OneDimSchr (1)

# OneDimSchr (1) - Schrdingers Equation in 1-D: Some Examples...

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Schrödinger’s Equation in 1-D: Some Examples Michael Fowler, UVa. 9/24/07 Curvature of Wave Functions Schrödinger’s equation in the form 2 22 () 2 ( () ) () dx m V x E x dx ψ = = can be interpreted by saying that the left-hand side, the rate of change of slope, is the curvature – so the curvature of the function is proportional to ( ) ( ) () . Vx E x This means that if E > V ( x ), for x positive ( ) x is curving negatively, for x negative x is curving positively. In both cases, ( ) x is always curving towards the x-axis so, for E > V ( x ), x has a kind of stability: its curvature is always bringing it back towards the axis, and so generating oscillations. The simplest example is that of a constant potential V ( x ) = V 0 < E , for which the wave function is ( )( sin xA k x ) δ =+ with a constant and 2 0 2/ . km E V =− = On the other hand, for V ( x ) > E , the curvature is always away from the axis. This means that x tends to diverge to infinity. Only with exactly the right initial conditions will the curvature be just right to bring the wave function to zero as x goes to infinity. (This is possible because as x tends to zero, the curvature tends to zero, too.) For a constant potential V 0 > E , the wave function is ( ) x x x Ae Be α , with 2 0 . mV E = = Of course, this wave function will diverge in at least one direction! However, as we shall see below, there are situations with spatially varying potentials where this wave function is only relevant for positive x , and the coefficients A , B are functions of the energy—for certain energies it turns out that 0 A = , and the wave function converges. One Dimensional Infinite Depth Square Well In an earlier lecture, we considered in some detail the allowed wave functions and energies for a particle trapped in an infinitely deep square well, that is, between infinitely high walls a distance L apart. For that case, the potential between the walls is identically zero so the wave function has the form ( ) ( ) sin . k x The wave function x necessarily goes to zero right at the walls, since it cannot have a discontinuity, and must be zero just inside the wall. Even a quantum particle cannot penetrate an infinite wall!

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2 An immediate consequence is that the lowest state cannot have zero energy, since k = 0 gives a constant () x ψ . Rather, the lowest energy state must have the minimal amount of bending of the wave function necessary for it to be zero at both walls but nonzero in between—this corresponds to half a period of a sine or cosine (depending on the choice of origin), these functions being the solutions of Schrödinger’s equation in the zero potential region between the walls. The allowed wave functions (eigenstates) found as the energy increases have successively 0, 1, 2, … zeros (nodes) in the well. Parity of a Wave Function Notice that the allowed wave eigenfunctions of the Hamiltonian for the infinite well are symmetrical or antisymmetrical about the center, ( ) ( ) . x x = ±− We call the operator that reflects a function in the origin the parity operator P , so these eigenstates of the Hamiltonian are also
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## This note was uploaded on 12/07/2011 for the course PHYSICS 751 taught by Professor Michaelfowler during the Fall '07 term at UVA.

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OneDimSchr (1) - Schrdingers Equation in 1-D: Some Examples...

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