Appendix 4.1 Alternate Solution to the Infinite Square Well with Symmetric Boundary Conditions. V(x) I II III +L/2

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1 Appendix 4.1 Alternate Solution to the Infinite Square Well with Symmetric Boundary Conditions V(x) I II III -L/2 +L/2 x In this problem, a particle of mass 7 confined by a potential energy which has a position dependence described by the figure above. The potential energy is infinite for region I, where B PÎ for region III, where B PÎ, but is zero in region II, where PÎ Ÿ B Ÿ PÎÞ We must find the solution to Schrodinger's time-independent equation in these three regions. Schrodinger's time-independent equation can be written in the form h ` ÐBÑ ZÐBÑ ÐBÑ œ I ÐBÑ 7`B < < < (0.1) ` 7 < ÐBÑ œ ci ZÐBÑd < ÐBÑ (0.2) `B h ` `B < ÐBÑ œ 5 < ÐBÑ (0.3) where 7 5 œ Ê ci ZÐBÑd (0.4) h Now in region I III where ZÐBÑ I for any energy, 5 is obviously imaginary. To make the imaginary nature of 5 more explicit, we will write 5 œ 3,, 7 where, œ È czðbñ Id is real, the equation above becomes h ` < ÐBÑ œ, < ÐBÑ (0.5) `B

2 The general solution to this differential equation is simple it must be of the form so that we have <ÐBÑ œ E/ -B (0.6) -B -B - E/ œ, E/ (0.7) so that - œ,, giving the general solutionðthe sum of the two possible solutions with appropriate arbitrary constants) <ÐBÑ œ E/, B, B (0.8) The time-dependent solution of the Schrodinger equation for the case where ZÐBÑ I is, therefore,, B, B 3 = > GÐB>Ñ, œ ce/ d/ (0.9) In the case of the infinite square well, where ZÐBÑ œ ß, œ ß giving 3 = > GÐB>Ñ, œ ce/ d/ (0.10) The second term goes to zero, but the first term blows up unless we require E œ!x In such a case, we require the wave function to be zero everywhere in region I region III. This means that the solution inside the well must go to zero at the boundaries. Now let's look at the solution in the region where ZÐBÑ œ!, 5 œ É 7I h. Again the general solution to the differential equation must be of the form giving <ÐBÑ œ E/ -B (0.11) - - E/ B œ 5 - E/ B (0.12) so that - œ 35. The general solution to this equation must be <ÐBÑ œ E/ 35B 35B (0.13) The time-dependent solution is, therefore, GÐB>Ñ, œ 35B 35B E/ 3 = > / (0.14) or GÐBß>Ñ œ E/ 3Ð5B = >Ñ 3Ð5B = >Ñ (0.15) Now using Euler's relationship we can write this last equation in the form G ÐBß>Ñ œ EccosÐ5B = >Ñ 3 sinð5b = >Ñd FccosÐ5B = >Ñ 3 sin Ð5B = (0.16) >Ñd

3 where we have used the fact that -9=Ð BÑ œ -9=ÐBÑ =38Ð BÑ œ =38ÐBÑ. Simplifying, we obtain G ÐBß>Ñ œ E cosð5b = >Ñ F cos Ð5B = >Ñ (0.17) 3cE sinð5b = >Ñ F sinð5b = >Ñd In both the real imaginary parts of this equation, we have two sinusoidal waves moving in opposite directions. The first term represents a real cosine wave moving in the B direction, while the second term represents such a wave moving in the B direction. This can be seen by observing a point on the wavefront where the phase, 9 œ 5B = > is equal to zero. Thus, the solution to Schrodinger's equation in the well is the solution of sinusoidal waves which are traveling in opposite direction =, interfering with each other. The only wave functions which are allowed, however, are ones which go to zero at the boundaries of the well, i.e., at B œ PÎ. Both the real the imaginary parts of the wavefunction must go to zero at the boundary. At the boundary where B œ PÎ, we have E cosð5pî = >Ñ F cos Ð5PÎ = >Ñ œ! (0.18) E sinð5pî = >Ñ F sin Ð5PÎ = >Ñ œ! (0.19) While at the boundary where B œ PÎ, we have E cosð 5PÎ = >Ñ F cos Ð 5PÎ = >Ñ œ! (0.20) E sinð 5PÎ = >Ñ F sin Ð 5PÎ = >Ñ œ! (0.21) This can be more compactly written in terms of the equation GÐB>Ñ, œ 35B 35B E/ 3 = > / (0.22) Since the boundary conditions must be satisfied at all times, we require the term in brackets to go to zero when B œ PÎ, when B œ PÎ, or 35PÎ 35PÎ E/ œ! (0.23) 35PÎ 35PÎ E/ œ! (0.24) Solving for E in both equations gives two different solutions E œ 35P (0.25)

4 both of which must be valid, so 35P œ 35P (0.26) cosð5pñ 3 sinð5pñ œ cos( 5PÑ 3 sin Ð5PÑ (0.27) sinð5pñ œ sin( 5PÑ (0.28) which can only be true if sin Ð5PÑ œ! Ê 5P œ 8 Ê 5 œ (0.29) P Applying this condition to both of the equations implied by Equ gives E œ FccosÐ81Ñ 3 sinð81ñd œ FccosÐ81Ñ 3 sin Ð81Ñd (0.30) which, since sin Ð81Ñ œ!, reduces to E œ F cos Ð81Ñ (0.31) This equation gives two different results for E, depending upon the valuse of 8: for even 8, E œ F, for odd 8, E œ F. This means we have two sets of solutions: 1. For odd 8 GÐB>Ñ, œ E 35B 35B / / 3 = > w / œ E Ð>Ñ cosð5bñ G ÐBß>Ñ œ E Ð>Ñ cos Ð81BÎPÑ (0.32) 9..8 w 3 = > where E Ð>Ñ œ EÎ / 2. For even 8 GÐB>Ñ, œ E 35B 35B / / 3 = > w / œ E Ð>Ñ sinð5bñ w G ÐBß>Ñ œ E Ð>Ñ sin Ð81BÎPÑ (0.33) /@/88 where E w Ð>Ñ œ EÎ3 / 3 = >. Note that the solution where 8 œ! gives the null wavefunction, which we interpret as no solution. The energy of the particle of mass 7 confined to the well is found using Equ. 0.0 Equ. 0.0, to be w h 5 h 81ÎP h 1 I œ œ œ 8 œ 8 I 7 7 7P " (0.34) The solutions to the wave equation are shown in blue in Figure.1 for the first four values of the quantum number 8. These solutions are plotted on a line which represents the energy of that particular state in terms of I. Plotted in red i = the probability density "

5 TÐBß>Ñ œ TÐBÑ given by * TÐBß>Ñ œ G ÐBß>ÑGÐBß>Ñ œ ¹ GÐBß>Ñ¹ (0.35) cos Ð81BÎPÑ 8 odd TÐBß>Ñ œ TÐBÑ œ œ 2 sin Ð81BÎPÑ 8 even 2 (0.36) Since we define the probability in terms of the square of the absolute magnitude of the wavefunction, the time dependent terms of a particular eigenfunction cancel so that the probability density of an eigenfunction is time independent! This is a direct consequence of the fact that the solution to Schrodinger's equation can be written as a time-dependent part times a position-dependent part. ÒNeeds to have proper limits indicated]

2. Let 0 be as in problem 1. Find the Fourier coefficient,&. (In other words, find the constant,& the Fourier series for 0.)

2. Let 0 be as in problem 1. Find the Fourier coefficient,&. (In other words, find the constant,& the Fourier series for 0.) Engineering Mathematics (ESE 317) Exam 4 April 23, 200 This exam contains seven multiple-choice problems worth two points each, 11 true-false problems worth one point each, and some free-response problems