Position and Momentum Uncertainties of the Normal and Inverted Harmonic Oscillators under the Minimal Length Uncertainty Relation
aa r X i v : . [ h e p - t h ] S e p Position and Momentum Uncertainties of the Normal and Inverted HarmonicOscillators under the Minimal Length Uncertainty Relation
Zachary Lewis ∗ and Tatsu Takeuchi † Department of Physics, Virginia Tech, Blacksburg, VA 24061, USA (Dated: September 11, 2011)We analyze the position and momentum uncertainties of the energy eigenstates of the harmonicoscillator in the context of a deformed quantum mechanics, namely, that in which the commutatorbetween the position and momentum operators is given by [ˆ x, ˆ p ] = i ~ (1 + β ˆ p ). This deformedcommutation relation leads to the minimal length uncertainty relation ∆ x ≥ ( ~ / / ∆ p + β ∆ p ),which implies that ∆ x ∼ / ∆ p at small ∆ p while ∆ x ∼ ∆ p at large ∆ p . We find that theuncertainties of the energy eigenstates of the normal harmonic oscillator ( m > x ∼ / ∆ p branch. The other branch, ∆ x ∼ ∆ p , is found to be populatedby the energy eigenstates of the ‘inverted’ harmonic oscillator ( m < x min = ~ √ β > √ ~ /k | m | ] / . Correspondence with the classical limit is also discussed. PACS numbers: 03.65.-w,03.65.Ge
I. INTRODUCTION
The consequences of deforming the canonical commu-tation relation between the position and momentum op-erators to [ ˆ x, ˆ p ] = i ~ (1 + β ˆ p ) (1)have been studied in various contexts by many authors[1–8]. The main motivation behind this was to use suchdeformed quantum mechanical systems as models whichobey the minimal length uncertainty relation (MLUR)[9]: ∆ x ≥ ~ (cid:18) p + β ∆ p (cid:19) . (2)This relation is expected on fairly generic grounds inquantum gravity [10, 11], and has been observed in per-turbative string theory [12]. The MLUR implies the ex-istence of a minimal length∆ x min = ~ p β , (3)below which the uncertainty in position, ∆ x , cannotbe reduced. For quantum gravity, ∆ x min would be thePlanck scale, ℓ P = p ~ G N /c , while for string theory thiswould be the string length scale, ℓ s = √ α ′ . The investi-gation of said model systems could be expected to shedsome light on the nature of these, and other, theorieswhich possess a minimal length scale.Note that the uncertainty in position which saturatesthe MLUR bound behaves as ∆ x ∼ / ∆ p for ∆ p < / √ β , while ∆ x ∼ ∆ p for 1 / √ β < ∆ p , as illustrated ∗ Electronic address: [email protected] † Electronic address: [email protected] (cid:144) Β D p Ñ ΒD x FIG. 1: The lower bound on ∆ x as a function of ∆ p whenthey obey the minimal length uncertainty relation, Eq. (2),is shown in red. The blue line indicates ∆ x = ( ~ / / ∆ p )while the green line indicates ∆ x = ( ~ β/ p . in Fig. 1. While we are familiar with the ∆ x ∼ / ∆ p be-havior from canonical quantum mechanics, the ∆ x ∼ ∆ p behavior is quite novel. It behooves us to understandhow it can come about, and how it can coexist with thecanonical ∆ x ∼ / ∆ p behavior within a single quantummechanical system. To this end, we calculate ∆ x and ∆ p for the energy eigenstates of the harmonic oscillator,ˆ H = 12 k ˆ x + 12 m ˆ p , (4)the wave-functions of which were derived explicitly inRef. [1]. We find that all of the eigenstates of thisHamiltonian inhabit the ∆ x ∼ / ∆ p branch, and notthe other branch, as long as both the spring constant k and the mass m remain positive. The cross-over happenswhen the inverse of the mass, 1 /m , is allowed to decreasethrough zero into the negative, in which case all the en-ergy eigenstates will move smoothly over to the ∆ x ∼ ∆ p branch. The objective of this paper is to provide a de-tailed account of this result.In the following, we solve the Schr¨odinger equation forthe above Hamiltonian without assuming a specific signfor the mass m . The spring constant k is kept positivethroughout. We find that the ‘inverted’ harmonic oscil-lator with k > m < x min = ~ p β > √ a , (5)where a = (cid:20) ~ k | m | (cid:21) / (6)is the characteristic length scale of the harmonic oscilla-tor. The uncertainties ∆ x and ∆ p are calculated for theenergy eigenstates, and the above mentioned cross-overthrough 1 /m = 0 from the ∆ x ∼ / ∆ p branch to the∆ x ∼ ∆ p branch is demonstrated.We then take the classical limit of our deformed com-mutation relation and work out the evolution of the clas-sical harmonic oscillator for both the positive and nega-tive mass cases. It is found that for the ‘inverted’ m < x = −∞ to x = + ∞ is finite, demanding the compactifi-cation of x -space, and also rendering the classical prob-ability of finding the particle near the origin finite. Thisprovides a classical explanation of why ‘bound’ statesare possible for the ‘inverted’ harmonic oscillator in thismodified mechanics. II. QUANTUM STATES AND UNCERTAINTIESA. Eigenvalues and Eigenstates
The position and momentum operators obeying Eq. (1)can be represented in momentum space by [13]ˆ x = i ~ (1 + βp ) ∂∂p , ˆ p = p . (7)The inner product between two states is h φ | ψ i = Z ∞−∞ dp (1 + βp ) φ ∗ ( p ) ψ ( p ) . (8)This definition ensures the symmetricity of the operatorˆ x . The Schr¨odinger equation for the harmonic oscillatorin this representation is thus " − ~ k (cid:26) (1 + βp ) ∂∂p (cid:27) + 12 m p Ψ( p ) = E Ψ( p ) . (9)Here, we do not assume m > p to ρ ≡ √ β arctan( p βp ) , (10) maps the region −∞ < p < ∞ to − π √ β < ρ < π √ β , (11)and casts the ˆ x and ˆ p operators into the formsˆ x = i ~ ∂∂ρ , ˆ p = 1 √ β tan( p βρ ) , (12)with inner product given by h φ | ψ i = Z π/ √ β − π/ √ β dρ φ ∗ ( ρ ) ψ ( ρ ) . (13)Note that ˆ x is the wave-number operator in ρ -space, sothe Fourier coefficients of the wave-function in ρ -spacewill provide the probability amplitudes for a discretized x -space. Eq. (9) is thus transformed into: " − ~ k ∂ ∂ρ + 12 mβ tan p βρ Ψ( ρ ) = E Ψ( ρ ) . (14)In ρ -space, the potential energy term k ˆ x / p / m effectively becomes a tangent-squared potentialwhich is ‘inverted’ when 1 /m <
0. We next introducedimensionless parameters and a dimensionless variableby κ ≡ (cid:2) β ~ k | m | (cid:3) / = ∆ x min a ,ε ≡ | m | Eβκ ,ξ ≡ √ βρκ , (15)where the length-scale a was introduced in Eq. (6). Thedimensionless variable ξ is in the range − π κ < ξ < π κ , (16)and the inner product is h φ | ψ i = κ √ β Z π/ κ − π/ κ dξ φ ∗ ( ξ ) ψ ( ξ ) . (17)The dimension of the inner product has all been absorbedinto the prefactor 1 / √ β . The Schr¨odinger equation be-comes (cid:20) ∂ ∂ξ ∓ κ tan κξ + ε (cid:21) Ψ( ξ ) = 0 , (18)where the minus sign in front of the tangent-squared po-tential is for the case m >
0, and the plus sign for thecase m <
0. Let Ψ( ξ ) = c λ f ( s ), where s ≡ sin κξ , Λ + Λ - Κ Λ FIG. 2: Plots of λ + (blue) and λ − (red), Eq. (23), as functionsof κ = β ~ p k | m | . c ≡ cos κξ = √ − s , and λ is a constant to be de-termined. The variable s is in the range − < s < , (19)with inner product given by h φ | ψ i = 1 √ β Z − dsc φ ∗ ( s ) ψ ( s ) . (20)The equation for f ( s ) is(1 − s ) f ′′ − (2 λ + 1) s f ′ + (cid:20)(cid:26) εκ − λ (cid:27) + (cid:26) λ ( λ − ∓ κ (cid:27) s c (cid:21) f = 0 . (21)We fix λ by requiring the coefficient of the tangentsquared term to vanish: λ ( λ − ∓ κ = 0 . (22)The solutions are λ =
12 + r
14 + 1 κ ≡ λ + ( m > ,
12 + r − κ ≡ λ − ( m < , (23)where we have chosen the branches for which λ ≥ / < λ + while ≤ λ − <
1, and that λ ± → κ = β ~ p k | m | →∞ . The dependence of λ ± on κ is shown in Fig. 2. The λ − branch does not extend below κ = 2. Setting λ = λ + for m >
0, and λ = λ − for m < − s ) f ′′ − ( 2 λ + 1 ) s f ′ + (cid:16) εκ − λ (cid:17) f = 0 , (24)the sign of m being encoded in the value of λ . Since f ( s )should be non-singular at s = ±
1, we demand a polyno-mial solution to Eq. (24). This requirement imposes the following condition on the coefficient of f : εκ − λ = n ( n + 2 λ ) , (25)where n is a non–negative integer [14]. Eq. (24) becomes(1 − s ) f ′′ − ( 2 λ + 1 ) s f ′ + n ( n + 2 λ ) f = 0 , (26)the solution of which is given by the Gegenbauer polyno-mial: f ( s ) = C λn ( s ) . (27)The Gegenbauer polynomials satisfy the following or-thogonality relation: Z − c λ − C λn ( s ) C λm ( s ) ds = 2 π Γ( n + 2 λ )[ 2 λ Γ( λ ) ] n ! ( n + λ ) δ nm . (28)The energy eigenvalues follow from the conditionEq. (25). Replacing λ with λ ± , we find ε ( ± ) n = κ (cid:2) n + (2 n + 1) λ ± (cid:3) = n + (2 n + 1) λ ± q λ ± (cid:12)(cid:12) λ ± − (cid:12)(cid:12) = κ (cid:18) n + n + 12 (cid:19) + (2 n + 1) r κ ± , (29)or in the original dimensionful units, E ( ± ) n = 12 β | m | n + (2 n + 1) λ ± λ ± (cid:12)(cid:12) λ ± − (cid:12)(cid:12) = ~ ω (cid:20)(cid:18) n + 12 (cid:19) r β m ~ ω ± (cid:18) n + n + 12 (cid:19) β | m | ~ ω (cid:21) = k (cid:20)(cid:18) n + 12 (cid:19) p (∆ x min ) ± a + (cid:18) n + n + 12 (cid:19) (∆ x min ) (cid:21) , (30)where ω = p k/ | m | . For the m > x min = ~ √ β →
0, and we recoverlim ∆ x min → E (+) n = ka (cid:18) n + 12 (cid:19) = ~ ω (cid:18) n + 12 (cid:19) . (31)For the m < x min ≥√ a for the square-root in Eq. (30) to remain real. Thisis the condition we cited in Eq. (5). Therefore, the limit∆ x min = ~ √ β → a . The two cases converge when | m | → ∞ , at which a = 0, and we find that the energy levels in that limitare lim | m |→∞ E ( ± ) n = k x min ) ( n + 1) . (32)Thus, the 1 /m > /m < /m = 0.The normalized energy eigenfunctions are thus givenby: Ψ ( λ ) n ( p ) = N ( λ ) n c λ C λn ( s ) , (33)where N ( λ ) n = p β " λ Γ( λ ) s n ! ( n + λ )2 π Γ( n + 2 λ ) ,c = cos p βρ = 1 p βp ,s = sin p βρ = √ βp p βp . (34)The wave-functions for the first few energy eigenstates forseveral representative values of λ are shown in Figs. 3. B. Expectation Values and Uncertainties
Using the wave-functions derived above, and the for-mula provided in the appendix, the expectation values ofˆ x , ˆ p , ˆ x , and ˆ p for the energy eigenstates are found tobe h n, λ | ˆ x | n, λ i = 0 , h n, λ | ˆ p | n, λ i = 0 , h n, λ | ˆ x | n, λ i = ( ~ β ) ( λ + n ) (cid:2) (2 λ − n + λ (cid:3) (2 λ − , h n, λ | ˆ p | n, λ i = 1 β (cid:18) n + 12 λ − (cid:19) , (35)giving the uncertainties in h ˆ x i and h ˆ p i as∆ x n = ∆ x min s ( λ + n ) (cid:2) (2 λ − n + λ (cid:3) (2 λ − , ∆ p n = 1 √ β r n + 12 λ − . (36)For fixed n and fixed β , ∆ x n is a monotonically decreas-ing function of λ in the range < λ <
1, and a mono-tonically increasing one in the range 1 < λ . ∆ p n on theother hand is a monotonically decreasing function of λ throughout. Eliminating λ from the above expressions,we find∆ x n ∆ x min = 12 √ β ∆ p n rh β ∆ p n ih (2 n + 1) + β ∆ p n i ≥ (cid:18) √ β ∆ p n + p β ∆ p n (cid:19) , (37)where the equality in the second line is saturated for the n = 0 case only. The first line gives the curve on the∆ p -∆ x plane that the point (∆ p n , ∆ x n ) follows as λ isvaried. Differentiating with respect to ∆ p n we find dd (∆ p n ) (cid:20) ∆ x n ∆ x min (cid:21) Λ= Λ= Λ= Λ= (cid:144) - Π - Π Π Π Β Ρ- - Y (cid:144) Β Λ= Λ= Λ= Λ= (cid:144) - Π - Π Π Π Β Ρ- - Y (cid:144) Β Λ= Λ= Λ= Λ= (cid:144) - Π - Π Π Π Β Ρ- - Y (cid:144) Β FIG. 3: λ -dependence of the wave-functions of the first fewenergy eigenstates. The λ > m > ≤ λ < m <
0. The λ = 1case corresponds to the limit | m | → ∞ . = β ∆ p n − (2 n + 1) √ β ∆ p n rh β ∆ p n ih (2 n + 1) + β ∆ p n i , (38)indicating that the curve is flat at the λ = 1 point where∆ p n = p (2 n + 1) /β and ∆ x n reaches its minimum of∆ x min ( n + 1). Therefore, the λ = 1 point is the turn-around point where the uncertainties switch from the∆ x ∼ / ∆ p behavior to the ∆ x ∼ ∆ p behavior. Togo from one branch to another one must flip the sign ofthe mass m .Eliminating n from Eq. (36), we find∆ x n ∆ x min = 12 r (1 + β ∆ p n ) h λ − β ∆ p n i . (39) n = = = = = = Λ = Λ = Λ = Λ = . Λ = . Λ = . m > m <
00 2 4 6 8 10 12 D p024681012 D x FIG. 4: ∆ p versus ∆ x for the lowest six energy eigenstatesof the harmonic oscillator. ∆ x is in units of ∆ x min = ~ √ β ,while ∆ p is in units of 1 / √ β = ~ / ∆ x min . The location of thestate along the curves shown is determined by the value of λ defined in Eq. (23). The λ = 1 points correspond to the case1 /m = 0. As 1 /m is increased to the positive side, the value of λ will increase away from one and the state will move towardthe left along the ∆ x ∼ / ∆ p branch. If 1 /m is decreased intothe negative, the value of λ will decrease toward 1 /
2, and thestate will move toward the right along the ∆ x ∼ ∆ p branch.The n = 0 curve (shown in red) corresponds to the minimallength uncertainty bound, Eq. (2). This gives the curve on which the points (∆ p n , ∆ x n ) fallon for constant λ . In particular, for λ = 1 this reducesto ∆ x n ∆ x min = 1 + β ∆ p n , (40)and gives the 1 /m = 0 boundary between the 1 /m > /m < x -∆ p space. These propertiesof the uncertainties have been plotted in Fig. 4. C. Limiting Cases
As shown in Fig. 4, the value of λ determines wherethe uncertainties (∆ p n , ∆ x n ) are along their trajectoriesgiven by Eq. (37), with λ > x ∼ / ∆ p branch of the trajectory, while <λ < x ∼ ∆ p branch. Let usconsider a few limiting values of λ to see the behavior ofthe solutions there. λ → ∞ The β → m > λ = λ + . As β →
0, the parameter λ + diverges to infinityas λ = λ + ∼ m ~ ωβ β → −−−→ ∞ , (41)where ω = p k/m . In that limit, the Gegenbauer poly-nomials become Hermite polynomials:lim λ →∞ n ! λ − n/ C λn ( x/ √ λ ) = H n ( x ) . (42)Noting that as β →
0, we have s ∼ p βp ∼ p √ λm ~ ω , (43)we can conclude thatlim λ →∞ n ! λ − n/ C λn ( s ) = H n (cid:18) p √ m ~ ω (cid:19) . (44)Similarly,lim λ →∞ c λ = lim λ →∞ (cid:18) − p λm ~ ω (cid:19) λ/ = exp (cid:18) − p m ~ ω (cid:19) . (45)Using Stirling’s formulaΓ( z ) ∼ √ π e − z z z − (1 / + · · · (46)the normalization constant can be shown to converge tolim λ →∞ N ( λ ) n λ n/ n ! = 1 √ n n ! 1 √ m ~ ωπ . (47)Therefore,lim λ →∞ Ψ ( λ ) n ( p )= 1 √ n n ! 1 √ m ~ ωπ exp (cid:18) − p m ~ ω (cid:19) H n (cid:18) p √ m ~ ω (cid:19) , (48)which are just the usual harmonic oscillator wave-functions in momentum space. The energy eigenvaluesreduce to the usual ones given in Eq. (31). The uncer-tainties reduce to the usual ones as well∆ x n λ →∞ −−−−→ s ~ mω (cid:18) n + 12 (cid:19) , ∆ p n λ →∞ −−−−→ s ~ mω (cid:18) n + 12 (cid:19) , (49)which satisfy∆ x n ∆ p n λ →∞ −−−−→ ~ (cid:18) n + 12 (cid:19) ≥ ~ . (50) λ → The limit λ = 1 is reached when β and k are keptconstant while | m | is taken to infinity. When λ = 1, theGegenbauer polynomials become the Chebycheff polyno-mials of the second kind: C n ( s ) = U n ( s ) , (51)where U n (cos θ ) = sin( n + 1) θ sin θ , (52)while the normalization constant reduces to N ( λ ) n λ → −−−→ p β r π . (53)Therefore, Ψ (1) n ( p ) √ β = r π c U n ( s ) . (54)The orthonormality relation for the Chebycheff polyno-mials is Z − p − x U n ( x ) U m ( x ) dx = π δ nm , (55)and we can see that the correct normalization constant isobtained. Since the argument in our case is s = sin √ βρ ,it is more convenient to express the Chebycheff polyno-mials as U n (sin θ ) = ( − n cos[(2 n + 1) θ ]cos θ ,U n +1 (sin θ ) = ( − n sin[(2 n + 2) θ ]cos θ , (56)for n = 0 , , , · · · . This will allow us to writeΨ (1)2 n ( p ) √ β = ( − n r π cos h (2 n + 1) √ βρ i , Ψ (1)2 n +1 ( p ) √ β = ( − n r π sin h (2 n + 2) √ βρ i . (57)The energy eigenvalues in this limit were given inEq. (32). Here, our procedure of keeping the spring con-stant k fixed while taking | m | to infinity maintains thefiniteness of E n , while taking the kinetic energy contribu-tion to E n to zero. From the n -dependence of the ener-gies, we can see that, in this limit, the problem reduces tothat of an infinite square well potential, of width π/ √ β ,in ρ -space. Indeed, the effective potential in ρ -space was12 mβ tan (cid:0)p βρ (cid:1) m →∞ −−−−→ ρ = ± π √ β , ∞ at ρ = ± π √ β . (58) This can also be seen from the form of the energy eigen-functions, which have reduced to simple sines and cosines.We will see in the next section that the classical solutionalso behaves as that of a particle in an infinite squarewell potential in ρ -space in the same limit.The uncertainties become∆ x n λ → −−−→ ∆ x min (cid:0) n + 1 (cid:1) , ∆ p n λ → −−−→ s (2 n + 1) β , (59)as was shown in Fig. 4. Note that ~ (cid:20) p n + β ∆ p n (cid:21) = ∆ x min n + 1 √ n + 1 ≤ ∆ x min (cid:0) n + 1 (cid:1) = ∆ x n , (60)the bound being saturated only for the ground state n =0. λ → The λ → limit is reached as ∆ x min → √ a when m <
0. In this limit, the Gegenbauer polynomials be-come the Legendre polynomials, C / n ( s ) = P n ( s ), whilethe normalization constant reduces to N ( λ ) n λ → −−−→ p β r n + 12 . (61)The wavefunctions areΨ (1 / n ( p ) √ β = r n + 12 √ c P n ( s ) . (62)Note that the orthonormality relation for the Legendrepolynomials is Z − P n ( x ) P m ( x ) dx = 22 n + 1 δ nm , (63)so these wave-functions are properly normalized. The in-tegrals for h ˆ x i and h ˆ p i diverge in this limit, so both ∆ x n and ∆ p n are divergent for all n . However, the energy,which is the difference between k h ˆ x i / h ˆ p i / | m | ,stays finite: E − n = k x min ) (cid:18) n + n + 12 (cid:19) . (64) III. CLASSICAL STATES ANDUNCERTAINTIES
As we have seen above, for values of β which maintainthe inequality ∆ x min = ~ √ β > √ a , the harmonic os-cillator Hamiltonian admits an infinite ladder of positiveenergy eigenstates even when m <
0. Furthermore, theseare states with finite ∆ x and ∆ p , implying that the par-ticle is ‘bound’ close to the phase space origin, just as inthe m > A. The Classical Equations of Motion
We assume that the classical limit of our commutationrelation, Eq. (1), is obtained by the usual correspondencebetween commutators and Poisson brackets:1 i ~ [ ˆ A, ˆ B ] → { A, B } . (65)Therefore, we assume { x, x } = 0 , { p, p } = 0 , { x, p } = (1 + βp ) . (66)Then, the equations of motion for the harmonic oscillatorwith Hamiltonian given by H = 12 kx + 12 m p , (67)are ˙ x = { x, H } = 1 m (1 + βp ) p , ˙ p = { p, H } = − k (1 + βp ) x . (68)We allow m to take on either sign: if m >
0, then ˙ x and p will have the same sign; if m < x and p have changed, the total energy willstill be conserved. Consequently, the time-evolution of x and p in phase space will be along the trajectory givenby H = constant. For the m > m < p to ρ ,which was introduced in Eq. (10) for the quantum case.Then, the equations become˙ x = 1 m √ β (cid:20) tan( √ βρ )cos ( √ βρ ) (cid:21) = 12 mβ ddρ h tan ( p βρ ) i , ˙ ρ = − kx . (69)Therefore,¨ ρ = − k ˙ x = − k mβ ddρ h tan ( p βρ ) i , (70)which integrates to˙ ρ = − kmβ (cid:20) tan (cid:0)p βρ (cid:1) − C (cid:21) , (71)where C is the integration constant. Since we must have˙ ρ >
0, the range of allowed values of C will depend onwhether m > m <
0. We will consider the two casesseparately. Ω t (cid:144) Π- Π - Π Π Π Β Ρ H t L Ω t (cid:144) Π- - p H t L(cid:144) m Ω A Ω t (cid:144) Π- - x H t L(cid:144) A FIG. 5: The dependence of the classical behavior of a pos-itive mass particle in a harmonic oscillator potential on theparameter C = 2 mEβ = βm ω A , where E is the particle’senergy, and A is the amplitude of the oscillation in x . Theundeformed β = 0 case is shown in red. The other four casesare C = 1 (orange), C = 2 (green), C = 4 (blue), and C = 8(purple). B. m > When m >
0, we introduce the angular frequency ω = r km (72)as usual. Then, Eq. (71) becomes β ˙ ρ = ω (cid:20) C − tan (cid:0)p βρ (cid:1)(cid:21) , (73)and taking the square-root, we obtain p β ˙ ρ = ± ω q C − tan ( p βρ ) . (74)In this case, we must have C > Ω t (cid:144) Π- Π - Π Π Π Β Ρ H t L Ω t (cid:144) Π- - - Β p H t L Ω t (cid:144) Π- - x H t L(cid:144) A FIG. 6: The classical behavior of a positive mass particle in aharmonic oscillator potential with modified Poisson brackets,Eq. (66), in the limit m → ∞ with C = 2 mEβ where E and β are kept fixed. ρ ( t ) and x ( t ) take on the behavior ofthe position and momentum of a particle in an infinite squarewell. obtain √ β dρ q C − tan (cid:0) √ βρ (cid:1) = ± ω dt . (75)The left-hand side integrates to Z √ β dρ p C − tan ( √ βρ )= 1 √ C arcsin "r CC sin (cid:0)p βρ (cid:1) . (76)Therefore, p βρ ( t )= arcsin "r C C sin n ±√ C ω ( t − t ) o (77)where t is the integration constant. Without loss of gen-erality, we can choose the sign inside the curly bracketsto be plus. Setting the clock so that t = 0, we obtain: p βmk x ( t ) = − √ βω ˙ ρ = − p C (1 + C ) cos (cid:0) √ C ωt (cid:1)q C cos (cid:0) √ C ωt (cid:1) , p β p ( t ) = tan( p βρ )= √ C sin (cid:0) √ C ωt (cid:1)q C cos (cid:0) √ C ωt (cid:1) , (78)and the energy is given by E = k x ( t ) + p ( t ) m = 12 βm hp βmk x ( t ) i + 12 βm hp β p ( t ) i = C βm > . (79)The period of oscillation T is no longer equal to 2 π/ω when β = 0. It is now T = 2 πω √ C . (80)Let A ≡ s Cβmk . (81)Then E = 12 kA , (82)and we can identify A as the oscillation amplitude in x .If we take the limit β → A fixed, we find: x ( t ) β → −−−→ − A cos( ωt ) ,p ( t ) β → −−−→ Amω sin( ωt ) , (83)which shows that the canonical behavior is recovered inthis limit. The behavior of the solution when β = 0 iscompared with the β = 0 limit for several representativevalues of C in Fig. 5.Another interesting limit is obtained by setting C =2 mEβ and letting m → ∞ while keeping E and β fixed.In that limit, √ C ω m →∞ −−−−→ p Eβk = p βkA ≡ ω , (84)and we find p β ρ ( t ) m →∞ −−−−→ arcsin h sin( ω t ) i , p β p ( t ) m →∞ −−−−→ + sin( ω t ) | cos( ω t ) | ,x ( t ) /A m →∞ −−−−→ − cos( ω t ) | cos( ω t ) | . (85)The behavior of the solution in this limit is shown inFig. 6. The motion of a particle in an infinite square wellpotential (in ρ -space) is reproduced, in correspondenceto the quantum λ → C. m < For the m < ω = s k | m | . (86)Then, Eq. (71) becomes β ˙ ρ = ω (cid:20) tan (cid:0)p βρ (cid:1) − C (cid:21) . (87)The integration constant C can have either sign in thiscase. We will consider the three cases C < C >
0, and C = 0 separately. C < (positive energy) case For the
C < p β ˙ ρ = ± ω q tan ( p βρ ) + | C | . (88)Therefore, √ β dρ q tan (cid:0) √ βρ (cid:1) + | C | = ± ω dt . (89)The left-hand side integrates to Z √ β dρ p tan ( √ βρ ) + | C | = p | C | − "s | C | − | C | sin (cid:0) √ βρ (cid:1) | C | > (cid:1) sin (cid:0) √ βρ (cid:1) (cid:0) | C | = 1 (cid:1) p − | C | arcsinh "s − | C || C | sin (cid:0) √ βρ (cid:1) | C | < (cid:1) (90)Therefore,sin (cid:0)p βρ ( t ) (cid:1) = s | C || C | − h ± p | C | − ω ( t − t ) i (cid:0) | C | > (cid:1) ± ω ( t − t ) (cid:0) | C | = 1 (cid:1)s | C | − | C | sinh h ± p − | C | ω ( t − t ) i (cid:0) | C | < (cid:1) (91)where t is the integration constant, which we will set tozero in the following. From this, we find: p β | m | k x ( t ) = − √ βω ˙ ρ - - - Ω t - Π - Π Π Π Β Ρ H t L - - - Ω t - - p H t L (cid:144) m ¤ k B - - - Ω t - - - - - x H t L(cid:144) B FIG. 7: The dependence of the classical behavior of a neg-ative mass particle in a harmonic oscillator potential on theparameter C = − µEβ = − β | m | kB , where E is the parti-cle’s energy, and B is the distance of closest approach to theorigin in x -space. Note that due to the negative mass, p ( t )is negative when ˙ x ( t ) is positive, and vice versa. The unde-formed β = 0 case is shown in red. The other four cases are C = − (orange), C = − (green), C = − C = − = ∓ p | C | ( | C | −
1) cos (cid:0) ± p | C | − ω t (cid:1)q | C | cos (cid:0) ± p | C | − ω t (cid:1) − (cid:0) | C | > (cid:1) ∓ p − ( ω t ) (cid:0) | C | = 1 (cid:1) ∓ p | C | (1 − | C | ) cosh (cid:0) ± p − | C | ω t (cid:1)q − | C | cosh (cid:0) ± p − | C | ω t (cid:1) (cid:0) | C | < (cid:1) p β p ( t ) = tan( p βρ )= p | C | sin (cid:0) ± p | C | − ω t (cid:1)q | C | cos (cid:0) ± p | C | − ω t (cid:1) − (cid:0) | C | > (cid:1) ± ω t p − ( ω t ) (cid:0) | C | = 1 (cid:1)p | C | sinh (cid:0) ± p − | C | ω t (cid:1)q − | C | cosh (cid:0) ± p − | C | ω t (cid:1) (cid:0) | C | < (cid:1) (92)In all three cases, we have E = k x ( t ) − p ( t ) | m | = 12 β | m | hp β | m | k x ( t ) i − β | m | hp β p ( t ) i = | C | β | m | > . (93)Let B ≡ s | C | β | m | k . (94)Then E = 12 kB , (95)and we can identify B as the distance of closest approachto the origin (aka impact parameter). Taking the limit β → B fixed, we find: x ( t ) β → −−−→ ∓ B cosh( ± ωt ) ,p ( t ) β → −−−→ B | m | ω sinh( ± ωt ) , (96)which recovers the canonical solution. This behavior of x ( t ) and p ( t ) for the β = 0 case is compared with that inthe β = 0 case for several representative values of C inFig. 7.It should be noted that for any finite value of C < x, p ) = ( ±∞ , ±∞ ) to ( x, p ) = ( ±∞ , ∓∞ ), orequivalently, for √ βρ to evolve from ∓ π/ ± π/
2. Wewill call this time
T / T is given by: T = 4 ω × p | C | − p | C | (cid:0) | C | > (cid:1) (cid:0) | C | = 1 (cid:1) p − | C | arccosh 1 p | C | (cid:0) | C | < (cid:1) (97)This dependence on C < - - - Ω t - Π - Π Π Π Β Ρ H t L - - - Ω t - - - - - p H t L(cid:144) p min - - - Ω t - - m ¤ k x H t L(cid:144) p min FIG. 8: The dependence of the classical behavior of a neg-ative mass particle in a harmonic oscillator potential on theparameter C = − | m | Eβ = βp , where E is the particle’senergy, and p min is the momentum of the particle at the origin x = 0. The undeformed β = 0 case is shown in red. The otherfour cases are C = + (orange), C = + (green), C = +1(blue), and C = +4 (purple). C > (negative energy) case For the
C > p β ˙ ρ = ± ω q tan (cid:0)p βρ (cid:1) − C . (98)1Therefore, √ β dρ q tan (cid:0) √ βρ (cid:1) − C = ± ω dt . (99)The left-hand side integrates to Z √ β dρ p tan ( √ βρ ) − C = 1 √ C arccosh (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)r CC sin (cid:0)p βρ (cid:1)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (100)Therefore,sin (cid:0)p βρ ( t ) (cid:1) = ± r C C cosh h ±√ C ω ( t − t ) i (101)where t is the integration constant, which we will setto zero in the following. The sign on the argument ofthe hyperbolic cosine is also irrelevant so we will set it toplus. From this, we find: p β | m | k x ( t ) = − √ βω ˙ ρ = ∓ p C (1 + C ) sinh (cid:0) √ C ω t (cid:1)q − C sinh (cid:0) √ C ω t (cid:1)p β p ( t ) = tan( p βρ )= ± √ C cosh (cid:0) √ C ω t (cid:1)q − C sinh (cid:0) √ C ω t (cid:1) (102)and E = k x ( t ) − p ( t ) | m | = 12 β | m | hp β | m | k x ( t ) i − β | m | hp β p ( t ) i E > E < - - - C Ω T FIG. 9: ωT as a function of − C = 2 | m | Eβ , where ω = p k/ | m | , and E is the particle’s energy. T / = − C β | m | < . (103)Let p min ≡ s Cβ . (104)Then E = − p | m | , (105)and we can identify p min as the magnitude of the momen-tum that the particle has at the origin x = 0. Taking thelimit β → p min constant, we find x ( t ) β → −−−→ ∓ p min | m | ω sinh( ± ωt ) ,p ( t ) β → −−−→ ± p min cosh( ± ωt ) . (106)As in the C <
0, the time
T / x = ∓∞ to x = ±∞ is finite. T is given by: T = 4 ω √ C arcsinh 1 √ C . (107)This dependence on
C > C = 0 (zero energy) case For C = 0, Eq. (87) leads to p β ˙ ρ = ± ω tan (cid:0)p βρ (cid:1) , (108)or √ β dρ tan (cid:0) √ βρ (cid:1) = ± ω dt , (109)which can be integrated easily to yieldln (cid:12)(cid:12)(cid:12) sin (cid:0)p βρ (cid:1)(cid:12)(cid:12)(cid:12) = ± ω ( t − t ) , (110)or sin (cid:0)p βρ (cid:1) = ± e ± ω ( t − t ) , (111)with all combinations of signs allowed. Set the clock sothat t = 0. The solutions for the t > p β | m | k x ( t ) = − √ βω ˙ ρ = ± √ e ωt − , √ β p ( t ) = tan (cid:0) √ βρ (cid:1) = ± √ e ωt − . (112)The particle starts out at ( x, p ) = ( ±∞ , ±∞ ) and asymp-totically approaches the origin, taking an infinite amountof time to get there. This behavior is show in Fig. 10.2 Ω t - Π - Π Π Π Β Ρ H t L Ω t - - - Β p H t L Ω t - - - Β m ¤ k x H t L FIG. 10: The classical behavior of a zero-energy particle withnegative mass in a harmonic oscillator potential. The particlestarts out at x = −∞ at time t = 0 and asymptoticallyapproaches the origin. D. Compactification
As we have seen, when m < β = 0, it only takesa finite amount of time for the particle to traverse theentire classical trajectory as long as the energy of theparticle is non-zero. This means that we must specifywhat happens to the particle after it reaches infinity. Forthis, we could either compactify x -space so that the par-ticle which reaches x = ±∞ will return from x = ∓∞ ,in which case the momentum of the particle will bounceback from infinite walls at p = ±∞ , or we could compact-ify p -space so that the particle which reaches p = ±∞ willreturn from p = ∓∞ , in which case the position of theparticle will bounce back from infinite walls at x = ±∞ .Here, we choose to compactify x -space so that the | m | → ∞ limit of the m < C < m → ∞ limit of the m > C > - - Ω t - Π - Π Π Π Β Ρ H t L - - Ω t - - Β p H t L - - Ω t - - x H t L(cid:144) B FIG. 11: The periodic behavior of the negative-mass parti-cle in a harmonic oscillator potential in compactified x -space.The example shown is for C = − posed in the quantum case, in which the wave-functionin ρ was demanded to vanish at the domain boundaries ρ = ± π/ √ β , which corresponds to placing infinite po-tential walls there. The m > C > | m | → ∞ limit of Eq. (92) whilekeeping B fixed, we find p | C | − ω m →∞ −−−−→ p Eβk = p βkB ≡ ω , (113)and p β ρ ( t ) | m |→∞ −−−−−→ arcsin h sin( ± ω t ) i , p β p ( t ) | m |→∞ −−−−−→ + sin( ± ω t ) | cos( ± ω t ) | ,x ( t ) /B | m |→∞ −−−−−→ ∓ cos( ± ω t ) | cos( ± ω t ) | . (114)which formally agrees with Eq. (85), and if graphed willlead to a figure similar to Fig. 6. The one significant3difference is, however, that when the particle jumps from x = ± A to x = ∓ A in the m > x = 0, while when it jumps from x = ± B to x = ∓ B inthe m < x = ∞ .By compactifying x -space, all motion when E = 0 willbecome oscillatory through x = ∞ , and the T calculatedabove becomes the oscillatory period. As an example, weplot the x -compactified solution for C = − T = 4 /ω . Note that the periodfor the m < C < | m | → ∞ becomeslim | m |→∞ T = lim | m |→∞ ω p | C | − p | C | = 2 πω , (115)the arccosine providing a π/ E. Classical Probablities
Consider the m < C = − | m | Eβ < √ βρ evolves from − π/ π/ T /
2, that is: T Z T/ − T/ dt = Z π/ √ β − π/ √ β dtdρ dρ = Z π/ √ β − π/ √ β dρ ˙ ρ = 1 ω Z π/ √ β − π/ √ β √ β dρ q tan (cid:0) √ βρ (cid:1) + | C | . (116)Thus, we can identify P ( ρ ) = 2 ωT √ β q tan (cid:0) √ βρ (cid:1) + | C | (117)as the classical probability density of the particle in ρ -space. The classical probablity in p - and x -spaces can bedefined in a similar manner: P ( p ) = P ( ρ ) dρdp = 2 ωT √ β (1 + βp ) p | C | + βp ,P ( x ) = P ( ρ ) dρdx = 2 T √ β | m | p β | m | kx − | C | (cid:16) β | m | k x − | C | + 1 (cid:17) = 2 ωT B √ x − B h | C | ( x − B ) + B i . (118)These probability functions are plotted in Fig. 12 for thecase C = − | m | Eβ = − − C = n + (2 n + 1) λλ (1 − λ ) , < λ < . (119) - Π - Π Π Π Β Ρ H Β Ρ L - - - Β p H Β p L - - - x (cid:144) B H x (cid:144) B L FIG. 12: Classical probabilities in ρ -, p -, and x -spaces of anegative mass particle in a harmonic oscillator potential forthe case C = − | m | Eβ = −
1. The time dependence of thissolution was shown in Fig. 11. Though the trajectory of theparticle is not confined to a finite region of phase space, theclassical probabilities of finding the particle near the phase-space-origin is still finite due to the finiteness of the time ittakes for the particle to traverse the entire trajectory.
We expect the quantum and classical probabilities tomatch for large n . As an example, we take λ = and n = 30, which correspond to: C = − ,κ = ∆ x min a = 1 p λ (1 − λ ) = 2 √ ≈ . ,B = p | C | κ ∆ x min 1 ≪ n ≈ n ∆ x min = 30 ∆ x min . (120)The comparison of the quantum and classical probabil-ities for this case in ρ -, p -, and x -spaces are shown inFig. 13. If we average out the bumps in the quantumcase, it is clear that the distributions agree, up to the typ-ical quantum mechanical phenomenon of seepage of theprobability into energetically forbidden regions. Thus,the existence of ‘bound’ states with a finite ∆ x and ∆ p - Π - Π Π Π Β Ρ H Β Ρ L - - - Β p H Β p L - - x (cid:144) B H x (cid:144) B L FIG. 13: Comparison of quantum and classical probabilitiesin ρ -, p -, and x -spaces for a negative mass particle in a har-monic oscillator potential for the case λ = and n = 30,which corresponds to the classical C = − | m | Eβ = − x -space is discrete due to theexistence of the minimal length. There is also some seepageof the probability into classically forbidden regions in x -spaceas is expected of quantum probabilities. in the quantum case can be associated with the fact thatthe particle spends a finite amount of time near the phasespace origin in the classical limit. IV. SUMMARY AND DISCUSSION
We have solved for the eigenstates of the harmonicoscillator hamiltonian under the assumption of the de-formed commutation relation between ˆ x and ˆ p as givenin Eq. (1), with the objective of calculating their uncer-tainties in position and momentum.For the normal harmonic oscillator with positive mass(1 /m > x ∼ / ∆ p branch of the MLUR, where decreasing 1 /m leads tolarger ∆ p , and thus smaller ∆ x . Somewhat surpris-ingly, 1 /m can be decreased through zero into the neg-ative, thereby ‘inverting’ the harmonic oscillator, whilestill maintaining an infinite ladder of positive energy eigenstates so long as the condition ∆ x min / √ > a ≡ [ ~ /k | m | ] / is satisfied. There, the eigenstates arefound on the ∆ x ∼ ∆ p branch of the MLUR, wheredecreasing 1 /m away from zero further into the negativeleads to larger ∆ p , and thus larger ∆ x , with both diverg-ing as a approaches the above bound from below. The1 /m = 0 line separating the ∆ x ∼ / ∆ p and ∆ x ∼ ∆ p regions is given by Eq. (40).Taking the classical limit by replacing our deformedcommutator with a deformed Poisson bracket, we solvethe corresponding classical equations of motion and findthat the solutions for the ‘inverted’ harmonic oscillatorare such that the particle only takes a finite amount oftime to traverse its entire trajectory. This leads to a fi-nite classical probability density of finding the particlenear the phase space origin, and provides and explana-tion of why ‘bound’ states with discrete energy levels arepossible in the quantum case.One significant difference between the classical andquantum cases is, however, that the classical system hasno restriction on the sign of the energy, whereas the quan-tum system only allows for positive energy eigenstates.The latter is guaranteed by the above mentioned condi-tion on ∆ x min and a . Indeed, the condition is equivalentto E = k x ) − | m | (∆ p ) > , (121)when one assumes ∆ x ∼ ( ~ β/ p .We have found that the energy eigenstates of the har-monic oscillator only populate the ∆ x ∼ / ∆ p branchof the MLUR when the mass is positive, and only the∆ x ∼ ∆ p branch when the mass is negative. A natu-ral question to ask is whether some superposition of theenergy eigenstates could cross over the 1 /m = 0 line tothe other side. For the free particle, which can be con-sidered the k = 0 limit of the harmonic oscillator, theanswer is in the affirmative. In that case, the uncertain-ties of the energy eigenstates are ∆ p = 0 and ∆ x = ∞ .Superpositions of these states, with uncertainties posi-tioned anywhere along the MLUR bound, can be easilyconstructed, as will be discussed in detail in a subsequentpaper [15]. Is a similar ‘cross-over’ possible for the k = 0case?Another interesting question is whether it is possibleto construct classically behaving ‘coherent states’ in ei-ther of these mass sectors, and whether their uncertain-ties are contained as they evolve in time. In particu-lar, when the mass is negative, the classical equationsof motion calls for the particle to move at arbitrarilylarge speeds. What is the corresponding quantum phe-nomenon? We are also assuming that Eq. (1) embodiesthe non-relativistic limit of some relativistic theory witha minimal length. From that perspective, the infinitespeed that the negative-mass particle attains seems prob-lematic. Would such states not exist in the relativistictheory, or will they metamorphose into imaginary masstachyons? These, and various other related questions will5be addressed in future works. Acknowledgments
We would like to thank Lay Nam Chang, George Hage-dorn, and Djordje Minic for helpful discussions. Thiswork is supported by the U.S. Department of Energy,grant DE-FG05-92ER40709, Task A.
Appendix A: An Integral Formula for GegenbauerPolynomials
The following integral formula is necessary in calculat-ing the expectation value of ˆ p . For two non-negativeintegers m and n such that m ≤ n , and λ > , we have Z − c λ − C λm ( s ) C λn ( s ) ds = π Γ (cid:0) m + 2 λ (cid:1) (2 λ − (cid:2) λ Γ (cid:0) λ (cid:1) (cid:3) m ! if n − m = even,0 if n − m = odd,(A1)where c = √ − s . We were unable to find this resultin any of the standard tables of integrals [14] though Mathematica seems to be aware of it. Here we present aproof.We start from recursion relations which can be foundin Ref. [14]. In section 8.933 of Gradshteyn and Ryzhik,we have:( n + 2 λ ) C λn ( x ) = 2 λ h C λ +1 n ( x ) − x C λ +1 n − ( x ) i ,n C λn ( x ) = 2 λ h x C λ +1 n − ( x ) − C λ +1 n − ( x ) i . (A2)Eliminating C λ +1 n − ( x ) and then shifting λ by one unit, weobtain C λn ( x ) − C λn − ( x ) = (cid:18) n + λ − λ − (cid:19) C λ − n ( x ) , (A3)which is Equation 22.7.23 of Abramowitz and Stegun.Iterating this relation, we deduce that C λ k ( x ) = C λ ( x ) + k X i =1 (cid:18) i + λ − λ − (cid:19) C λ − i ( x ) ,C λ k +1 ( x ) = C λ ( x ) + k X i =1 (cid:18) i + λλ − (cid:19) C λ − i +1 ( x ) . (A4)Since C λ ( x ) = 1 = C λ − ( x ) ,C λ ( x ) = 2 λ x = (cid:18) λλ − (cid:19) C λ − ( x ) , (A5) we can write C λ k ( x ) = k X i =0 (cid:18) i + λ − λ − (cid:19) C λ − i ( x ) ,C λ k +1 ( x ) = k X i =0 (cid:18) i + λλ − (cid:19) C λ − i +1 ( x ) . (A6)Thus, the even C λn ’s can be expressed as a sum of the even C λ − n ’s, and the odd C λn ’s as a sum of the odd C λ − n ’s.Invoking the orthogonality relation, Eq. (28), which isvalid when λ > − , it is clear that Z − c λ − C λ k ( s ) C λ ℓ +1 ( s ) ds = 0 (A7)for λ > . So for the integral of Eq. (A1) to be non-zero, m and n must be both even, or both odd. For twonon-negative integers k and ℓ such that k ≤ ℓ , we find Z − c λ − C λ k ( s ) C λ ℓ ( s ) ds = Z − c λ − " k X i =0 (cid:18) i + λ − λ − (cid:19) C λ − i ( s ) × " ℓ X j =0 (cid:18) j + λ − λ − (cid:19) C λ − j ( s ) ds = k X i =0 ℓ X j =0 (cid:18) i + λ − λ − (cid:19) (cid:18) j + λ − λ − (cid:19) × Z − c λ − C λ − i ( s ) C λ − j ( s ) ds = k X i =0 (cid:18) i + λ − λ − (cid:19) π Γ(2 i + 2 λ − i + λ − (cid:2) λ − Γ( λ − (cid:3) (2 i )!= 2 π (cid:2) λ − Γ( λ ) (cid:3) k X i =0 (2 i + λ −
1) Γ(2 i + 2 λ − i )! , Z − c λ − C λ k +1 ( s ) C λ ℓ +1 ( s ) ds = Z − c λ − " k X i =0 (cid:18) i + λλ − (cid:19) C λ − i +1 ( s ) × " ℓ X j =0 (cid:18) j + λλ − (cid:19) C λ − j +1 ( s ) ds = k X i =0 ℓ X j =0 (cid:18) i + λλ − (cid:19) (cid:18) j + λλ − (cid:19) × Z − c λ − C λ − i +1 ( s ) C λ − j +1 ( s ) ds = k X i =0 (cid:18) i + λλ − (cid:19) π Γ(2 i + 2 λ − i + λ ) (cid:2) λ − Γ( λ − (cid:3) (2 i + 1)!= 2 π (cid:2) λ − Γ( λ ) (cid:3) k X i =0 (2 i + λ ) Γ(2 i + 2 λ − i + 1)! . (A8)6The sums in the above expressions are given by k X i =0 (2 i + λ −
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