Eigenstates of the Hamiltonian and the One-Dimensional Stationary Schr¨Odinger Equation

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MATH3385/5385. Quantum Mechanics. Handout # 5: Eigenstates of the Hamiltonian and the One-dimensional Stationary Schr¨odinger Equation

Time-independent potentials

In the case of a structureless particle subject to a potential V (r,t) (possibly time-dependent!) we have seen in Handout # 3 that the Hamiltonian can be represented by the diﬀerential operator: ~2 H = 2 + V (r,t) −2m∇ (i.e. from the replacement of the momentum vector p by the operator i~∇ in the classical b − Hamiltonian function H(r, p; t)). If now the potential V is actually time-independent, i.e. V (r,t)= V (r) only depending on the position vector r, then we can look for solutions of the Schr¨odinger (SE) equation in the separated form:

ψ(r,t)= χ(t)φ(r) which inserted in the SE leads to 1 dχ 1 i~ = Hφ = E χ dt φ where the l.h.s. of the equation depends only onb t, whereas the r.h.s. depends only on r. Thus, both sides must be equal to a constant E, and we can thus separate the equations into

Hφ = Eφ the time independent SE (1a) − dχ i~ = Eχ (1b) bdt The second equation (1b) can be readily solved to give:

iEt/~ χ(t)= e− χ(0) leading to a simple time-evolution1. The ﬁrst equation (1a), i.e. the stationary or time- independent Schr¨odinger equation, which for a single structureless particle reads:

~2 2φ + V (r)φ = Eφ (2) −2m∇ has the form of an eigenvalue problem –with eigenvalue E, the energy– of the diﬀeren- tial operator H. Solving this eigenvalue (or “spectral”) problem under certain boundary conditions will lead to a possible set of values that E can take, i.e. the eigenvalues of the b 1Obviously, the separated solution ψ = χφ is not the general solution of the time-dependent Schr¨odinger equation, but only a particular solution: see Handout # 4.

1 operator under the boundary condtions, and this set of values is called the spectrum2. From now on we will concentrate, in given problems, in developing the techniques to calculate the spectrum of the Hamiltonian: once we know this and the corresponding eigenfunctions φ = φE(r), the quantum mechanical problem is practically solved. All physical quantities (such as expectation values of observables) can then in principle be calculated.

Since the Hamiltonian operator is Hermitian H† = H, the eigenfunctions for diﬀerent eigenvalues are orthogonal w.r.t. the inner product: b b (φ , φ ′ )= dr φ∗ (r)φ ′ (r) = 0 provided E = E′ . E E E E 6 Z Recalling further that if φE(r) is a solution of the eigenvalue problem (1a), e.g. (2), with eigenvalue E, then so is any constant multiple cφE(r). We can encounter then the following situations:

the spectrum is discrete: the eigenvalues of the stationary SE take on a discontinuous • set of values En, n = 1, 2,... . In this case the corresponding eigenfunctions φn(r)=

φEn (r) can be normalised, i.e.

φ 2= dr φ (r) 2 = 1 k E k | E | Z The corresponding physical states of the system are called bound states. the spectrum is continuous: the eigenvalues can take on a continuous range of values, • say in an interval E0

dr φ∗ (r)φ ′ (r)= δ(E E′) E E − Z The corresponding physical states of the system are called scattering states. The completeness relation in this case can be written as:

E1 φ (r)φ∗ (r′) dE = δ(r r′) . E E − ZE0 the spectrum is partly discrete and partly continuous. In this mixed situation there • are in the system bound states as well as scattering states.

It can happen that we have diﬀerent (i.e. independent) eigenfunctions for one and the same eigenvalue E: in that case we say that the corresponding “energy level” is degenerate. The corresponding eigenspace is then no longer one-dimensional as would be the case if the eigenvalue were nondegenerate (i.e. when the eigenfunction is unique up to a multiplicative constant). In the degnerate case we need additional “quantum numbers” to characterise the eigenstate uniquely. As we have noted in Handout # 4, this means that we need to look for other physical observables whose operators commute with the Hamiltonian operator H, and consequently for which we can then impose an additional eigenvalue problem consistent with the one for the Hamiltonian. 2 b The name, which has become customary in mathematics, is clearly inspired by physics: it is the set of values of E that corresponds to the measured spectra of atoms in experiments.

2 Example 1: particle in an inﬁnite potential well

As a first example of a system with a discrete spectrum, i.e. physically representing bound states, we consider the case of a one-dimensional particle in an infinite potential well given by: 0 , a x a V (x)= − ≤ ≤ otherwise ∞ In view of the fact that outside the interval a x a the potential is infinite (and − ≤ ≤ thus we have a potential “barrier” inhibiting the particle from being found anywhere else than within the mentioned interval) we solve the eigenvalue problem by imposing: ~2 d2φ = Eφ , a x a −2m dx2 − ≤ ≤ φ(x) = 0 , x a or x a ≤− ≥ implying that the particle has zero probability of being found outside the mentioned interval. So, concentrating on the problem within the interval [ a, a], we note that we have − a simple linear homogenous differential equation with constant coefficients to solve. We distinguish two cases: the case that E > 0 and the one when E 0. ≤ E > 0

In this case the stationary Schr¨odinger equation takes the form: 2 d φ 2 iκx iκx + κ φ = 0 φ(x)= Ae + Be− , (3) dx2 ⇒ with κ = √2mE/~ (which is real in this case). imposing continuity of the eigenfunction we have that at the boundaries of the interval, at x = a the wave function vanishes: ± Aeiκa + Be iκa = 0 φ( a) = 0 − ± ⇒ Ae iκa + Beiκa = 0 − This coupled set of equations has only a nontrivial solution for the coeﬃcients A and B if the following condition holds:

2iκa 2iκa e e− = 2i sin(κa) = 0 2κa = nπ (n N) . − ⇒ ∈ Thus, we ﬁnd the following energy levels: ~2 n2π2 E = E = (4) n 2m 4a2 associated with the following eigenfunctions: 2iA sin( nπx ) , n even φ (x)= n 2a (5) n 2A cos( nπx ) , n odd n 2a The coeﬃcients An are determined by the normalisation of the eigenfunctions: a a 2 2 nπx 2 2 2 nπx 2 4 An sin ( ) dx =1 resp. 4 An cos ( ) dx = 1 | | a 2a | | a 2a Z− Z− 3 leading in both cases to A 2 = 1/(4a) . Thus, we can take for the normalised eigenfunc- | n| tions: 1 nπx √a sin( 2a ) , n even φn(x)= 1 nπx (6) ( √a cos( 2a ) , n odd

Exercise: Check that these eigenfunctions are orthogonal, i.e. that we have: a φn∗ (x)φm(x) dx = 0 , n = m . a 6 Z−

E 0 ≤ In this case κ is purely imaginary, and we have instead to introduce the real quantity:

√ 2mE k − ≡ ~ in which case the stationary Schr¨odinger equation takes the form:

2 d φ 2 kx kx k φ = 0 φ(x)= αe + βe− . (7) dx2 − ⇒ Again imposing continuity of the eigenfunction at the boundaries of the interval, the wave function vanishes: ka ka αe + βe− = 0 φ( a) = 0 ka ka ± ⇒ αe− + βe = 0 which only leads to a nontrivial solution for α, β if

e4ka = 1 k = 0 . ⇒ Thus, in this case only the eigenvalue E = 0 is allowed, from which we conclude that there are no negative eigenstates! Finally, we have also to reject the eigenvalue E = 0, because it is easy to see that if k = 0 the above boundary conditions only lead to the case α = β = 0 which would lead to the trivial function φ(x) = 0 which is to be rejected as an eigenfunction.

Thus, we can draw the following conclusions about the spectrum for the inﬁnite potential well:

~2 n2π2 Eigenvalues: En = 2m 4a2 , (n positive integer)

1 nπx √a sin( 2a ) , n even 1 nπx √a cos( 2a ) , n odd Eigenfunctions: φn(x)=   if : a

In Example 1 we have investigated a siuation where the potential is unbounded, namely V = at the boundaries of the interval [ a, a]. In that case the derivatives of the solution ∞ − φ at the boundaries of the inﬁnite potential well do not play a role. In the examples that follow now, V (x) is discontinuous but bounded in a neighborhood of the discontinuity. In that case we have to consider also the boundary values for the derivative φ′(x). In fact, from the stationary SE we have:

ǫ d2φ dφ dφ ǫ V (x) E dx = = φ(x) dx =0 as ǫ 0 , 2 ~2 − ǫ dx dx x=ǫ − dx x= ǫ ǫ /(2m) → Z− − Z− whenever V (x) is bounded around x = 0 and since φ(x) in the last integrand is continuous, implying continuity of φ′(x) at x = 0.

Example 2: Finite one-dimensional well

In example 1, we have seen a situation where the spectrum is discrete leading to bound states only, whereas in example 2 we have seen the other extreme situation, where the spectrum is fully continuous leading to scattering states characterised by the reflection and transmission coefficients. In the last example of this handout we consider a mixture of both situations, namely the case in which the spectrum consists of both a discrete part as well as a continuous part. However, the two parts of the spectrum have to be dealt with separately. We consider again a one-dimensional particle, now in a finite potential well given by: 0 x > a V (x)= | | V a x a − 0 − ≤ ≤ with fixed V0 > 0. Again the analysis of this case proceeds in different steps according to whether E V or 0 E < V , (it can easily be shown that the value E < V does not ≥ 0 ≤ 0 − 0 lead to nontrivial physical solutions.

V E < 0 − 0 ≤ In this case it is convenient to introduce the wave numbers:

~κ = √ 2mE , ~k = 2m(E + V ) , − 0 pertinent to the regimes x > a respectively a

2 φ′′ = k φ x < a , −2 | | φ′′ = κ φ x > a , | |

5 and these are readily solved in the various areas as follows αeκx , x< a ikx ikx − φ(x)= Ae + Be− , aa where we have already taken into account the requirement that the solution should be bounded as x . Imposing φ(x) and φ (x) to be continuous at x = a leads to the → ±∞ ′ ± following matching conditions: αe κa = Ae ika + Beika continuity at x = a : − − − καe κa = ik Ae ika Beika − − − βe κa = Aeika + Be ika continuity at x = a : − − κβe κa = ik Aeika Be ika − − − − Eliminating α from the ﬁrst set of equations and β from the second we obtain two diﬀerent expressions for the fraction B/A leading to the relation: 2 B ik + κ 2ika ik κ 2ika ik + κ 2ika = e = − e− e = 1 , A ik κ ik + κ ⇒ ik κ − − and thus B ik + κ = e2ika = 1 , (9) A ik κ ± − where depending on whether we have the + sign or the sign we will have either one of − the following relations: (+) k tan(ka)= κ resp. ( ) k cot(ka)= κ , (10) − − which follow after some algebra. These relations between k and κ actually determine the allowed values of E (i.e. the eigenvalues of the SE), although these equations being transcendental we cannot give explicit algebraic expressions for the solutions for the eigenvalues E. However, as we shall show below, there is a nice countable discrete of solutions of these equations which we can visualise through a graphical proceduere. Then, having obtained numerical values for the eigenvalues En, and hence also of the corresponding wave numbers kn and κn, the corresponding eigenfunctions are immediately obtained from the relations (8) and (9) to yield:

iknx iknx 2An cos(knx) , (+) φn = An e e− = (11) ± 2iAn sin(knx) , ( ) − where An may be determined by normalisation (one can absorb the i in the second row in the An if necessary). Thus, the solutions going with the (+) are even functions, whilst the ones determined by the ( ) equation of the spectrum are odd3. − To analyse the equations (10) for the spectrum further we note that from the deﬁnition of k and κ we get: 2mV ~2k2 + ~2κ2 = 2m(E + V ) 2mE = 2mV (κa)2 + ξ2 = γ2 0 a2 , 0 − 0 ⇒ ≡ ~2 3Note that in this problem all the diﬃculty stems from the determination of the spectrum through the equations (10), not so much from the eigenfunctions themselves!

6 where we have introduced as new variable ξ = ka. Using the relations for the spectrum, we get then: tan2 ξ ξ = γ cos ξ , (+) ξ2 1+ = γ2 | | cot2 ξ ⇒ ξ = γ sin ξ , ( ) | | − where (as a consequence of (10) noting that κ/k is positive) the equations for ξ should be solved subject to the additional conditions that: (+) tan(ξ) > 0 , ( ) tan(ξ) < 0 . − The solutions can be graphically found from the intersection points of the graphs (in the (ξ, η)-plane) of the line η = ξ/γ and the function η = cos ξ respectively η = sin ξ , as | | | | indicated in the ﬁgure.

0.8

0.6

0.4

0.2

0 2 4 6 8 10 12 x

Although we cannot give analytic expressions for the intersection points indicated in 1 the graph, it is clear that for given value of V0 (which through γ− determines the slope of the line in the figure) we will have a finite number of discrete values for ξ and hence for E. If γ is large enough and, hence, the slope is very small (indeed the slope is of the order of ~!) then approximately the lowest eigenvalues correspond to π2~2n2 ξ nπ/2 E V + ≈ ⇒ n ≈− 0 8ma2 (using the definitions of ξ respectively k).

7 E > 0

In this case we will have, as in Example 2, only scattering states that can be characterised through the reﬂection and transmission coeﬃcients R and T . The solution can be analysed in a similar manner as the next example. We will, therefore not treat this case separately here.

Example 3. The quantum harmonic oscillator

The quantum mechanical system of the one-dimensional harmonic oscillator is one of the simplest nontrivial quantum models. In the classical model the potential is given by 1 2 V (x)= 2 kx leading to the linear equations of motion mx¨ + kx = 0 i.e. the Newton equation for a particle on a spring (following Hooke’s law). Obviously, the classical equation is easily solved in trigonometric funcions, taking k = mω2. The model is of key importance, since it represents the basic ﬁrst step in studying the quantum mechanics of systems of vibrating particles. In principle any smooth potential V (x) in the neighborhood of a local minimum x0 can be approximated by a harmonic oscillator, in view of the Taylor expansion:

1 2 V (x)= V (x )+ V ′′(x )(x x ) + . . . 0 2 0 − 0 so any one-dimensional potential behaves approximately as a harmonic oscillators for low enough energy values. We will study the model here in detail , because it connects quantum mechanics with another interesting area of mathematics: that of special functions and orthogonal polynomials. a) Eigenvalue Problem for the harmonic Oscillator

The classical Hamiltonian of the one-dimensional harmonic oscillator reads p2 1 H(x,p)= + mω2x2 2m 2 which is quantised by taking for the quantum Hamiltonian the form: p2 1 H = + mω2x2 (12) 2m 2 b leading to the eigenvalue problem (i.e.b the stationaryb Schr¨odinger equation): ~2 d2φ 1 + mω2x2φ = Eφ . (13) −2m dx2 2 To analyse this equation it is convenient to go to some dimensionless variables: ~ x = λξ , λ = (14) rmω 8 and rewrite the eigenfunctions as:

1 1 ξ2 φ(x)= e− 2 H(ξ) , (15) √λ where the new function H(ξ) should not be confused with the notation for the Hamiltonian! (It is for historical reasons that we use the symbol H for these functions). Implementing the change of variables we are led to the equation :

d2H dH 2E 2ξ + 1 H = 0 , (16) dξ2 − dξ ~ω − which is a well-known differential equation called the Hermite equation. Note that this is a linear homogeneous second order equation, but the coefficients are not constant (otherwise we could readily solve the equation). If the original eigenfunctions are normalised it is easy to check that through this change of variables the new functions H(ξ) are normalised as well: ∞ 2 ∞ 2 ξ2 φ(x) dx = 1 H(ξ) e− dξ = 1 . (17) | | ⇒ | | Z−∞ Z−∞ It turns out (cf. Appendix D) that the solutions of the Hermite equation (16) that are of physical interest are only those for which the coefficient in the last term takes on integer values, i.e. from which: 2E 1 = 2n , n N , (18) ~ω − ∈ in which case remarkably the solutions are actually polynomials: the so-called Hermite polynomials. Thus, associated with the values of the energy given by (18) we have poly- nomial Hn(ξ) of order n solving the equation:

H′′ 2ξH′ + 2nH = 0 , (19) n − n n in which the primes denote diﬀerentiation w.r.t. the variable ξ.

To summarise our ﬁndings from the above investigation of the the quantum eigenvalue problem associated with the harmonic oscillator:

1 eigenvalues: E = En = (n + )~ω , n = 0, 1, 2,... 2 2 − 1 x e 2 ( λ ) eigenfunctions: φ(x)= φ (x)= 1 2 H (x/λ) n (2n n!√πλ) / n ~ where λ = mω , Hn(ξ) are the standard Hermite polynomials q Thus, we have for the quantum harmonic oscillator a discrete spectrum, where the eigenvalues En represent the (non-degenerate) energy-levels. Note that the spacing between two adjacent levels is always the same (i.e. they are equidistant), namely given by ~ 1 ~ quantum ω, the lowest energy-level being given by E0 = 2 ω. The latter corresponds to the so-called ground state of the model, and the fact that we have a non-zero energy for this ground state is referred to as zero-point energy. We mention also that the oc- currence of discrete eigenvalues of the form given above is in agreement with the basic

9 assumption made in 1905 by A. Einstein to explain the behaviour of speciﬁc heat as a function of temperature in crystals: the vibrational modes of atoms in the crystal can be eﬀectively modelled by quantum harmonic oscillators. The normalised eigenstates (given above, where the Hermite polynomials Hn are described in the next section) form an or- thonormal basis of the Hilbert space of the model. The completeness relation follows from the following property of the Hermite polynomials:

∞ Hn(ξ)Hn(η) 1 2 2 = δ(ξ η)e 2 (ξ +η ) (20) 2nn! √π − n=0 X (see Appendix D).

10 A plot of the probability densities φ (x) 2 for the ﬁrst few energy levels is given in | n | Figures 1-4. 4 2 x 0 0.4 0.2 0.1 0.5 0.3 -2 4 -4 2 x

Figure 1: 0 0.4 0.2 0.1 0.3 -2 -4

Figure 2: 4 2 x 0 0.2 0.1 0.3 0.35 0.25 0.15 0.05 -2 4 -4 2 x

Figure 3: 0 0.2 0.1 0.3 11 0.35 0.25 0.15 0.05 -2 -4

Figure 4: b) Operator Solution ∗ Once we have established the eigenfunctions of the Hamiltonian and its spectrum for the harmonic oscillator we can, in principle, calculate all physically interesting quantities, in particular expectation values and matrix elements of operators. However, in practice, these calculations can become technically diﬃcult requiring the evaluation of inﬁnite sums (namely sums over all eigenvalues). In order to tackle this problem we will now reformulate the quantum harmonic oscillator in terms of some new operators which allows us to deal with these calculations in an algebraic way. Let us introduce the following operators (called ladder operators, or lowering resp. raising operators for reasons that will become clear):

mω 1 mω 1 a = x + ip , a† = x ip , (21) 2~ mω 2~ − mω r r which are eachother’sb Hermitianb conjugateb inb view of the herbmiticityb of x and p: x† = x , p† = p . The commutation relations for these operators are easily checked: b b b b [ x , p ]= i~1 [ a , a† ]= 1 . (22) b b ⇒ We can now write the Hamiltonian in the following suggestive way b b b b b b 1 2 1 2 2 1 H = p + mω x = ~ω(n + ) , where n = a†a (23) 2m 2 2 where the operatorb n isb referred tob as the numberb operator. This operatorb hasb b the following commutation relations with a and a†: b [ n , a ] = [ a†a , a ] = [ a† , a ]a = a b b − [ n , a† ] = [ a†a , a† ]= a†[ a , a† ]= a† b b b b b b b b b which can also be written as: b b b b b b b b b

n a = a (n 1) , n a† = a† (n + 1) (24) − More generally, for k integer, we have: b b b b b b b b b b k k k k n a = a (n k1) , n (a†) = (a†) (n + k1) (25) − or equivalently b b b b bk b k b b k b b k [ n , a ]= ka , [ n , (a†) ]= k(a†) . (26) −

Exercise: Verify theb aboveb commutationb relations.b b b

Now, we make the assumption (justiﬁed by the conclusions of section 1) that there is a distinguished nonzero quantum state, which we denote by φ0(x) , called the vacuum state, which is annihilated by the operator a, i.e.

aφ = 0 normalised as φ = 1 . (27) 0 b k 0k

b 12 (Note: this vacuum state is not the zero vector in the Hilbert space). In fact, the vacuum state can be obtained by solving (27) as a differential equation. From (21) we have 1 ~ dφ mω x + ip φ = 0 + xφ = 0 φ (x)= C exp x2 mω 0 mω dx 0 ⇒ 0 2~ which correspondsb tob the exponential prefactor in (15). Determining C by normalisation subsequently leads to the solution φ0 given in part (a). From the algebra given above, we then have that 1 nφ = 0 Hφ = ~ωφ , 0 ⇒ 0 2 0 using (23). Thus, the vacuum state is an eigenstateb (with eigenvalue 0) of the number operator n, and hence alsob an eigenstate of the Hamiltonian with eigenvalue (energy) 1 ~ E0 = 2 ω. To obtainb the other eigenstates of the Hamiltonian we make the observation that from (25) we have k k k n(a†) φ0 = (a†) (n + k1)φ0 = k(a†) φ0 , k implying that the state (a†) φ0 is an eigenstate of n with integer eigenvalue k, and hence from (23) it is anb eigenstateb ofb theb Hamiltonianb associb ated with the energy level 1 ~ Ek = (k + 2 ) ω . What remainsb to be done is to normaliseb the eigenstate by including a multiplicative factor. Setting now k φk(x)= αk(a†) φ0(x) , to be this normalised eigenstate, i.e. imposing: φk = 1 , we can easily find this b k k multiplicative factor αm. In fact, from the commutation relations (22) it is not difficult to demonstrate the following identity: k k [ a , (a†) ] φ0 = k!φ0 , (e.g. by induction), and then we can assert that: b b 2 k k 2 k k 1 = (φk, φk)= αk ((a†) φ0, (a†) φ0)= αk (φ0, a (a†) φ0) k | k| 2 | | 2 = (φ0 , [ a , (a†) ] φ0)= k! αk (φ0, φ0)= k! αk , b | b | | |b b and thus αk = 1/√k! up to a complex phase factor which we can neglect. Thus, we find that the normalised eigenstatesb b of the quantum harmonic oscillator are given by:

1 k φk(x)= (a†) φ0(x) . (28) √k! Using the commutation relations b k k 1 k k 1 [a, (a†) ]= k(a†) − , [a†, a ]= ka − , − one can obtain recursion formulae following from the action of a and a† on the eigenstates as follows: b b b b b b

1 k 1 k b b aφk = a(a†) φ0 = [ a, (a†) ] φ0 √k! √k!

1 k 1 k (k 1)! b = bk(ba†) − φ0 = b b− φk 1 √k! p√k! −

b 13 whilst 1 k+1 (k + 1)! a†φk = (a†) φ0 = φk+1 . √k! p √k! Thus, we obtain the following formulae: b b

aφk = √k φk 1 − a†φk = √k + 1 φk+1 b b Thus, it has now become clear why we call a† and a a raising- resp. lowering operator: they are the operators that (mathematically) bring you from one energy state into the next higher (resp. lower) energy state! b b

One merit of the operator description is that it facilitates the calculation of “physical” quantities such as expectation values. For this we need to consider the matrix elements of the operators a and a†, which are easily calculated from the above formulae:

(φk, aφl) = √l(φk, φl 1)= √l δk,l 1 (29a) b b − − (φk, a†φl) = √l + 1(φk, φl+1)= √l + 1 δk,l+1 (29b) b and from these we can calculate the matrix elements of x and p by inverting the relations (21), i.e. b ~ bmω~ b x = (a + a†) , p = i (a† a) . (30) r2mω r 2 − We can actually present the matrix elements X k x l and P k p l (k, l = kl ≡ h | | i kl ≡ h | | i 0, 1, 2,... ) of the positionb and momentumb b operatorsb as semi-inﬁniteb b matrices as follows: 0 √1 0 0 . . . b 0 √1b 0 0 . . . − √1 0 √2 0 . . . √1 0 √2 0 . . . ~   mω~  −  X = 0 √2 0 √3 . . . , P = i 0 √2 0 √3 . . . . 2mω  √  2  √ −  r  0 0 3 0 . . .  r  0 0 3 0 . . .   . . . . .   . . . . .   ......   ......         (31)  Expectation values of the observables x and p and its powers can now be simply calculated by using combinations of these semi-inﬁnite matrices, which multiply according to the rule:

∞ (XP )kl = XkjPjl (k, l = 0, 1, 2,... ) . Xj=0

Example: To calculate the expectation values x and p in the eigenstate k we need h i h i | i only to ﬁnd the corresponding diagonal elements of the matrices X resp. P : x = X =0 and p = P = 0 h i kk h i kk whilst the spreading is calculated from ~ mω~ x2 = (X2) = (2k + 1) and p2 = (P 2) = (2k + 1) . h i kk 2mω h i kk 2