Enegy Levels Of Hydogen Atom Using Ladde Opeatos Ava Khamseh Supeviso: D. Bian Pendleton The Univesity of Edinbugh August 11 1
Abstact The aim of this pape is to fist use the Schödinge wavefunction methods and then the ladde opeato methods in ode to study the enegy levels of both the quantum Simple Hamonic Oscillato SHO and the Hydogen atom. Intoduction Stating fom the Simple Hamonic Oscillato[1 we fist wok though the solution of the bound-state wavefunctions fo the hamonic oscillato in ode to obtain the enegy eigenvalues, E =n + 1 ω, enegy eigenfunctions and the Hemite polynomials by solving diffeential equations. Then, we will use algebaic ladde opeato methods in ode to find the enegy eigenvalues which should esult in the same answe. In the second section the aim is to find the the enegy levels of the Hydogen atom[1 using the two methods[ exploed in the pevious section. Howeve, this diffeence between the SHO and the Hydogen atom is that in the case of the latte we need to conside thee dimensional space. Finally the esults of both methods will be compaed in the end. 1 The Hamonic Oscillato 1.1 Diffeential Equations Method Fo the SHO the Hamiltonian is of the fom Ĥ = ˆp m + 1 kx 1.1 So the time independent Schödinge equation 1.1 becomes d ψx m dx + 1 kx ψx =Eψx 1. Now, intoducing the fequency of the oscillato k ω = 1.3 m witing ε = E ω 1.4 and substituting fo k and E into equation 1. we get m d ψx dx + 1 ω mx ψx = ε ω ψ x d ψx dx + ωm [ ε ωm x ψ x = 1.5 3
Changing vaiables to equation 1.5 can then be witten as y = mω 1.6 d ψ dy + ε y ψ = 1.7 As y,εbecomes negligible in compaison with y and we can ewite equation 1.7 as d ψ y dy y ψ y = 1.8 Now if we multiply the equation by dψ dy and use the chain ule we get [ d dψ y d dψ ψ d x dy dy dy = y ψ = yψ 1.9 dy dy In ode to make this equation simple, we assume that the tem on the ight side of the equation can be neglected. Howeve, we will check late that this assumption is coect. So we have d dy [ dψ x dy y ψ = 1.1 Integating both side of the above equation with espect to y and eaanging, Both ψ and dψ dy dψ x dy =C + y ψ 1/ 1.11 must vanish at infinity, so C =so dψ x dy and the acceptable solution, which vanishes at infinity is ψ y =e y / = ±yψ 1.1 If we now obseve the tem d dy y ψ which appeas in equation 1.9, 1.13 d dy y ψ= d dy y e y y 3 e y +ye y 1.14 it can be said that yψ =ye y is indeed negligible compaed with y 3 e y fo lage y so the assumption made above is coect. Hence, we can make an ansatz that the geneal solution fo ψy takes the fom ψy =hye y / 1.15 NB Note that this is a geneal solution fo ψy not ψ y which is only defined fo lage y. 4
It is impotant to mention that the solution is exact fo the gound state whee h y =1see equation 1.17, fo m =, h y =a =1 We inset this back into equation 1.7 and devide both sides by e y / to get d hy dy y dhy dy +ε 1hy = 1.16 Now that we have consideed the behavio at infinity, we should obseve the behavio nea y =by stating fom the powe seies expansion hy = a m y m 1.17 m= which when substituted into equation 1.16, eaanged and devided by y m esults in the ecusion elation m + 1m + a m+ =m ε + 1a m 1.18 Fo lage m, ie m>n whee N is a lage numbe, we have m a m ma m a m+ m a m 1.19 which implies that the geneal solution to hy is hy = solution fo m<n + solution fo m N so hy = a polynomial in y +a N [y N + N yn+ + NN + yn+4 + 3 NN + N + 4 yn+6 +... 1. whee the fist tem just comes fom the powe seies expansion and last tems is deived as follows 1. m = N a N y N. Accoding to equation 1.19, a N+ = N a N N a N y N+ 3. Applying 1.19 again, a N+4 = N+ a N+ = N+ N a N = NN+ yn+4 NN+ a N and so on. Note that, fo simplicity, we have only consideed the even solution. The second tem in equation 1. can, fo easons that will soon become clea, be witten in the following fom a N y [y N + 1 y N 1 + N N +... +1yN+ N = a N y 1 [! y N 1 N 1 + 1 N! N y N 1 + N N N +1 +... +1y 5
N = a N y! y N 1 1 N 1! + y N N + y N +1! N +1 1.1! We intoduce k such that N =k so the equation above takes the fom [ y y k 1 y k y k+1 k 1! k 1! + + k! k + 1! +... [ [ y = y k 1! e y 1+y y k + +...! k! 1. Whee we have made use of the Taylo seies fo the exponential function e x = 1+x + x! + x3 3! +... In othe wods, equation 1. can be intepeted as hy = a polynomail + a constant y e y. When this is substituted back into equation 1.15, ψy =a polynomial e y / +a constant y e y / 1.3 which does not vanish at infinity so it is not an acceptable solution. Theefoe, the ecusion elation 1.18 must teminate fo some n, ie, n ε +1= ε =n +1 1.4 If we know go back to the definition of ε on page 3, E = ε ω, E = ωn + 1 ; n =, 1,,... 1.5 which is the enegy eigenvalue fo the Simple Hamonic Oscillato. The polynomials hy apat fom the nomalisation constants ae known as the Hemite polynomials H n y and some of thei impotant popeties ae stated below, d H n y dy and the ecusion elations y dh ny dy +nh n y = 1.6 H n+1 yh n +nh n 1 = 1.7 H n+1 + dh n dy yh n = 1.8 Both of the above elations can be poved by induction and it is left as an execise fo the eade. The fist few polynomials ae listed below: H y =1 H 1 y =y 6
H y =4y 1. Opeato Methods H 3 y =8y 3 1y The Hamiltonian fo the SHO is of the fom Ĥ = ˆp m + 1 mωˆx 1.9 whee the opeatos fo momentum and position ae ˆp = i x and ˆx = x espectively. Note that both of these opeatos ae Hemitian, ie, they have eal eigenvalues. It is useful to pove that these opeatos ae hemitian: Fo any linea opeato Ô, the hemitian conjugate Ô is defined by In othe wods, = dxψ Ô x Ôψ x dx ψ x ψ x 1.3 ψ O ψ = Ô ψ ψ 1.31 Hence, fo the momentum opeato ˆp we have dxψ x ˆpψ x = dxψ i d ψ = i dxψ dψ dx dx Using integation by pats = i {[ψ ψ dx dψ dx } ψ = i dx dψ andassuming that ψ and ψ vanish at ±, = dx i dψ ψ = dx ˆpψ x ψ x 1.3 dx Theefoe, ˆp =ˆp = i d dx as equied. The same agument holds fo x. The commutation elation between the two opeatos is dx ψ [ˆp, ˆx =ˆpˆx ˆxˆp = i 1.33 One might think that the Hamiltonian 1.9 can be witten in the fom mω Ĥ = ω ˆx i ˆp mω mω ˆx + i ˆp 1.34 mω Howeve, since ˆp and ˆx do not commute, equation 1.34 is not coect. 7
In fact, mω ω ˆx i ˆp mω mω ˆx + i ˆp = mω mω ˆx + iω ˆxˆp iω = mω ˆx + ˆp m iω ˆpˆx ˆxˆp =mω ˆx + ˆp m iω i ˆpˆx + ˆp m = Ĥ 1 ω 1.35 Now, we intoduce the opeatos  and  such that mω  = ˆx + i ˆp 1.36 mω mω  = ˆx + i ˆp 1.37 mω 1 whee the facto of is intoduced in ode to make the opeatos dimensionless. Note that since ˆx and ˆp ae both Hemitian,  is the Hemitian conjugate of Â. Since [ ˆx commutes with itself, and so does ˆp, in ode to obtain the elation A, A we only need to conside [ [Â,  [ mω ˆp ˆp mω = ˆx, i + i, mω mω ˆx = i i [ˆx, ˆp+ [ˆp, ˆx =1 1.38 Equation 1.35 implies that Ĥ = ω   + 1 1.39 We also have [Ĥ,  [ = ω Â,  [ = ω  Â  = ωâ 1.4 [Ĥ,  [ [ = ω Â,  = ωâ    = ωâ 1.41 The eigenvalue equation fo the Hamiltonian is Now, let us conside the state  E. Using the commutation elation 1.4 Ĥ E = E E 1.4 Ĥ E = ÂĤ E ωâ E =E ωâ E 1.43 Theefoe, it can be said that  E is also an eigenstate of Ĥ but with the enegy loweed by ω. Each time we apply  again, the enegy will be loweed by ω but thee is a point whee this pocess must teminate since the expectation 8
value of Ĥ is positive. The state of lowest enegy is called the gound state and is denoted by. So fo the gound state  = 1.44 whee the enegy cannot be loweed any moe. Using equation 1.39 the enegy of the gound state is Ĥ = ω   + 1 = 1 ω 1.45 Let us now conside the enegy fo the state Â,  1 Ĥ = Ĥ + ωâ = ω +1  1.46 This time the enegy is ω moe than that of the gound state. When  is applied again, the enegy will incease by a futhe ω, ie two units of enegy above the gound state, and so on. In this context,  and  ae known as the ladde opeatos. The figue below [1 indicates the aising and loweing opeatos whee ε = ω. Hence it can be concluded that E = n + 1 ω 1.47 which is exactly the same as the enegy eigenvalue equation 1.5 deived in the pevious section using the diffeential equations method. So we have deived the enegy levels fo the SHO using two diffeent methods. The Hydogen atom.1 Diffeential equations method We stat fom the 3-D Schödinge equation ˆp Ĥψ = µ + V ψ =Eψ.1 9
whee ˆp =ˆp x +ˆp y +ˆp z and ˆp =ˆp x, ˆp y, ˆp z = i x, i y, i z and µ is the educed mass 1 and the Schödinge equation is witten as a patial diffeential equation µ ψ x, y, z+v x, y, z, ψ x, y, z =Eψ x, y, z. With = x + y + z In spheical pola coodinates[3 this becomes = + + 1 which can be witten in the fom 3 θ + cot θ θ + 1 sin θ ϕ.3 = + L.4 whee L is the obital angula momentum. We will only conside the case of a cental potential so that the potential V = V only depends on the distance fom the oigin. So the Schödinge equation becomes µ + ψ + L ψ +V ψ =Eψ.5 µ The wavefunction ψ can be split into adial and angula components ψ =R nl Y lm θ, ϕ.6 which is then substituted back into equation.5 to get µ + l l + 1 R nl +V R nl =E nl R nl.7 Note: we have used the fact that Y lm θ, ϕ ae eigenfunctions of L with eigenvalue l l + 1. 4 Fo the case of Hydogen atom we have the attactive Coulomb potential in SI units V = Ze 4πε so the Schödinge equation.7 becomes.8 1 Fo a two-body system we need to conside the educed mass µ = mem m e +M whee m e is the mass of the electon and M is the mass of the poton. These ae all fully deived in Mathematics fo Physics 4: Fields lectue notes. Altenatively efe to Suppliment 7-B of the Quantum Physics Ref [1 on www.wiley.com/college/gasioowicz 3 Refee to equation 7B-9 of the Suppliment. 4 Refe to Chapte 7 of Quantum Physics by S.Gasioowicz. 1
[ d d + d d + µ E + Ze 4πε l l + 1 µ R =.9 We only conside the solutions with E<, ie the bound states. To make the above equation simple, let us define the dimensionless paametes ρ and λ such that 8µ E ρ =.1 and with α = e 4πε c d R ρ dρ λ = Ze µ 4πε E = Zα µc E =1/137. Now, equation.9 takes the fom + dr ρ l l + 1 λ ρ dρ ρ R ρ+ ρ 1 4.11 R ρ =.1 In ode to solve this we will fist conside the behavio fo lage ρ, ie d R dρ 1 R =.13 4 the acceptable solution that behaves popely at infinity, ie goes to zeo, is R e ρ/.14 theefoe, fo all ρ, we make an ansatz fo the geneal solution to be of the fom R ρ =e ρ/ G ρ.15 fo some function G ρ. Substituting back into equation.1 we get d G dρ 1 [ dg λ 1 ρ dρ + l l + 1 ρ ρ G =.16 In ode to be able so solve this, fist conside vey small ρ, so the equation becomes d G dρ + dg l l + 1 ρ dρ ρ G =.17 By obseving the equation one can conclude that G ρ ρ l. So the geneal solution will be of the fom thefoe G ρ =ρ l H ρ.18 d H l + dh dρ + 1 ρ dρ + λ l 1 H =.19 ρ which is implied by substituting equation.18 into.16. In ode to solve this diffeential equation, we use the same method used in pat 1 of section1, ie the powe seies expansion 11
H ρ = a k ρ k. which can be substituted into.19 and eaanged to yield l + a k [k k 1 ρ k + k 1 ρ k 1 +λ l 1 ρ k 1 = ρ k= k= ρ k 1 [k + 1 k +l + a k+1 +λ l 1 a k =.1 k= So the coefficient of ρ k 1 must be zeo which implies the ecusion elation below a k+1 k + l +1 λ = a k k +l + k + 1. Again, the geneal solution of H ρ= solution fo k < N + solution fo k>n whee N is a lage numbe. Fo small k, ie k<n, we have a polynomial in ρ. Howeve fo lage k>n the above elation can be witten as a k+1 k a k k = 1.3 k It can easily be shown that fo vey lage k a k+1 = 1 k!.4 Hence H ρ = a polynomial in ρ +e ρ which is not acceptable since R ρ will have the tem ρ l e ρ/ e ρ which means that R ρ will gow as e ρ/ which does not vanish at infinity. So the ecusion elation. must teminate, ie k + l +1 λ = It can be said that fo a given l thee exists an intege k, denoted by n, such that In fact, the pincipal quantum numbe n is defined as Fom λ = n and equation.11, λ = n + l +1.5 n = n + l +1.6 E = 1 Zα µc n.7 Also, since n, one can conclude that n is an intege and n l +1. The enegy can also be witten in the following fom using.11 and substituing fo α and λ, Z E = µe4 4πε n.8 Fo the case of Hydogen, Z =1. 1
. Opeato Method Finally, we use opeato methods to deive the Boh levels of the hydogen atom. This is an extension and genealisation of the techniques used in pat of section 1. The Hamiltonian can be witten as whee Ĥ = p µ + L µ e p = i + 1.9.3 and the potential is in cgs units. It can be checked that this is the same as the Hamiltonian used in section.1 compae with equations.4 and.5, in fact [ ψ p [ψ x = p [p ψ = i p + ψ = i + 1 [ ψ + ψ = ψ + ψ + 1 ψ + ψ = ψ + ψ.31 and the potential V = e. The only diffeence is that the potential in the peviouse section was taken to be V = Ze 4πε which is diffeent by a facto of 1 4πε fo hydogen Z=1. This is meely due to use of diffeent units. As it can be seen, the Hamiltonian is only witten in a diffeent way. The opeatos, Ĥ, L and L z the component of angula momentum in the z-dieciton satisfy the eigenvalue equations H E,l,m = E E,l,m.3 L E,l,m = l l + 1 E,l,m.33 L z E,l,m = m E,l,m.34 whee l takes non-negative intege values and m = l, l +1,...,, 1,...,l. The states E,l,m ae in the domain of definition of opeatos Ĥ, L and L z. It can be obseved that these opeatos commute. Equations.3 and.33 imply that whee H E,l,m = H l E,l,m = E E,l,m.35 H l = p µ + l l + 1 µ e Now, we define the opeato D l such that.36 13
l + 1 D l = p i µe l + 1.37 Then we define the adjoint of D l, D l, extending equations 1.3 and 1.31 to 3-D space. In othe wods, we integate ove d 3 x by consideing abitay functions f x, g x such that f x and g x vanish fo both and +. So the aim is to find D l such that g Dl f = D lg f.38 The ight hand side is ˆ g D l fdx 3 Fo simplicity, we assume that f and g ae functions of only, ie, they do not depend on θ and φ. It is left to the eade to pove the esults fo the geneal case, f x and g x. Using pola coodinates =4π = ˆ π ˆ π l + 1 g [p i this can be boken into two steps: 1. We find p, g p f =4π g D l f sin θ d dθ dφ µe f d l + 1 g [ i + 1 [ f d =4π i g f d + then we integate by pats to get { [g = i 4π f g f d g f + the fist tem goes is zeo by applying bounday conditions, so [ g ˆ π ˆ π =i 4π f d + g f d = ˆ π ˆ π [ = i + 1 Hence, p is hemitian. Using the same method it can be shown that 1 } g f d i g + i g g f sin θ d dθ dφ = p g f g p f = p g f = p = p.39 is also hemitian. g f d f sin θ d dθ dφ Theefoe, 1 and imply that l + 1 D = p + i µe l + 1 Befoe we poceed, it would be useful to pove the following elations:.4 14
p = + aleady shown above, see equation.31 [p,= i Poof: Take some function f, then [p,f = i { + [ 1 [f + } 1 f = i f + f + f f f = i f [ p, 1 = i Poof: [ { p, 1 f = i + [ [ } 1 f 1 f + f = i i f [ 1 p, = i 3 Poof: [ p, 1 i 3 { f = i + 1 f [ p, 1 = ip Poof: [ p, 1 f = { + [ 1 [ [ } f 1 f + f = i [ [ f 1 f + 1 f f + f + f 3 f 1 f 3 f f } f f + f 1 f f = [ = Using what we have, the following fou elations can be established Poof: D l H l = [ D l,h l +H l D l = l l + 1 µ [ i = l l + 1 µ = H l D l + l + 1 µ 3 e [ i 1 f f + f 3 1 f 3 + f = 3 f = ip f H l+1 D l = D l H l.41 [ p, i l + 1 + µ i l iµe l + 1 + i + p Poof: Using the above esults, we have [ 1 e p, 1 i l + 1 + µ = ip + H l D l [ H l + l + 1 µ [ 1,p +H l D l D l = H l+1d l H l D l = D l H l+1.4 15
D l H l = H l+1 D l = H l D l = D l H l+1 but the Hamiltonian is hemitian, ie H = H, so H l D l = D l H l+1 D l D l H =µ µe 4 l + l + 1.43 Poof: Stating fom the definition of D l and D l and applying them on an abitay function f we have D l D l [f { = i + 1 } [ l + 1 i µe f i l + 1 + f l + 1 + i =µ f + = f + l + 1 µe l + 1 = D l D l = p l + 1 + l + 1 f µe f l + 1 µe f l + 1 f + f l + 1 + µe [f l + 1 l + 1 µe l + 1 µe 4 =µ H l + l + 1 p µ + l + 1 µ l + 1 µ e + µe 4 l + 1 D l D µe 4 l =µ H l+1 + l + 1.44 Poof: Using.43 D l D l D l=µ µe 4 µe 4 H l + l + 1 D l =µ H l D l + l + 1 D l But.4 implies that D l D l D µe 4 l =µ D l H l+1 + l + 1 D l D l D l D µe l = D l [µ 4 H l+1 + l + 1 D l D µe 4 l =µ H l+1 + l + 1 f µe l + 1 f 16
Theefoe, we have E,l,m D l D l E E,l,m =µ µe 4 µe 4 + l + 1 E,l,m E,l,m =µ E + l + 1 which is always positive since.45 E,l,m D l D l E,l,m = D l E,l,m.46 and the length of the vecto is always positive. In othe wods µe 4 E l + 1.47 The minimum possible value fo the enegy is whenl =, ie E = µe4.48 fo l lage than this, the positivity is violated. Howeve, we will show that this is also an eigenvalue of the Hamiltonian Ĥ. Using equations.35 and.41, we have in geneal, H l+1 D l E,l,m = D l H l E,l,m =E D l E,l,m.49 which implies that D l E,l,m is an eigenstate of H l+1 with the same enegy E, so we can wite D l E,l,m = c l E,l +1,m.5 [ 1/ c µe 4 = i µ E + l + 1.51 whee the facto i is chosen fo convenience. It can be said that D l+1 D l E,l,m is an eigenstate of H l+ and so on. Howeve, due to the positivity of D l D l, the sequence must teminate fo some l = l max fo E E<. Hence, fo E + µe4 to be geate than zeo, thee exists l+1 an l max fo an eigenvalue E<such that D l max E,l max,m is the zeo vecto. As obseved above, fo the gound state enegy, l max =. We can wite l max = n 1 whee n =1,, 3,... since l max it is a non negative intege. Hee n =1coesponds to the gound state level. Finally, since c l max = c n 1 must vanish, we conclude that E n = µe4 n.5 NB This implies that the enegy is independent of l. 17
This is ou final answe fo the enegy eigenvalue of Hydogen atom using opeato methods. Refeing to equations.43 and.46, the gound state enegy level we must satisfy D l E,,m =.53 Theefoe, the gound state enegy is E = µe4.54 which means that the minimum possible value of enegy equation.48 is in fact an eigenvalue of the Hamiltonian Ĥ. Compaing the esults with those obtained in Section.1, equation.8, it is 1 clea that they ae the same apat fom the facto 4πε which, as mentioned befoe, is due to use of diffeent units fo the potential. 18
Refeences [1 Gasioowicz, Stephen. Quantum Physics. New Jesey: Wiley & Sons, 3 [ Manoukian, Edwad. B. Quantum Theoy: A Wide Spectum. Dodecht, The Nethelands: Spinge, 6. [3 Pendleton, Bian. Mathematics fo Physics 4: Fields lectue notes, Edinbugh Univesity [4 www.wiley.com/college/gasioowicz [5 Giffiths, David. j. Intoduction to Quantum Mechanics. Uppe Saddle Rive, N.J. : Peason/Pentice Hall, 5. 19