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Chapter 5: Q67E (page 191)

Prove that the transitional-state wave function (5.33) does not have a well-defined energy.

Short Answer

Expert verified

The obtained solution for the energy still hasa wave function, so the original wave function does not have a defined energy.

Step by step solution

01

Identification of given data 

The given data from equation 5.33 is:

  • The sum of the two different solutions is Aψnxe-iEn/ℏt+Bψme-iEm/ℏt
02

Concept/Significance of wave function

A wave function is a function that plots the probability of a particle's existence in a quantum system as a function of position, momentum, duration, and/or rotation.

03

Proof of the transitional state wave function does not have defined energy 

The transitional state wave function is:

ψx,t=ψnxe-iEn/ℏt+ψme-iEm/ℏt

Apply the energy operator on the above equation.

E^ψx,t=iℏ∂∂xψnxe-iEn/ℏt+ψme-iEm/ℏt=iℏ∂ψnxe-iEn/ℏt∂x+iℏ∂ψme-iEm/ℏt∂x=Enψnxe-iEn/ℏt+Emψme-iEm/ℏt.

The obtained solution for the energy still hasa wave function;therefore, the original wave function does not have defined energy.

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Most popular questions from this chapter

Determine the expectation value of the momentum of the particle. Explain.

equation (5-33). The twosolutionsare added in equal amounts. Show that if we instead added a different percentage of the two solutions. It would not change the important conclusion related to the oscillation frequency of the charge density.

Show that the uncertainty in the position of a ground state harmonic oscillator is 1/2ℏ2/mk1/4.

Sketchψ(x) . Would you expect this wave function to be the ground state? Why or why not?

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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