Building Confidence in the Dirac δ -function
BBuilding Confidence in the Dirac δ -function Asim Gangopadhyaya ∗ and Constantin Rasinariu † Department of Physics, Loyola University Chicago
Abstract
In this note we present an example from undergraduate quantummechanics designed to highlight the versatility of the Dirac δ -function.Namely, we compute the expectation value of the Hamiltonian of afree-particle in a state described by a triangular wave function ψ ( x ) .Since the first derivative of ψ ( x ) is piecewise constant, and becausethis Hamiltonian is proportional to the second order spatial derivative,students often end up finding the expectation value to be zero –anunphysical answer. This problem provides a pedagogical applicationof the Dirac δ -function. By arriving at the same result via alternatepathways, this exercise reinforces students’ confidence in the Dirac δ -function and highlights its efficiency and elegance. Introduction
Physics majors at most liberal arts colleges in United States generally seethe Dirac δ -function during their sophomore or junior year. Its introductionis many times associated with a disclaimer that the name δ -function is amisnomer because it is not a function; it should be seen as a distribution[1, 2]. However, as all practitioners know the essential role δ -function playsin physics, it is important for the students to not only know its definition[3–6] and learn how it use it [7–9], but also to understand its subtlety [10],and verify the results of using the δ -function by alternate means, and thusincrementally build confidence in the robustness of the procedure.Here, we present an example from undergraduate quantum mechanics [9]in which students are asked to determine the expectation value of a free par-ticle Hamiltonian ˆ H in a particular state ψ ( x ) by several methods, includingone that employs the Dirac δ -function. ∗ [email protected] † [email protected] a r X i v : . [ phy s i c s . g e n - ph ] J u l escription of the Problem Consider a free particle, which at t = 0 , is in a state described by thefollowing normalized triangular wave function ψ ( x ) = (cid:113) a ( x + a ) − a < x < (cid:113) a ( a − x ) 0 ≤ x < a | x | ≥ a (1)The isosceles triangle with its base from − a/ to + a/ , illustrated in Fig.(1), is a square-integrable function, and hence is a valid wave function forany potential defined over a segment of the real-axis (or the entire real-axis)containing [ − a/ , + a/ . It is interesting to note that such a function is also -1.0 -0.5 0.5 1.00.51.01.5 Figure 1: An isosceles triangular wave function of height (cid:113) a and extendingfrom − a/ to + a/ for a = 1 .the groundstate for an infinite square well potential with an attractive δ -potential in the middle [11], and for such a system its shape does not evolvewith time.The Hamiltonian for the system is ˆ H = − ¯ h m ∂ ∂x . Our objective is todetermine the expectation value of the Hamiltonian in this state.In our experience, when presented with this problem, many times stu-dents rush to the naive conclusion that (cid:104) H (cid:105) = − ¯ h m (cid:82) ψ ∗ ( x ) ∂ ∂x ψ ( x ) d x = 0 ,because the first derivative of the wave function is piecewise constant. Theywould soon realize that, because of the discontinuities of ψ (cid:48) ( x ) , the secondderivative of the wave function must introduce several Dirac δ -functions.This allows them to calculate the integral in a straightforward way. However,2e think that students’ understanding improves when alternate, equivalentsolution pathways are presented. Therefore, in this paper we will use threedifferent methods to compute the expectation value of (cid:104) H (cid:105) . First, we con-sider that the free particle is inside an infinite square well of width a . Thus, ψ ( x ) can be expressed in terms of the eigenfunctions φ n ( x ) of the squarewell Hamiltonian. This is equivalent to the Fourier series representation of ψ ( x ) . The second method makes use of the Fourier transform of ψ ( x ) which,up to a constant ¯ h , gives the momentum wave function representation φ ( p ) .The third method uses the Hermiticity of the momentum operator [12] al-lowing us to cast (cid:104) ψ | ˆ p | ψ (cid:105) as (cid:104) ˆ pψ | ˆ pψ (cid:105) . This turns out to be the easiest wayto compute the integral. Finally, the fourth method employs the Dirac δ -function mentioned above, and is by far the most efficient way of getting theresult. We believe that by solving the same problem via several equivalentways, students become more comfortable in using the Dirac δ -function and,equally important, learn to appreciate the elegance and effectiveness of suchtechniques. Method 1: Fourier Series
In this section we assume that the particle is free within the domain ( − a , a ) and is blocked from the beyond by infinite barriers at x = ± a . The eigenfunc-tions are φ n ( x ) = (cid:113) a cos( nπx/a ) for odd n ’s and φ n ( x ) = (cid:113) a sin( nπx/a ) for even n ’s, where n = 1 , , , · · · . As expected, they vanish at x = ± a .Since φ n ( x ) form a complete set, we can expand the triangular function ψ ( x ) as the linear combination ψ ( x ) = ∞ (cid:88) n =1 c n φ n . Since ψ ( x ) is an even function, only the cosine functions remain, with oddcoefficients c n +1 . Hence, ψ ( x ) = ∞ (cid:88) n =0 c n +1 (cid:32)(cid:114) a cos (cid:20) (2 n + 1) πxa (cid:21)(cid:33) . The coefficients c n +1 are given by c n +1 = (cid:90) a/ − a/ ( φ n ) ∗ ψ ( x ) dx = 2 (cid:90) − a/ (cid:32) (cid:114) a · (cid:18) x + a (cid:19)(cid:33) (cid:32)(cid:114) a cos (cid:20) (2 n + 1) πxa (cid:21)(cid:33) dx √ π (2 n + 1) . The expectation value of the Hamiltonian is (cid:42) ˆ p m (cid:43) = − ¯ h m (cid:90) a/ − a/ ( ψ ( x )) ∗ (cid:32) d ψdx (cid:33) dx = ¯ h m ∞ (cid:88) k,n =0 c n +1 c k +1 (cid:18) (2 k + 1) πa (cid:19) × (cid:90) a/ − a/ (cid:114) a cos (cid:18) (2 n + 1) π xa (cid:19) · (cid:114) a cos (cid:18) (2 k + 1) π xa (cid:19) dx (cid:124) (cid:123)(cid:122) (cid:125) δ n,k = ¯ h m ∞ (cid:88) n =0 (cid:32) √ π (2 n + 1) (cid:33) (cid:18) (2 n + 1) πa (cid:19) = 96¯ h mπ a · ∞ (cid:88) n =0 n + 1) = 96¯ h mπ a · π h ma , (2)where we used Mathematica [13] to compute the infinite sum ∞ (cid:88) n =0 n + 1) = π . Students can arrive at the same result using the formula 0.234.2 on page 7of [14].
Method 2: Fourier Transform
In this section we will use the momentum wave function representation φ ( p ) which, up to a constant ¯ h , is exactly the inverse Fourier transform of ψ ( x ) : φ ( p ) = 1 √ π ¯ h (cid:90) + ∞−∞ ψ ( x ) e − ipx/ ¯ h dx . (3)For the triangular wave function given in Eq. (1), we obtain φ ( p ) = 2 √ π ¯ h (cid:114) a (cid:32)(cid:90) − a/ ( x + a/ e − ipx/ ¯ h dx + (cid:90) a/ ( a/ − x ) e − ipx/ ¯ h dx (cid:33) = 1 p (cid:115) h πa (cid:16) − e − iap/ h − e iap/ h (cid:17) = 4 (cid:115) h πa × p sin (cid:18) ap h (cid:19) . (4)4
40 -20 20 400.100.200.30
Figure 2: The momentum space wave function for a = 1 and ¯ h = 1 .In Fig. (2), we depict the momentum space wave function φ ( p ) . In thisrepresentation, the expectation value of the Hamiltonian is (cid:42) (cid:98) p m (cid:43) = 12 m (cid:90) + ∞−∞ p | φ ( p ) | dp . (5)Now, inserting φ ( p ) from Eq. (4) into the above equation, we get (cid:42) (cid:98) p m (cid:43) = 96¯ h mπa (cid:90) + ∞−∞ p sin (cid:18) ap h (cid:19) dp = 6¯ h ma (6)Again, we resorted to Mathematica [13] to calculate the integral (cid:90) + ∞−∞ sin ( x ) x d x = π . Equivalently, one can use the formula 3.821.10 from page 446 of [14].
Method 3: Hermiticity of the Momentum Operator
Since the momentum operator ˆ p is Hermitian, we can write (cid:104) ψ | ˆ p | ψ (cid:105) as (cid:104) ˆ pψ | ˆ pψ (cid:105) . Thus, in coordinate representation, we have (cid:42) (cid:98) p m (cid:43) = 12 m (cid:104) ˆ pψ | ˆ pψ (cid:105) = 12 m (cid:90) a/ − a/ (cid:18) − i ¯ h dψdx (cid:19) ∗ (cid:18) − i ¯ h dψdx (cid:19) dx, ¯ h m (cid:32) (cid:114) a (cid:33) (cid:90) a/ − a/ dx = 6¯ h ma (7)This method, used in Ref. [12], turns out to be the easiest of the four methodsconsidered here. Method 4: Dirac δ -function Since our objective is to see an application of the δ -function, in this sectionwe will derive the expectation value by directly applying ˆ H to the triangularwave function of Eq. (1). (cid:42) (cid:98) p m (cid:43) = − ¯ h m (cid:90) a/ − a/ ( ψ ( x )) ∗ (cid:32) d ψdx (cid:33) dx . (8)To compute this integral, we first find dψdx and d ψdx . The first derivative isgiven by dψ ( x ) dx = (cid:113) a − a < x < − (cid:113) a < x < a | x | ≥ a (9) -1.0 -0.5 0.5 1.0-3-2-1 123 -1.0 -0.5 0.5 1.0-3-2-1123(a) (b) Figure 3: The first derivative (a) and second derivative (b) of ψ ( x ) for a = 1 .The δ -functions are represented by vertical arrows.6ue to the discontinuities of dψdx at x = − a/ , x = 0 and x = a/ , the secondderivative will be a sum of three δ -functions centered on these points: d ψdx = 2 (cid:114) a δ (cid:18) x + a (cid:19) − (cid:114) a δ ( x ) + 2 (cid:114) a δ (cid:18) x − a (cid:19) . (10)Note that the coefficients in front of each δ -function are equal to the differ-ences in the derivatives of ψ across each point of discontinuity.The first and second derivatives are depicted in Fig. (3) for a = 1 .Inserting Eq. (10) into Eq. (8), and taking into accont that ψ (cid:0) ± a (cid:1) = 0 , weget (cid:42) (cid:98) p m (cid:43) = − ¯ h m (cid:90) a/ − a/ ψ ∗ ( x ) (cid:34) (cid:114) a δ (cid:18) x + a (cid:19) − (cid:114) a δ ( x ) + 2 (cid:114) a δ (cid:18) x − a (cid:19)(cid:35) dx = − ¯ h m (cid:34) ψ ∗ (cid:18) − a (cid:19) (cid:32) (cid:114) a (cid:33) − ( ψ (0)) ∗ (cid:32) − (cid:114) a (cid:33) − ψ ∗ (cid:18) a (cid:19) (cid:32) (cid:114) a (cid:33)(cid:35) = − ¯ h m · (cid:32) (cid:114) a · a (cid:33) (cid:32) − (cid:114) a (cid:33) = 6¯ h ma . (11)Note that this derivation is also straightforward and, unlike the first twomethods, doesn’t require any use of Mathematica. Conclusion
In this note we computed the expectation value of a free-particle Hamil-tonian in a state described by a triangular wave function ψ ( x ) . Since thefirst derivative of ψ ( x ) is piecewise constant, and because the Hamiltonianis proportional to a second order spatial derivative, students often end upfinding the expectation value to be zero –an unphysical answer. This prob-lem provides a natural setting for the introduction of Dirac δ -function asthe derivative of a step function. By computing the expectation value ofthe Hamiltonian via several different methods we attempt to bolster theirconfidence in δ -function, which appears in many areas of physics. Acknowledgement
We would like to thank the reviewers for their valuable comments and expressour gratitude to the editors for the opportunity to improve our manuscript.7 eferences [1] Aguirregabiria J M, Hernández A, and Rivas M 2002, δ -function con-verging sequences, Am. Jour. of Phys. , 180; doi: 10.1119/1.1427087[2] Boykin T B 2003, Derivatives of the Dirac delta function by ex-plicit construction of sequences, Am. Jour. of Phys. , 462; doi:10.1119/1.1557302[3] Arfken G B and Weber H J 2001, Mathematical Methods for Physicists , 281; doi: 10.1119/1.16058[5] Muller F A 1994, Definitions of Delta Distributions, Am. Jour. of Phys. , 11; doi: 10.1119/1.17729[6] Uginčius P 1972, An Integral Representation for the Dirac Delta Func-tion, Am. Jour. of Phys. , 1690; doi: 10.1119/1.1987016[7] Atkinson D A, and Crater H W 1975, An exact treatment of the Diracdelta function potential in the Schrödinger equation, Am. Jour. of Phys. , 301; doi: 10.1119/1.9857[8] Shankar R 1994, Principles of Quantum Mechanics