Transcript Introduction to Quantum Mechanics AEP3610 Professor Scott

Review of statistics and probability I
• consider first a discrete distribution of possible outcomes
of some measurement: each measurement has one value hi
and can be placed in one bin, bins labeled by the index i
• the total number of possible different bins is H
• the number of measurements that fall into bin i is ni
• Example: measurements of human height, to nearest cm
• the set of ni values is the discrete distribution H
• the total number of measurements is N:
N :  ni
i 1
• rescale ni by 1/N to get probability pi of hi:
• pi distribution is said to be normalized:
H
ni
pi :
N
 pi  1
i 1
• will shortly convert this discrete stuff to a continuous distribution
by replacing summations (over heights) by integrations
Review of statistics and probability II
H
H
• if we lined up all the people end to end
L   ni hi  N  pi hi
we would obtain the total length L:
i 1
i 1
• the average height of a person is <h>:
L 1
 h  
N N
H
 ni hi 
i 1
H
 ni hi
H
i 1

H
 ni
i 1
i 1
 pi hi
• and this is a
paradigm for
calculating
ANY average!
• How narrow or broad is the distribution?
• the deviationi of a measurement hi from the average := hi – <h>
• what is the average deviation? What does it tell us?
 hi   h    h    h    h    h   0
• it tells us NOTHING about the “spread” of the distribution
Review of statistics and probability III
• consider the squared deviationi:= (hi - <h>)2
• this is always ≥ 0
• now consider the average of this quantity: the variance
1
variance :
N

H
i 1
H

i 1
H
 ni hi   h    pi hi   h 2
2
pi hi2
2
H

pi hi  h  
i 1
i 1
H

pi  h  2
i 1
2
  h 2  2  h  h    h    h 2    h  2
• the square root of this quantity is the standard deviation,
with the same units as the quantity in question
• it will turn out to be the uncertainty in a measurable
standard deviation  : variance 
   h2    h 2
1
N
H

i 1
ni hi   h   
2
H

i 1
pi hi   h  2
Now convert to a continuous distribution
• now the bin index is a continuous variable: i  x
• now the sums from i =1 to i = H become integrals over the
domain of x (say, –∞ ≤ x ≤ ∞)
• define the probability density r(x) as follows: the probability
that a measurement of x occurs between x and x + dx is r(x) dx
• the dimensions of r(x) are the dimensions of 1/x
b
• probability that measurement x is between a and b: Pab :  r ( x)dx
• r(x) distribution is said to be normalized:

a
 r ( x)dx
1
b 
• average value of x in that interval a < x < b:  x   xr ( x)dx
a
b
• average value of f(x) in that interval a < x < b:  f   f ( x) r ( x)dx
a
Application of these ideas to the black body radiator
• we assume that a cavity with a small hole acts as a perfect
black body, since any radiation (light) that strikes the hole is
‘absorbed’… so whatever light comes out of the hole is
characteristic of black body radiation
• if one ‘counts standing wave modes’ in a cavity of volume V,
it may be shown that the number of modes in frequency range
df is expressed as a probability-like distribution
8V 2
 number of modes in tiny frequency interval df is dN :  ( f ) df 
f df
3
c
• for a system to possess an energy E when in an environment at
temperature T, the probability that its energy is within dE is
expressed by the Boltzmann factor: r ( E ) dE  Ae  E kT dE
• When this is normalized, it is easy to show A = 1/kT
b
• average energy is therefore
1
 E 
Ee  E / kT dE  kT
kT

a
The wrong black body theory and its repair
• Now we make an incorrect but reasonable assumption:
assume that each mode can have any energy E at all, as long
as the average in the mode is that average energy kT (thus,
consistent with the Boltzmann probability), so that the amount
of energy in frequency range df is (f) kT df:
8V
 e ( f ) df  3 kTf 2 df
c
• this ‘energy-per-frequency’ function e(f), unfortunately, diverges to
infinity as f  ∞, and so does its integral over all frequencies, which
would be the total energy in the cavity!
• both problems are collectively termed the ultraviolet catastrophe!
• to fix the problem, Planck proposed that the mode energy E is
restricted to only be an integer times a constant times the frequency:
E = nhf, and the value of h will be determined by the data
• thus, the mode energy is quantized so now are statistics are those of
a discrete distribution: we say that each mode contains an integer #
of photons appropriate to its frequency
Implications of quantizing a mode’s energy
• again we invoke Boltzmann, and say that the probability for
 nhf / kT
where E  nhf
a mode to have energy E is p E  Ae
• of course, we must normalize this discrete probability distribution:

 pE  1 A e
 since 0  e
 nhf / kT
 e

1 A
n 0
 hf / kT

 hf / kT n
1
n 0
 1 for any f , this is a common and well - behaved sum :



1
xn 
for | x |  1 
e  hf / kT
1 x
n 0
n 0


n

1
1  e  hf / kT

 A  1  e  hf / kT

 hf / kT  nhf / kT
p

1

e
e
• so the probability to have energy E is E
• now we can find the average energy of a mode as usual:

E


Ep E 
n 0

 hf  1  e
 hf / kT
 nhf / kT





nhf
1

e
e


n 0
 hf / kT
 ne

n 0
 nhf / kT
The UV catastrophe has vanished!
 this too is a common and well - behaved sum :



ne
n 0

 hf / kT n

e  hf / kT
1  e

 e
 hf / kT 2

 nx  1  x 2 for | x |  1
n 0
1
hf / kT
x
n

 1 1  e hf / kT

• assembling all the pieces we arrive at the Planck distribution

8V 2
1
 hf / kT 

 df
 e ( f ) df  3 f hf  1  e
hf / kT
 hf / kT 

c
1 1  e
 e


8Vh 
f3
 df
 3
hf
/
kT
c  e
 1 
• drops to zero for small f
like f3
• drops to zero for large f
like hf / kT

e


