Download Solutions to Schrodinger's Equation for a Particle in a 1D Square Well Potential and more Study notes Quantum Physics in PDF only on Docsity! Phys 560 Fall 2007, Particle in a Well reminders C. Thompson (Dated: October 8, 2007) Particle in a Well solutions reminders I. SOLUTION Consider a particle of mass m in a square well potential (one dimension). V (x) = 0, |x| ≤ aV0, |x| ≥ a Enumerate the three regions I, II, III are where x < −a, −a < x < a, and x > a, respectively. Look for eigenfunction ψE(x) that satisfy Schrodinger’s equation. Those that describe a bound state will decay exponentially as x→ ±∞. In the different regions of the potential, it is convenient to define k2 = 2mE~2 and κ 2 = 2m(V0−E)~2 and rewrite the Schrodinger’s equations with those variables. Solutions in each of the regions that will satisfy the requirements of a bound state will have E < V0, (otherwise, in regions I, and III, solutions will also be traveling wave solutions, and although normalizable, are not localized). The pieces of a solution in each region are ψ(x) = Ae+κx x < −a B cos kx+ C sin kx |x| < a De−κx x > a where we have already eliminated terms that would not go to 0 at ±∞. So, any particular ψE(x) will be described with the three pieces as above. The k and κ are fixed by the energy eigenvalue E. At x = −a and x = +a, there should not be any discontinuities of the wavefunction or its first deriviative. And overall normalization should be one. This will put requirements on the allowed k and κ, which puts constraints on the possible E eigenvalues that will work. Continuity will also put constraints on the relationship of the constants. 2 For the Even solution, A = D, and C = 0. ψ(x) = Ae+κx x < −a B cos kx |x| < a Ae−κx x > a Find the boundary conditions on ψ and dψ dx and divide those conditions to give the compact relation between κ and k for the Even solutions, Ae−κx = B cos kaκAe−κa = Bk sin ka → κ = k tan ka Similarly, for the Odd solutions, A = −D and B = 0. ψ(x) = Ae+κx x < −a C sin kx |x| < a −Ae−κx x > a The boundary conditions to give a compact relation for κ and k for the Odd solutions, Ae−κx = C sin kaκAe−κa = −Bk cos ka → κ = −k cot ka Using the definitions of κ and k in the initial simplified notations of Schrodingers equations gives the relation, for ψ(x) at the boundaries, gives k2 + κ2 = 2mV0/~2 Note that as V0 tends to ∞ we find that the wavefunction and energies become identical with the ‘particle in the box’ solutions. As V0 goes to ∞, so does κ. And then the outside regions of the solutions becomes, eκ|x| → 0 and the interior regions remain sin or cos. Figure I plots the various relationships for κ(k). The x axis is in dimensionless units ka. The y axis is in dimensionless units, κa. The solid lines are the conditions y = x tanx and the dotted lines are −x cotx that any valid solutions for any well must satisfy. The circle shows the condition (in dimensionless units x2 + y2 = U2) k2a2 + κ2a2 = 2mV0a 2/~2 = U2 for a particular well. Therefore, the radius U is determined by the system parameters, V0, m, and a. The intersection of this circle with the other conditions gives the ka of the valid bound solutions for that well, and thus the E of the bound eigenfunctions. Changing any of the a, or V0 or m will change those intersections by changing the radius, U .
