The Hydrogen Atom - Reed College

1 Physics 342 Lecture 24 The Hydrogen AtomLecture 24 Physics 342 Quantum Mechanics IMonday, March 29th, 2010We now begin our discussion of the Hydrogen atom. Operationally, thisis just another choice for spherically symmetric potential ( Coulomb).Morally, of course, this is one the great triumphs of our time (technically,the time two before ours). We already know the angular solutions, the usualYm`( , ), so all we need to do is establish the radial portion of the wave-function, and put it all together. Notice that we are following ProfessorGriffiths treatment here, and he uses a different initial dimensionless lengththan you did for your homework.

2 This is no problem, in the end, the spec-trum has to be the same no matter which choice one makes. You will see aslight difference in the recursion relation, and since the recursion relation inthis case is more directly related to the associated Laguerre definition, it issomewhat easier to get the actual radial wavefunctions Radial WavefunctionThe potential, in this case, represents the electrostatic field set up by thenucleus of the Hydrogen atom, as felt by the electron:U(r) = e24 0r.( )This goes into the usual (withu(r) =rR(r) as before) ~22md2udr2+[U(r) +~22m`(`+ 1)r2 E]u= 0( )where we are associatingmwith the mass of the electron.

3 We just made apretty dramatic approximation. We know that the two-particle problem can1 of RADIAL WAVEFUNCTIONL ecture 24be reduced to a stationary center, provided we use the reduced mass of thesystem. On the one had, that is fine but on the other: What do we meanby a two-particle problem in quantum mechanics? For now, just imaginethe nucleus doesn t have much kinetic energy, so that it remains prettymuch fixed (what about the energy associated with having it around at all?Its relativistic rest energy is still there, but we are not doing relativisticquantum mechanics yet).

4 If we write the above out, we have: ~22md2udr2+[ e24 0r+~22m`(`+ 1)r2 E]u= 0.( )As with the infinite square well, it makes sense to let = 2mE~(negativeinside the square root, now bound states will haveE <0 and we wantto make real). We want to define a new coordinate r. Theadvantage is to render the coordinate variable itself unitless. Whenever wewant to consider limiting cases of an equation or more generally, a physicalsetting, we need a point of does it mean to be far away from a distribution of charge, forexample?

5 That clearly depends on how large the distribution itself is. Byre-parametrizing using a fundamental length in the problem, we have allowedfor easier classification of limits. For example, on the E&M side, suppose wehave a dipole moment with a certain lengthd. Then far away means thata field point at a distancerfrom the origin is large compared tod:r suppose we wrote everything in our problem in terms of the new length r r/d. We have eliminated the explicit comparison withdand can referto small r unambiguously as r 0, making for easier Taylor expansion, point is, has units of 1/length and involves the fundamental (and asyet unknown) energy scale, it is a natural choice for constructing = r, aunitless quantity.

6 In the above, we just replacer / , andddr dd .Performing this simple change of variables, multiplying by2m~2in the process,we have d2u( )d 2+[ me22 0~2 +`(`+ 1) + 1]u( ) = 0.( )We have another scale defined by 0 me22 0~2 (there are, evidently, twoenergy scales of interest to us here, hence two lengths we could have written2 of RADIAL WAVEFUNCTIONL ecture 24 in terms of = ( 0 )r), and with this, we can write the final form:d2ud 2=[1 0 +`(`+ 1) 2]u.( )As for limiting cases, we can take , which gives us growing anddecaying exponentials as solutions:d2ud 2=u u( ) =Ae ( )where we have thrown out the growing exponential, that will not be the other hand, when the barrier-term dominates, for small , we have(using uto distinguish from the actual solution)d2 ud 2=`(`+ 1) 2 u,( )and we can solve this by consider a generic polynomial (always a good ansatzfor ODE s of the above flavor).

7 U( ) =a p, thenap(p 1) p 2=a(`(`+ 1)) 2a p( )and then we have a solution forp(p 1) =`(`+ 1), orp= `,p=`+ general solution is a linear combination: u( ) =a `+b `+1( )and we seta= 0, for near zero, this will blow , we will use these two regimes to factor the full solution takeu( ) = `+1e v( ),( )this is naturally dominated by the polynomial near 0, and the exponen-tial will help with integration at infinity. If we input this into our differentialequation, we get d2vd 2+ 2 (`+ 1 )dvd + ( 0 2 (`+ 1))v= 0.( )Letx 2 , then in terms ofx, the above isxd2vdx2+ (2 (`+ 1) x)dvdx+(12 0 (`+ 1))v= 0.

8 ( )3 of ASSOCIATED LAGUERRE POLYNOMIALSL ecture 24 Now, the differential equation:xd2dx2 Lkn(x) + (k+ 1 x)dLkn(x)dx+nLkn(x) = 0( )has solutionsLkn(x), the associated Laguerre polynomials , for is almost the above, if we setk+ 1 = 2(`+ 1) andn= (12 0 (`+ 1)) andwe assume thatnis an integer. In that case, the solution to our problemis just:v(x) =L2`+112 0 (`+1)(x),( )This pre-supposes that12 0 nis an integer, but we can return to thatlater on. For now, this is the source of the quantization of energy, since wehave:2 n= 0=me22 0~2 =me22 2 0~ E m E= me432 20~2 2n2,( )or in more standard form, labelled usingnthe principal quantum number :E n= (m2~2(e24 0)2)1 n2 E1 n2.

9 ( )This is the energy spectrum of Hydrogen we shall return to it in a Associated Laguerre PolynomialsThe associated Laguerre polynomials are defined as the solution to the abovedifferential equation ( ). Most special functions arise as solutions to difficult ODEs, meaning ones not solvable by exponentials or solutions usually proceed by series expansion (Frobenius method), andinvolve points at which we remove certain elements of the solution withbehaviour we do not want to allow. The associated Laguerre polynomialshave Rodrigues formula1 Lkn(x) =exx kn!

10 Dndxn(e xxn+k)( )and can be related to the Laguerre polynomials viaLkn(x) = ( 1)kdkdxkLn+k(x)( )1 For a further compilation of properties, see Arfken and Weber, p. of ASSOCIATED LAGUERRE POLYNOMIALSL ecture 24 Because it is the integerization of 0that quantizes the energy in theHydrogen atom, it is worthwhile to generate the series solution and seehow this appears effectively as aboundary condition(vanishing of the radialwavefunction at infinity). We ll return to the direct ODE, d2vd 2+ 2 (`+ 1 )dvd + ( 0 2 (`+ 1))v= 0,( )and make the series ansatz:v( ) = j=0cj jv ( ) = j=0cjj j 1v ( ) = j=0cjj(j 1) j 2.