Research: Science and Education
Compounding Selection Rules in Quantum Mechanics K. R. Brownstein* and K. S. Capelle Department of Physics and Astronomy, 5709 Bennett Hall, University of Maine, Orono, ME 04469-5709; *
[email protected] Selection rules in quantum mechanics provide a useful algorithm for deducing that certain wide classes of matrix elements are zero (1). By way of illustration, we shall consider the one-dimensional harmonic oscillator matrix elements for the particle coordinate x, kn j|x|n kl. The harmonic oscillator hamiltonian * is
mω 2 * = 1 p2 + x 2 2m
(1)
Here m is the mass of the particle, ω is the classical oscillator frequency [ω = (k/m)1/2, where k is the restoring force constant], and (x,p) are the (coordinate, linear momentum) operators, which obey the fundamental commutation rule (2) (3)
where the energy eigenvalues E n are given by a quantum number n, E n = h⁄ ω (n + 1 ⁄2 ),
n = 0, 1, 2, 3, …
∆n = ± 2
orthonormal: kn j |n kl = δj,k
(4)
knj |x 3 |nk l = knj |x1x1x|nk l = knj |x
Σ |nj lknj | = 1
nj =0
(5) (6)
This can be rearranged to the form of a sum of products
knj |x 3 |nk l = Σ Σ knj |x|ns l kns |x|nt l knt |x|nk l
(7)
(8)
We see that if the selection rule is not satisfied, then the matrix element kn j|x|n k l vanishes. To put it another way, the vanishing of the quantity f (n j,n k ) is a necessary (but not sufficient) condition for the nonvanishing of the matrix element kn j|x|n k l. For our example (matrix element kn j |x|n kl), the selection rule is (see Appendix A) f (nj,nk) = (n j + 1⁄ 2)2 – 2(nj + 1⁄2)(nk + 1⁄2) + (nk + 1⁄2 )2 – 1 = 0 (9) which is usually written in the abbreviated form nj = nk ± 1
(10)
or even more compactly as ∆n = ± 1
(11)
In this article we shall also make use of the selection rule for matrix elements kn j|+|n k l where the hermitian operator + (which can be interpreted as the lagrangian) is given by 992
(15)
ns nt
Now the first term, kn j|x|n s l, vanishes by eq 10, unless (16a)
Similarly, the second term, kn s|x|n t l, vanishes unless (16b)
and the last term, kn t|x|n k l, vanishes unless nk = nt ± 1
We then say that a selection rule is f (n j,n k) = 0
t
nt = ns ± 1
Suppose one can obtain a relation of the form f (n j,n k )kn j|x|n k l = 0
Σn |nslkns| x Σn |nt lknt | x|nk l (14)
ns = nj ± 1
∞
complete:
(13)
Suppose we are interested in the selection rule for a product of operators, assuming that we know the selection rules for the individual factors. As a concrete example, we consider finding the selection rule for kn j|x 3|n kl. Making use of the completeness relation (eq 6),
These energy eigenvectors |nl are and
(12)
This selection rule turns out to be (see Appendix B)
s
The eigenvalue problem for * is *|nl = E n|nl
2
Compounding Selection Rules
2
[x,p] = i h⁄
mω 2 + = 1 p2 – x 2 2m
(16c)
Because these three terms are multiplied together, the condition (selection rule) for a non-zero matrix element kn j|x 3 |n kl is that (for at least one of the products in eq 15) all three of the above selection rules (eqs 16a, 16b, 16c), are obeyed. That is, the overall selection rule is or more compactly
nk = nj ± 1, nj ± 3
(17)
∆n = ±1, ±3
(18)
We see that the above analysis can be graphically interpreted as indicated in Figure 1. One starts by considering state nj in the left energy level diagram and then follows the individual selection rules through the intermediate diagrams ns and nt to the right diagram n k. We shall refer to the indicated arrows as “transitions”, although these are merely graphical aids rather than actual physical energy level transitions. We say that we have compounded the selection rules of the individual operators (in this case three x operators) to obtain the selection rule for their product (in this case x3). Ghost States and Ghost Transitions One must be careful in applying the above compounding rules to a situation in which the spectrum terminates. In our case (the harmonic oscillator) the energy spectrum terminates
Journal of Chemical Education • Vol. 76 No. 7 July 1999 • JChemEd.chem.wisc.edu
Research: Science and Education
these nonexisting n < 0 levels are shown by dashed arrows (“ghost” transitions). Strictly speaking, one should not have any n < 0 levels present in the construction. The question then arises, should we—— Choice I: initially “prune” all n < 0 states from all the spectra
or, is it sufficient to simply Choice II: finally “prune” the nk < 0 final states yielded by the construction.
Figure 1. Energy level diagrams to illustrate the compounding of selection rules for matrix elements of x to obtain the selection rule for matrix elements of x3. Transitions obeying the selection rules ∆n = ±1 for matrix elements of x are shown by the arrows. The initial state is n j and the possible final states are n k = n j ± 1, n j ± 3.
Of course choice I is the correct procedure but the simpler choice II might prove to be sufficient. In our case, choice II would prune the nk = {1, {3 final states and give as the (correct) result n k = 1, 3, agreeing with what we would obtain using choice I. A little consideration of Figure 2 should convince the reader that choice II will always yield the correct result for matrix elements of all powers of x (kn j |x m|n k l; m = 1, 2, 3, …) and for all initial states nj (nj = 0, 1, 2, …). At this point one is tempted to believe that choice II will always be sufficient to replace choice I. The purpose of this article is to show, by a simple counter-example, that this conjecture is false. Consider the (hermitian1) operator γ = x+x
(19)
The selection rule for matrix elements of x is ∆n = ±1 and that for matrix elements of + is ∆n = ±2. By compounding these individual selection rules, we see that the selection rule for matrix elements of γ = x+x is ∆n = 0, ±2, ±4 Figure 2. Same as Figure 1 except that the initial state is n j = 0. Ghost states (with n < 0) are indicated by dashed lines; ghost transitions are indicated by dashed arrows. The physically possible final states are n k = 1, 3.
(20)
Consider the matrix element knj = 0| γ |n k l. Applying the above selection rule yields the possible values of nk = {4, {2, 0, 2, 4 as shown in the graphical construction in Figure 3. Pruning the nk < 0 final states (choice II) leaves the possibilities nk = 0, 2, 4 (choice II)
(21)
However, pruning all (including the intermediate ns and nt) n < 0 states (choice I), yields only the possibilities n k = 2, 4 (choice I)
(22)
We see that the final state nk = 0 is allowed by naive application of choice II but is not allowed by application of choice I. Note 1. It is not necessary to restrict ourselves to hermitian operators. However, hermitian operators do correspond to observables.
Literature Cited Figure 3. Same as Figure 2 except that the selection rule for matrix elements of γ = x + x (rather than x3) is obtained. Ghost states (with n < 0) are indicated by dashed lines; ghost transitions are indicated by dashed arrows. The initial state is n j = 0. The physically possible final states are n k = 2, 4 (and not n k = 0).
1. Dirac, P. A. M. The Principles of Quantum Mechanics, 4th ed.; Oxford University Press: London, 1957; pp 159–165.
Appendix A Selection Rule for kn j|x|n kl Using the known commutator [x,p] = i h⁄
from below, n = 0 being the lowest possible state. Consider the selection rule for the matrix elements kn j|x 3 |n k l with nj = 0, that is, k0|x 3|n k l. The (naive) construction for this, analogous to that in Figure 1, is shown in Figure 2. The dashed energy levels corresponding to negative values of n (“ghost” states) do not, of course, exist. Transitions associated with
(A1)
we find that the commutator of * with x is ⁄ [*,x] = { (i h/m )p
(A2)
Because this involves more than just * and x, we try taking the commutator of * with eq A2, (A3) [*,[*,x]] = (h⁄ ω)2 x
JChemEd.chem.wisc.edu • Vol. 76 No. 7 July 1999 • Journal of Chemical Education
993
Research: Science and Education Writing out this double commutator (A3), * 2 x – 2*x* + x* 2 – (h⁄ ω) 2x = 0
(A4)
This involves only * and x and furthermore is linear in x. We now take the matrix element kn j |(A4)|n k l of this equation. Using the eigenvalue problem *|nl = En|nl, this yields (Ej2
– 2EjEk +
Ek2
–
(h⁄ ω)2 ) kn
j|(x)|n k l
=0
(A5)
Thus the selection rule is, in terms of the quantum number n, f (n j ,nk) = (nj + 1⁄2)2 – 2(n j + 1⁄2)(n k + 1⁄2) + (nk + 1⁄2)2 – 1 = 0 (A6)
(B4) (B5)
Since this is not of the correct form to furnish a selection rule for matrix elements of +, we take the commutator of * with eq B5. After a little algebra we have [*,[*,+]] – 4h⁄ 2 ω 2+ = 0 (B6) This is of the correct form to furnish a selection rule for matrix elements of +. We form the matrix element kn j |(B6)|nk l and find that (E j2 – 2E jE k + E k2 – 4h⁄ 2 ω 2 )kn j |+|nk l = 0 (B7) Thus the selection rule, in terms of the quantum number n, is
Appendix B Selection Rule for kn j|+|nkl Starting with * = (1/2m) p 2 + (m ω2/2) x2
(B1)
⁄ ⁄ ω2x p and [*,p] = i hm [*,x] = { (i h/m)
(B2)
we find that
We then have ⁄ ⁄ ω 2 (xp + px) (B3) [*,x 2 ] = { (i h/m)(xp + px) and [*,p 2 ] = i hm
994
The commutator of * with the quantity + = (1/2m) p 2 – (m ω2/2) x 2 is then [*,+] = i h⁄ ω 2 (xp + px)
f (nj ,nk) = (nj + 1⁄2)2 – 2(nj + 1⁄2)(nk + 1⁄2) + (nk + 1⁄2)2 – 4 = 0 (B8) Note that the selection rule for matrix elements of x2 (i.e., k n j |x 2|n k l), is ∆n = 0,±2, as is also the rule for matrix elements of p 2 (i.e., kn j | p 2 |n k l). Therefore the selection rule for a general linear combination of x2 and p2 would also be ∆n = 0,±2. However, the particular linear combination of x 2 and p 2 which constitutes the hamiltonian * (eq 1) has a selection rule of simply ∆n = 0, while another linear combination of x2 and p 2, namely the lagrangian + (eq 12), has the “complementary” selection rule ∆n = ±2.
Journal of Chemical Education • Vol. 76 No. 7 July 1999 • JChemEd.chem.wisc.edu
