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Chapter 2: Problem 19

Demonstrate that accidental degeneracies can exist in a rectangular infinite square-well potential provided that the lengths of the sides are in a rational proportion, that is when \(L_{1}=\lambda L_{2},\) with \(\lambda\) an integer. What is the degeneracy?

Short Answer

Expert verified
Degeneracy can occur when the lengths of the sides of the well are in a rational proportion, specifically when \(L_{1}=\lambda L_{2}\), where \(\lambda\) is an integer. The degeneracy for a given value of \(n\) is given by \(D(n)=\sum_{m=1}^{n}\)gcd\((m,n)\).

Step by step solution

01

Define the wavefunctions

The wavefunctions in a rectangular infinite potential well are given by \(\phi_{n1,n2}(x_{1},x_{2})=A\sin(n_{1}\pi x_{1}/L_{1})\sin(n_{2}\pi x_{2}/L_{2})\) where \(n_{1}\) and \(n_{2}\) are the quantum numbers, \(L_{1}\) and \(L_{2}\) are the lengths of the sides, and \(A\) is the normalization constant.
02

Calculate the energy eigenvalues

The energy eigenvalues for these wavefunctions are given by \(E_{n1,n2}=h^{2}(n_{1}^{2}/2mL_{1}^{2}+n_{2}^{2}/2mL_{2}^{2})\) where \(h\) is the Planck constant and \(m\) is the mass of the particle.
03

Find conditions for degeneracy

We can see that for accidental degeneracy to occur, the energy eigenvalues of two wavefunctions need to be the same. This can be achieved if \(L_{1}=\lambda L_{2}\), where \(\lambda\) is an integer. To prove this, we set the energy eigenvalues equal to each other and solve for \(L_{1}\) and \(L_{2}\). After simplifying, we find that \(L_{1}/L_{2}=\sqrt{n_{2}^{2}/n_{1}^{2}}\), which can only be satisfied if \(n_{1}\) and \(n_{2}\) are integers and \(L_{1}=\lambda L_{2}\) with \(\lambda\) an integer.
04

Solve for degeneracy

The degeneracy of energy levels is the number of wavefunctions that have the same energy. This is given by \(D(n)=\sum_{m=1}^{n}\)gcd\((m,n)\), where gcd\((m,n)\) is the greatest common divisor of \(m\) and \(n\). Therefore, for a given value of \(n\), the degeneracy is the sum of the greatest common divisors of all numbers less than or equal to \(n\) and \(n\) itself.

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Most popular questions from this chapter

Calculate the energies and wavefunctions for a particle in a one-dimensional square well in which the potential energy rises to a finite value \(V\) at each end, and is zero inside the well; that is \\[ \begin{array}{ll} V(x)=V & x \leq 0 \text { and } x \geq L \\ V(x)=0 & 0

Consider a harmonic oscillator of mass \(m\) undergoing harmonic motion in two dimensions \(x\) and \(y .\) The potential energy is given by \(V(x, y)=\frac{1}{2} k_{x} x^{2}+\frac{1}{2} k_{y} y^{2}\). (a) Write down the expression for the hamiltonian operator for such a system. (b) What is the general expression for the allowable energy levels of the two- dimensional harmonic oscillator? (c) What is the energy of the ground state (the lowest energy state)? Hint. The hamiltonian operator can be written as a sum of operators.

The root mean square deviation of the particle from its mean position is \(\Delta x=\left\\{\left\langle x^{2}\right\rangle-\langle x\rangle^{2}\right\\}^{1 / 2} .\) Evaluate this quantity for a particle in a well and show that it approaches its classical value as \(n \rightarrow \infty\). Hint. Evaluate \(\left\langle x^{2}\right\rangle=\int_{0}^{L} x^{2} \psi^{2}(x) \mathrm{d} x\) In the classical case the distribution is uniform across the box, and so in effect \(\psi(x)=1 / L^{1 / 2}.\)

(a) Show that the variables in the Schrödinger equation for a cubic box may be separated and the overall wavefunctions expressed as \(X(x) Y(y) Z(z)\) (b) Deduce the energy levels and wavefunctions. (c) Show that the function .are orthonormal. (d) What is the degeneracy of the level with \(E=14\left(h^{2} / 8 m L^{2}\right) ?\)

Calculate the values of (a) \(\langle x\rangle\) (b) \(\left\langle x^{2}\right\rangle\) \(,(\mathrm{c})\left\langle p_{x}\right\rangle,(\mathrm{d})\left\langle p_{x}^{2}\right\rangle\) for a harmonic oscillator in its ground state by evaluation of the appropriate integrals (as in Problems \(2.13-2.15\) ). Examine the value of \(\Delta x \Delta p_{x}\) in the light of the uncertainty principle. Hint. Use the integrals \\[ \begin{array}{l} \int_{-\infty}^{\infty} \mathrm{e}^{-\alpha x^{2}} \mathrm{d} x=\left(\frac{\pi}{\alpha}\right)^{1 / 2} \\ \int_{0}^{\infty} x \mathrm{e}^{-\alpha x^{2}} \mathrm{d} x=\frac{1}{2 \alpha} \\\ \int_{-\infty}^{\infty} x^{2} \mathrm{e}^{-\alpha x^{2}} \mathrm{d} x=\frac{1}{2}\left(\frac{\pi}{\alpha^{3}}\right)^{1 / 2} \end{array} \\]
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