Download 1 Planck`s black body radiation formula

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Planck’s black body radiation formula
The point of the …rst three lectures was to contrast classical physics and
quantum physics. The work of Planck, Einstein and de Broglie was emphasized. The work of Planck gave a “correct” formula for the distribution of
black body radiation at a …xed temperature, T , as a function of the wave
length, , of the light,
c
=
f
with f the frequency. The breakthrough involved the “ansatz” that the
energy of a photon is E = hf . Thus n photons would have energy nhf . His
derivation of his formula involved methods of statistical mechanics. In class
we gave an “explanation”of Planck’s constant based on the correspondence
principle. The starting point is the Raleigh-Jeans formula for black body
radiation distribution
2ckT
4 jd j
with c the speed of light and k Boltzman’s constant. This formula is essentially correct for long wavelengths but indicates that as the wavelength goes
to 0 the intensity goes to in…nity. To counter this Wien gave a formula that
was graphically more like the actual measurements
C1
5
C2
T
e
jd j
with C1 and C2 to be chosen to …t the data. This is also related to Wien’s
Law which says that the maximum of the distribution function depends only
on the temperature and in fact the value of the maximum determines the
temperature. Taking his density
C1
5
C2
T
e
and di¤erentiating one …nds
d C1
e
dt 5
C2
T
=
C1 C2
5C1 T
T
which has a unique 0 at
=
C2
:
5T
1
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e
C2
T
In our “bogus history”we suggested that Planck observed that a formula
like
C1 jd j
5
C2
e T 1
was close to the Raleigh-Jeans formula for long and close to the Wien
suggestion for short. To see these assertions set u = 1 then Planck’s
formula is
u
C1 u4 C2
e( T )u 1
the factor
u
C2
e( T )u 1
is (according to freshman calculus) for
C2
T
long so u small
T
1
=
(1 + O(u))
C2
+ O(u)
thus Planck’s formula for
long is
C1 T
1
4 (1 + O( )):
C2
Thus to …t the classical formula we need
C1
= 2kc:
C2
If
is small we have
C1
5
1
e
C2
T
=
1
C1
5
e
1
C2
T
1
e
C2
T
=
C1
5
e
C2
T
(1 + O(e
C2
T
)):
Which is progressively closer to Wien’s suggestion when is progressively
shorter. How do we give a good reason that Planck’s formula is physically
reasonable? First we observe that if we set
C1 = C10 c2
and
c
C2 = C20 ;
k
2
Then to agree with the classical formula at long wave lengths we need
C10
= 2:
C20
Thus if we set h = C20 we have Planck’s formula
2hc2
5
e
jd j
hc
kT
1
:
and Wien’s wave length at the maximum is
=
hc
:
5kT
This can be used to calculate h, Planck’s constant which is approximately
6 10 34 joules sec :Thus if the wave length is long then we can think of h
as essentially 0: If we do this and look at Planck’s formula it is
2c2
5
h
e
hc
kT
1
=
2c2
5
kT
2ckT
(1 + O(h)) =
4 (1 + O(h)):
c
Thus we have the …rst instance of the correspondence principle that if we
can consider Planck’s constant is 0 then quantum formulas should be good
approximations to classical ones.
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Einstein’s explanation of the photoelectric
e¤ect and de Broglie’s extension
Due to the success of Maxwell’s theory of electricity and magnetism and
the overwhelming evidence of wave phenomena in light and electromagnetic
phenomena the prevailing idea was that light is propagated in waves. We
all know that if we consider sound waves then the intensity of the waves
(how painful it can be to the ears) is proportional to the amplitude of the
wave. However, many researchers including Hertz found that if a plate that is
negatively charged has light of very short wavelength (high frequency) shine
on it then it will lose the charge. However, nothing happens if the wave
length is su¢ ciently long. The prevailing theory was that the loss of charge
(and hence the current) was caused by the light waves adding energy to the
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electrons which caused them to escape. This contradicted the idea that the
amplitude of a wave should be the cause of its intensity.
Einstein’s solution to the problem was to take Planck’s “ansatz”seriously
that is a photon is a particle with energy equal to hf Planck’s constant times
the frequency. He published his paper on the photoelectric e¤ect in his great
year of 1905. It was for this insight that he received the Nobel prize in
physics in 1921. His theory, in fact, …t perfectly with the measurements and
led to the actual beginning of quantum mechanics in the hands of de Broglie,
Schrödinger, Heisenberg, etc. However, there is now a new problem what
does it mean for a particle to have a wave length? de Broglie came up with
a novel idea: every particle has both wave and particle behavior.
de Broglie (in his thesis) considers a plane wave
(t; !
x ) = exp
!
with ! = 2 f and k =
!
c
i !t
! !
k x
!
and k is the vector of that magnitude in the
direction of the wave. Thus E = hf = ~! with ~ =
h
2
. Hence
@
(t; !
x ) = ~! (t; !
x ) = E (t; !
x ):
@t
Now if !
p is the momentum vector that is m!
v (m the mass and the velocity
!
v vector) then
2
j!
pj
E=
+ V (x)
2m
the …rst term is kinetic energy and the second is the potential energy. From
!
the above we see that !
p =~ k and
i~
!
(t; !
x)= k
2
2
j!
pj
!
(t; x ) = 2 (t; !
x ):
~
The upshot is that
i~
@
(t; !
x)=
@t
~2
2m
+ V (!
x)
(t; !
x ):
Schrödinger to this equation as being the basis of the dynamics of the new
quantum mechanics. In particular, it should be applicable to any quantum
mechanical wave function. Before we go into this in more detail we should
take a brief view of the symplectic formulation of classical mechanics.
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Conservation of energy
Recall that Newton’s equations are given by
!
d
F = !
p:
dt
!
!
!
x
) and so the
The force F is usually given by a formula (e.g. F = jC!
x j3
force law is a second order ordinary di¤erential equation. If the force has a
!
C
potential, that is it is the gradient of a function (e.g. F = r j!
) that is
x j2
P @V
!
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F = rV =
j @xj ej (e1 ; e2 ; e3 ) the standard basis of R then Newton’s
equations become
d
@V
=
pi
@xi
dt
along the ‡ow. We then have that along the ‡ow
P 2
X @V dxj
X d
pj
pj
d
d
V =
=
pj
=
:
dt
@xj dt
dt
m
dt 2m
j
j
This implies that the quantity
2
j!
pj
+ V (!
x)
E=
2m
is constant along the solutions to Newton’s equation when the force has the
potential V . This conserved quantity is called the energy.
This leads to Hamilton’s version of Newton’s equations when there is a
potential. We will phrase them in the language of Poisson brackets. If f; g
are functions on an open subset of R6 with coordinates p1 ; p2 :p3 ; q1 ; q2 ; q3 then
we set
X
@f
@g
@f
@g
ff; gg =
:
@p
@q
@q
@p
j
j
j
j
j
Then writing H = E and qi = pi Newton’s equations become (the dot
corresponds to derivitive in t)
q_i = fH; qi g
p_i = fH; pi g:
H is called the Hamiltonian.
Exercise. Show that if f is a function of the p’s and the q’s then along
the solutions to the ODE we have f_ = fH; f g:This gives another proof that
H_ = 0.
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Schrödinger’s equation
We saw above in the case when the force …eld has a potential Newton’s
equations become
f_ = fH; f g
for f a function of position and momentum. If we now write the quantum
Hamiltonian as
~2
+ V (q)
H=
2m
then the quantum analogue to the above equation is
i~
@
=H :
@t
This is Schrödinger’s equation and H is called the time independent equation.
More generally we can consider the operator H to be a densely de…ned
symmetric operator on a Hilbert space H. This means that H is de…ned for
vectors v 2 D H a dense subspace such that if v; w 2 D then hHvjwi =
hvjHwi : It is generally assumed that H is self adjoint (or at least desired).
This means that the closure of H; H, is also symmetric and H = H . To de…ne
the terms.The domain of the closure of H is the subspace of H consisting of
the elements, v, such that
v (w)
= hvjHwi
initially de…ned for w 2 D extends to a continuous linear functional on H.
e and since it contains D it is dense.
This subspace will be denoted denoted D
The Riesz representation theorem implies that there exists u 2 H such that
if
v (w) = hujwi
we de…ne the closure of H; H by u = Hv:If H is self adjoint we will assume
that it is equal to its closure.
If H is self adjoint then there exists a function from the Borel sets of R to
the orthogonal projections on H (P 2 = P; P = P ). that has the properties
of a vector valued measure. That is, P (R) is the identity and if S \ T = ;
then P (S)P (T ) = 0 and if Sj are Borel subsets that are mutually disjoint
then
X
P (Sj ) = P ([Sj ):
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This implies that if v; w 2 D then S 7 ! hP (S)vjwi is a complex Borel
measure. Denoted v;w . The spectral theorem implies that this P can be
chosen so that
Z
hvjHwi =
d v;w ( ):
R
This is also written as
H=
Z
dP ( )
R
Using the spectral theorem we can de…ne
Z
itH
eit dP ( ):
e =
R
We therefore see that we can solve the Schrödinger equation
i~
@
=H
@t
by setting
i
tH
~
(t) = e
7
(0):