Transcript lecture 17

PHYSICS 420
SPRING 2006
Dennis Papadopoulos
LECTURE 17
THE PARTICLE IN A BOX
CLASSICAL STANDING WAVES
1 ( x, t )  Bsin(kx  t )
 2 ( x, t )  Bsin(kx  t )
 ( x, t )  1 ( x, t )   2 ( x, t )  Asin(kx) cos(t )
A  2B
Nodes at nl/2
1
0.75
0.5
0.25
Nodes
2
-0.25
-0.5
-0.75
-1
4
6
l/2
8
10
String clamped between
two points separated by
a distance L. What
possible standing waves
can fit it?
Only the ones that have
nodes at the end l2L/n.
Quantum mechanical versionthe particle is confined by an
infinite potential on either side.
The boundary condition- the
probability of finding the particle
outside of the box is ZERO!
l=2L
l=L
l=3L/2
 (0, t )  0 at x  0
 ( L, t )  0 at x  L
For the quantum mechanical case:
For the guitar string:
Assume a general case:
1( x, t )   2( x, t )
 A1ei ( kx t )  A2ei ( kx t )
 A1ei (
2 mE x / h  Et / h )
2 mE x / h  Et / h )
 A2ei (
at x  0
A1ei (  Et / h )  A2ei (  Et / h )  0
at
xL
 A1ei (
hk  p
2 mEL / h  Et / h )

2 mEL / h  Et / h )
 A2ei (
E  p / 2m  (h k ) / 2m
 Ae  iEt / h e  i
k  2mE / h
 2iA sin( 2mE L / h )  0
2
2
2
2 mEL / h
 ei
2 mEL / h
 2mE L / h  n
n 2 2h 2
 En 
2mL2
n  1, 2,3...
0
0
Fig. 6-7, p. 202
TIME
INDEPENDENT
SE
 ( x, t )   ( x) (t )
d
ih
 E (t )
dt
 (t )  exp(iEt / h )
h 2 d 2

 U ( x) ( x)  E ( x)
2
2m dx
Free particle U(x)=0
 ( x)  exp(ikx)
E  h k / 2m
2
2
Constant E is the total particle energy
Notice that
( x, t )   ( x) exp(iEt / h )
 ( x, t )   ( x)
2
2
FOR STATIONARY STATES
ALL PROBABILITIES ARE
STATIC AND CAN BE
CALCULATED BY THE TIME
INDEPENDENT SE
Quantum mechanical versionthe particle is confined by an
infinite potential on either side.
The boundary condition- the
probability of finding the particle
outside of the box is ZERO!
here, we
assumed the
walls were
infinite—there
was no
possibility the
particle could
escape.
ODD AND
EVEN
 (0, t )  0 at x  0
 ( L, t )  0 at x  L
For the quantum mechanical case:
d 2
2
 k  ( x)
2
dx
2mE
2
k  2
h
 ( x)  A exp(ikx)  B exp(ikx)
 (0)  0
A  B
 ( L)  0
A[exp(ikL)  exp(ikL)]  2 AiSin(kL)  0
kL  n
h 2 kn2 n 2 2h 2
En 

2m
2mL2
ALTERNATIVE SOLUTION
 ( x)  A sin(kx)  B cos( kx)
 (0)  0
B0
 ( L)  0
ASin(kL)  0
kL  n
h 2 kn2 n 2 2h 2
En 

2m
2mL2
n x
 n ( x)  A sin(
)
L
0 xL
n  1, 2,3,..
n x
1  A  dx sin (
)
L
0
L
2
A  2/ L
2
Note odd and even
solutions
What is the probability that a particle will be found between L/4 and 3L/4 in
the ground state?
  2m(U  E ) / h
2
2
 2  2m(U  E ) / h 2
Fig. 6-15, p. 209
I
II
III
U
E
d 2  ( x)
2m



( E  U ) ( x)
2
2
dx
h
Since E<U:
d 2  ( x ) 2m
 2 (U  E ) ( x)
2
dx
h
Generally, solutions are then:
0
Setting
2m
  2 (U  E )
h
2
L
This begins to look like the
familiar undamped SHO equation:
d 2  ( x)
2


 ( x)
2
dx
( x)  C1e x  C2e x  C1e
But remember the conditions
imposed on wave-functions so that
they make physical sense:
2 m (U  E ) x / h
 C2e
2 m (U  E ) x / h
I
II
( x)  C1e x  C2e x  C1e
III
 C2e
2 m (U  E ) x / h
C1 must be 0 in Region III and C2 must be zero
in Region I, otherwise, the probabilities would
be infinite in those regions.
U
E
-L/2
2 m (U  E ) x / h
L/2
Note that the wavefunction is not necessarily 0 in Regions I and II.
(It is 0 in the limit of an infinite well.) How is this possible when
U>E?? The uncertainty principle.
Now we must use the condition of continuity (the wavefunction must be
continuous at the boundary, and so must its first derivative).
Suppose we had
a discontinuous
function…
(x)
L/2
x
Here, the
acceleration
would be
infinite. Uh-oh!
d/dx
L/2
x

1


h
2m(U  E )
n 2 2h 2
En 
2m( L  2 ) 2
Fig. 6-16, p. 210
Even solutions
Odd solutions
Note that the particle has negative kinetic energy
outside of the well: K=E-U
The wave-function decreases rapidly- 1/e in a space of 1/
1/ is a penetration depth. To observe the particle in this region, you
must measure the the position with an accuracy of less than 1/
h2
x  1/  
2m(U  E )
From the
uncertainty
principle:
p  h / x  h
p 

K 
2m
2m(U  E )
h2
2
 8 2 m(U  E )
The uncertainty in the measurement is larger
than the negative kinetic energy.
A “real world example” of a finite box would be a neutron in a nucleus.
Note that the allowed
energies are inversely
proportional to the length
of the box.
The energy levels of a
finite box are lower than
for an infinite well
because the box is
effectively larger. There
is less confinement
energy.
At x=0, all of a particle’s energy is potential
energy, as it approaches the boundary, its
kinetic energy becomes less and less until it
all of the particles energy is potential energyit stops and is reflected back. An example
would be a vibrating diatomic molecule.
This is analogous to a classical system,
such as a spring, where potential energy is
being exchanged for kinetic energy.
In contrast to the square well, where the particle moves
with constant kinetic energy until it hits a wall and is
reflected back, in the parabolic potential well of the
harmonic oscillator, the kinetic energy decreases
(wavelength increases) as the boundary is approached.
p2
 E  U  p  2 m( E  U )
Kinetic energy: K 
2m
p
h
l
 2 m( E  U )
The wavelength is position dependent:
Making an approximationassuming small penetration
depth and high frequency, the
condition for an infinite
number of half wavelengths
as in an infinite well must be
recast as an integral to
account for a variable
wavelength:

b
a
Boundary conditions:
  0 at x  
1 3
2m[ E  U ( x)]dx  ,1, ...
2 2
Note that the width of the well is
greater for higher energies. As
the energy increases, the
“confinement energy” decreases.
The levels are evenly spaced.
We have Planck’s quantization
condition!
d 2 2m 1
2 2

(
m

x  E ) ( x)
2
2
dx
h 2
  m / 2h
1
E  (n  )h 
2
n  1, 2,3,...
In classical physics,
the “block on a
spring” has the
greatest probability
of being observed
near the endpoints
of its motion where it
has the least kinetic
energy. (It is
moving slowly here.)
This is in sharp
contrast to the quantum
case for small n.
In the limit of large
n, the probabilities
start to resemble
each other more
closely.
Fig. 6-18, p. 214
Fig. 6-19, p. 215
Fig. 6-20, p. 216
•We can find allowed wave-functions.
•We can find allowed energy levels by plugging those wavefunctions into
the Schrodinger equation and solving for the energy.
•We know that the particle’s position cannot be determined precisely, but
that the probability of a particle being found at a particular point can be
calculated from the wave-function.
•Okay, we can’t calculate the position (or other position dependent
variables) precisely but given a large number of events, can we predict
what the average value will be? (If you roll a dice once, you can only
guess that the number rolled will be between 1 and 6, but if you roll a
dice many times, you can say with certainty that the fraction of times you
rolled a three will converge on 1 in 6…)
If you roll a dice 600 times, you can
average the results as follows:
1  2  4  6  5  5  3  4  ...
600
Alternatively, you can count the number of
times you rolled a particular number and
weight each number by the the number of
times it was rolled, divided by the total
number of rolls of the dice:
99
97
104
(1) 
(2) 
(3)  ...
600
600
600
After a large number of rolls, these ratios
converge on the probability for rolling a
particular value, and the average value
can then be written:
x   xPx
This works any time you have discreet values.
What do you do if you have a continuous variable,
such as the probability density for you particle?




2
 x   xPx dx   x  ( x, t ) dx
It becomes an integral….
2.5  3.7  1.4  ....  5.3
 5.46
18
1.4(1/18)  2.5(1/18)  ...  5.4(3 /18)  6.2(2 /18)  ...  8.8(1/18)  5.46
x
x   xPx
Expectation, value

x 


x  ( x, t ) dx
2
Table 6-1, p. 217
The expectation value can be interpreted as the average value of x that we
would expect to obtain from a large number of measurements. Alternatively it
could be viewed as the average value of position for a large number of
particles which are described by the same wave-function.
We have calculated the expectation value for the position x, but this can be
extended to any function of positions, f(x).
For example, if the potential is a function of x, then:

U   U ( x) ( x, t ) dx

2
expression for
kinetic energy
p2
KE 
;
2m
the potential
kinetic plus potential
energy gives the total
energy
p  hk
x
p
x
potential energy
U
U(x)
kinetic energy
K
h 2 2

2m x 2
total energy
E
h 
i x
ih

t
operator
observable
position
momentum
In general to calculate the expectation value of some observable quantity:
Q   * Qdx


We’ve learned how to calculate the observable of a value that is simply a function of x:
U    U dx    U ( x)dx   U ( x)  dx


*


*

2

But in general, the operator “operates on” the wave-function and the
exact order of the expression becomes important:
 h 2 2 
K     K dx     
dx
2 


 2m x 

*

*
Table 6-2, p. 222