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1 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 7 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. 2 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 3 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. 4 3 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 ! 3 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. 5 4 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 ): 6 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):