A comment on Path integral for the quantum harmonic oscillator

Transcription

1 A comment on Path integral for the quantum harmonic oscillator Kiyoto Hira (Dated: May 8, 203) An elementary derivation of quantum harmonic oscillator propagator using the path integral have been reconsidered to modify the calculation, which used to be, for more pedagogical significance.

2 I. INTRODUCTION As is well known, the brilliant accomplishment of path integral ranks one of the most significant in the domain of modern quantum physics. Nowadays it is important for the students who major in physics to master the path integral in quantum theory. However, familiar problems that can be solved rigorously by using the path integral are only a few. The harmonic oscillator problem is one of them. We come across it in the early stage of period when we start to learn the path integral. The solutions 8 for this problem are available in textbooks and journals and online. One of two solutions we now refer to is the solution of Itzykson and Zuber in the text book 2 and the other is Cohen s solution in the journal 5. In order to make the multiple integration integrable in these approaches we need to diagonalize the quadratic form in the argument of the exponential appearing in the path integral formula. There we make use of the diagonalization of the symmetric matrix. The calculations 2 of Itzykson and Zuber seem to be too concise for the beginners to understand, and also those of Cohen 5 seem to be logically clear,mathematically meaningful, and beneficial to the students. However the details of Cohen s calculations are involved and not self-contained in themselves. Accordingly those might be drawbacks from the pedagogical point of view.that is why we are prompted to write this note. Therefore, for more pedagogical significance, in this note we try to modify their calculations to remove their drawbacks. Our calculations for the modification consist in the simple computations for the determinants, somewhat different from the others. We believe our approach may make their calculations simpler, more transparent, and more pedagogical to those novice students who start to learn the another method of quantum mechanics. In this note we shall do our calculations along Cohen s lines. II. FORMULATION FOR QUANTUM HARMONIC OSCILLATOR PROPAGA- TOR According to the discretization recipe for Feynman path integral, the harmonic oscillator propagator K(x, x) from the position x at the time t to the position x at the time t is given as in Cohen s paper 5, with the harmonic oscillator Lagrangian L(x(t), ẋ(t)) = 2 mẋ2 2

3 2 mω2 x 2, ( m ) K(x 2, x) = lim N dx dx 2 dx N N 2π iε e im/2 ε N n= {(x j x j ) 2 ε 2 ω 2 x 2 j }, () where we divide the time interval T = t t into N intervals of width ε each such that ε = T/N and we denote t j = t+jε (j = 0,,, N). For each point (x,, x N ) in (N ) dimensional real space R N, a so-called path function x(t) is defined by corresponding every t j (j = 0,, N) to x j = x(t j ) with x(t 0 ) = x and x(t N ) = x fixed, thereby we get a possible path by joining the successive points (x j, t j ) on the x t plane with the segments. The argument of the exponential contains the quadratic form N Q = [(x j x j ) 2 ε 2 ω 2 x 2 j] n= = x x 2 N ε 2 ω 2 x 2 N 2x x 0 2x N x N + Q, (2) where we may write, Q = x T A x. Here x T = (x x 2 x N ) (3) is the tranpose of x, and 2 ε 2 ω ε 2 ω A = 0 2 ε 2 ω (4) ε 2 ω Now we will find a transformation of variables, x = O y, (5) such that O T AO = Λ, (6) 3

4 where Λ is a diagonal matrix, such that Λ ij = δ ij, and O is a orthogonal since A is symmetric and real. Note that A and Λ have the following properties: Then we may rewrite deta = detλ, A = N j=, (7) (A ) T = A, (A ) i,j = (A ) j,i, (8) OΛ O T = A, Completing the squares, N k= N Q =x x 2 N ε 2 ω 2 x 2 N 2x N N 2x 0 j= O,j y j + N j= O i,k λ k O j,k = (A ) i,j. (9) j= O N,j y j y 2 j. (0) Q = x x 2 N ε 2 ω 2 x 2 N + N j= ( y j O N,jx N + O,j x 0 = x x 2 N ε 2 ω 2 x 2 N + x 2 N N j= N j= N O N,j O N,j x 2 0 = x x 2 N ε 2 ω 2 x 2 N + N j= ) 2 N j= ( y j O N,jx N + O,j x 0 j= (O N,j x N + O,j x 0 ) 2 ) 2 N O,j O,j 2x 0 x N ( y j O N,jx N + O,j x 0 ) 2 j= O N,j O,j x 2 N(A ) N,N x 2 0(A ), 2x 0 x N (A ) N,, () where we have used Eq.( 9 ) to get the last line. Here we change variables once again to yielding z j = y j O N,jx N + O,j x 0, (2) Q = N j= z 2 j + x x 2 N ε 2 ω 2 x 2 N x 2 N(A ) N,N x 2 0(A ), 2x 0 x N (A ) N,. (3) 4

5 Since O is orthogonal, deto =, and the Jacobian of both transformations, Eqs.(5) and (2) is unity. Hence, in the multiple integral Eq.(), along with the transformation of variables,we have the replacement We obtain dx dx 2 dx N = dz dz 2 dz N. (4) ( m ) K(x 2, x) = lim N exp{ im N 2πi ε 2 ε (x2 0 + x 2 N ε 2 ω 2 x 2 N x 2 N(A ) N,N x 2 0(A ), 2x 0 x N (A ) N, )} N j= = lim N ( m 2πi ε exp( im 2 ε z 2 j )dz j ) 2 exp{ im 2 ε (x2 0 + x 2 N ε 2 ω 2 x 2 N x 2 N(A ) N,N x 2 0(A ), 2x 0 x N (A ) N, )} (deta) /2, (5) where to get the last line, we have used the well known Gaussian Integral exp( im 2 ε zj 2 2πi ε )dz j =, m along with Eq.(7). III. DETERMINANTS, RECURSION RELATIONS We shall denote the determinant of the following type matrix with n rows and n columns as deta n+ : 2 ε 2 ω ε 2 ω deta n+ = 0 2 ε 2 ω (6) ε 2 ω If we expand this determinant along the first row, we obtain the recursion relations deta n+ = (2 ε 2 ω 2 )deta n deta n for n =, 2, 3,, (7) 5

6 where we set deta 0 = 0 and deta =. Here to solve this recursion relations for deta n,we deform this equation with the constants α and β as deta n+ α deta n = β(deta n α deta n ) for n =, 2, 3,. (8) Now by comparing Eq.(7) with Eq.(8) we must have α + β = 2 ε 2 ω 2, and αβ =. (9) From these equations α and β turn out to be the solutions of From Eq.(8) we find successive equations t 2 (2 ε 2 ω 2 )t + = 0. (20) deta n+ α deta n = β(deta n α deta n ) = = β n (deta α deta 0 ). Then we obtain with deta 0 = 0 and deta = Swapping α for β in Eq.(8), similarly we also obtain If α β, we obtain from these two equations deta n+ α deta n = β n. (2) deta n+ β deta n = α n. (22) deta n = βn α n β α. (23) Note that, from Eq.(20), since (2 ε 2 ω 2 ) 2 4 < 0, t 2 (2 ε 2 ω 2 )t + = 0 yields t = (2 ε2 ω 2 ) ± i 4 (2 ε 2 ω 2 ) 2 2 = e ±iφ, (24) where we set cos φ = (2 ε 2 ω 2 )/2, (25) sin φ = 4 (2 ε 2 ω 2 ) 2 /2. (26) Then, if we adopt β = e iφ and α = e iφ, by eq.(23) we obtain an important equation deta n = βn α n β α = einφ e inφ e iφ e iφ 6 = sin(nφ) sin(φ). (27)

7 From this expression, we obtain the determinant of the matrix A previously defined by Eq.(4) deta = deta N = sin(nφ) sin(φ). (28) Next, we shall compute the relevant matrix elements of A in Eq.(5 ). We shall denote the cofactor of (i, j) element of matrix A as D i,j, which is the product of ( ) i+j and the determinant of the submatrix formed by deleting the i-th row and j-th column from the matrix A. Thanks to Eq.(27), (A ) N,N = D N,N deta N = ( )N +N deta N deta N = sin(n )φ, (29) sin Nφ (A ), = (A ) N,N, (30) (A ) N, = (A ),N = D N, deta N = ( )N + ( ) N 2 deta N = sin φ sin Nφ. (3) IV. EXACT PROPAGATOR FOR N DISCRETE TIME INTERVALS Using the above results Eqs.(28)-(3), we can obtain the propagator for any number N of partitions of the time interval T. From Eq.(5), we obtain ( m ) K(x 2, x) = lim exp{ im N 2πi ε 2 ε (x2 0 + x 2 N ε 2 ω 2 x 2 N x 2 N(A ) N,N x 2 0(A ), 2x 0 x N (A ) N, )} (deta) /2 ( ) m sin φ 2 im sin(n )φ = lim exp{ [( )x 2 0 N 2πi ε sin Nφ 2 ε sin Nφ + ( ε 2 ω 2 sin(n )φ )x 2 sin φ N 2x 0 x N ( = mω 2πi sin ωt sin Nφ sin Nφ ]} ) /2 imω exp{ 2 sin ωt [(x2 0 + x 2 N) cos ωt 2x 0 x N ]}, (32) where to get the last line, we have used the fact that for large N φ = ωε = ω T N. (33) Finally we can reproduce the exact harmonic oscillator propagator. Furthermore we can recognize that our results concide with those of Cohen perfectly although our mathematical symbols are somewhat different from Cohen s ones. 7

8 V. CONCLUDING REMARK As you can see in our calculations,our analytical approach may be transparent, relatively simple and straightforward, and our calculation is done in a concise manner. The essential point of this note consists in the simple computations for the determinant Eq.(6), which yields Eq.(27). Moreover, it is an advantage of our technique that we can compute the relevant matrix elements of A in Eq.(5 ) very easily thanks to Eq.(27). We may emphasize that our technique is also applicable to the other calculations similar to Cohen s one and may make them simpler and more transparent. We believe the most undergraduates can easily follow our approach within the elementary mathematical manipulations and our approach may be of interest to those instructors who want to introduce path integral into their courses. That is why we stress the accessibility of our calculation. R.P.Feynman and A.R.Hibbs Quantum Mechanics and Path Integrals, McGraw-Hill,New York p63, p7-p73(965) 2 Claude Itykson and Jean-Bernard Zuber Quantum Field Theory, International Student Edition,McGraw-Hill, (985). 3 Hitoshi Murayama 22A Lecture Notes Path Integral 4 L. Q. English and R. R. Winters Continued fractions and the harmonic oscillator using Feynman path integral Am. J. Phys, 65, (997). 5 S. M. Cohen Path integral for the quantum harmonic oscillator using elementary methods Am. J. Phys, 66, (997). 6 B. R. Holstein The harmonic oscillator propagator Am. J. Phys. 66, (998). 7 L. Moriconi An elementary derivation of the harmonic oscillator propagator Am. J. Phys. 72, 258 (2004). 8 E. Wikberg Path Integral in Quantum Mechanics 8

9 This is a perfect supplementary note of the calculation of Itykson and Zuber project work,4p, Department of Physics,Stockholm University, emma/pathintegrals.pdf 9