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A Ladder Operator Solution to the Particle in a Box Problem Carl W

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A Ladder Operator Solution to the Particle in a Box Problem Carl W University of Connecticut OpenCommons@UConn Chemistry Education Materials Department of Chemistry September 2006 A Ladder Operator Solution to the Particle in a Box Problem Carl W. David University of Connecticut, [email protected] Follow this and additional works at: https://opencommons.uconn.edu/chem_educ Recommended Citation David, Carl W., "A Ladder Operator Solution to the Particle in a Box Problem" (2006). Chemistry Education Materials. 25. https://opencommons.uconn.edu/chem_educ/25 A Ladder Operator Solution to the Particle in a Box Problem C. W. David Department of Chemistry University of Connecticut Storrs, Connecticut 06269-3060 (Dated: September 21, 2006) I. SYNOPSIS construct our ladder operators. Symbolically, we define the up-ladder operator through the statement A ladder operator solution to the particle in a box M +|n >→ |n + 1 > problem of elementary quantum mechanics is presented, although the pedagogical use of this method for this prob- and we will require the inverse (the down-ladder opera- lem is questioned. tor), i.e., M −|n >→ |n − 1 > II. INTRODUCTION where the down-ladder operator is lowering us from n to n − 1 as usual. Recent interest in ladder operators [1], and the curious One knows that there is a lower bound to the ladder, omission from all current texts of the ladder operator i.e., that one can not ladder down from the lowest state solution to the particle in a box problem of elementary (n = 1), i.e., quantum mechanics makes one suspect that hidden in notes in various laboratories scattered all over the world M −|lowest >= 0 are the details of why this particular approach, when ap- plied to this particular problem, is not particularly ped- agogically valid. A. Form of the Up-Ladder and Down-Ladder Were it otherwise, it would imply that no one has Operators thought of applying this wonderful technique to what is the simplest of all quantum mechanical problems. Since What must the form of M + be, in order that it func- it is inconceivable that persons teaching the ladder op- tion properly? The up-ladder operator, M +, is defined erator technique for harmonic oscillators [2], rigid rotors according to and angular momentum [3, 4], and the H-atom [5–7], etc., nπx (n + 1)πx have not thought of this application, it must be that there M + sin → sin (3.2) is something curious about the problem that makes it of L L less than paramount interest. which, using Equation 3.1 and α = nπx and β = πx may For the precocious student, the one who asks radi- L L be expressed as cally different questions from the norm, the existence of a ladder operator in this particular problem should excite πx nπx πx nπx M + = cos sin + sin cos some interest. Here then is an approach to the ladder L L L L operator problem for the particle in a box in the domain (3.3) 0 ≤ x ≤ L which possibly illustrates why the method i.e., has not gained universal acceptance for this problem. It πx L πx ∂|n > appears to be related to that of Kais and Levine [8]. M +|n >→ cos |n > + sin = |n+1 > L nπ L ∂x The down operator is obtained in an analogous way, III. CONSTRUCTING THE LADDER OPERATOR (n − 1)πx πx nπx πx nπx sin = cos sin −sin cos L L L L L We start by remembering the trigonometric definition making use of the even and odd properties of cosines and of the sine of the sum of two angles, i.e., sines, respectively. sin(α + β) = sin α cos β + cos α sin β (3.1) πx L πx ∂|n > M −|n >→ cos |n > − sin → |n−1 > L nπ L ∂x We seek an operator, M + which takes a state |n > and ladders it up to the next state, i.e., |n + 1 >. Since the particle in a box problem is already solved, i.e., we know − πx L πx ∂ nπx M = cos − sin that |n > is sin L we can employ this knowledge to L nπ L ∂x Typeset by REVTEX 2 + − 2 πx The operators M and M can be rewritten in and [px, cos L ] (and its sine equivalent): co¨ordinate-momentum language as πx π πx π 2 πx 2 2 + πx L πx [px, cos ] = 2ı¯h sin px + ¯h cos M = cos − sin px L L L L L L ı¯hnπ L πx L πx M − = cos + sin p (3.4) L ı¯hnπ L x πx π πx π 2 πx [p2 , sin ] = −2ı¯h cos p +¯h2 sin x L L L x L L IV. COMMUTATION OF LADDER OPERATORS WITH THE HAMILTONIAN + A. Elementary Commutators B. The Commutator of M with Hop πx πx We need [px, cos L ] and [px, sin L ]: Remembering that πx π πx πx πx 2 [px, cos ] = ı¯h sin = px cos − cos px p2 ¯h ∂2 L L L L L H = = − op 2m 2m ∂x2 πx π πx πx πx [p , sin ] = −ı¯h cos = p sin −sin p after some effort, one has x L L L x L L x ¯h2 π 2 π 2 πx 2 πx ¯h2 ∂2 [H, M +] = − − (2n + 1) M + + 2n cos − cos (4.1) 2m L L L n L 2m ∂x2 V. LADDERING and since [H, M +] = HM + − M +H, using Equation 4.1 one has Then, if + + HM |n > −[H, M ]|n >= n|n + 1 > H|n >= n|n > + + operating from the left with M one has H|n + 1 > −[H, M ]|n >= n|n + 1 > + + M H|n >= nM |n >= n|n + 1 > and then, after a tedious amount of algebra, one obtains π 2 4m πx ¯h2 π 2 H|n + 1 > + 2n − n cos |n >= n|n + 1 > + (2n + 1) |n + 1 > L n¯h2 L 2m L From this we conclude that, if the elements in square then brackets cancelled perfectly, M +|n + 1 > would be an eigenfunction, i.e., if π 2 4m 2 2 2n − 2 n = 0 ¯h π L n¯h H|n + 1 >= n + (2n + 1) |n + 1 > 2m L i.e., n2¯h2π2 = n 2mL2 3 ¯h2π2 ¯h2π2 = n2 + 2n + 1 |n + 1 >= (n + 1)2 |n + 1 >= |n + 1 > 2mL2 2mL2 n+1 This represents a deluge of results, since we have obtained ladder operator, is that the trigonometric formula for the the redundant result of the form of the eigenvalue and sin of the sum of two angles makes the form possible. the form of the next (higher) eigenvalue. Generally, we This implies that similar “addition” formulae lie at the have to work a bit harder in ladder operator solutions fundament of the ladder operator approach in more com- to quantum mechanical problems, than we have had to mon cases, which is, perhaps, enlightening. here. In all, however, this ladder operator approach to the particle in a box problems probably deserves its own ob- scurity (or absence). No simplification occurs in carrying VI. DISCUSSION out this derivation which warrants expending the effort to lay the ground work for the derivation. Is it worthwhile to note the existence of this ladder One concludes that the ladder operator exists, is inter- operator approach? That curious students might worry esting, but that its relevance to the teaching and learning about this obvious omission (or absence) prompts these processes lies somewhere in the venue of exercises. remarks. But there is no question that the presenta- tion implies a circularity of reasoning which, although it might exist in other problems, is never so self-evident. We are presupposing the existence and form of the eigen- VII. ACKNOWLEDGEMENTS functions prior to attacking the problem. Had we started with the operators as postulated in Equation 3.2, the cir- It is a pleasure to acknowledge that the material pre- cularity of the argument would not have been so evident. sented here arose out of a discussion with Professor Jef- What is worth keeping here? It is clear that the current frey Bocarsly concerning Honors Freshman Chemistry in- repertoire of exercises of evaluating commutators can be struction. That no magic scheme for teaching this ma- n m expanded to include examples other than x and px e.g., terial to the mathematically unsophisticated absent tra- [px, sin x]. ditional methods was found does not detract from the Also, what becomes clear in constructing the original pleasure of having had the discussion. [1] D. D. Fitts, J. Chem. Ed. 72, 1066 (1995). [6] J. B. Boyling, Am. J. Phys. 56, 943 (1988). [2] L. U. R. Montemayor, Am. J. Phys. 51, 641 (1983). [7] C. W. David, Am. J. Phys. 34, 984 (1966). [3] R. E. R. O. L. deLange, Am. J. Phys. 55, 913 (1987). [8] R. D. L. S. Kais, Phys. Rev. A 34, 4615 (1986). [4] R. E. R. O. L. deLange, Am. J. Phys. 55, 950 (1987). [5] R. M. G. J. D. Newmarch, Am. J. Phys. 46, 658 (1978)..
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