Kramers Relation Douglas H. Laurence Department of Physical Sciences, Browar College, Davie, FL 333 Introuction Kramers relation, name after the Dutch physicist Hans Kramers, is a relationship between expectation values of nearby powers of r for the hyrogen atom: s + n r s (s + )a r s + s [(l + ) s ]a r s = 0 The relation is very important when computing, specifically, perturbative corrections to the hyrogen spectrum, as those computations require expectation values of the raial Hamiltonian (which has powers of r such as r an r ) an potentially perturbative corrections that have higher (negative) powers of r. Having this relation hany makes a lot of those calculations go more quickly. This erivation is going to require the solution to the hyrogen atom to be known: the Hamiltonian, the wavefunctions, the energy spectrum, etc. To solve Kramers relation for all s > is straightforwar, an will be shown, but to solve for any s requires one of these s values to alreay be known, for instance r. The Feynman-Hellmann theorem allows for an easy computation of this, which I erive in the appenix, as well as apply to compute r explicitly, allowing for r s to be compute for all values of s using Kramers relation. The Derivation of the Relation The raial wave equation for the hyrogen atom is typically written as: where the energy is: H r R nl (r) = [ m r + m l(l + ) r ke r ] R nl (r) = E n R nl (r) () E n = m n a () where a is the Bohr raius, efine as /kme. A more useful form of the above equation, however, is to make the typical substitution u(r) = rr(r), an then express the raial wave equation as: [ l(l + ) u = r ar + ] n a u (3) Note that I ve roppe the labels of nl; these are implie by their presence in the above equation.
Notice the terms we have in the above equation: r, r, an r 0. If we multiply both sies by r s, then we ll get exactly the powers of r that we nee for Kramer s relation: [ r s u = l(l + )r s a rs + ] n a rs u Now what we nee to o is multiply both sies by u = u an integrate. Then, we ll have: ur s u r = l(l + ) r s a rs + n a rs () What we nee to o is to figure out how to eal with the integral on the left-han-sie. Luckily, this integral is set up so that it inclues a factor of ur s an a factor of u r. If we integrate by parts, it will take two steps to convert the u to a u, resulting in an integral like uf(r)ur i.e. an expectation value an uring those two steps, the power of r will rop by one or two, so we ll have powers like r s, r s, an r s, exactly as we like! Integration by parts makes perfect sense to use, then, to reuce the integral on the left-han-sie of the above equation into aitional factors of expectation values of these powers of r. First, note that the integral goes from r = 0 to r =, an u equals zero at both limits (because R ecreases exponentially as r, beating the factor of r that increases linearly), so the terms in the integration by parts that are evaluate at these limits will always result in zero, since they will have some epenence on u. Note that I ll be using the notation: wv = vw for the integration by parts; this way, my choice in how to perform the integration will be clear. For the first step of the integration process, I ll choose w = ur s an v = u r, so integration by parts yiels: ur s u r = u r s u r } {{ } I s ur s u r } {{ } I where I have calle the first integral on the right-han-sie I an the secon integral I for brevity. Each of these integrals will also nee to be evaluate. For I, I ll make the seemingly-o choice of w = (u ) an v = r s r; if I ha chosen something similar to before, like w = u r s an v = u r, I get a recursion relationship that wouln t be helpful. So, the integration by parts results in: I = u r s u r = s + (5) u r s+ u r (6) I alreay have an equation for u in terms of u (the raial wave equation), so ultimately I m going to be solving integrals like u r k u, so let s figure out what those integrals yiel before plugging u into the above equation. Choosing w = ur k an v = u r, integration by parts yiels: u r k ur = u r k ur k ur k ur Pulling the first integral to the left-han-sie an noting the secon integral is just the expectation value of r k, we fin: u r k ur = k rk
Now we want to plug our raial wave equation into I, equation (6), an use the above result to evaluate it: [ l(l + ) u r s+ u r = u r s+ r ar + ] n a ur = l(l + ) u r s ur u r s ur + a n a u r s+ ur Thus, the integral I equals: I = l(l + )(s ) = r s + s a rs s + n a rs l(l + )(s ) (s + ) r s s a(s + ) rs + n a rs We can evaluate the integral I with the formula above: I = s ur s u s(s ) r = r s So, our original integral, given by equation (5), is: ur s u l(l + )(s ) r = I I = r s s + (s + ) a(s + ) rs s(s ) n a rs + r s Now, going all the way back to the start, we see the above result is only the left-han-sie of equation (), so we can set it equal to the left-han-sie an solve: l(l + )(s ) (s + )... + r s + s(s ) s a(s + ) rs n a rs +... r s = l(l + ) r s a rs + n a rs If we multiply both sies by (s + ) an group everything to the left-han-sie, we arrive at: l(l + )(s ) r s + s (s + ) a rs n a r s + s(s )(s + ) r s...... l(l + )(s + ) r s + (s + ) a Notice that combining the r s terms gives a coefficient of: r s (s + ) n a r s = 0 l(l + )(s ) l(l + )(s ) + s(s )(s + ) = l(l + )s + s(s ) = s(l + l s + ) = s [ (l + ) s ] The coefficient of the r s term becomes (s + )/a, an the coefficient of the r s term becomes (s + )/n a. Putting this all together, our above equation becomes: s [ (l + ) s ] r s + (s + ) a r s (s + ) n a r s = 0 Finally, multiplying the above equation by a /, an re-orering from r s to r s, we have Kramers relation: s + n r s (s + )a r s + s [ (l + ) s ] a r s = 0 (7) 3
3 Solutions of the Relation The main application of Kramers relation is that of computing perturbative corrections to the hyrogen atom, such as for the fine structure or hyperfine structure of hyrogen. In orer to see this in action, let s compute some particular solutions to Kramers relation. First, the s = 0 solution yiels: n r0 a r = 0 So, re-arranging, we fin the expectation value of /r: = r n a (8) Notice also the s = solution yiels: n r 3a r 0 + [ (l + ) ] a r = 0 Since we just foun r, we can solve the above for r. We ll see that setting s = will allow us to solve for r, an so on. However, we run into a problem when trying to go to larger negative powers of r s, such as trying to fin r (by setting s =, for instance): the best we ll be able to o is to fin an equation for r in terms of r 3. Going to progressively larger (negative) powers keeps the problem going unless we can solve for one along the chain, e.g. r or r 3, then Kramers relation will never allow us to solve for any value of s below. This is seen by checking the s = case: a r [ (l + ) ] a r 3 = 0 As we can see, without knowing either r or r 3, we can t solve for the other. A common way to circumvent this problem is to use the Feynman-Hellmann theorem to compute r, which allows for all subsequent solutions for s < to be foun using Kramers relation. The Feynman- Hellmann theorem, an the computation of r itself using the theorem, are presente in the appenix; I ll merely present the result here: r = (l + )n3 a (9) Knowing r, we can solve the above s = solution to Kramers relation for r 3 : r 3 = a[(l + ) ] r = [(l + ) l](l + )n3 a 3 Notice that (l + ) = l + l + = l(l + ), so the above equation results in: r 3 = l(l + )(l + (0) )n3 a 3 Now that the negative s solutions have been starte, one coul compute r s for all values of s using Kramers relation.
A Feynman-Hellmann Theorem Imagine we ha a Hamiltonian H with normalize eigenstates ψ such that: H ψ = E ψ If we consiere some parameter λ that we coul express the Hamiltonian, the states, an the energies in terms of, then the above equation woul become: This, for normalize states, means that: H(λ) ψ(λ) = E(λ) ψ(λ) () E(λ) = ψ(λ) H(λ) ψ(λ) Now we want to consier taking the erivative of the energy with respect to this parameter λ. Applying the prouct rule to the right-han-sie, we see that: ( ) ( ) E(λ) = λ λ ψ(λ) H(λ) ψ(λ) + ψ(λ) H(λ) λ ψ(λ) + ψ(λ) H(λ) λ ψ(λ) For the first an final terms on the right-han-sie of the above equation, we can apply H(λ) right-war an left-war, respectively, yieling: ( ) ( ) E(λ) = E(λ) λ λ ψ(λ) ψ(λ) + ψ(λ) H(λ) λ ψ(λ) + E(λ) ψ(λ) λ ψ(λ) Notice, then, that the first an thir terms above are simply an example of the prouct rule: ( ) ( ) λ ψ(λ) ψ(λ) + ψ(λ) λ ψ(λ) = λ ψ(λ) ψ(λ) Here s the final step of the erivation: recall that we require that the states be normalize. Thus, for any parameter λ, ψ(λ) ψ(λ) =, an so the above term equals zero. Therefore, our above equation for E/λ simply results in: E(λ) λ = ψ(λ) H(λ) λ ψ(λ) This equation is known as the Feynmann-Hellmann theorem. Using the Feynman-Hellmann theorem, I can erive many interesting results. Most important to this note is the erivation of r for the hyrogen atom. To o so, we ll express our (raial) Hamiltonian in the usual way: H = m r + m l(l + ) r e πɛ 0 r an we ll express the energy of a state ψ nlm in terms of n = j max + l +. This is because, in orer to compute r, we nee to choose λ = l in the Feynman-Hellmann theorem, so E nees to be expresse in terms of its epenence on l, which is hien in its epenence on n. In this respect, the energy is given by: E n = ma (j max + l + ) Technically the claim that this equation is true is the theorem. The woring that everyone uses just bothers me. () 5
As I sai above, we want to apply the Feynman-Hellmann theorem in the case of λ = l. So, first we ll calculate the erivative of E: E(l) l = ma (j max + l + ) 3 = mn 3 a where I have converte back from l epenence to n epenence because we no longer nee the l epenence to be explicit. Next, we ll calculate the erivative of H with respect to l: H(l) l = l + m r Finally, we ll compute the expectation value of this erivative: H(l) = (l + ) l m r = m ( l + ) r So, plugging our results into the Feynman-Hellmann theorem, we have: ( mn 3 a = l + ) m r from which r is very easily solve for: r = (l + )n3 a (3) 6