A particle of mass m is described by a stationary state (i.e. it is a solution t

Question

A particle of mass m is described by a stationary state (i.e. it is a solution to the time-independent Schroedinger equation):

= 0 for x < 2nm and x > 8 nm.

(x) = C (x _ 2nm) for 2 nm ? x < 5 nm

(x) = C (8nm _ x) for 5 nm ? x ? 8 nm.

(Note: Making a sketch is always an advisable first step. Very often, a good sketch will help you avoid doing needless calculations. At the very least, it may help point out how to start a calculation.)

? First of all, explain why this wavefunction could not truly describes a real particle.

? Normalize the wavefunction.

? What is the probability of finding the particle in the region 2 nm ? x < 5 nm? What is the probability of finding the particle in the region 5 nm ? x ? 8 nm.

? What is the expectation value of position ?

? What is the expectation value of the momentum

?

Explanation / Answer

Consider, now, a particle of mass trapped in a one-dimensional square potential well of width and finite depth . In fact, suppose that the potential takes the form (869) Here, we have adopted the standard convention that as . This convention is useful because, just like in classical mechanics, a particle whose overall energy, , is negative is bound in the well (i.e., it cannot escape to infinity), whereas a particle whose overall energy is positive is unbound. Since we are interested in bound particles, we shall assume that . We shall also assume that , in order to allow the particle to have a positive kinetic energy inside the well. Let us search for a stationary state (870) whose stationary wavefunction, , satisfies the time independent Schrödinger equation, (832). Now, it is easily appreciated that the solutions to (832) in the symmetric [i.e., ] potential (870) must be either totally symmetric [i.e., ] or totally antisymmetric [i.e., ]. Moreover, the solutions must satisfy the boundary condition (871) otherwise they would not correspond to bound states. Let us, first of all, search for a totally symmetric solution. In the region to the left of the well (i.e., ), the solution of the time independent Schrödinger equation which satisfies the boundary condition as is (872) where (873) and is a constant. By symmetry, the solution in the region to the right of the well (i.e., ) is (874) The solution inside the well (i.e., ) which satisfies the symmetry constraint is (875) where (876) and is a constant. The appropriate matching conditions at the edges of the well (i.e., ) are that and both be continuous [since a discontinuity in the wavefunction, or its first derivative, would generate a singular term in the time independent Schrödinger equation (i.e., the term involving ) which could not be balanced]. The matching conditions yield (877) Let . It follows that (878) where (879) Moreover, Equation (878) becomes (880) with (881) Here, must lie in the range , in order to ensure that lies in the range . Figure: The curves (solid) and (dashed), calculated for . The latter curve takes the value when . Now, the solutions of Equation (881) correspond to the intersection of the curve with the curve . Figure 62 shows these two curves plotted for a particular value of . In this case, the curves intersect twice, indicating the existence of two totally symmetric bound states in the well. Moreover, it is clear, from the figure, that as increases (i.e., as the well becomes deeper) there are more and more bound states. However, it is also apparent that there is always at least one totally symmetric bound state, no matter how small becomes (i.e., no matter how shallow the well becomes). In the limit (i.e., the limit in which the well is very deep), the solutions to Equation (881) asymptote to the roots of . This gives , where is a positive integer, or (882)

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