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Chapter 5: Q31E (page 188)

Verify that solution (5-19) satisfies the Schrodinger equation in form (5.18).

Short Answer

Expert verified

The solution of the Schrodinger equation is

d2ψ(x)dx2=Aα2exp±αxd2ψ(x)dx2=α2ψ(x)

Step by step solution

01

Time-dependent equation of Schrodinger.

Time-dependent equation of Schrodinger

d2ψ(x)dx=2m(Uo(x)-E)h2ψ(x)………………(1)

and the wave function is provided

ψ(x)=Aexp±αx,α=2m(Uo(x)-E)h2………………(2)

02

Schrodinger equation

Differentiate the wave function twice and replace into the Schrodinger equation to ensure that it satisfies the Schrodinger equation (1).

dψ(x)dx=±Aαexp±αx=±αψ(x)d2ψ(x)dx2=Aα2exp±αx=α2ψ(x)

So, the Schrodinger equation isd2ψ(x)dx2=Aα2exp±αxd2ψ(x)dx2=α2ψ(x).

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Most popular questions from this chapter

What is the probability that a particle in the first excited (n=2) state of an infinite well would be found in the nuddle third of the well? How does this compare with the classical expectation? Why?

A study of classical waves tells us that a standing wave can be expressed as a sum of two travelling waves. Quantum-Mechanical travelling waves, discussed in Chapter 4, is of the form ψ(x,t)=Aei(kx=ωt). Show that the infinite well’s standing wave function can be expressed as a sum of two traveling waves.

Sketch the wave function. Is it smooth?

ψ(x)={2a3xe-axX>00X<0

Consider a particle bound in a infinite well, where the potential inside is not constant but a linearly varying function. Suppose the particle is in a fairly high energy state, so that its wave function stretches across the entire well; that is isn’t caught in the “low spot”. Decide how ,if at all, its wavelength should vary. Then sketch a plausible wave function.

It is possible to take the finite well wave functions further than (21) without approximation, eliminating all but one normalization constant C . First, use the continuity/smoothness conditions to eliminate A, B , andG in favor of Cin (21). Then make the change of variables z=x-L/2 and use the trigonometric relations

sin(a+b)=sinacosb+cosasinband

cos(a+b)=cosacosb-sinasinbon the

functions in region I, -L/2<z<L/2. The change of variables shifts the problem so that it is symmetric about z=0, which requires that the probability density be symmetric and thus that ψ(z)be either an odd or even function of z. By comparing the region II and region III functions, argue that this in turn demands that (α/k)sinkL+coskL must be either +1 (even) or -1 (odd). Next, show that these conditions can be expressed, respectively, as αk=tankL2 and αk=-cotkL2. Finally, plug these separately back into the region I solutions and show that

ψ(z)=C×{eα(z+L/2)          z<L/2coskzcoskL2          -L/2<z<L/2e-α(z-L/2)          z>L/2


or

ψ(z)=C×{eα(z+L/2)          z<L/2-sinkzsinkL2          -L/2<z<L/2e-α(z-L/2)          z>L/2

Note that Cis now a standard multiplicative normalization constant. Setting the integral of |ψ(z)|2 over all space to 1 would give it in terms of kand α , but because we can’t solve (22) exactly for k(or E), neither can we obtain an exact value for C.
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