Some degeneracies are easy to understand on the basis of symmetry in the physical situation. Others are surprising, or “accidental”. In the states given in Table 7.1, which degeneracies, if any, would you call accidental and why?

Short Answer

Expert verified

The way that the initial three have similar energy is grasped by the cubic balance. Nonetheless, the way that (3,3,3) has a similar energy to the initial three is not perceived as evenness, so it is viewed as an "accidental" degeneracy.

Step by step solution

01

Step 1:Define degeneracy

"Degeneracy" signifies more than one quantum state with the equivalent strongly characterized energy.

The permitted energies for a molecule in the cubic boundless potential are.

Enx,ny,nz=(nx2+ny2+nz2)π2h22mL2

Where nx,ny,nz are integers, m is the mass of the particle, and L is the lengthin which we are studying the behavior of the particle.

02

The accidental degeneracy in Table 7.1.

Because of cubic evenness, all potential blends of any three numbers nx,ny,nzhave similar energy.

For instance (2,1,1) , (1,2,1) and (1,1,2) structure a three-crease degenerate state.

When we take a gander at Table 7.1 cautiously, see that (5,1,1) , (1,5,1) , (1,1,5) and (3,3,3) structure a four-overlap degenerate state.

The way that the initial three have similar energy is grasped by the cubic balance. Nonetheless, the way that (3,3,3) has a similar energy to the initial three is not perceived as evenness, so it is viewed as an "accidental" degeneracy.

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Most popular questions from this chapter

Here we Pursue the more rigorous approach to the claim that the property quantized according to ml is Lz,

(a) Starting with a straightforward application of the chain rule,

φ=xφ/x+yφy+zφz

Use the transformations given in Table 7.2 to show that

φ=-yx+xy

(b) Recall that L = r x p. From the z-component of this famous formula and the definition of operators for px and py, argue that the operator for Lz is -ihφ..

(c) What now allows us to say that our azimuthal solutioneimlφ has a well-defined z-component of angular momentum and that is value mlh.

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

For an electron in the(n,l,ml)=(2,0,0) state in a hydrogen atom, (a) write the solution of the time-independent Schrodinger equation,

(b) verify explicitly that it is a solution with the expected angular momentum and energy.

For states where l = n - t the radial probability assumes the general form given in Exercise 54. The proportionality constant that normalizes this radial probability is given in Exercise 64.

(a) Show that the expectation value of the hydrogen atom potential energy is exactly twice the total energy. (It turns out that this holds no matter what l may be)

(b) Argue that the expectation value of the kinetic energy must be the negative of the total energy.

Prove that if the functioneiDφis to meet itself smoothly whenφchanges by 2π, D must be an integer.

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