Generalized raising and lowering operators for supersymmetric quantum mechanics

0.12 MB·English

by  Mark W. Coffey

Generalized raising and lowering operators for
supersymmetric quantum mechanics
Mark W. Coffey
Department of Physics
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1
Colorado School of Mines
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2 Golden, CO 80401

n
USA
a
J
mcoff[email protected]

7
2
January 9, 2015

]
h
p
-
h Abstract
t
a
m Supersymmetric quantum mechanics has many applications, and typically
uses a raising and lowering operator formalism. For one dimensional problems,
[
we show how such raising and lowering operators may begeneralized to include
1
an arbitrary function. As a result, the usual Rodrigues’ formula of the theory
v
9 of orthogonal polynomials may be recovered in special cases, and it may other-
4
wise be generalized to incorporate an arbitrary function. We provide example
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6 generalized operators for several important classical orthogonal polynomials,
0 including Chebyshev, Gegenbauer, and other polynomials. In particular, as
.
1 concerns Legendre polynomials and associated Legendre functions, we supple-
0 ment and generalize results of Bazeia and Das.
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1
:
v Key words and phrases
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X
raising operator, lowering operator, supersymmetry, Rodrigues’formula, Legendre
r
a polynomial, Laguerre polynomial, Chebyshev polynomial, Gegenbauer polynomial
2010 PACS codes
02.30.Gp, 03.65.-w, 02.30.Hq
1
Introduction
The concepts of supersymmetry from quantum field theory have led to diverse ap-
plications in many other areas of physics including statistical, condensed matter, and
nuclear physics, and quantum mechanics [7, 10, 12]. The supersymmetric scheme per-
mits to combine fermionic and bosonic degrees of freedom. Supersymmetry presents
a hierarchy of Hamiltonians, and for nonrelativistic quantum mechanics provides fac-
torizations of the Schro¨dinger differential equation. Factorization of the Hamiltonian
is done with operators incorporating the superpotential W, and W satisfies Ricatti
equations with the partner potentials being the inhomogeneous term.
The factorization of Hamiltonians is typically developed via raising and lowering
operators, each being of first order. In this paper we focus on the one dimensional
Schro¨dinger equation and describe how an arbitrary function, denoted h(x), may be
incorporated into these operators. This procedure allows known Rodrigues’ formulae
to be recovered in short order for even the simplest choices of h(x), and otherwise
gives a way to develop generalized Rodrigues’ formulae and other identities. In this
Introduction we provide illustrations for the quantum simple harmonic oscillator, and
the associated Hermite and Laguerre polynomials. We then concentrate on the Leg-
endre differential equation, thereby generalizing many of the results of [4]. We also
supplement [4] by giving proofs of certain identities. Furthermore, we provide a gen-
eralization to Gegenbauer polynomials with parameter λ. Our method is applicable
in a great many contexts, and we briefly mention other examples at the end.
We let Lα(x) denote the nth Laguerre polynomial, H (x) the nth Hermite poly-
n n
nomial, (z) = z(z+1) (z+n 1) = Γ(z+n)/Γ(z) the Pochhammer symbol, with
n
··· −
Γ the Gamma function (e.g., [1, 2, 11, 14]). We recall that Hermite polynomials are
special cases of Laguerre polynomials, such that
H (x) = ( 1)n22nn!L−1/2(x2) and H (x) = ( 1)n22n+1n!xL1/2(x2) (1.1)
2n − n 2n+1 − n
2
(e.g., [11], p. 1037). Both of these types of polynomials are instances of the confluent
hypergeometric function F .
1 1
We work with dimensionless quantities throughout, so that for the quantum har-
monic oscillator the raising and lowering operators may be taken as a± = x d . For
∓ dx
the raising operator, we make the ansatz
d d
a+ = x = f g +h,
1 2
− dx dx
whereinthefunctionsf andg aretobedetermined intermsofthearbitraryfunction
1 2
h. Letting a+ act on a differentiable function f, and then equating coefficients of f
′
and f , we find that, aside from multiplicative constants,
g (x) = exp (x h)dx ,
2
− −
(cid:20) Z (cid:21)
and
1
f (x) = = exp (x h)dx .
1
−g (x) − −
2 (cid:20)Z (cid:21)
−
It goes very similarly for generalizing a , so that with a sign change the roles of f
1
±
and g are reversed. The original operators a do not depend upon the energy level n
2
and accordingly neither do the generalized operators with h in general different from
x. The well known equal energy spacings follow.
As usual, the ground state obtains from aψ = 0 and ψ (x) exp( x2/2). Acting
0 0
∝ −
successively on the ground state wavefunction with the raising operator provides the
Hermite functions. If h = 0, then a+ = ex2/2 d e−x2/2. By multiplying by ex2/2
− dx
and imposing a sign convention, even in the simplest case of taking h = 0, the usual
Rodrigues’ formula for Hermite polynomials is recovered.
For the Laguerre differential equation
′′ ′
xu +(α x+1)u +nu = 0, (1.2)
−
3
raising and lowering operators may be taken as
d d
A+ = x x+α+n, A− = x +n, (1.3)
dx − − dx
so that the differential equation assumes the forms
A+A−Lαn(x) = n(n+α)Lαn(x)
and
A−A+Lαn−1(x) = n(n+α)Lαn−1(x).
Here we have applied recurrence relations ([11], 8.971.3) so that A Lα (x) = nLα(x)
+ n−1 n
andA−Lαn(x) = (n+α)Lαn−1(x). IfweputA+ = f1ddxg2+h, thenwefindthat,omitting
a multiplicative constant,
h(x)
g (x) = xα+ne−xexp dx ,
2
− x
(cid:20) Z (cid:21)
and
x h(x)
f (x) = = x1−α−nexexp dx .
1
g (x) x
2 (cid:20)Z (cid:21)
ThecorrespondencetotheusualRodrigues’formulaLα(x) = 1exx−α d n(e−xxn+α)
n n! dx
is then evident again for h = 0. The Laguerre polynomials appear,(cid:0)for(cid:1)instance, in
the wave function for the hydrogen atom.
Generalized raising and lowering operators for the Legendre equations
We now focus on the Legendre differential equation (x = cosθ, 1 x 1)
− ≤ ≤
d d
(x2 1) n(n+1) P (x) = 0, (2.1)
n
dx − dx −
(cid:20) (cid:21)
and the associated Legendre differential equation
d d m2
(x2 1) + n(n+1) Pm(x) = 0, (2.2)
dx − dx 1 x2 − n
(cid:20) − (cid:21)
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with m < n. The associated Legendre functions are related to the Legendre poly-
| |
nomials via
dmPm(x)
Pm(x) = (1 x2)m/2 n .
n − dxm
Note that for m odd, Pm(x) is algebraic, not polynomial, in x.
n
Taking the raising operator for (2.1) to be
d d
R = (x2 1) +nx = f g +h, (2.3)
n 1 2
− dx dx
we find, multiplicative constants henceforth being omitted,
h(x)
g (x) = (x2 1)n/2exp dx ,
2 − − x2 1
(cid:20) Z − (cid:21)
and
x2 1 h(x)
f (x) = − = (x2 1)1−n/2exp dx .
1
g (x) − x2 1
2 (cid:20)Z − (cid:21)
As a brief example, when h(x) = ax2 +bx+c, so that
ax2 +bx+c 1
dx = ax+ [(a+b+c)ln(1 x)+(b a c)ln(1+x),
x2 1 2 − − −
Z −
the exponential factor becomes
h(x)
exp dx = eax(1 x)(a+b+c)/2(1+x)(b−a−c)/2.
x2 1 −
(cid:20)Z − (cid:21)
Proceeding similarly with the lowering operator for (2.1),
d d
L = (x2 1)+nx = g f +h,
n 2 1
−dx − − dx
we determine that
h(x)
f (x) = (x2 1)1−n/2exp dx ,
1 − x2 1
(cid:20)Z − (cid:21)
and
x2 1 h(x)
g (x) = − = (x2 1)n/2exp dx .
2 f (x) − − x2 1
1 (cid:20) Z − (cid:21)
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Generalizing the m-raising operator for the associated Legendre equation (2.2),
d mx d
R = (1 x2)1/2 + = f g +h,
m − dx (1 x2)1/2 1dx 2
−
we have f (x) = (1 x2)1/2/g (x) and
1 2
−
h(x)
g (x) = (1 x2)−m/2exp dx .
2 − − (1 x2)1/2
(cid:20) Z − (cid:21)
Similarly for the m-lowering operator
d mx d
L = (1 x2)1/2 + = g f +h,
m −dx − (1 x2)1/2 − 2dx 1
−
we determine that
h(x)
f (x) = (1 x2)(m+1)/2exp dx ,
1
− (1 x2)1/2
(cid:20)Z − (cid:21)
with g (x) = (1 x2)1/2/f (x).
2 1
−
TheseoperatorssatisfyRnPn−1(x) = nPn(x),Ln+2Pn(x) = nPn−1(x), RmPnm(x) =
Pm+1(x), and L Pm+1(x) = [n(n+1) m(m+1)]Pm(x) = (n m)(m+n+1)Pm(x).
n m n − n − n
For h = 0 these results reduce to the raising and lowering operators of (10) and (27)
of [4].
From the generalized raising operator (2.3) we find that
(x2 1)−n/2+1 h(x) d h(x) dn−1
P (x) = − exp (x2 1)n/2exp (x2 1)n−1
n 2n−1n! x2 1 dx − − x2 1 dxn−1 −
(cid:20)Z − (cid:21) (cid:20) Z − (cid:21)
dn−1
+h(x) (x2 1)n−1. (2.4)
dxn−1 −
This leads to the equation 2n−1n!Pn(x) = Rn2n−1(n 1)!Pn−1(x). The relation (2.4)
−
is a generalization of (16) of [4] for nonzero functions h(x).
We let
h(x)
e(x) exp . (2.5)
≡ x2 1
(cid:20)Z − (cid:21)
Then we may arrive at families of generalized Rodrigues’ formulae in the form
d d n−1 (x2 1)1/2
n!P (x) = (x2 1)1−n/2[e(x)]n−1 (x2 1)3/2 − +F[h],
n
− dx − dx e(x)
(cid:20) (cid:21)
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wherein F, being 0 for h = 0, depends upon powers of h and its derivatives with re-
specttox. ThetermofF withthehighestorderderivativeofhisx(x2 1)n−2h(n−2)(x)
−
and the term of the highest power of h is ( 1)n(n 1)!xhn−1(x).
− −
Generalization to Gegenbauer polynomials
One of the generalizations of Legendre polynomials is the Gegenbauer (or ultra-
spherical) polynomials Cλ, in turn a special case of the Jacobi polynomials P(α,β),
n n
[2, 11]
(2λ)
Cλ(x) = n P(λ−1/2,λ−1/2)(x).
n (λ+1/2) n
n
These polynomials satisfy the ordinary differential equation
(x2 1)y′′+(2λ+1)xy′ n(2λ+n)y = 0,
− −
and the case λ = 1/2 recovers the Legendre polynomials. A pair of lowering and
raising operators in n for the Gegenbauer polynomials is given by
d
(1 x2) +nx Cλ(x) = (n+2λ 1)Cλ (x),
− dx n − n−1
(cid:20) (cid:21)
d
(1 x2) (n 1+2λ)x Cλ (x) = nCλ(x).
− dx − − n−1 − n
(cid:20) (cid:21)
Putting
d d
C+ = (1 x2) (n 1+2λ)x = f g +h,
n − dx − − 1dx 2
we find that
h(x)
g (x) = (1 x2)(n−1+2λ)/2exp dx ,
2 − x2 1
(cid:20)Z − (cid:21)
and
1 x2 h(x)
f (x) = − = (1 x2)(3−n)/2−λexp dx .
1 g (x) − − x2 1
2 (cid:20) Z − (cid:21)
With Cλ(x) = 1, via iteration of Cλ(x) = 1C+Cλ (x) for h = 0 we determine that
0 n −n n n−1
( 1)n d d n−1
Cλ(x) = − (1 x2)(3−n)/2−λ (1 x2)3/2 (1 x2)λ.
n n! − dx − dx −
(cid:20) (cid:21)
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For comparison, the usual Rodrigues’ formula for Gegenbauer polynomials is
( 2)nΓ(n+λ)Γ(n+2λ) dn
Cλ(x) = − (1 x2)1/2−λ (1 x2)n+λ−1/2.
n n! Γ(λ)Γ(2n+2λ) − dxn −
From the latter formula we determine that
( 2)n−1Γ(n+λ 1)Γ(n+2λ 1)
Cλ(x) = − − −
n − n! Γ(λ)Γ(2n+2λ 2)
−
d dn−1
(1 x2)(3−n)/2−λ (1 x2)n/2 (1 x2)n+λ−3/2.
× − dx − dxn−1 −
A Legendre polynomial identity
The following identity, useful for deriving a Rodrigues’ type formula, was given
without proof in [4]. We present three proofs of this identity, the first using corre-
spondences with the Legendre polynomials, and the other two employing the product
(Leibniz) rule. For n 2,
≥
dn dn−2
(x2 1) (x2 1)n−1 = (n 1)n (x2 1)n−1. (3.1)
− dxn − − dxn−2 −
For the first proof, we begin by making various identifications of the left side of
the identity (3.1) in terms of Legendre polynomials P (x) and associated Legendre
n
functions Pm(x) [1, 5, 11]. From the usual Rodrigues’ formula
n
1 dℓ
P (x) = (x2 1)ℓ, (3.2)
ℓ 2ℓℓ!dxℓ −
the left side of (3.1) is
d
2n−1(n−1)!(x2 −1)dxPn−1(x) = −2n−1(n−1)!(1−x2)1/2Pn1−1(x),
where we have used P1(x) = (1 x2)1/2 d P (x). From recurrence 12.5.8(f) of [5] we
ℓ − dx ℓ
also have
d
2n−1(n 1)!(x2 1) Pn−1(x) = 2n−1n![Pn(x) xPn−1(x)]. (3.3)
− − dx −
8
The right side of (3.1) is the integral of the polynomial
dn−1
(n−1)ndxn−1(x2 −1)n−1 = 2n−1(n−1)n!Pn−1(x).
By integrating the recurrence 12.5.8(b) of [5] by parts we obtain
(n+1) Pn(x)dx = Pn−1(x)+xPn(x), (3.4)
−
Z
and
x
(n+1) Pn(x)dx = Pn−1(x)+xPn(x)+Pn−1(0).
−
Z0
The right side of (3.1) is, using (3.4),
dn−1
(n−1)n dxn−1(x2 −1)n−1dx = 2n−1(n−1)n! Pn−1(x)dx
Z Z
= (n 1)!(n 1)2n−1[ Pn−2(x)+xPn−1(x)]
− − −
= n!2n−1[Pn(x) xPn−1(x)]. (3.5)
−
In the last step we have used recurrence 12.5.8(a) of [5] in order to eliminate the
Pn−2(x) term. The equality of (3.3) and (3.5) verifies identity (3.1).
Remark. The expressions (3.3) or (3.5) verify that (3.1) evaluates to 0 when
x = 1.
±
A second proof of (3.1) may be based upon the product rule and the factorization
x2 1 = (x 1)(x+1). We then obtain for the left side of (3.1)
− −
dn n n
(x2 1) (x2 1)n−1 = ( 1)n (1 n)ℓ(1 n)n−ℓ(x 1)ℓ(x+1)n−ℓ, (3.6)
− dxn − − ℓ − − −
ℓ=0 (cid:18) (cid:19)
X
while the right side is given by
n−2
n 2
n(n 1)( 1)n − (1 n)n−ℓ−2(1 n)ℓ(x 1)ℓ+1(x+1)n−ℓ−1
− − ℓ − − −
ℓ=0 (cid:18) (cid:19)
X
n−1
n 2
= n(n 1)( 1)n − (1 n)n−ℓ−1(1 n)ℓ−1(x 1)ℓ(x+1)n−ℓ. (3.7)
− − ℓ 1 − − −
ℓ=1 (cid:18) − (cid:19)
X
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Since in (3.6), (1 n) = 0 unless n = 0, the ℓ = 0 and ℓ = n terms evaluate to 0.
n
−
Comparing (3.6) and (3.7) we need to verify the equality
n 2 n
n(n 1) − (1 n)n−ℓ−1(1 n)ℓ−1 = (1 n)ℓ(1 n)n−ℓ.
− ℓ 1 − − ℓ − −
(cid:18) − (cid:19) (cid:18) (cid:19)
With the ratios (1 n)n−ℓ/(1 n)n−ℓ−1 = ℓ and (1 n)ℓ/(1 n)ℓ−1 = ℓ n, this
− − − − − −
equality reduces to the readily confirmed statement that
n 2 n n!
n(n 1) − = (n ℓ)ℓ = .
− ℓ 1 ℓ − (ℓ 1)!(n ℓ 1)!
(cid:18) − (cid:19) (cid:18) (cid:19) − − −
For a third proof, we first binomially expand each side of (3.1), and then differ-
entiate term by term:
dn−2 dn−2 n−1 n 1
(x2 1)n−1 = − ( 1)n−ℓ−1x2ℓ
dxn−2 − dxn−2 ℓ −
ℓ=0 (cid:18) (cid:19)
X
n−1
n 1
= 2( 1)n − ( 1)n−ℓ−1ℓ(1 2ℓ)n−3x2ℓ−n+2,
− − ℓ − −
ℓ=0 (cid:18) (cid:19)
X
and
dn dn n−1 n 1
(x2 1)n−1 = − ( 1)n−k−1x2k
dxn − dxn k −
k=0(cid:18) (cid:19)
X
n−1
n 1
= 2( 1)n − ( 1)n−k−1k(1 2k)n−1x2k−n.
− − k − −
k=0(cid:18) (cid:19)
X
Therefore, the left side of (3.1) is given by
n−1
n 1
2( 1)n − ( 1)n−k−1k(1 2k)n−1(x2k−n+2 x2k−n)
− − k − − −
k=0(cid:18) (cid:19)
X
n−1
n 1
= 2( 1)n − ( 1)n−k−1k(1 2k)n−1x2k−n+2
− − k − −
"
k=0(cid:18) (cid:19)
X
n−2
n 1
− ( 1)n−k(k +1)( 1 2k)n−1x2k−n+2 .
− k +1 − − −
#
k=0(cid:18) (cid:19)
X
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