On the recursive solution of the quantum harmonic

European Journal of Physics Eur. J. Phys. 39 (2018) 015402 (17pp)

https://doi.org/10.1088/1361-6404/aa9584

On the recursive solution of the quantum harmonic oscillator José M Pérez-Jordá Departament de Química Física, Universitat d’Alacant, E-03080, Alacant, Spain E-mail: [email protected] Received 6 September 2017, revised 13 October 2017 Accepted for publication 24 October 2017 Published 18 December 2017 Abstract

This paper advocates writing the eigenfunctions of the harmonic oscillator without Hermite polynomials, by means of the following recurrence relation: Yn = 2a n x Yn - 1 - (n - 1) n Yn - 2 (n > 0), with Y0 = (a p )1 4 2 e-ax 2 . It is easy to prove that these functions are eigenfunctions of the harmonic oscillator. It is shown that many of the exercises related to the harmonic oscillator found in textbooks can be solved in a simple and elegant way by using this recurrence relation. Keywords: quantum mechanics, harmonic oscillator, recursion

1. Introduction The one-dimensional harmonic oscillator is an unavoidable model system in any quantum mechanics course, either introductory or advanced. There are several reasons for its popularity. First, it is an exactly solvable system. Second, students, once in possession of its solutions, can gain insight into the many concepts of the quantum world by practicing the myriad of exercises that pack textbooks [1–9] and journals [10, 11]. Third, despite its simplicity, the harmonic oscillator model is able to explain ‘real-world’ phenomena such as infrared spectroscopy or black-body radiation. The eigenfunctions of a Hamiltonian Hˆ are the set of functions Yn that fulfil the timeindependent Schrödingerʼs equation:

Hˆ Yn = En Yn,

(1)

where En is the eigenvalue corresponding to Yn . For the one-dimensional harmonic oscillator, the Hamiltonian takes the form 0143-0807/18/015402+17$33.00 © 2017 European Physical Society Printed in the UK

1

Eur. J. Phys. 39 (2018) 015402

J M Pérez-Jordá

1 2 1 Hˆ = pˆ + kxˆ 2 , 2m 2 where xˆ and pˆ are respectively the position and momentum operators, xˆ = x,

(2)

(3)

d , (4) dx m is the particleʼs mass and k the oscillatorʼs force constant. It is known from the early days of quantum mechanics that the eigenvalues and eigenfunctions of the harmonic oscillator are given by pˆ = - i

⎛ 1⎞ En = hn ⎜n + ⎟ , ⎝ 2⎠ Yn(x ) =

n = 0, 1, 2  ,

(5)

1 ⎛⎜ a ⎞⎟1 4 2 Hn (x a ) e-ax 2 , n ⎝ ⎠ 2 n! p

(6)

where ν is the oscillatorʼs frequency,

n=

1 2p

k m

(7)

and

2pnm km = . (8)   The symbol Hn represents the Hermite polynomial of degree n. Obtaining these eigenfunctions is not trivial. Two different approaches can be employed. One is the power-series method [1–3, 6–9], which requires lengthy manipulations of the equation, some of them controversial [12–14]. As an elegant alternative, but lengthy nonetheless, we can use the ladder or creation/annihilation operators [2, 4, 5, 7, 8] method. In both cases, it is shown that the solutions fulfil a certain recurrence relation, out of which, and after some effort, the Hermite polynomials emerge. One may wonder if the Hermite polynomials are really unavoidable. When solving this eigenproblem from scratch, one may consider stopping at the recurrence relation and circumvent the Hermite polynomials. Even if this proof from scratch is omitted, as may be the case in freshman courses, working with Hermite polynomials is tedious and error-prone, and the students have to be exposed to yet another concept, the properties of which are probably unknown to them and are often given without proof, increasing the feeling that quantum mechanics is some sort of revealed truth that cannot be understood, only believed. This paper advocates to write the harmonic oscillator eigenfunctions by means of the recurrence relation a=

Yn =

2a x Yn - 1 n

n-1 Yn - 2 (n > 0) , n

Y0 =

⎛⎜ a ⎞⎟1 4 -ax 2 2 e . ⎝p⎠

( 9)

This expression can be worked out from the recursion property of the Hermite polynomials [15], or from the properties of ladder operators [2]. We will prove in section 2 that the functions generated from equation (9) are indeed eigenfunctions of the harmonic oscillator, and then we will use this relation to solve some of the exercises usually found in textbooks. The manipulations needed to achieve these goals are simple enough to be suitable for nonadvanced courses and do not require Hermite polynomials, nor solving differential equations, 2

Eur. J. Phys. 39 (2018) 015402

J M Pérez-Jordá

nor applying ladder operators. Some conclusions will be gathered in section 3 (in an advanced course, it may be advisable to prove that no function other than those generated from equation (9) can be an eigenfunction of the harmonic oscillator. This is done, to make the paper self-contained, in appendix C). 2. Properties and applications In this section we will study some properties of the harmonic oscillator eigenfunctions, and then use these properties to solve a selection of the exercises often found in textbooks. All this will be achieved entirely from equation (9), without requiring the use of Hermite polynomials nor any other advanced skill or concept. 2.1. Position operator

The result of acting on an eigenfunction Yn with the position operator can be obtained by means of equations (3) and(9),

xˆ Yn = x Yn =

n Yn - 1 + 2a

n+1 Yn + 1, 2a

n  0.

(10)

Note that this equation, unlike the recursion relation in equation (9), is also valid for n=0. 2.2. Momentum operator

According to equation (4), to get the effect of pˆ on any Yn obtained from equation (9), we need the first derivative of Yn , which is given by

dYn an a ( n + 1) (11) = Yn - 1 Yn + 1, n  0 dx 2 2 (see appendix A for the proof of this expression). Therefore, the effect of the momentum operator on Yn will be Y ¢n =

pˆ Yn = - iY ¢n = - i

an a ( n + 1) Yn - 1 + i Yn + 1, 2 2

n  0.

(12)

b 2.3. Eigenfunctions of H

Let us check now that any function Yn from equation (9) is indeed an eigenfunction ofHˆ with eigenvaluehn (n + 1 2). We computexˆ 2 Yn by applying equation (10) twice,

⎛ n ⎞ n+1 Yn - 1 + Yn + 1⎟ xˆ 2 Yn = xˆ(xˆ Yn) = xˆ ⎜ 2a ⎝ 2a ⎠ ⎞ n ⎛ n-1 n = Yn - 2 + Yn⎟ ⎜ 2a ⎝ 2a 2a ⎠ +

n+1⎛ n+1 Yn + ⎜ 2a ⎝ 2a

3

⎞ n+2 Yn + 2⎟ , 2a ⎠

(13)

Eur. J. Phys. 39 (2018) 015402

J M Pérez-Jordá

which simplifies to

xˆ 2 Yn =

n ( n - 1) 2n + 1 Yn - 2 + Yn + 2a 2a

(n + 1)(n + 2) Yn + 2. 2a

(14)

Similarly, for pˆ 2 Yn , if we apply equation (12) twice we get

⎛ ⎞ an a (n + 1) Yn - 1 + Yn + 1⎟ pˆ 2 Yn = ipˆ ⎜ 2 2 ⎝ ⎠ ⎛ an ⎞ ⎛ a ( n - 1) an ⎞ ⎟ ⎜= -  2⎜ Yn - 2 + Yn⎟ ⎝ 2 ⎠⎝ 2 2 ⎠ ⎞ a (n + 1) ⎛ a (n + 1) a (n + 2) - 2 Yn + Yn + 2⎟ , ⎜2 2 2 ⎝ ⎠

(15)

which gives

 2a ( n (n - 1) Yn - 2 - (2n + 1) Yn + (n + 1)(n + 2) Yn + 2). 2 Now we use these results to apply Hˆ to Yn . We get, according to equation (2), pˆ 2 Yn = -

1 2 1 Hˆ Yn = pˆ Yn + kxˆ 2 Yn 2m 2 ⎛  2a k ⎞ = ⎜+ ⎟ n (n - 1) Yn - 2 ⎝ 4m 4a ⎠ ⎛  2a k ⎞ +⎜ + ⎟ (2n + 1) Yn ⎝ 4m 4a ⎠ ⎛  2a k ⎞ + ⎜+ ⎟ (n + 1)(n + 2) Yn + 2. ⎝ 4m 4a ⎠

(16)

(17)

It turns out from the definition of α in equation (8), however, that

 2a  2 2pnm = = hn , m m 

(18)

2a 2

k ( k )2  2a = = m = = hn , m a a a so that we can simplify the expression of Hˆ Yn to ⎛ 1⎞ Hˆ Yn = hn ⎜n + ⎟ Yn, ⎝ 2⎠

(19)

(20)

which proves that any Yn from equation (9) is an eigenfunction of Hˆ with the correct eigenvalue given by equation (5). Note, however, that the above discussion does not exclude the possibility that some functions other than those from equation (9) are eigenfunctions of Hˆ too. See appendix C for a rigorous proof. 2.4. Parity

A function f is said to be even if f (x ) = f (-x ), and odd if f (x ) = -f (-x ). Let us consider the eigenfunctions defined by the recurrence relation in equation (9). It is clear that Y0 is even. 4

Eur. J. Phys. 39 (2018) 015402

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Y1 is given by

Y1 =

2a x Y0

(21)

and therefore is odd. It is easy to go on from here by inspecting equation (9): Y2 is even, Y3 odd, and so on. We conclude that an eigenfunction Yn is even if n is even, but odd if n is odd, which can be expressed by means of this formula:

Yn(x ) = ( - 1)n Yn( - x ).

(22)

2.5. Orthonormality

The set of eigenfunctions Yn form an orthonormal set,

⟨ Yn∣Ym⟩ = dnm.

(23)

Let us prove it by means of equation (9). Two eigenfunctions Yn and Ym corresponding to different eigenvalues are orthogonal,

⟨ Yn∣Ym⟩ = 0,

n ¹ m.

(24)

This is true not only for the harmonic oscillator, but for the eigenfunctions of any Hermitian operator. The proof is well known but we include it here for completeness. By the properties of Hermitian operators,

⟨ Yn∣Hˆ ∣Ym⟩ = ⟨ Ym∣Hˆ ∣Yn⟩*

(25)

and, as Yn and Ym are eigenfunctions of Hˆ with respective (and different) real eigenvalues En and Em,

E m ⟨ Yn∣Ym⟩ = En ⟨ Ym∣Yn⟩* .

(26)

We know, in addition, that

⟨ Ym∣Yn⟩* = ⟨ Yn∣Ym⟩ ,

(27)

E m ⟨ Yn∣Ym⟩ = En ⟨ Yn∣Ym⟩

(28)

(E m - En)⟨ Yn∣Ym⟩ = 0.

(29)

so that and therefore As Em and En are different, we must conclude that

⟨ Yn∣Ym⟩ = 0.

(30)

Let us prove now that any Yn is normalised. We use equation (9) to work on ⟨ Yn∣Yn⟩,

⟨ Yn∣Yn⟩ =

2a ⟨ Yn∣x Yn - 1⟩ n

n-1 ⟨ Yn∣Yn - 2⟩ , n

(31)

2a ⟨ x Yn∣Yn - 1⟩. n

(32)

but, as Yn and Yn - 2 are orthogonal,

⟨ Yn∣Yn⟩ =

2a ⟨ Yn∣x Yn - 1⟩ = n

5

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J M Pérez-Jordá

Now we use equation (10):

2a ⎛ n ⟨ Yn - 1∣Yn - 1⟩ + ⎜ n ⎝ 2a

⟨ Yn∣Yn⟩ =

⎞ n+1 ⟨ Yn + 1∣Yn - 1⟩ ⎟. 2a ⎠

(33)

Again, as Yn + 1 and Yn - 1 are orthogonal,

⟨ Yn∣Yn⟩ = ⟨ Yn - 1∣Yn - 1⟩ ,

(34)

so that, by repeated application of this equation, we arrive at

⟨ Yn∣Yn⟩ = ⟨ Y0∣Y0⟩.

(35)

The ground-state eigenfunction is normalised, as proved in appendix B, and therefore

⟨ Yn∣Yn⟩ = 1,

(36)

which concludes the proof that the eigenfunctions of the harmonic oscillator given in equation (9) form an orthonormal set. 2.6. Uncertainty relation

Heisenbergʼs uncertainty relation for the position and momentum operators states that

Dx Dp 

 , 2

(37)

where

Dx =

⟨ x 2⟩ - ⟨ x⟩2 ,

(38)

Dp =

⟨ p 2 ⟩ - ⟨ p ⟩2 .

(39)

The expectations values ⟨ x⟩, K,⟨ p 2 ⟩ are computed from the normalised wave function Ψ describing the system,

⟨ x⟩ = ⟨ Y∣xˆ∣Y⟩ = ⟨ Y∣xˆ Y⟩ ,

(40)

⟨ x 2⟩ = ⟨ Y∣xˆ 2∣Y⟩ = ⟨ Y∣xˆ 2 Y⟩ ,

(41)

⟨ p ⟩ = ⟨ Y∣ pˆ∣Y⟩ = ⟨ Y∣ pˆ Y⟩ ,

(42)

⟨ p 2 ⟩ = ⟨ Y∣pˆ 2 ∣Y⟩ = ⟨ Y∣pˆ 2 Y⟩.

(43)

Next, we check the uncertainty relation for the eigenfunctions of the harmonic oscillator, Y = Yn . We begin by computing ⟨ x⟩ by means of equation (10),

n n+1 ⟨ Yn∣Yn - 1⟩ + ⟨ Yn∣Yn + 1⟩ 2a 2a and, recalling the orthonormality of Yn , we get For get

⟨ x 2⟩,

⟨ x⟩ =

(44)

⟨ x⟩ = 0.

(45)

by using equation (14) and taking into account again the orthonormality of Yn , we 6

Eur. J. Phys. 39 (2018) 015402

J M Pérez-Jordá

2n + 1 . 2a Similarly, by using equations (12) and(16), we obtain ⟨ x 2⟩ =

(46)

⟨ p ⟩ = 0,

(47)

⟨ p2 ⟩ =

 2a (2n + 1) . 2

(48)

We conclude that

⎛ 1⎞  DxDp =  ⎜n + ⎟  , ⎝ ⎠ 2 2

(49)

which agrees with the uncertainty relation. 2.7. Virial theorem

The virial theorem [9] states that

x

dV dx

= 2⟨T ⟩,

(50)

where V represents the potential energy of the system and T its kinetic energy. For the harmonic oscillator,

T=

1 2 p , 2m

(51)

V=

1 2 kx 2

(52)

and as

x

dV dx

= ⟨ x kx⟩ = 2 ⟨ V ⟩ ,

(53)

the virial theorem takes the form

⟨ V ⟩ = ⟨ T ⟩.

(54)

The expectations values of V and T for any eigenfunction Yn are readily computed from equations (46) and(48):

⟨V ⟩ =

1 k k ⟨ x 2⟩ = (2n + 1) , 2 4a

1  2a ⟨ p2 ⟩ = (2n + 1) 2m 4m and taking into account equations (18) and (19) we obtain ⟨T ⟩ =

⟨V ⟩ = ⟨T ⟩ =

hn ⎛⎜ 1⎞ n + ⎟, 2 ⎝ 2⎠

(55)

(56)

(57)

which agrees with the virial theorem. 7

Eur. J. Phys. 39 (2018) 015402

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2.8. Selection rules

The harmonic oscillator eigenfunctions Yn can be used to model the vibrational states of diatomic molecules [15, 16]. Assuming that the change of the moleculeʼs dipole moment with internuclear distance is linear, a spectroscopic transition from state Y n¢¢ to state Yn¢ is forbidden unless [16]

⟨ Yn∣x∣Yn¢⟩ ¹ 0.

(58)

This bracket is non-zero when the following selection rule is fulfilled [16]:

Dn = n¢ - n =  1.

(59)

The proof is immediate from equation (10),

⟨ Yn∣x∣Yn¢⟩ =

n¢ ⟨ Yn∣Yn¢- 1⟩ + 2a

n¢ + 1 ⟨ Yn∣Yn¢+ 1⟩ , 2a

(60)

which, by the orthogonality of the eigenvectors, is zero unless n = n¢ - 1 (Dn = +1) or n = n¢ + 1 (Dn = -1). 3. Conclusions The following conclusions can be drawn from this work. • It is possible to write the eigenfunctions of the harmonic oscillator without having to use Hermite polynomials, by means of the recurrence relation shown in equation (9). • It is easy (see section 2.3) to prove that these recursive functions are indeed eigenfunctions of the harmonic oscillator. • It is simple and elegant, using the recursive formulation proposed in this work, to solve many of the typical exercises related to the harmonic oscillator often found in textbooks. • Equation (9) can be proved from scratch by means of ladder operators (to make the paper self-contained, the proof is given in appendix C). Appendix A. Derivative of Ψn In order to prove equation (11) we will use the relation

⎛ d ⎞⎟ ax Yn + Y ¢n = ⎜ax + Yn = ⎝ dx ⎠

2an Yn - 1,

n  0,

(61)

which is proved next. For n=0, we operate on the expression of Y0 given in equation (9),

⎛ d ⎞⎟ ⎡⎛⎜ a ⎞⎟1 4 -ax 2 2⎤ e ax Y0 + Y ¢0 = ⎜ax + ⎢ ⎥ = ax Y0 - ax Y0 = 0, ⎝ dx ⎠ ⎣⎝ p ⎠ ⎦ which agrees with equation (61). 8

(62)

Eur. J. Phys. 39 (2018) 015402

J M Pérez-Jordá

For n > 0 , we operate on the expression of Yn from equation (9),

⎛ d ⎞⎟ ⎛ 2a ax Yn + Y ¢n = ⎜ax + x Yn - 1 ⎜ ⎝ dx ⎠ ⎝ n ⎛ 2a x Yn - 1 = ax ⎜ ⎝ n

⎞ n-1 Yn - 2⎟ n ⎠

⎞ n-1 Yn - 2⎟ n ⎠

2a 2a n-1 x Y ¢n - 1 Yn - 1 + Y ¢n - 2 n n n 2a x (ax Yn - 1 + Y ¢n - 1) = n n-1 2a (ax Yn - 2 + Y ¢n - 2) + Yn - 1. n n +

(63)

The terms in parenthesis are the n - 1 and n - 2 counterparts of ax Yn + Y ¢n . This suggests the use of mathematical induction: we assume that equation (61) have been proved for n - 1 and n - 2, and then replace them in equation (63) (note that for n=1, only ax Y0 + Y ¢0 is needed),

ax Yn + Y ¢n =

4a 2 ( n - 1) x Yn - 2 n 2a (n - 1)(n - 2) Yn - 3 + n

2a Yn - 1. n

(64)

Next we replace the value of x Yn - 2 , which can obtained from equation (10),

4a 2 ( n - 1 ) ⎛ n - 2 Yn - 3 + ⎜ n 2a ⎝

ax Yn + Y ¢n =

⎞ n-1 Yn - 1⎟ 2a ⎠

2a (n - 1)(n - 2) 2a Yn - 3 + Yn - 1 n n 2a 2a = ( n - 1) Yn - 1 + Yn - 1 = 2an Yn - 1, n n -

(65)

which agrees with equation (61). We conclude, by the logic of mathematical induction, that equation (61) is true for any n  0 : from the case n=0 we can prove the case n=1, from n=0 and n=1 we can prove the case n=2, and so on. We use equation (61) to get the derivative of Yn ,

Y ¢n =

2an Yn - 1 - ax Yn

(66)

and, after replacing the value of x Yn from equation (10), we obtain equation (11),

⎛ n Y ¢n = 2an Yn - 1 - a ⎜ Yn - 1 + ⎝ 2a =

an Yn - 1 2

a (n + 1) Yn + 1. 2

9

⎞ n+1 Yn + 1⎟ 2a ⎠ (67)

Eur. J. Phys. 39 (2018) 015402

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Appendix B. Normalisation of Ψ0 We will check here that the ground-state eigenfunction Y0 given in equation (9) is normalised. Let us compute the bracket ⟨ Y0∣Y0⟩,

a p

⟨ Y0∣Y0⟩ =

¥

ò-¥ e-ax

2

dx.

(68)

The integral above can be readily found in many tables of integrals [17], but, for the sake of completeness, we will compute it here. We express it in terms of a two-dimensional integral, ¥

ò-¥ e-ax

(ò

2

dx =

¥

(ò

¥

-¥

¥

e-ax dx 2

)(ò

¥

-¥

e-ay dy 2

)

(69)

ò e-a (x +y ) dx dy) , -¥ -¥

=

2

2

(70)

which we can simplify, first by using polar coordinates, ¥

ò-¥

e-ax dx = 2

¥

ò0 ò0

2p

e-ar r dr dj = 2

2p

ò0

¥

e-ar r dr ,

(71)

p . a

(72)

2

then by the change of variable t = ar 2 , ¥

ò-¥ e-ax

2

dx =

p a

ò0

¥

e - t dt , =

p [ - e-t]¥ 0 = a

We conclude that

⟨ Y0∣Y0⟩ = 1.

(73)

Appendix C. Equation (9) gives all possible eigenfunctions In this section we will prove that any function given by equation (9) is a normalised eigenfunction of the harmonic oscillator, and, conversely, that any normalised eigenfunction can be obtained from equation (9). We will use the ladder operators method [2]. b and a b† operators C.1. a

The ladder or creation/annihilation operators can be defined as

aˆ =

aˆ † =

a xˆ + 2m a xˆ 2m

i pˆ , 2m

(74)

i pˆ , 2m

(75)

(note that aˆ and aˆ† are usually defined to be dimensionless operators, and differ from the above expressions by a constant). Operators aˆ and aˆ† are each other adjoints. 10

Eur. J. Phys. 39 (2018) 015402

J M Pérez-Jordá

ba b† and a b† a b operators C.2. a

Let us apply aa ˆ ˆ† to an arbitrary wave function Ψ (no necessarily an eigenfunction of Hˆ ):

⎛ a aa xˆ + ˆ ˆ †Y = ⎜ ⎝ 2m a 2 2 2 = xˆ Y 2m a 2 2 2 = xˆ Y + 2m

⎞ ⎛ a ⎞ i i pˆ ⎟ ⎜ xˆ pˆ ⎟ Y 2m ⎠ ⎝ 2 m 2m ⎠ a a 1 2 i xp px pˆ Y ˆˆ Y + i ˆˆ Y + 2m 2m 2m a 1 2 pˆ Y - i [xˆ , pˆ] Y , 2m 2m

(76)

which, from the definition of α in equation (8) and the expression of Hˆ in equation (2), becomes

ˆ ˆ †Y = aa

a a 1 2 1 2 [xˆ , pˆ] Y = Hˆ Y - i [xˆ , pˆ] Y. kxˆ Y + pˆ Y - i 2 2m 2m 2m

(77)

We can compute the commutator [xˆ , pˆ] from the definition of the operators xˆ and pˆ in equations (3) and (4),

[xˆ , pˆ] Y = xp ˆ ˆ Y - px ˆ ˆ Y = - ix

dY d (x Y ) + i = iY , dx dx

(78)

so that

a 2 ˆ ˆ †Y = Hˆ Y + Y aa 2m

(79)

and, from equation (18),

hn ˆ ˆ †Y = Hˆ Y + Y. aa 2

(80)

As Ψ is an arbitrary function, we conclude that

hn ˆ ˆ † = Hˆ + aa . 2

(81)

Similarly, we can prove that

hn aˆ †aˆ = Hˆ . 2

(82)

b Ψ ¼ EΨΨ bΨ, H C.3. Properties of a

Let us assume that Ψ is an eigenfunction of Hˆ with eigenvalue EΨ,

Hˆ Y = E Y Y.

(83)

Let us study how the function aˆ Y is transformed by operator Hˆ : we have, from equations (81) and (82), that 11

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J M Pérez-Jordá

⎛ † hn ⎞⎟ hn ˆˆ (aˆ Y) = aˆ (aˆ †aˆ) Y Hˆ (aˆ Y) = ⎜aa aˆ Y ⎝ 2 ⎠ 2 ⎛ hn ⎞⎟ hn ˆ ˆ Y - hn (aˆ Y) , = aˆ ⎜Hˆ Yaˆ Y = aH ⎝ 2 ⎠ 2

(84)

so that, from equation (83), we get

Hˆ (aˆ Y) = (E Y - hn )(aˆ Y).

(85) ˆ This means that aˆ Y is either zero or an eigenfunction of H with eigenvalue E Y - hn . Let us denote the ground-state eigenfunction by Y0 and its eigenvalue byE Y0 . The function aˆ Y0 cannot be an eigenfunction of Hˆ because, according to equation (85), it would have E Y0 - hn as eigenvalue, which is below the ground-state energy. Therefore, aˆ Y0 = 0.

(86)

It is easy to obtain, from this equation and equation (82), the ground-state eigenvalue:

⎛ hn ⎞⎟ hn hn Y0 = aˆ † (aˆ Y0) + Y0 = Y0. Hˆ Y0 = ⎜aˆ †aˆ + ⎝ ⎠ 2 2 2

(87)

We conclude that E Y0 = hn 2, which agrees with E0 in equation (5). Next, we will prove that for any excited state Ψ, aˆ Y cannot be zero. We proceed by contradiction, and assume that for some excited Ψ, aˆ Y = 0 . Then, by repeating the manipulations carried out in equation (87), we find out that the eigenvalue of Ψ should be equal to the eigenvalue E0 of the ground state, which is in contradiction with Ψ being an excited state. We conclude, according to equations (85) and(86), that aˆ Y is zero when Ψ is the ground state, but it is and eigenfunction of Hˆ with eigenvalue E Y - hn when Ψ is an excited state,

Hˆ Y = E Y Y

⟶

⎧ if Y = Y0: aˆ Y = 0 ⎨ ⎩ if Y ¹ Y0: aˆ Y ¹ 0,

Hˆ (aˆ Y) = (E Y - hn )(aˆ Y).

(88)

b Ψ ¼ EΨΨ b† Ψ, H C.4. Properties of a

Given any eigenfunction Ψ of Hˆ , let us study now how aˆ†Y is transformed byHˆ : we have, from equations (81) and (82), that

⎛ hn ⎞⎟ † hn † (aˆ Y) = aˆ † (aa ˆ ˆ †) Y + Hˆ (aˆ †Y) = ⎜aˆ †aˆ + aˆ Y ⎝ ⎠ 2 2 ⎛ hn ⎞⎟ hn † = aˆ † ⎜Hˆ + Y+ aˆ Y = aˆ †Hˆ Y + hn (aˆ †Y) , ⎝ ⎠ 2 2

(89)

so that, from equation (83), we get

Hˆ (aˆ †Y) = (E Y + hn )(aˆ †Y).

(90) ˆ This means that is either zero or an eigenfunction of H with eigenvalue E Y + hn . We will prove next that aˆ†Y cannot be zero, and therefore it has to be an eigenfunction of Hˆ . We will proceed by contradiction, and assume that, for some eigenfunctionΨ, aˆ†Y is zero, aˆ†Y

aˆ †Y = 0.

(91)

12

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Then, from equation (81),

⎛ † n ⎞⎟ n n ˆˆ Y = aˆ (aˆ †Y) Y = - Y, Hˆ Y = ⎜aa ⎝ 2 ⎠ 2 2

(92)

that is, Ψ would be an eigenfunction of Hˆ with eigenvalue -hn 2 , but this is impossible because it would be below the ground-state eigenvalue obtained in equation (87). We conclude that aˆ†Y is an eigenfunction of Hˆ with eigenvalue E Y + hn ,

Hˆ Y = E Y Y

aˆ †Y ¹ 0,

⟶

Hˆ (aˆ †Y) = (E Y + hn )(aˆ †Y).

(93)

C.5. Ground-state eigenfunction

To obtain the ground-state eigenfunction, we have to solve equation (86). From the definition ofaˆ in equation (74) we get

ipˆ Y0 = - axˆ Y0,

(94)

so that, taking into account the expression of xˆ and pˆ in equations (3) and (4),

dY0 = - ax Y0. dx Solving this ordinary differential equation is simple,

dY0 = - a x dx , Y0

(95)

(96)

which we integrate to Y0

òC

dY0 = -a Y0

ò0

x

x dx ,

(97)

where C is an integration constant. After computing the integrals we get

ln Y0 - ln C = -

ax 2 , 2

(98)

so that

Y0 = C e-ax

2

2.

(99)

We can fix the value of C by normalisation. The normalised expression of Y0 is (see appendix B)

Y0 =

⎛⎜ a ⎞⎟1 4 -ax 2 2 e , ⎝p⎠

(100)

which agrees with the expression given in equation (9). It is important to note that Y0 is, up to a normalization constant, the only possible solution of equation (86). This means that the ground state is non-degenerate. C.6. Excited-state eigenvalues

We have seen, in equation (87), that the ground-state eigenvalue is E0 = hn 2 , and we have proved in appendix C.5 that it is non-degenerate. Let us extend this study to excited-state eigenvalues. 13

Eur. J. Phys. 39 (2018) 015402

J M Pérez-Jordá

Let us begin with the ground-state eigenfunction Y0 , which fulfils

hn Hˆ Y0 = Y0. (101) 2 Then, according to equation (93), aˆ†Y0 will be an excited eigenfunction of Hˆ with eigenvalue 3hn 2,

3hn † (aˆ Y0) Hˆ (aˆ †Y0) = 2 and, by n successive applications of operator aˆ† over Y0 , we will arrive at

(102)

⎛ 1⎞ (103) En = hn ⎜n + ⎟ , Yn = (aˆ †)n Y0, n = 0, 1, , ⎝ 2⎠ that is, Yn = (aˆ†)n Y0 is an eigenfunction of Hˆ with eigenvalue hn (n + 1 2). This agrees with the well-known eigenvalues of the harmonic oscillator given in equation (5). Having proved that hn (n + 1 2), with n a non-negative integer, is an eigenvalue of Hˆ , now we have to prove that no value other than hn (n + 1 2) is an eigenvalue. Negative integer values of n are out of the question, because they will yield eigenvalues below the ground state, which is absurd. Let us try now non-integer values of n, and assume that there is an excited-state eigenfunctionΨ with an eigenvalue EΨ located between En and En + 1, Hˆ Yn = En Yn,

En < E Y < En + 1.

(104)

Now we apply the aˆ operator to Ψ n times, to get, according to equation (88), an eigenfunction aˆ n Y with eigenvalue

E aˆn Y = E Y - nhn ,

(105)

which is still above the ground state,

E 0 < E aˆn Y < E1.

(106) aˆ n Y

As Eaˆ n Y is above the ground state, the eigenfunction is an excited state, and, according to equation (88), if we apply aˆ one more time, we will get an eigenfunction aˆ n + 1Y with an eigenvalue below the ground state, which is impossible. Therefore, the only possible eigenvalues of the harmonic oscillator are those given by equations (5). We have seen in appendix C.5 that the ground state is non-degenerate. We will prove next that excited states are also non-degenerate. We will proceed by contradiction, and assume that the first excited state is degenerate, with two eigenfunctions Y1 and Y¯ 1 with the same eigenvalue E1 = 3hn 2. Y1 and Y¯ 1 must be linearly independent,

Y¯ 1 ¹ k Y1,

(107) ¯ for any constant k. Next, we apply the aˆ operator to both Y1 and Y1, and obtain, according to equation (88), the respective eigenfunctions aˆ Y1 and aˆ Y¯ 1, both with eigenvalue E0 = hn 2 , which correspond to the ground state. However, as the ground state is non-degenerate, aˆ Y1 and aˆ Y¯ 1 must be linearly dependent, that is, aˆ Y¯ 1 = kaˆ Y1 for some constant k. Now we apply the

(108) aˆ†

operator to both sides of this relation,

aˆ †aˆ Y¯ 1 = kaˆ †aˆ Y1,

(109)

14

Eur. J. Phys. 39 (2018) 015402

J M Pérez-Jordá

use equation (82),

⎛ˆ ⎛ˆ hn ⎞ ¯ hn ⎞ ⎜H ⎟ Y1 = k ⎜H ⎟ Y1, (110) ⎝ ⎠ ⎝ 2 2 ⎠ take into account that both Y¯ 1 and Y1 are eigenfunctions of Hˆ with eigenvalue3hn 2 , ⎛ 3hn ⎛ 3hn hn ⎞ ¯ hn ⎞ ⎜ ⎟ Y1 = k ⎜ ⎟ Y1 ⎝ 2 ⎝ 2 2 ⎠ 2 ⎠

(111)

Y¯ 1 = k Y1,

(112)

and arrive at which is in contradiction with equation (107). Therefore, the first excited state Y1 must be non-degenerate. We can now repeat the proof for Y2 , then for Y3, and so on. We conclude that all the states of the harmonic oscillator are non-degenerate. C.7. Normalisation

Given an eigenfunction Yn , we know, from equations (88) and(93), that aˆ Yn and aˆ†Yn are the eigenfunctions corresponding to the eigenvalues En - 1 and En + 1, and therefore should be respectively proportional to Yn - 1 and Yn + 1,

aˆ Yn = Kn Yn - 1,

(113)

aˆ †Yn = Kn† Yn + 1,

(114)

where Kn and are constants (note that is not the ‘adjoint’ of Kn, but the constant corresponding to aˆ†). We can find out the values of these constants by assuming that all three Yn , Yn - 1, and Yn + 1 are normalised. Let us compute ⟨ aˆ Yn∣aˆ Yn⟩ with the help of equations (82) and(5),

Kn†

Kn†

⟨ aˆ Yn∣aˆ Yn⟩ = ⟨ Yn∣aˆ †aˆ∣Yn⟩ =

hn Yn∣Hˆ ∣Yn 2

= hnn ,

(115)

where we have used the fact that Yn is normalised. On the other hand, we can also compute this bracket by means of equation (113),

⟨ aˆ Yn∣aˆ Yn⟩ = Kn2 ⟨ Yn - 1∣Yn - 1⟩ = Kn2,

(116)

so that, by equating the two previous equations, we get

Kn =

hn n .

We proceed similarly for

Kn†,

(117) but using equations (81) and(5),

⟨ aˆ †Yn∣aˆ †Yn⟩ = ⟨ Yn∣(aˆ †)†aˆ †∣Yn⟩ = Yn∣Hˆ +

= ⟨ Yn∣aa ˆ ˆ †∣Yn⟩

hn ∣Yn 2

= hn (n + 1)

(118)

and, from equation (114),

⟨ aˆ †Yn∣aˆ †Yn⟩ = (Kn† )2 ,

(119)

which gives us

Kn† =

hn (n + 1) .

(120)

15

Eur. J. Phys. 39 (2018) 015402

J M Pérez-Jordá

Summing up,

aˆ Yn =

hnn Yn - 1,

aˆ †Yn =

(121)

hn (n + 1) Yn + 1.

(122)

Note that for the ground state Y0 , and according to equation (88), aˆ Y0 should be zero, which agrees with equation (121). C.8. Proof of the recurrence relation

To complete the derivation of the harmonic oscillator solutions, we have to prove the recurrence relation for Yn given in equation (9). First we express the position operator xˆ in terms of the operators aˆ and aˆ†, which is easily done by taking into account their definition in equations (74) and (75),

xˆ =

m (aˆ + aˆ †) a 2

(123)

and then, by means of equations (121) and (122), we apply it to Yn - 1,

xˆ Yn - 1 =

m ( hn (n - 1) Yn - 2 + a 2

hnn Yn) ,

n > 0.

(124)

Next, from the definition of α in equation (8), 2pnm

m

hn =

a 2

2pnm a  = = = a 2 a 2 a 2

1 , 2a

(125)

so that

xˆ Yn - 1 =

1 ( n - 1 Yn - 2 + 2a

n Yn) ,

n > 0.

(126)

Noting that, according to equation (3), xˆ = x , we finally get

2a x Yn - 1 n which proves equation (9). Yn =

n-1 Yn - 2, n

n > 0,

(127)

ORCID iDs José M Pérez-Jordá

https://orcid.org/0000-0002-7968-1499

References [1] Schiff L I 1968 Quantum Mechanics 3rd edn (Singapore: McGraw-Hill) [2] Cohen-Tannoudji C, Diu B and Laloë F 1977 Quantum Mechanics vol 1 (New York: Wiley) [3] Pauling L and Wilson E B Jr 1985 Introduction to Quantum Mechanics with Applications to Chemistry (New York: Dover) [4] Bohm D 1989 Quantum Theory (New York: Dover) [5] Pilar F L 1990 Elementary Quantum Chemistry 2nd edn (Mineola, NY: Dover) [6] Lowe J P 1993 Quantum Chemistry 2nd edn (San Diego, CA: Academic) [7] Shankar R 1994 Principles of Quantum Mechanics 2nd edn (New York: Kluwer Academic) [8] Zettili N 2009 Quantum Mechanics 2nd edn (New York: Wiley) 16

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[9] Levine I N 2014 Quantum Chemistry 7th edn (Boston, MA: Pearson) [10] Nogueira P H F and de Castro A S 2016 Revisiting the quantum harmonic oscillator via unilateral Fourier transforms Eur. J. Phys. 37 015402 [11] Borghi R 2017 Quantum harmonic oscillator: an elementary derivation of the energy spectrum Eur. J. Phys. 38 025404 [12] Buchdahl H A 1974 Remark on the solutions of the harmonic oscillator equation Am. J. Phys. 42 47–50 [13] Bowen M and Coster J 1980 Methods of establishing the asymptotic behavior of the harmonic oscillator wave functions Am. J. Phys. 48 307–8 [14] Solt G and Amiet J-P 1985 On the ‘series method’ to solve Schrödinger’s equation for the linear oscillator Am. J. Phys. 53 180–1 [15] Levine I N 1975 Molecular Spectroscopy (New York: Wiley) [16] Herzberg G 1950 Molecular Spectra and Molecular Structure: I. Spectra of Diatomic Molecules 2nd edn (Princeton, NJ: Van Nostrand) [17] Gradshteyn I S and Ryzhik I M 1994 Table of Integrals, Series, and Products 5th edn (San Diego, CA: Academic)

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