Atomic Units, Revisited

Transcription

1 Atomic Units, evisited C. W. David (Dated: September, 004)

2 I. INTODUCTION We assume that the reader has gone through an initial discussion of the quantum mechanics of atoms and molecules, and seeks an alternative treatment which looks at the same material backwards, i.e., motivated by the question of why we have done what we ve done? The raison d être for changing units to atomic units is usually treated laconically as a side issue. Therefore, we here explore atomic units in a manner different than textbook manner. II. EIGENFUNCTIONALITY The world of atomic physics is so small that the unit systems we use in the macroscopic world you and I inhabit (by default) are inappropriate. To create a system of units which makes sense, one starts with the H-atom s associated Schrödinger Equation: h m ψ Ae ψ = Eψ where we have used m for the mass of an electron (normally m e ), A is the atomic number of the nucleus (A= means H atom, A= means He +, A=3 means Li +, where Ze is the charge on the nucleus and e is the charge on the electron). This is called the infinite nuclear mass approximation, and if one carried out the more exact two body problem of the nucleus and extra nuclear electron moving about the common center of gravity, this would have become µ i.e., m P + m e = µ and m e µ, the reduced mass of the entire H-atom. We make only one notational change, defining the Cartesian and spherical polar forms of the Laplacian: = + sin ϑ X + = sin ϑ ( sin ϑ ϑ ϑ Y + ) Z φ + i.e., we use capital letters to indicate coördinates in centimeters (meters might be preferred, but since both are absurd from an atomic/molecular point of view, let s not go crazy), where

3 = X + Y + Z. One has, for the ground (s) state, ψ s ([cm], ϑ[rad], φ[rad]) = e α[cm] where α is a (to be determined) constant and whose units must be cm. (This one time the units have been indicated in square brackets.) Substituting this s solution into the Schrödinger Equation one has which is h m e α Ae e α = Ee α h e α m Ae e α = Ee α where, of course, we only need the radial part of the Laplacian. Carrying out the first (inner) partial derivative, we obtain ( h ( αe α ) ) m Ae e α = Ee α and, after carrying out the second (outer) partial derivative, we have which simplifies to which becomes e-writing this in a suggestive form: h m (α α)e α Ae e α = Ee α h m (α α) Ae = E h α h m α + m Ae = E h α h m α + m Ae E = 0 = g() () we see that if the left hand side, called temporarily g(), is to equal zero for arbitrary values of α, then we have an equation here for a special radius,, at which g( ) = 0. But we are charged with finding solutions to the Schrödinger Equation for all, not just some special one. Therefore we are forced to choose a value of α which makes it possible for the equation g() to equal zero for all values of. 3

4 If the coëfficient of / were to vanish then E (a constant) would be identified, i.e., if then Equation would become ( α h m Ae ) = 0 () h m α + (0) E = 0 = g() (3) This is certainly possible (i.e., not a function of anymore)! To achieve this, we need to choose α: and, squaring, α = mae h α = m A e 4 h 4 so that, left multiplying both sides by h, we then obtain m h m α = h m A e 4 m h 4 = E arriving finally at which is a well known result! E = A e 4 m h (4) III. WHAT WOULD CHANGE IF WE CHANGE THE SCALE OF X, Y AND Z Knowing α s value, we return to the Schrödinger Equation: h m X,Y,Zψ(X, Y, Z) Ae ψ(x, Y, Z) = Eψ(X, Y, Z) (again, we use our non-standard notation to remind the reader that the Laplacian is still in terms of the original (cm) coördinate scheme) where we know that h is in erg-seconds, m (the mass of the electron) is in grams, e, the charge on the electron (4.8x0 0 ) is in statcoulomb, and E, the energy, is in ergs. This is the cgs system. Let us define a dimensionless radius, r, so that r = β ; = r/β which means X = x/β ; Y = y/β ; Z = z/β 4

5 where β is a constant (to be determined) whose units are cm, and x, y, and z are to be forced dimensionless. Then = r using the chain rule. Then, we have ( ) e α ( ) h β r β β e αr/β r m r r β or so r = β r = (r/β) β (r/β) βe αr/β r r Ae r/β e αr/β = Ee αr/β ( h βα r e αr/β) Ae m r r r/β e αr/β = Ee αr/β h βα (re αr/β r αβ ) m r e αr/β Ae r/β e αr/β = Ee αr/β + h βα mr h α m Ae r/β = E (5) and with A=, ( h ) βα m e β r h α ( h ) m Eβ = 0 = β α m e r h α m (which shows that the argument concerning α is independent of the value of β, since the coefficient of r is the same as in Equation ) i.e., β factors out.. We chose to set A= momentarily, so that later we can handle the entire isoelectronic series of one electron atoms and ions in one fell swoop. The fact that β ultimately factors out of the first term (above) h α m e 0 indicates that the argument about choosing β will not influence the form of the Schrödinger Equation, i.e., we are free to choose it as we wish, and we choose to make it so that the exponential has no constants in it, i.e., we choose α β = as an appropriate form. This makes the exponential (since = r/β) e α e αr/β e r = e x +y +z (said another way (substituting for α), e me h e me h r/β ) where, or course, r = x + y + z, and x, y, z and r all are now dimensionless.ψ(x, y, z) is a clean function if ever there was one. For the isoelectronic one electron series the s orbital would now be e Ar. 5

6 i.e., With A = we have β = gram(dyne/ cm) (erg sec) = β = α = me h (6) gram dyne cm erg sec = gram cm erg sec (which has the expected units reciprocal centimeters in cgs units). Further β = ( ( 3.459) ) ( ) cm This happens to be approximately 0.5Å, the Bohr radius. = cm IV. FOMING THE DIMENSIONLESS SCHÖDINGE EQUATION Again, returning to the Schrödinger Equation, we had We re-write it as since h m Xψ(X, Y, Z) Ae ψ(x, Y, Z) = Eψ(X, Y, Z) { } h m X + Y + ψ(x, Y, Z) Ae ψ(x, Y, Z) = Eψ(X, Y, Z) Z we can write it in components using = βr ; r = /β X = x/β ; Y = y/β ; Z = z/β where x, y, &z are the new coordinates. the squared sum of which lead to r. From the chain rule, we obtain X = x X x = β x = β X x so (with similar terms for y and z) we transform the Laplacian from the set X,Y,Z to x,y,z. β h Ae β m x,y,zψ(x, y, z) ψ(x, y, z) = Eψ(x, y, z) x + y + z 6

7 which becomes (using Equation 6) so, with A=, choosing, one recovers Equation 6: x,y,zψ mae h βr ψ = me β h ψ me h β = β = me h where we pull out the A so that later we can treat more than just H Continuing, we have which makes rψ r ψ = E = ɛ me 4 h me E h h me ψ = ψ = ɛψ (7) me4 h ; gram dyne cm 4 erg sec = erg (8) (ɛ is unit-less, hereafter to be called a.u. for atomic unit) which resembles the ydberg we ve come to know and love (Equation 4): y = me4 h With A= (Hydrogen) and n= (the ground state), one then has E H = y = me4 h = ( ) which, since it comes out in ergs, can be converted to the easier to remember ev value using ev Thus in this case, rearranging Equation 8 we have erg and therefore and y h e 4 m = ( me 4 h E h e 4 m = ɛ ) ) ( h = e 4 m When E is (about) -3.6 ev for Z= we have ɛ = au so au = 3.6eV au = 3.6eV 7 = ɛ(in au)

8 V. VEIFYING EIGENFUNCTIONALITY Once one has a dimensionless form for the Schrödinger Equation, 7, it becomes simple to verify whether or not a function is an eigenfunction of the Hamiltonian, i.e., solves the Schrödinger Equation. Consider, again, the s case, ψ s = e αr We ask the question, what value of α is required to solve the Schrödinger Equation? We had which is, for s-states rψ r ψ = ɛψ r r ψ r r r ψ = ɛψ Here is a MAPLE example for showing the above: psi := exp(-alpha*r); t := - (()/())*(/r^)*diff(r^*diff(psi,r),r)-(/r)*psi; t := expand(t/psi);#this lines shows why the alpha= choice is needed E_s := subs(alpha=,t); The resultant answer (-/) validates the assigning of the value of the a.u. previously. It is significantly more difficult to deal with a s orbital. In MAPLE, we have: psi_ = (+gamma*r)*exp(-(alpha/)*r);# we ve made some assumtions here t := - (()/())*(/r^)*diff(r^*diff(psi,r),r)-(/r)*psi; t := expand(t/psi); ta := coeff(t,(+gammma*r)^(-)); tb := coeff(ta,/r); t := normal(subs(alpha=,t)); E_s := normal(subs(gamma=-/,t)); carried out The answer here can be rewritten which is the correct value. = 8 8

9 VI. ESTOING THE A VALUE We had in Equation 5 + h βα mr h α m Ae r β where we set A equal to in order to proceed. Now, we allow A to continue having a value = E greater than or equal to, so ( h ) βα + m βae r h α m = E and now we set the coëfficient of r values of r, i.e., equal to zero so that this equation can be true for all +β ( h ) α m Ae = 0 and since β isn t zero, the part in parenthesis must be, so + h α m = Ae which means that the A dependent value of α should be α = Ae m h and since β = me h e α e αr/β e Ar = e A x +y +z 9