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Bound States in One Dimension Bound States in One Dimension In this chapter we will concern ourselves with obtaining stationary state so- lution of the time independent Schr¨odinger Equation: ¯h2 ∂2 − ψ(x)+V (x)ψ(x)=Eψ(x)(1) 2m ∂x2 for particles (such as electrons) bound in one dimensional potential wells. In general these solutions can be chosen to be real rather than complex functions which can often be considerable simplification. To see this lets take the complex conjugate of the time independent SE given in Eq. (1): ¯h2 ∂2 − ψ∗(x)+V (x)ψ∗(x)=Eψ∗(x)(2) 2m ∂x2 Because all the factors which operate on ψ(x) are real, they don’t change under complex conjugation and hence ψ(x)andψ∗(x) are equally valid solutions of the same potential V (x) with the same energy E. Since the SE is a homogenous linear differential equation, we can always form new valid solutions by taking linear combinations of valid solutions. In particular, the solution ψr(x)whichis constructed out of a general solutions ψ(x)andψ∗(x) according to ∗ ψr(x) ∝ ψ(x)+ψ (x)(3) is an intrinsically real. Of course exp (iφ) × ψr(x) is an equally valid solution with any complex phase φ that one cares to choose. We will generally choose real bound state solutions to the problems posed in this chapter. Some of the problems which we will address are a bit artificial but they are easily solvable and give considerable insight into perhaps more realistic but less tractable problems. I begin by describing the general “potential well” illustrated below in hopefully familiar classical terms. 1 Classically Allowed V turning turning point point E m KE Classically Classically Forbidden Forbidden PE X By a potential well, we mean a graph of potential energy as a function of coordinate x. In this well picture, we indicate a constant energy level (total potential plus kinetic energy) for the particle of mass m by the horizontal “dotted line”. The picture is meant to evoke conservation of energy ie the particle has the same total energy E = KE +PE independent of x. As a particle of mass m slides back and forth in x along the well much like a mass sliding down a frictionless valley, it constantly exchanges potential energy for kinetic energy while keeping a constant E. The (energy) distance between the energy level and the potential curve rep- resents the kinetic energy since KE = E − V (x). Classically the kinetic energy is an intrinsically positive quantity because KE = mv2/2. The classical turning points are the two intersections of the energy level and the potential well where the particle has zero kinetic energy. In the region between the turning points E>V(x)andKE=E − V>0. Since the kinetic energy is positive this is a perfectly reasonable or “allowed” region for the particle to exist. In the reason outside of the turning points, E<V(x)andKE=E − V<0whichisaclassi- cally “forbidden” condition. Classically the particle can never be found outside of the interval between the classical turning points. 2 Think of the particle as released from rest down at the left turning point on the frictionless gravitational valley represented by th potential, the particle picks up maximum speed at the bottom of the well, slows down as it approaches the right turning point. It stops at the right turning point, turns around and heads back down the well valley. Turning points where the particle turns around and reverses are aptly named. Stepwise constant potentials Perhaps the simplest solvable systems are constructed out of constant me- chanical potentials. To make life interesting we can use different potentials steps in different x regions. A physical system that approaches the ideal of a stepwise potential is provided by the ion trap depicted below: Φ Φ1 3 x V3 =e Φ3 V1 = e Φ1 classically classically allowed classically forbidden forbidden E V2 = 0 -a/2 0 a/2 x We think of the motion of the charged electron (serving as the ion) as con- strained to small diameter metal tubes which serve as “Faraday cages” . When the electron is within the electrostatic environment of one of the tubes comprising the trap, it has the constant electrical potential of the tube. In the asymmetric trap depicted above, the center tube is grounded and the outside tubes ride at a negative voltage with respect to the central tube. To the negatively charged 3 electron, the outside tubes are at a higher mechanical potential than the zero potential central tube. We’ve given the electron a total energy of E which traps in the center tube by making it impossible for the electron to classically travel to the outside tube. In this idealized system the only electrical forces act at the tube gaps where the potential is changing since since the electrical field is the negative of the potential gradient. If the mechanical potential actually makes discrete jumps as depicted above, the forces and electrical fields at the tube gaps become infinite. The potential is maintained by the batteries and we’ve drawn it for the general case of unequal voltages. Particle in the box A particularly simple case to consider involves taking the limit where the potential on the outer tubes in the ion trap depicted above approaches infinity. This is a way of insuring that the electron is constrained in the region of the central tube which serves as a confining “box”. We will consider the case where the electron has zero potential (V = 0) inside the box and is always found within the walls of the impenetrable box. We can place the origin of our x coordinate system anywhere we choose for convenience. One convenient choice for the particle in the box problem is to put the origin on the left side of the box (which has a width a) as shown below: m x 0 a Within the walls of the box, we have a free particle and we know from the previous chapter that valid solutions of the free particle problem are complex travelling waves of the form ψ(x)=exp(ikx) where we have suppressed the time dependence given by exp (−iωt). Without regard to the the boundary condition (the box walls) any combination of k’s is possible. We get some guidance by 4 restricting ourselves to real ψ(x) which can be obtained by exp (ikx)±exp (−ikx). The real combinations of these complex traveling waves are just sines and cosines: exp (ikx)+exp(−ikx) exp (ikx) − exp (−ikx) cos (kx)= , sin (kx)= 2 2i ψ(x)=b cos (kx)+c sin (kx)(4) and the real wave function for the free particle can be constructed as a linear combination of sines and cosines with unknown coefficients b and c. Like any problem involving differential equations, once the general solution is written down, we need to match the boundary conditions to obtain a fully explicit solution. What are the boundary conditions for the particle in the box? To answer this question we need to return to our interpretation of the wave function as a PDF amplitude PDF(x)=ψ∗(x)ψ(x)=ψ2(x). Because (for real wave functions) the square of the wave function gives the probability density for finding the electron from x → x + dx, we must insist that ψ(x) is well defined at every point x, eg ψ(x) must be continuous in the sense Continuity of the Wave Function ψ(x +∆)=ψ(x − ∆) as ∆ → 0 In particular this means that the wave function cannot make discrete jumps at boundaries. In the particle in the box problem, we assume that the electron has no probability of being found outside the box. Hence ψ(x)=0forx<0 and x>a. By continuity requirement since the wave function vanishes outside the box it must vanish on the walls of the box as well and thus: ψ(0) = ψ(a)=0 (5) The condition that ψ(0) = 0, tells us that our wave function of the general form ψ(x)=b cos (kx)+c sin (kx) cannot include a cos (kx) piece since: 5 0=ψ(0) = b cos (k0) + c sin (k0) = b. This limits us to the form ψ(x)= c sin (kx). We can limit this further by invoking the ψ(a) = 0 condition: nπ ψ a c ka → ka nπ → k n , , ... 0= ( )= sin ( ) = = a where =1 2 3 (6) The condition sin (ka) = 0 occurs when the argument of the sin () vanishes which occurs at an integral multiple of π radians. We thus have decided that the particle in the box wave function is of the form: nπx ψ x c ( )= sin a (7) There is also a far simpler and less formal way of obtaining Eq. (7). Since the wave function vanishes outside the box, it must vanish on the walls of the box at x =0andx = a. Since we know that the wave function is comprised of sinusoidal functions we are searching for a wave function with a nodes at x =0 and x = a. Only a pure sine function has a node at the origin. The distance between nodes of a sinusoidal function is λ/2. We need to cram an integral number of 1/2 wavelengths in the box width a. ψ (x) n = 3 λ/2 x λ/2 λ/2 a Therefore we require λ 2a n = a → λ = 2 n πx nπx ψ x c 2 → c ( )= sin λ sin a (8) 6 It is of interest to find the energy levels.
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