Hermite Functions and the Quantum Oscillator
Background Differentials Equations handout The Laplace equation – solution and application Concepts of primary interest: Separation constants Sets of orthogonal functions Discrete and continuous eigenvalue spectra Generating Functions Rodrigues formula representation Recurrence relations Raising and lowering operators Sturm-Liouville Problems Sample calculations: Tools of the trade:
The Quantum Simple Harmonic Oscillator is one of the problems that motivate the study of the Hermite polynomials, the Hn(x). Q.M.S. (Quantum Mechanics says.): 2 du2 n ()(1 kx2 E u x)0 [Hn.1] 2mdx2 2 nn This equation is to be attacked and solved by the numbers STEP ONE: Convert the problem from one in physics to one in mathematics. The equation as written has units of energy. The constant has units of energy * time, m 2 has units of mass, and x has units of length. Combining, has units of (length)4 so a mk
1/4 mk -1 dimensioned scaling constant = 2 is defined with units (length) as a step
toward defining a dimensionless variable z = x and a dimensionless constant n = 2 E n which represents the scaled energy or the energy in natural units. The (/km )1/2
Contact: [email protected] k ½ equation itself is divided by the natural energy ½ ( /m) to reach a dimensionless
(mathspeak) form: du2 n ()() zuz2 0. [Hn.2] (**See problem 2 for important details.) dz2 nn The advantage of this form is that one does not need to write down as many symbols.
STEP TWO: Identify and factor out the large z behavior. For large z, the n un term is
2 2 dun 2 negligible compared to z un. Study: zu() x. To leading order in z, u(z) ≈ A dz 2 n
2 2 e z /2 + B ez /2for |z| very large. As the function u(z) must be normalized, the second
2 behavior e z /2 is discarded.
z 2/2 STEP THREE: Propose that the full solution is: un(z) = g(z) e Nn where g(z) is a
2 more slowly-varying, well-behaved function that is to multiply e z /2 to yield the full solution to the differential equation [Hn.2].
dd22g 2 gz() ezz/2 e/2 zgz() ez/2 dz dz
22 dd22gdg 2 2 gz() ezz/2 e /22 z e z /2 z 2 gz() ez /2 dz22dz dz
z2 /2 After some substitutions and division by e , it follows that un(z) will satisfy dg2 dg equation [Hn.2] if g(z) satisfies: 2(1)0zg [Hn.3] dz2 dz n This equation is the DE for the Hermite polynomials (See problems 5 and 6.). The ratio and comparison tests indicate that the series solution to equation [Hn.3] diverges and that it diverges faster than ez2 /2 converges. The series must be terminated after a finite number of terms if the overall solution functions are to remain finite. Therefore the functions g(z) are the Hermite polynomials, the Hn(z) to within a multiplicative normalization constant. The conclusions flow forth as series termination requires that
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-2 1/2 n = 2 n + 1 leading to energy eigenvalues En = (n +½) (/km ) = (n + ½) o and
n 1/2 -1/2 z 2/2 spatial and temporal eigenfunctions: un(z) = hn(z) = [2 n! ] Hn(z) e and
n 1/2 -1/2 z 2/2 in(1/2) t (x,t) = [2 n! ] Hn(z) e e . Hn(z) is the Hermite polynomial of order n,
and the hn(z) are the spatial parts of the normalized wavefunctions for the quantum harmonic oscillator.
Comment/Exercise: Consider the original form of the equation du2 () zu2 ()0z. dz2 Insert a template power series solution and isolate the coefficient of the zs terms. Show that the recurrence relation becomes (s + 2)(s + 1) as+2 + as – as-2 = 0. A two-term recurrence relation is a good thing. A three-term relation is an indication that you should factor out an asymptotic or other behavior out of the solution and try again.
Charles Hermite 1822-1901
Hermite made important contributions to number theory, algebra, orthogonal polynomials, and elliptic functions. He discovered his most significant mathematical results over the ten years following his appointment to the École Polytechnique. In 1848 he proved that doubly periodic functions can be represented as quotients of periodic entire functions. In 1849 Hermite submitted a memoir to the Académie des Sciences which applied Cauchy's residue techniques to doubly periodic functions. Sturm and Cauchy gave a good report on this memoir in 1851 but a priority dispute with Liouville seems to have prevented its publication.
MacTutor Archive on Mathematics History; University of St. Andrews Scotland
Properties of the Hermite Polynomials: dH2 dH The differential equation: nn 22znH0 dz2 dz n n n n – 2 n – 4 Normalization condition: Hn(z) = 2 x + (terms with powers of x , x , …) Hermite polynomials are alternately even and odd
xn2 Orthogonality Relation: eHxHxdxnnm() () 2 ! mn
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-3 22 (1) nzeed n z Rodriques Formula: Hn(z) = dzn
n (2zh h2 ) h Generating Function: e Hzn () n0 n!
d Hz() 2 nHz () Recurrence Relations: dz nn 1
d Hz() 2 zHzH () () z dz nnn 1
2()2()(zHnnn z nH11 z H z)
2 Sample Hermite Polynomials: H0(z) = 1 H1(z) = 2 z H2(z) = 4 z - 2 3 4 2 5 3 H3(z) = 8 z - 12 z H4(z) = 16 z - 48 z + 12 H5(z) = 32 z - 160 z + 120 z
Mathematica syntax: HermiteH[n,z] Hn(z)
nn(1) nn (1)(2)(3) n n Hz( ) (2 z )nn (2 z )24 (2z )n .... n 1! 2!
Int[/2] n k (1) n ! nk2 Hzn () (2) z k 0 kn!( 2 k )!
z2 1 2 Properties of the Harmonic Oscillator Wavefunctions: hznn() Hze () 2!n n
dh2 The differential equation: n ()()0;2zhx2 n1 dz2 nn n
Normalization condition: hz() hzdz () (The hn(z) are real-valued.) nm mn
2 z n 2 [2nnne ! ]1/2 ( 1) 2 d ez Rodriques Formula: hn(z) = dzn
Recurrence Relations: d hz()nn h () z1 h () z dz nn2211n
zh() znn h () z1 h () z nn2211n
d ||1|nn nn1 n 1 Abstract Space Notation: dz 22
zn||1| nn n1 n 1 22
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-4 z2 z2 2 z2 1 2 2 z 2 21z 2 Sample Wavefunctions: h0(z) = e h1(z) = e h2(z) = e 4 2
3 z2 42 z2 53 z2 23zz 2 412zz 3 2 42015zzz 2 h3(z) = e h4(z) = e h5(z) = e 3 4! 60
Notation Alert: The use of the zero vector notation 0 representing the additive identity is sometimes preempted by another convention. For example, in quantum mechanics, |0 represents the ground state (lowest energy state) for problems such as the quantum harmonic oscillator. In these cases the additive identity is to
be represented as |ZERO or |NULL. Stay alert; this conflict will arise when you study the quantum oscillator. That is 0 is the ground state, a state vector with non-zero magnitude, for the QSHO problem. Lowering 1 yields 0, the ground state; it does not annihilate the state. Lowering 0 results in annihilation.
Interlaced Zeros Property
The eigenfunctions of a Sturm-Liouville operator (includes most Hamiltonians studied in the first year of quantum) have the properties that they can be chosen to be real valued and that the zeros of functions with adjacent eigenvalues are interlaced. This discussion assumes a discrete eigenvalue spectrum. Other than perhaps endpoint zeroes, a zero for the function with the higher eigenvalue will appear between any two zeros of the function with the lower eigenvalue.
SHOwavefn[n_,z_]:= HermiteH[n,z] Exp[-(z^2)/2]*(1/Sqrt[2^n n! Sqrt[Pi]]) Plot[{SHOwavefn[5,z], SHOwavefn[6,z]},{z,-5,5}, AspectRatio0.4]
This result does not mean that a function with a higher eigenvalue can never share an interior zero location with a lower eigenvalue function. It means that adjacent-
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-5 eigenvalue functions will not share an interior zero location and that the wavefunctions gain a node for each step up in eigenvalue.
0.4
0.2
-4 -2 2 4 -0.2
-0.4
Plots of the n = 5 and 6 QHO wavefunctions
0.4
0.2
-10 -5 5 10 -0.2
-0.4
Plots of the n = 30 and 31 QHO wavefunctions
Given an eigenfunction, every higher eigenfunction with a higher eigenvalue will have at least one zero between any two zeroes of that eigenfunction. Exercise: Give an example in which a SHO wavefunction shares an interior zero location with a function with a lower eigenvalue.
The interlace behavior suggests that the higher eigenfunctions wiggle more rapidly and have smaller and smaller separations between their zero locations indicating that they might represent finer and finer spatial variations. The case that the eigenvalue spectrum is unbounded above might then suggest that the eigenfunctions can wiggle as
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-6 fast as necessary to characterize any reasonable functions and hence that the set of eigenfunctions is a complete basis for all reasonable functions defined on the domain.
Completeness: The SHO wavefunctions hn(x) are orthonormal and so one can attempt to use them as an expansion set for a function f(x).
hxn () fx () x2 f ()x ahxnn ()hx n () hxfxedxhx n () () n () nn00hxhxnn() () n0
Recall that the hn(x) are normalized so that hxhxnn() ()= 1. The set is complete if:
N 2 Limit[|fx() ahx () |2 ex dx ]0 reasonablef ( x ) N nn n0 Basically if the total integrated value of the square of the error in the series representation goes to zero, then no important parts of the behavior of f(x) are absent. All our functions are faithfully represented so expansion set is adequate to provide a complete representation.
Tools of the Trade
Ladder Operations and the Solution of Differential Equations: The form of the differential operator for the Hermite functions suggests an alternative approach to solving the equation for the Hermite functions. du2 n zu2 () z u () z [Hn.4] dz2 nnn that has the eigenvalue form: d 2 Luˆˆ() z u () z where L z2 [Hn.5] nnn dz2 Recalling the great algebraic identity: x2 – y2 = (x – y) (x + y), one is ˆ ˆ ˆ ˆ ˆ ˆ d ˆ d tempted to express L as either AB or BA where Az dz and Bz dz .
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-7 Unfortunately, neither representation is correct due to the appearance of product rule terms in the derivative.
ABˆ ˆˆ u() z zdd z u d2 z2 u u L 1 u 1 u nnndz dz dz2 nnnn [Hn.6]
BˆˆAuˆ () z zdd z u d2 z2 u u L 1 u 1 u nnndz dz dz2 nnnn [Hn.7] Note: The process seems to produce multipliers differing in value by 2. At this point, it is postulated that the spectrum for the eigenvalues n is discrete and consists of values
separated by intervals of magnitude 2. Thus n+1 is assumed to be n + 2. This problem arose in the study of the quantum harmonic oscillator and Lˆ is the dimensionless form of the Hamiltonian, the operator for the sum of the kinetic energy 2 and the positive potential energy ½ k x . All the eigenvalues n should be positive. The spectrum of eigenvalues should start with a minimum positive value, and include values generated by incrementing that value by +2 multiple times. (Rather than a minimum positive value, a more generally applicable condition is that there should be a minimum or lowest eigenvalue such as – 13.6 eV for the hydrogen atom problem.)
Exercise: What combination of the operators Aˆ and Bˆ would equal Lˆ ?
Summary of the notation and results: d 2 Luˆˆ() z u () z where L z2 nnn dz 2 ˆ d ˆ d Azdz Bz dz [Hn.8]
ABuˆ ˆˆ() z L 1 u 1 u BˆˆAuˆ () z L 1 u 1 u nn nn nn nn A short calculation verifies that the multiplication of the linear differential operators
Aˆ , Bˆ and Lˆ is associative, but not commutative. Aˆ ( Bˆ Aˆ ) = ( Aˆ Bˆ ) Aˆ , but Bˆ Aˆ Aˆ Bˆ .
In fact, Aˆ Bˆ - Bˆ Aˆ
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-8 Exercise: Verify that Aˆ ( Bˆ Aˆ ) = ( Aˆ Bˆ ) Aˆ and that the operator Aˆ Bˆ - Bˆ Aˆ is equivalent to multiplication by
Consider:
Aˆˆ BAuˆˆ A ˆ(1) u AB ˆˆ Au (1) Au ˆ [Hn.9] nnn nnn ˆ ˆ Compare this result for A B on an eigenfunction uk (See [Hn.7]). ˆ ˆ AB ukkkn (1)(1) u ukwhich requires that k = n - 2.
The conclusion is that starting with the eigenfunction un with eigenvalue n, the action ˆ ˆ of A on un is to return a new eigenfunction with eigenvalue k = n - 2. The operator A takes the function one rung lower on the eigenfunction ladder. Such operators are called lowering operators.
For the analogous raising result, begin by considering:
Bˆˆ ABuˆˆ B ˆ(1) u B ˆˆ ABu (1) Bu [Hn.10] nnn nnn ˆ ˆ Compare this result for B A on an eigenfunction uk (See [Hn.6]). ˆ ˆ BA ukkk(1)(1) u n uk which requires that k = n + 2.
The conclusion is that starting with the eigenfunction un with eigenvalue n, the action ˆ of B on un is to return a new eigenfunction with eigenvalue k = n + 2. The operator Bˆ takes the function one rung higher on the eigenfunction ladder. Such operators are called raising operators.
Aˆ lowers the eigenfunction to the one with the next lowest eigenvalue and Bˆ raises the eigenfunction to the one with the next highest eigenvalue. These operators seem to be
joined together or associated. The operator Bˆ is to be relabeled Aˆt and dubbed the
adjoint of Aˆ and read as ‘A-dagger’. Nothing has been shown about the normalization and any multiple of an eigenfunction is a solution to the eigenvalue equation. The
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-9 results are restated allowing for a loss of normalization when the operators raise or lower a function.
Auˆˆ() z c u () z A n c n 1 nnn1 n [Hn.11] ˆˆtt Aunn() z d un1 () z A n dn n 1
Recalling previous result and converting them to abstract vector space notation, ˆˆttˆˆ AAunnn() z 1 u () z AA nn1 n [Hn.12] ˆˆttˆˆ A Aunnn() z 1 u () z A A n n1 n [Hn.13]
As the n are positive and separated from one another by 2, there must be a least value of Assume that the lowest value is o and that the corresponding eigenfunction is uo(z). It is impossible to lower the lowest eigenfunction as there is none lower. The ˆ 1 action of A uo(z) should be to annihilate it (return zero as the result).
ˆ d dun ()z A uo(z) = 0 or zuzdz nn() 0 dz zu
z2 2 This equation can be integrating to yield uo(z) = c e where c is a scalar constant. With the lowest eigenfunction in hand, its eigenvalue can by computed. d 2 Luˆˆ() z u () z where L z2 Luˆ () z u () z nnn dz 2 000
2 22 2 d zce2½zz ce ½ 1 ce ½z dz2
The conclusion is that the lowest eigenvalue o is 1. The next task is to raise uo(z) to
find u1(z).
ˆt d ½ z2 A uo(z) = dn un(z) zuzduzzedz 01()n () (2)
½ z2 The conclusion is that u1(z) is a multiple of (2ze ) . The overall constant multiplier would be set by a normalization procedure. Even without normalization, we can find
the eigenvalue for u1(z).
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-10 22 2 d2 zze2½(2)zz (6)ze½ 3(2)ze ½ z dz2
The eigenvalue 1 = 3 which is 0 + 2 supporting the previous speculation about the
spacing between eigenvalues and leading to the final result that n = 2 n + 1.
The energy eigenvalues for the quantum harmonic oscillator follows from the values for the . 2E 21nEnn (½) k nnm k m
Combining the result n = 2 n + 1 with equations: [Hn.11], [Hn.12] and [Hn.13], the identifications ½ ½ cn = (2 n) and dn = (2 [n + 1]) are possible. In an effort to stomp out the root 2’s, new operators are defined.
aˆt1Aˆt1 zd and aAˆ 11ˆ z d [Hn.14] 22dz 22 dz New Action Summary:
anˆ nn11aˆt n n n 1 [Hn.15]
n (annuzzuˆt )n 0 ! ()1 d ()z [Hn.16] n 2!n n dz 0
It follows that aanˆˆt nnand aaˆˆt n n1 n leading to the identification of aaˆˆt as the
ˆˆtt ˆ ˆ ˆ ˆ t number operator and of the relation aa a a a,1 a . The action of this combination of aaˆˆt and is to multiply the wavefunction by 1. Hence that operator
ˆˆttˆ ˆ ˆ ˆt combination, aaa a a, a , is completely equivalent to multiplication by 1.
ˆˆtt ˆˆ ˆtˆ ˆˆt aa, aa aa is called the commutator of aaand .
ˆ ˆˆtt ˆ ˆ With a little effort, it follows that Laaaa½ . Comparing with the eigenvalue equation,
Lnˆ ½( aaˆˆtt aa ˆ ˆ n n2 n1) n n
1 This behavior leads to the alternative name set of annihilation and creation for the lowering and raising operators.
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-11 ˆ k ½ ˆ ˆ k ½ The dimensionful Hamiltonian is H = ½ ( /m) L leading to H n = (2 n + 1)½ ( /m) .
ˆ d The dimensionless momentum operator is pz i dz .
azˆt 1 d and azˆ 1 d become aziˆˆt 1 pand azipˆˆ1 2 dz 2 dz 2 z 2 z Inverting the equations,
zaa1 ˆˆt and pˆˆ i aat ˆ 2 z 2 [Hn.17] The commutator of the position and momentum can now be computed in terms of the commutator
ˆˆttt ˆˆ ˆˆ aa, aa aa.
zp,,ˆˆˆˆˆ 1 att ai a a z 22
z,,,,,0(1)(1)(0pˆˆˆˆˆˆˆˆˆii aatt aa t aa t aa )i z 22 This result is valid for our dimensionless operators. The dimensionful form is:x, piˆ x . The value of the commutator is deduced by allowing the operator to act on an arbitrary function f(x).
xp,()()()()()ˆ fx x idd fx i xfx x dx dx df df ix x fxifx() () dx dx
The action of the commutator on an arbitrary function is to multiply it by i so the commutator is equivalent to multiplication by i.
Developing the Rodriques Formula:
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-12 The machinery developed in the previous section provides a basis for developing the
2 z n 2 uz() e2 d e z Rodriques formula for the n dzn . Consider an arbitrary function f(z).
22 22 eefzeezfzzfz½½zzdd() ½½ zz() d () dz dz dz Next the induction step. It is assumed that:
n 22 ddzgze() ½½zzn e gz() dz dzn This assumption is used to establish the result for n + 1.
nn1 22 dddddzgz() z zgz() ze½½zzn egz () dz dz dz dz dzn
n1 22 22 ddddzgze() ½½zznn egzzeegz() ½½ zz () dz dz dznn dz 22 2 2 22 ze½½zzddnn e g() z e ½ z1 e ½ z g() z z e ½½ zz d n e g() z dznn dz1 dz n
n1 22 or ddzgze() ½½zzn1 egz () dz dzn1 With the n = 1 anchor step and the induction (n n + 1) step validated, the result is true for all n.
n 22 ddzgze() ½½zzn e gz( ) n and well- behavedgz ( ) [Hn.18] dz dzn
To this point, g(z) has been an arbitrary function; now g(z) is identified with uo(z) = c e 2 - ½ z . The operation begins with the lowest eigenfunction, the ground state and raises it n times to generate a multiple of un(z).
n n 222 ddzuz() At uzduz () () e½½½zzzn e ce dz 00 n dzn
22 d uz() e½ zzd n e c n dzn d The factor /c provides for normalization of needed. It is claimed that the process has generated a multiple of the nth eigenfunction. Developments covered in the problem section establish that:
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-13 22 2 2 ( 1)nzeeezbzbzeHz½½2d n z z(2 ) n n n 4½... z () dzn nn24 n
22 22 It follows that e½½zn(1)ez Hz() Hz ()orthat(1)nzz ed n e Hz(). Rodriques nn dzn n got it right.
Sturm – Liouville Problems Almost all of our DEs
A homogeneous linear differential equation for y(x) of the form:
ddy px() wx () qx () yx () 0 [SL.19] dx dx where p(x), q(x) and w(x) are real-valued functions, w(x) 0 and some rather general boundary conditions are met at a and b, the ends of the interval over which the solutions are to be defined. The equation is found to possess physically meaningful solutions only if the parameter is one of a certain set of characteristic values call eigenvalues. The permissible values of are known as its characteristic values (or
eigenvalues) k, and the corresponding functions yk(x), which then satisfy the conditions of the problem when = k, are known as the characteristic functions (or eigenfunctions)1. In most physics applications, the functions p(x) and w(x) are positive
definite in the interval [a,b], except possibly at one or both of the end points.
Consider the differential equation and its complex conjugate.
1 Eigen is German for characteristic property. eigen, eigene, eigenes: own or its own; Eigenschaft (f): property
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-14 d dym px()mm wx () qx () y () x 0 dx dx [SL.20] d dy* m ** px() mm wx() qx () y () x 0 dx dx
Multiply the second equation by yn(x).
d dy* m ** yxnm() px () wxqxyxyx() ()n ()m () 0 dx dx Prepare a copy with m and n interchanged and subtract it from the one above.
dddy* dy mn*** ynm() x px () y () x px () nmwx() yn () x ym () x 0 dx dx dx dx Integrate both sides over the range a to b. Use integration by parts on the left-hand side:
**b * dy dyb dy dy dy dy p()xyx ()mn yx* () mnmnpx() px () dx nm a dx dxa dx dx dx dx [SL.21] b **y ()xy () xwxdx () nma m n It is to be assumed that the following boundary condition is met:
* b dymn* dy px()ynm () x y () x 0 [SL.22] dx dx a and it is clear that the second integral vanishes so the result is:
b **yxyxwxdx() () () 0 [SL.23] nma m n
Note that this procedure has generated a natural inner product for the functions associated with this differential equation. The function w(x) is everywhere positive and is called the weight function.
b f ()xgx () fx ()* gxwxdx () () [SL.24] a
In the case that m = n,
b **yxyxwxdx() () () 0 nna n n
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-15 b * and, as y ()xy () xwxdx () 0 for non-trivial functions yn(x), a nn
n =n* [SL.25] If a complex number is equal to its complex conjugate, it is a real number. The eigenvalues are real, and, as all the eigenvalues are real for Sturm-Liouville problems, the general case becomes:
b yxyxwxdx* () () () 0 nma m n
Solution functions yn(x) and ym(x) with distinct eigenvalues (n m) are necessarily orthogonal.
b yy yxyxwxdx* () () () 0 for [SL.26] mna m n n m Eigenfunctions with distinct eigenvalues are orthogonal. In the cases that an eigenvalue is repeated, the several functions corresponding to the eigenvalue can be used to build a set of mutually orthogonal functions by applying the Gram-Schmidt procedure. COMPLETE BASIS: The collection of the eigenfunctions that solve the differential equation is a complete set of basis functions for the problem. A well-behaved, general function defined over same interval can be expanded in a generalized Fourier series as:
ym f fx() aymm ()where x a m [SL.27] m1 ymmy
Exercise: Show that the Hermite polynomial equation can be written as:
22 d xxdHn en 2()0eHxn dx dx Compare this with the general Sturm-Liouville form to identify each function in the general form. What values are the allowed eigenvalues? What is the interval (a, b)?
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-16 ddy px() wx () qx () yx () 0 dx dx Give the form of the natural inner product for the space spanned by the Hermite polynomials.
* b dymn* dy Is the condition px() ynm () x y () x 0 met for the particulars of the Hermite dx dx a problem?
The Generating Function: What is it good for?
Consider the generating function relation for the Hermite polynomials:
n nm 2 h 222 hh e(2zh h ) Hz() and hence ee(2zh h ) (4zh 2 h ) Hz() Hz () . n nm n0 n! nm00nm!! Note that: Sums start at zero for the Hermite problem. Exercise (what else) the inner product.
nm 2 22hh e(4zh 2 h ) edzzz Hz() Hzedz () nm nm00nm!!
nm 2 2 hh 2 ee2hz(2)zh dz H() z H () z e dz nm mn,0 nm!! We have shown that that the Hermite polynomials form a mutually orthogonal set.
22 HzHze() ()zz dz HzHze() () dz nm mnnn
2n 22hu22 2 h h z2 eedu e HzHzenn() () dz 2 n0 (!)n Expanding the left-hand side:
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-17 2 nn2 2n 2 e2h 2 h h HzHze() ()z dz n! 2 nn nn00(!)n Each power of h is an independent function so the coefficient must match left to right.
22 2(n 1 HzHze)()ozz dzr( HzHze)()2 dzn n n! (!n )2 nn nn
After incorporating the normalization and splitting the weight function between um(x)*
and un(x), the wavefunctions for the harmonic oscillator become:
2 2 1 ½ z 1 ½ z in(½)0 t uznn() Hze () nn()zHz ()ee 2!n n 2!n n CONCLUSION: The generating function may have application beyond being a difficult method to generate the eigenfunctions.
Warm Up Problems
1/4 mk WUP1. The constant = 2 has dimensions of length inverse. It is used to define the dimensionless coordinate z = x. a.) Provide an equation for x given z. b.) The potential energy of a SHO is ½ k x2. Give the expression for the potential energy as a function of z. d d c.) Express /dx in terms of /dz. d.) Express piˆ in terms of a derivative with respect to z. x 22 e.) Express the kinetic energy operator using derivatives with respect to z. 2mx 2 f.) What was identified as the natural energy unit for the SHO problem?
2 1/4 22 1 2 1 k 2 Partial Answers: x = zz kin. en. op. = 2 2 m 2 mk 2mz z
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-18 WUP2. The Hermite polynomials are alternately even and odd and are normalized n n-2 n-4 such that they have the form: Hn(x) = (2x) + an-2 x + an-4 x + … . dHnn() x d11 H () x d n H () x Give the values of: nn;;n dxnn dx11 dx n
Problems 1.) Find the location of all the zeros of the seventh and eighth eigenfunctions for the harmonic oscillator in terms of their dimensionless argument. Locate the zeros to three significant figures.
z2 1 2 hznn() Hze () 2!n n Do the zeros obey the interlaced-zeros hypothesis? Give a specific example in which a wavefunction has exactly the same location of an interior zero as does another wavefunction with a lower eigenvalue. Use Mathematica: NSolve[HermiteH[n,x]0,x,6]
2.) Begin with the Shroedinger equation for the harmonic oscillator. 2 du2 n ()(1 kx2 E u x)0 2mdx2 2 nn 2 As each term must have the same dimensions, argue that has units of (length)4 and mk
1/4 1/2 mk that (/km ) has the dimensions of energy. Define = 2 and use the change of variable z = x and division by ½ (/km )1/2 to convert the Shroedinger equation a dimensionless form, a simple problem in mathematics. The DE becomes the dimensionless (mathspeak) form:
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-19 dU2 n ()() zUz2 0 dz2 nn
Note that if one wants to be pure, Un(z) = Un( x) = un(x). The variable x has dimensions while z is dimensionless. In the mathform, functions and their arguments du2 are dimensionless. We are sloppy and often just write n ()() zuz2 0 even dz2 nn though we should be more careful. !This course is SP35x, not SM35x.
3.) Identify and factor out the large z behavior. For large z, the n un term is negligible
2 2 2 dun 2 z /2 compared to z un. Study: zu() x. To leading order in z, u(z) ≈ A e for |z| dz 2 n very large. Show that:
2 d 22 ezezzz/2 22/2 1() where (z2 )represents “and terms of order that vanish dz2
2 as z-2 or faster”. Why is the behavior u(z) ≈ A e z /2 , which works to the same order, banished for the list of possibilities? Propose that the full solution is: un(z) = Nn Hn(z)
z 2/2 e where Nn is a normalization constant, and show that Hn(z) must satisfy
2 dHnn dH 2(1)z H0 if un(z) is to satisfy the original equation . The equation dz2 dz nn dH2 dH nn2(1)z H0 is the DE for the Hermite polynomials (See problems 5 dz2 dz nn and 6.) The ratio and comparison tests indicate that the series diverges and that it
2 diverges faster than e z /2 converges. The series must terminate to yield polynomials if the overall solution functions are to remain finite. The conclusions flow forth as series 1 termination requires that n = 2 n + 1 leading to energy eigenvalues: En = (n + /2)
1/2 1 (/km ) =(n + /2) o and spatial and temporal eigenfunctions:
2 2 n 1/2 -1/2 z /2 n 1/2 -1/2 z /2 -i(n+½)t un(z) = hn(z) = [2 n! ] Hn(z) e and (x,t) = [2 n! ] Hn(z) e e
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-20
1 2 x 4.) The normalized oscillator wave functions are: ux() 2m m ! Hxe () 2 . mm
Verify this normalization for uo(x) and u2(x) and show that uo(x) and u2(x) are
orthogonal. The quantum mechanics inner product: f gfxgx [()]*()dx
5.) The Hermite polynomials are a family of polynomials central to the study of the harmonic oscillator in quantum mechanics and the transverse intensity pattern of laser beams. The governing differential equation is: dH2 dH nn 21xH 0 dx2 dx nn
Apply the power series method to find the indicial equation. If the series for Hn is allowed to be infinite, it diverges by the ratio test. Identify the spectrum of values for
n that terminate the series after a finite number of terms. Generate the form of the four lowest order polynomials using the recurrence relations for the coefficients in the
power series trial solution and setting the lowest index coefficients ao and a1 alternately to 1 and 0 or to 0 and 1. Renormalize the mth order polynomial by multiplying by whatever value makes the coefficient of the highest (mth) power of x equal to 2m. Compare with the forms given below and comment.
23 Hx01() 1; Hx () 2 xHx ; 2 () 4 x 2; Hx 3 () 8 x 12 x
1/2 Quantum mechanics identifies n as (2En) /[(k/m) ] where the En are the allowed
energy levels for the harmonic oscillator. What are the allowed energy values?
6.) Hermite Polynomials: There are many ways to develop the Hermite polynomials. One method begins with the initial set of polynomials { 1, x, x2, x3, … , xn, … } and
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-21 uses the Gram-Schmidt procedure to generate an orthogonal set over the interval (- ,) using the inner product:
x2 HxHxe() () dxHH 0 for m mm
The weight function ex2 is necessary to assure that the integrals over the infinite range 2 converge. Assume that H0(x) = 1 and that H1(x) = 2 x is orthogonal to it. Start with x and use the Gram-Schmidt procedure to get an H2(x) that is orthogonal to H0(x) and n n 0 H1(x) and that has 2 s the coefficient of the x . Continue to find H3(x). Starting at x and proceeding up the list of powers, the Gram-Schmidt procedure guarantees that each new polynomial is orthogonal to all polynomials of lower order.
2 2 x 2 n x 13 1 Use: edx and xe dx n22 n ...2 for n = 1,2, … 4 2 Partial Answer: H4(x) = 16 x – 48x + 12.
22 (1) nxeed n x 7.) Use the Rodriques representation Hn(x) = dxn for the Hermite
d Hx() 2 xHxH () () x polynomials to show that dx nnn1 . Use the first derivative of this result and the differential equation satisfied by the Hermite polynomials dH2 dH nn22xnH0 to establish the recurrence relation d Hx() 2 nHx (). dx2 dx n dx nn1
Continue to show that: 2x Hxnnn () 2 nHx11 () H () x. Use recurrence properties of the
n d Hx() 2nn nHx ! () 2 n ! Hermite polynomials to show that: dxn n 0 .
8.) a.) Generate the first four Hermite polynomials be exercising the Rodriques formula. Show that your results are consistent with
2 22 1 nnd n ex 2 x axnn24 ax... e x Hxe ( ) x. That is the exponential dxn nn24 n n n times a polynomial with leading term 2 x followed by terms that drop the power of x
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-22 1 0 n nn24 by two as they step down to x or x . Assume pn(x) = 2 xaxax nn24... and
22 show that ddp ()xex 2 xp () x ( p ()) x ex leads to: dxnnn dx
2 22 1 nn11d n1 exaxax 2 nn13xeHxe... x() x dxn1 nn13 n1 Proving the result for n = 0 anchors the hypothesis. Assuming the result for the case n and then proving it for the case n + 1 establishes the result for all cases. The Hn(x) are alternately even and odd polynomials. Continue to argue that: d n Hx() 2n nHx ! () 2n n ! dxn n 0 . Use this result in the problem below.
d Hx() 2 nHx () d n Hx() 2nn nHx ! () 2 n! b.) Use the identity dx nn1 to show that dxn n 0 .
9.) Use the information in the previous two problems to evaluate: (Assume m n)
xm22xd m x2x2n HxHxedx() () Hx()(1) em e e dx2 n ! nm n dx mn Explain why one can assume m n without loss of generality. (The Hermite polynomial of higher order is represented by its Rodriques formula.) Note that the results of this problem identify the orthonormal harmonic oscillator wavefunctions
x2 1 2 asuxnn() Hxe () . 2!n n
n (2xh h2 ) h 10.) The Hermite generating function relation is: e Hxn (). Expand the left- n0 n! hand side keeping all terms up to and including order h4. Identify the Hermite polynomials Ho to H4.
n d Hx() 2nn nHx ! () 2 n ! 11.) Give the value of H0(x). Prepare an argue that: dxn n 0 based on the recurrence relations. (short form of #8)
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-23 z2 1 2 12.) Given the Hermite function definition: nhznn() Hze () . Develop 2!n n
d recurrence like expressions for z |n and ( /dz) |n. Use the recurrence relations for the Hermite polynomials; do not assume those for the Hermite functions (SHO d wavefunctions). What states are linked to |n by z and dz ?
13.) Combine the Hermite function recurrence relations to find: 1 zh d ()z. 2 dz n Recast your results in the abstract vector space KET-style notation. What states are linked to |n by azˆ 1 d and azˆ† 1 d ? 2 dz 2 dz
14.) Develop the recurrence relations based on d 2 hz() and zh2 () z. What states are dz2 n n linked to |n by z2 and the second derivative with respect to z?
x2 1 2 15.) Given: nn()zhx () Hxen () , use the recurrence relations for the 2!n n oscillator wave functions to evaluate:
*2 *2 * d * d 31()x xxd ()x mn()x xxd ()x 13()x ()xdx mn()x ()xdx dx dx
d 2 d 2 * ()x ()xdx and *2()x x ()xdx mn2 mn2 dx dx
x2 1 2 16.) Given: nn()zhx () Hxen () Use the recurrence relations to 2!n n
* d * d evaluate: mn()x xx ()dx, mn()x xxd () x and dx dx
* dd mn()x xx () xdx. dx dx
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-24 x2 x2 x2 17. Starting with the linearly independent set of functions {1 e 2 , x e 2 , x2 e 2 , x2 x3 e 2 , …. } over the domain (-,) construct the first three members of a set of orthonormal functions using the Gramm-Schmidt method with the inner product: f gfxgx [()]*()dx.
Hermite Functions: ho(x) = ______; h1(x) = ______; h2(x) = ______
Companion problem for #6.
18. Identify and factor out the large z behavior. For large z, the n un term is negligible
2 2 2 dun 2 z /2 compared to z un. Study: zu() x. To leading order in z, u(z) ≈ A e for |z| dz 2 n very large. Show that:
2 d 22 ezezzz/2 22/2 1() where (z2 )represents “and terms of order that vanish dz2
2 as z-2 or faster”. Why is the behavior u(z) ≈ A e z /2 which works to the same order banished for the list of possibilities? Propose that the full solution is: un(z) = Nn Hn(z)
z 2/2 e where Nn is a normalization constant, and find the equation satisfied by Hn(z is dH2 dH nn2(1)zH 0. dz2 dz nn This equation is the DE for the Hermite polynomials (See problems 5 and 6.) The ratio and comparison tests indicate that the series diverges and that it diverges faster than
2 e z /2 converges. The series must terminate to yield polynomials if the overall solution functions are to remain finite. The conclusions flow forth as series termination requires
1 1/2 1 that n = 2 n + 1 leading to energy eigenvalues: En = (n + /2) (/km ) =(n + /2) o and spatial and temporal eigenfunctions:
n 1/2 -1/2 z 2/2 n 1/2 -1/2 z 2/2 un(z) = hn(z) = [2 n! ] Hn(z) e and (x,t) = [2 n! ] Hn(z) e
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-25 z2 1 2 19.) The eigenvalue problem for the QSHO: nhznn() Hze () 2!n n
Find the recurrence relations for d hz() , z h (z) and d 2 hz() and zh2 () z. Use then dz n n dz2 n n to evaluate d 2 zhz2 (). Conclude that h (z) is an eigenfunction of the operator dz2 n n
d 2 z2 with eigenvalue 2 n +1. Recall that d 2 z2 is the dimensionless form of dz2 dz2
k ½ the operator for the energy in terms of the energy unit ½ ( /m) .
20.) The eigenfunctions of the QSHO are known to be a complete set so any time
independent solution can be represented as uz() amn h () z where the hm(z) are ortho- m0 normal. The general form for a time dependent wave function
in(½)0 t is (,)zt amn h () ze . Use this form to compute the normalization integral m0
[ (,)]zt (,) zt dz. Be sure to use distinct dummy indices in the representations of the wavefunction and its complex conjugate. Use the orthogonality relation,
hzhzdz() () . Express the probability that the particle is in the state hk(z) in mn mn terms of the expansion coefficients in the series representation of .
21.) The eigenfunctions of the QSHO are known to be a complete set so any time
independent solution can be represented as uz() amn h () z where the hm(z) are ortho- m0 normal. The general form for a time dependent wave function
in(½)0 t in(½) 0 t is (,)zt amn h () ze . Each state hzen () has an energy eigenvalue of 2 n m0
½ 2 + 1 (in units of ½ (k/ ) ) where d z2 is the operator for the energy. Compute m dz2
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-26 the expectation value of the operator [(,)]zt d 2 z2 (,) zt dzusing series dz2 representations for the wave function and its complex conjugate. Also use the
orthogonality relation, hzhzdz() () . Discuss the result and the interpretation mn mn that ak*ak is the probability that the system is in the quantum state k which has the (dimensionless) energy eigenvalue 2 k + 1.
22. The eigenfunctions of the QSHO are known to be a complete set so any time
independent solution can be represented as uz() amm h () z where the hm(z) are ortho- m0 normal. The general form for a time dependent wave function
im(½)0 t is (,)zt amm h () ze . Find the expectation value of the coordinate z in a m0 general state of the QSHO. Compute that expectation value of position as:
[(,)]zt z (,) zt dz using series representations for the wave function and its
complex conjugate. Also use the orthogonality relation, hzhzdz() () . Make mn mn use of the recurrence relation: zh()z nnh() z1 h() z . The result has two nn221 n1 sums. Redefine the summation index in one sum as needed to make the summation ranges identical and show that the sums are the complex conjugates of one another. The result can be written as:
Ce* it00Ce it || Ceiieit0|| Ce e it 0 Acos[ t ] z 0
i ½ where CCe|| (2) aakkk1 1 , A = 2 |C| (Answer not verified) k0 That is the expectation value of position has the same time dependence as does the position of a classical oscillator with the same m and k. Note that the expectation value of the classical observable z is real-valued.
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-27 This answer is for the dimensionless form of the problem. x = -1z
1/4 2 x= 2 |C| cos[0t + ] mk 23. The eigenfunctions of the QSHO are known to be a complete set so any time
independent solution can be represented as uz() amm h () z where the hm(z) are ortho- m0 normal. The general form for a time dependent wave function
im(½)0 t is (,)zt amm h () ze . Find the expectation value of the dimensionless m0 momentum in a general state of the QSHO using its operator i . Compute that z
expectation value of position as: [(,)]zt i (,) zt dz using series z representations for the wave function and its complex conjugate. Also use the
orthogonality relation, hz()h()zdz . Make use of the recurrence relation: mn mn d hz()n h() z n1 h ()z. The result has two sums. Redefine the summation dz nn212n1 index in one sum as needed to make the summation ranges identical and show that the sums are the complex conjugates of one another. The result can be cast in the form:
pB sin[o t + ] = 2 |C| sin[o t + ] (see previous problem) That is the expectation value of position has the same time dependence as does the momentum of a classical oscillator with the same m and k and a position zcos[o t + ]. Conclude that the expectation values of the quantum system obey the classical equations of motion. Note that the expectation value of the classical observable p is real-valued. Compare with the previous problem.
1/4 mk ˆ With the dimensioned constants reasserted, pi i2 . x z
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-28 d Show that in the dimensioned form p = m /dt x. Expectation values obey the classical equations of motion.
24.) The most general form for a time dependent wave function for a quantum
im(½)00 t im(½) t harmonic oscillator is (,)zt cmm h () z e cm me . Compute: mm00
* (,)zt (,) zt dz. Use the orthogonality statement n|m = nm.
* Your answer should be: cckk. As it represents the normalization integral, k 0
* | = 1.What quantity is being normalized? Provide your best interpretation of cc55
* and cckk.
25.) Consider our lowering and raising operators: azˆ 1 d and azˆ† 1 d . 2 dz 2 dz
anˆˆ||1and|1| n n a† n n n1 The ground state cannot be lowered. Rather, it is annihilated by the lowering operator. anˆ|0 0 | 1 0. Moving from abstract vector to function notation, this becomes
1 z d u (z) = 0. 2 dz a.)Solve this equation to find the spatial form of the ground state wave function of a harmonic oscillator. Be sure to normalize your result. b.) anˆ† |1| n n 1 translated into 1 z d u (z) = (n + 1)½ u (z). Use the 2 dz n n+1 raising operator repetitively to generate u1(z) and u2(z) from your uo(z). Note that the process generates the properly normalized wave functions.
References:
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-29 1. Mary L. Boas, Mathematical Methods in the Physical Sciences, 2nd Edition, chapter 3, John Wiley & Sons (1983). 2. K. F. Riley, M. P. Hobson and S. J. Bence, Mathematical Methods for Physics and Engineering, 2nd Ed., Cambridge, Cambridge UK (2002). 3. The Wolfram web site: mathworld.wolfram.com/ 4. M. Abramowitz and J. A. Stegun. Handbook of Mathematical Functions, NIST (1964).
d 22dH 2 Hze() zz/2 n e/2 zHze() z/2 dznn dz
2 2 d 22dH dH 2 2 Hze() zz/2nn e /22 z e z /2 zHze 2() z /2 dz22n dz dz n dH2 dH nn 2(1)zH 0 dz2 dz nn
10/8/2009 Physics Handout Series.Tank:Hermite Functions - QM SHO Hn(x)-30