In section 7.5,eimlφis presented a sour preferred solution to the azimuthal equation, but there is more general one that need not violate the smoothness condition, and that in fact covers not only complex exponentials but also suitable redelinitions of multiplicative constants, sine, and cosine,

Φm1(Φ)=Ae+imlφ+Be+imlφ

(a) Show that the complex square of this function is not, in general, independent of φ.

(b) What conditions must be met by A and/or B for the probability density to be rotationally symmetric – that is, independent of φ ?

Short Answer

Expert verified

(a) In above equation, in general, the complex square of the azimuthal equation is not independent of φ.

(b) When the value of m1takes both positive and negative integral values, they will be equivalent.

Step by step solution

01

Complex Square of azimuthal equation:

The equation Φm1(Φ)=eimlφis the improved form of the solution of the equation governing azimuthal motion.

Consider the known equation as below.

Φm1(Φ)=Ae+imlφ+Be+imlφ

By taking the complex square of the azimuthal equation, we get,

localid="1659178898885" Φ*mΦΦmlΦ=A*e-imlφ+B*e+imlφAe+imlφ+Be-imlφΦ*mΦΦmlΦ=A2+B2+A*Be-2imlφ+B*Ae+2imlφ(1)

As it can be clearly shown in the above eq. (1) that in general, the complex square of the azimuthal equation is not independent of φ.

02

(b) Conclusion:

If we make A=0 and B=0 to make the equation independent of φ, the solution itself will be zero, hence, we cannot do that.

But when the value of mltakes both positive and negative integral values, they will be equivalent.

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Most popular questions from this chapter

A wave function with a non-infinite wavelength-however approximate it might be- has nonzero momentum and thus nonzero kinetic energy. Even a single "bump" has kinetic energy. In either case, we can say that the function has kinetic energy because it has curvature- a second derivative. Indeed, the kinetic energy operator in any coordinate system involves a second derivative. The only function without kinetic energy would be a straight line. As a special case, this includes a constant, which may be thought of as a function with an infinite wavelength. By looking at the curvature in the appropriate dimension(s). answer the following: For a givenn,isthe kinetic energy solely

(a) radial in the state of lowest l- that is, l=0; and

(b) rotational in the state of highest l-that is, l=n-1?

If the constantCxinequation(7-5)were positive, the general mathematical solution would be

Ae+cxx+Be-cxx

Show that this function cannot be 0 at two points. This makes it an unacceptable solution for the infinite well, since it cannot be continuous with the wave functions outside the walls, which are 0.

For a hydrogen atom in the ground state. determine (a) the most probable location at which to find the electron and (b) the most probable radius at which to find the electron, (c) Comment on the relationship between your answers in parts (a) and (b).

To conserve momentum, an atom emitting a photon must recoil, meaning that not all of the energy made available in the downward jump goes to the photon. (a) Find a hydrogen atom's recoil energy when it emits a photon in a n = 2 to n = 1 transition. (Note: The calculation is easiest to carry out if it is assumed that the photon carries essentially all the transition energy, which thus determines its momentum. The result justifies the assumption.) (b) What fraction of the transition energy is the recoil energy?

Question: Show that the angular normalization constant in Table 7.3 for the case (l,ml)=(1,0) is correct.

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