.
Solution of the Wave Equation
for a Free Particle
`Free particle’, means V=0 everywhere, such as
an electron in the absence of external forces.
Using the TISE,
 2 d 2y
d 2y
2mE
Hy  
 Ey , ie.
 2 y
2
2
2m dx
dx

Try a solution of the form, yA eax,
2
d ( Ae ax )
d
y
2mE
2
 a ( Ae ax ) ,

a
y


y
2
2
dx
dx

2mE
2mE
ie. a  i
,
thus
if,
k


,
2


y ( x)  Aeikx , which is a solution t o the TISE
The energy is given by Hy  Ey, thus
(k ) 2
E
2m
Since k (the wave number) is
real, we have shown that any
positive value of energy is
allowed for a free particle.
Particles (electrons) in a Box: Solving
the Wavefunction for Bound Electrons.
When electrons are acted on by a force, this
acts to restrict the allowed energies (and hence
other quantities e.g. momentum, spin..) This
force is specified by the potential, V(x,t).
Example: Electrons in a 1-D Box
Suppose an e- (in 1-d) is placed
in a `box’ with impenetrable
walls at x=0 and x=L. The
electron can move freely (in x
direction) in the box, such that:
V=0
V  0 0  x  L and
V   x  0 and  L
x=0
x=L
Note if V tends to infinity, the prob. of finding
the e- there must tend to zero.ie y0 outside box.
There are also boundary conditions that
y(x) = 0 for x=0 and x=L, for all times, t
1AMQ/2002 W.N.Catford
& P.H. Regan
9
Outside the box, y=0, now we solve for inside.
Since inside the box,
 2 d 2y
 Ey
V=0 for all t, we can Hy  
2
2m dx
use the T.I.S.E, i.e.
For a free particle, we saw that Aei( k x) was
(k ) 2
a solution for any value of k and that
E
2m
For a particle in a box, we consider the most
general solution for a given energy, and then
apply the boundary conditions.
General solution is of the form
y ( x)  A1e
ikx
+ A2 e
ikx
, with k 
2mE

Thus we can write,
y ( x)  ( A1 + A2 ) cos kx + i ( A1  A2 ) sin kx
 A sin kx + B cos kx, where A, B are consts.
One of the boundary conditions is y(0)=0, thus,
B=0 and hence, y(x)=Asin(kx).
We also have y(L)=0, so 0=Asin(kl), hence,
either, A=0 or sin(kL)=0.
If A=0, the wavefunction =0 everywhere! (ie.
there is no probability of the a particle having
this wavefunction!(solution can be disregarded).
npx
y ( x)  A sin(
)
L
Hence, kL = np,
where n=1,2,3….,
ie k=np/L and thus
Determine A using the normalisation condition,

npx
2 L
|y ( x) | dx  1 | A | 0 sin ( L )dx | A | 2
L
2
Thus
y ( x) 
2
2
npx
sin(
) and
L
L
2
npx iwt
Y ( x, t ) 
sin(
)e
L
L
2
BCs mean that only
certain modes are
allowed for Y(x,t).
A plot of the real
(or imag.) part
of Y(x,t) looks like
standing waves.
Energy Quantization
For particles (eg, electrons) in box, the
boundary conditions mean that only certain
modes are allowed for the wavefunction, ie.
Y( x, t ) 
2
npx iwt
sin(
)e
L
L
p
which arises from
kL

n
from the restriction,
But the particle energy is given by,
2
2
p
p
(k )
E
+V 

2m
2m
2m
2
The restriction on k, means only certain energies
are allowed, i.e. the energy is quantized.
Thus, by simple substitution, the energy levels
are given by,
np 2
(
)
2 2
(k ) 2
h
n
En 
 L

, with n  1,2,3...
2
2m
2m
8mL
Note the ZPO is consistent with the
uncertainty principle, i.e. particles with zero
momentum can not be localised.