A Relativistic Electron in a Coulomb Potential

Transcription

1 A Relativistic Electon in a Coulomb Potential Alfed Whitehead Physics 518, Fall 009 The Poblem Solve the Diac Equation fo an electon in a Coulomb potential. Identify the conseved quantum numbes. Specify the degeneacies. Compae with solutions of the Schödinge equation including elativistic and spin coections. Appoach My appoach follows that taken by Diac in [1] closely. A few modifications taken fom [] and [3] ae included, paticulaly in egads to the final quantum numbes chosen. The geneal stategy is to fist find a set of tansfomations which tun the Hamiltonian fo the system into a fom that depends only on the adial vaiables and p. Once this fom is found, I solve it to find the enegy eigenvalues and then discuss the enegy spectum. The Radial Diac Equation We begin with the electomagnetic Hamiltonian with H = p 0 cρ 1 σ p q A c + ρ 3 mc 1 ρ 1 = ρ 3 =

2 σ = , 0 i 0 0 i i 0 0 i 0, We note that, fo the Coulomb potential, we can set using cgs units: p o = eφ = Ze This leads us to this fom fo the Hamiltonian: A = 0 H = Ze cρ 1 σ p + ρ 3 mc 5 We need to get equation 5 into a fom which depends not on p, but only on the adial vaiables and p. We do this by looking fo quantities that commute with the tems of the Hamiltonian. In ode to find the fist, we conside the total angula momentum J = L + 1 σ 6 By assumption, we assume that we can wite L = ˆx ˆp 7 j = J 1 + J + J = J We have ceated the left-hand side abitaily. The vaiable j is a non-zeo intege, and we will find late is elated to but not the same ase the total angula momentum quantum numbe j. The ight-hand side is abitay at this point, but will late be shown to be of use. In geneal, if B and C ae opeatos that commute with σ, it can be shown that whee we have defined σ B σ C = B C + i σ B C σ B C = ɛ ijk σ k B i C j 10 i,j,k 9 which is something like a coss poduct fo ou opeatos. [L 1, L ] = L 3 and cyclic pemutations, it can be shown that In paticula, due to the fact that

3 We can thus combine equations 9 and 11 to find L L = i L 11 = L L + i L = L 1 13 By cleve eaangement, this is equivalent to = L + 1 σ 3 4 [ + ] = J = j 14 whee I have used equation 6 to simplify. Note that we could identify j with the quantity in the backet on the left. Unfotunately, this does not have the desied commutation popeties. We can make futhe headway by continued use of equation 9: σ p = L p + i p = i p σ p = p L + i σ p L = i σ p L whee we have used p L = L p = 0. Adding these two expessions: σ p + σ p = i σ 1 L p 3 L 3 p + p L 3 p 3 L 13 whee the summation 13 denotes cyclic pemuation of the indices 1, and 3 a notation used often by Diac. By e-odeing the tems in the backet inside the sum, we can see that they ae, in fact, twice the commutato [ L, p] = i p. That means σ p + σ p = σ p We can edistibute the ight hand side into the left hand side, giving + σ p + σ p + = 0 15 This poves that + anti-commutes with σ p, which is in some ways simila to the Runge-Lenz vecto p L mkˆ, up to the constant tem. Due to the fact that cρ 1 is a constant see equation, + also anti-commutes with cρ 1 σ p, which is a tem in the Hamiltonian, equation 5. 3

4 Futhemoe, it can be shown by diect computation that + also commutes with the othe two tems in tha Hamiltonian Ze and ρ 3 mc. We would vey much like this quantity to commute with all thee tems. This can be accomplished by noticing that ρ 1 ρ 3 = 0. Theefoe, ρ 3 + must commute with cρ 1 σ p, since the ρs enfoce a zeo commutato. We have only multiplied by a constant see equation 3, which does not change the existing commutation with the othe two tems. Now we attempt to elate this to j by squaing the quantity ρ 3 + : [ρ 3 ] + = ρ 3 + = + = J whee I have used ρ 3 = I 4. This is the same expession we had in equation 14. We can now define j as so: j = ρ and est assued that it is a constant of motion, commuting with eveything in the Hamiltonian. Note that j is fee to take any non-zeo integal value. We now have one constant of motion. Equation 9 can yield still moe esults, by application to the x and p opeatos: σ x σ p = x p + i σ x p = x p + i = p + i ρ 3 j 1 whee I have used equation 16 and the fact that ρ 3 is its own invese. Note that I have used the fact that whee x p = p p = 1 xp x + yp y + zp z = i 17 We now define a new linea opeato ɛ: The squae of ɛ has the convenient popety ɛ = ρ 1 σ x 18 ɛ = ρ 1 σ x = σ x = x x + i σ x x = x = = ɛ = 1 4

5 whee I have used ρ 1 = I 4, x x = and x x = 0. Note that ρ 1 σ p commutes with j. The angula momentum opeatos in j don t disciminate between p and x, so ρ 1 σ x must also commute with j. Theefoe, ɛ also commutes with p. This fact can be exploited: [ σ x, x p] = σ x x p x p σ x = σ [ x x p x p x] = i σ x whee I have used the fact that the inteio of the backet in the second line is itself a known commutato. We can wite this esult in tems of ɛ: Now looking at ɛp p ɛ = i ɛ 19 σ x σ p = ɛρ 1 σ p = p + i ρ 3 j 1 whee I have used a pevious esult. Dividing though by ɛ, we obtain: ρ 1 σ p = ɛp ɛi + ɛi ρ 3j 0 This expession is in tems only of the adial vaiables, and ɛ and j. It is also the only tem in the Hamiltonian equation 5 containing non-adial vaiables! By making this substitution, we ae left with H = Ze which is the adial fom of the Hamiltonian. + cɛ p i + ciɛρ 3j + ρ 3 mc 1 A quick check shows that ɛ and ρ 3 commute with all othe tems, and that they anti-commute with one anothe. This means that we can tansfom to a epesentation in which ρ 3 is a diagonal matix and ɛ has the simple fom shown below. ρ 3 = ɛ = [ ] [ ] 0 i i 0 3 This tansfomation is done by educing ρ 3 to a matix, and then finding ɛ via the commutation and anti-commutation elations. 5

6 In this fom, the Diac Equation fo stationay states is: [ Ze + mc icp c c j icp + c c j Ze mc ] [ ψa ψ b Hψ ] = Eψ = λ [ ψa ψ b ] 4 Solution fo the Coulomb Potential Unfotunately, we cannot simply take the deteminant of the Hamiltonian matix in equation 4, as the deivative tems bought in by the momentum coodinate ae pesent. We will instead educe the system to coupled diffeential equations, and solve them by substituting an unknown function in the fom of a potentially infinite seies. Rewiting H λi = 0 as a system of coupled equations and substituting in p fom equation 17, we have: λ Ze λ Ze + mc ψ a + c mc ψ b c This can be simplified in fom using thee substitutions: 1 j dd dd 1 + j ψ b = 0 ψ a = 0 α = e c a 1 = a = mc λ c mc + λ c α is identically the fine stuctue constant, as we ae woking in cgs units. This makes ou equations, with appopiate eaangement and factoing 1 Zα a Zα a d ψ a ψ b We now assume solutions exist of the fom d + j + 1 ψ b = 0 8 d d j 1 ψ a = 0 9 ψ a = 1 e a f 30 ψ b = 1 e a g 31 6

7 With f and g epesenting unknown functions. The vaiable a is defined as a = a 1 a = m c λ c 1 3 By substituting expessions 30 and 31 into equations 8 and 9 and simplifying, we find: 1 Zα a Zα a d f g Next we expand the unknown functions f and g as seies: d 1 a + j g = 0 33 d d 1 a j f = 0 34 f = s g = s f s s 35 g s s 36 These ae then substituted into ou system of equations. In ode fo the equation to go to zeo as equied, each tem in the esulting seies must sepaately go to zeo. The coefficient of the s tems ae: f s Zαf s+1 s j g s+1 + g s a 1 a g s Zαg s+1 s + 1 j f s+1 + f s a a = 0 37 = 0 38 These can be combined by multiplying equation 37 by a and equation 38 by a and then subtacting the fome fom the latte. This gives us an expession diectly elating the f s coefficients with the g s coefficients: [Zαa a s j ] f s + [Zαa + as + j ] g s = 0 39 We can begin to obtain the values of the coefficients by consideing the bounday conditions. The functions f and g must go to zeo at = 0, because the ψ functions would othewise divege thee due to the 1 tem. This means that thee is some smallest s below which the seies does not continue. We call this s 0, and it has the popety: Plugging this into equations 37 and 38, we find: f s0 1 = g s0 1 = 0 Zαf s0 + s 0 + j g s0 = 0 Zαg s0 s 0 j f s0 = 0 7

8 Combining these and equation 39, we can wite the value s 0 in a vey simple fom s 0 = j Z α 40 This places a lowe bound on the seies. Note that this bound becomes imaginay if Zα > j. This will be discussed in moe detail shotly. The uppe bound of the seies is also useful. It can be shown that the seies must teminate if the enegy eigenvalue λ is to be less than mc see, fo example, [1]. The implication of this esult is that if the seies teminates at index s 1 such that then, using equations 37, 38 and 39, we have f s1+1 = g s1+1 = 0 s 1 a = Zα = λ a a 1 c Zα whee we have used equations 6 and 7 to expand a 1 and a. Squaing this expession and expanding a using equation 3, we get m c λ = Z α λ This can be solved fo the enegy eigenvalues λ: s 1 c c λ = ±mc [ ] Zα s 1 41 Note that the negative enegy solution doesn t imply negative enegy. What we ae seeing ae positon enegy levels. Fom hee fowad, I dop the negative oot and look only at the electon solution. The two end points of the seies indices s 0 and s 1 ae sepaated by an intege numbe of steps. If we call this intege n, then we can wite s 1 = n + s 0 = n + j Z α Plugging this in gives a esult fo the enegy eigenvalues in tems of only the two quantum numbes n and j : E n j = mc 1 + Zα [n + ] j Z α 1 4 8

9 This is the final esult quoted fo the enegy eigenvalues of the hydogenic atom by Diac in [1]. It tuns out that late developments in the field such as [, 3] pefe to use an equivalent set of quantum numbes that maps moe closely to the familia ones we have used in ou class. The numbe j is closely elated to the total angula momentum quantum numbe j. j has the ange 1,, 3,... while j has the ange 1/, 3/, 5/,.... It is natual, and in fact coect, to make the identification j = j The pincipal quantum numbe n is a bit moe ticky. The value of n we have chosen is simply any abitay intege whose value is positive since it epesents the length of the non-zeo elements of the seies expansion of f and g. While this is coect, it does not deal well with how that seies obtains the length that it does. A moe caeful calculation, pesented by Geine in [], shows that we should identify n with the moe familia n thusly: n = n j = n j 1 44 This has the same ange of values as Diac s solution, but is moe caeful in handling the degeneacy of each level. Combining these two adjustments with equation 4, we get the Sommefeld Fine-Stuctue Fomula: E nj = mc Zα 1 + [ ] j n j Z α 1 45 The deivation of the fom of the actual wave functions ψ a and ψ b is vey tedious and unenlightening. Geine addesses this in detail in [], and the inteested eade is diected thee. It was ealie noted that equation 40 has no solution if Zα > j = j + 1. This means that the s states stat to be destoyed above Z = 137, and that the p states begin being destoyed above Z = 74. Note that this diffes fom the esult of the Klein-Godon equation, which pedicts s states being destoyed above Z = 68 and p states destoyed above Z = 8 see, fo example, poblem set #1 fom the fist quate of class. Enegy Spectum of the Diac Hydogen Atom We ae inteested in the spectum of photons emitted by a hydogen atom when an electon tansfes fom one enegy level to anothe. This means that the est mass of the electon is not of concen. I define the obseved enegy levels of the hydogen atom to be taking Z = 1: E nj = E nj mc = mc α 1 + [ j n j α ]

10 This can be expanded in α as a Taylo Seies: Using the numeical values E nj mc α { [ 1 n + α n n j ]} 47 this can be witten as mc = eV α = { [ 1 E nj eV n n n j + 1 The fist tem is clealy ecognisable as the enegy spectum of the non-elativistic hydogen atom, ignoing the fine stuctue. The second tem contains all of the detail of the fine stuctue. This expession can be found identically by taking the fist-ode petubation of the Schödinge hydogen atom solution with the petubing Hamiltonian 3 4 ]} p4 H 1 = 8m 3 c which is the fist elativistic coection to the kinetic enegy of the electon. An excellent deivation of this esult is given by Fitzpatick in [4]. We ae now in a position to compute the enegy spectum and the degeneacies. The pincipal quantum numbe n defines the lage-scale enegy levels. Within each lage-scale level, the obital angula momentum quantum numbe l can take on values of 0, 1,..., n and its azimuthal numbe m uns fom l, l + 1,..., l 1, l. Fo each of these values of l, we compute the total angula momentum j by combining l with the electon spin quantum numbe m s, which can be ±1/. I pefomed these computations using equation 46 and the numeical values of mc and α listed above. The esult is tabulated below: n l j E ev Label Degeneacy 1 0 1/ s 1/ 0 1/ s 1/ 1 1/ p 1/ 1 3/ p 3/ / s 1/ 3 1 1/ p 1/ 3 1 3/ p 3/ 4 3 3/ d 3/ 4 3 5/ d 5/ 6 This confims the peviously detemined degeneacies found by using only the fist-ode elativistic coection. Note that the Lamb Shift and the effect of spin-obit coupling ae not pesent in ou esult, and the hypefine stuctue thus does not appea. These effects equie the use of a moe sophisticated Hamiltonian such as one with A 0. 10

11 Refeences [1] P. Diac, The Pinciples Of Quantum Mechanics, 4th ed. Oxfod Univesity Pess, 198. [] W. Geine, Relativistic Quantum Mechanics: Wave Equations, nd ed. Spinge, [3] F. Goss, Relativistic Quantum Mechanics and Field Theoy. Wiley, [4] R. Fitzpatick. 006 The fine stuctue of hydogen. [Online]. Available: 11