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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition · 469 pages

ISBN: 9780131118928

Chapter 6: Q12P (page 270)

Question: Use the virial theorem (Problem 4.40) to prove Equation 6.55.

Short Answer

Expert verified

It is proved that $<\frac{1}{r}> = \frac{1}{a n^{2}}$ .

Step by step solution

01

Expression of thepotential energy and Bohr energy

The expression for potential energy is as follows,

$V = \frac{- e^{2}}{4 π {\dot{o}}_{0}} \frac{1}{r}$

The expression for Bohrenergy is as follows,

localid="1658214362022" $E_{n} = - [\frac{m}{2 h^{2}} {(\frac{- e^{2}}{4 π {\dot{o}}_{0}})}^{2}] \frac{1}{n^{2}}$

02

Verification of <1r>=1an2

Write the expression for the relation between potential energy and Bohr energy.

$<V> = 2 E_{n}$

Substitute all the values in the above expression.

localid="1658214341492" $<\frac{- e^{2}}{4 π {\dot{o}}_{0}} \frac{1}{r}> = - 2 [\frac{m}{2 h^{2}} {(\frac{- e^{2}}{4 π {\dot{o}}_{0}})}^{2}] \frac{1}{n^{2}} \frac{- e^{2}}{4 π {\dot{o}}_{0}} <\frac{1}{r}> = - 2 [\frac{m}{2 h^{2}} {(\frac{e^{2}}{4 π {\dot{o}}_{0}})}^{2}] \frac{1}{n^{2}} <\frac{1}{r}> = \frac{m}{2 h^{2}} {(\frac{e^{2}}{4 π {\dot{o}}_{0}})}^{2} (\frac{4 π {\dot{o}}_{0}}{e^{2}}) \frac{1}{n^{2}} = \frac{m}{2 h^{2}} \frac{e^{2}}{4 π {\dot{o}}_{0}} \frac{1}{n^{2}}$

Consider $a = \frac{4 π {\dot{o}}_{0} h^{2}}{{me}^{2}}$ , where, a is the Bohr radius.

Substitute $\frac{4 π {\dot{o}}_{0} h^{2}}{{me}^{2}}$ for a in the above expression.

$<\frac{1}{r}> = \frac{1}{a n^{2}}$

Thus, the equation 6.55 is verified.

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Most popular questions from this chapter

Question: In a crystal, the electric field of neighbouring ions perturbs the energy levels of an atom. As a crude model, imagine that a hydrogen atom is surrounded by three pairs of point charges, as shown in Figure 6.15. (Spin is irrelevant to this problem, so ignore it.)

(a) Assuming that $r ≪ d_{1}, r ≪ d_{2}, r ≪ d_{3}$ show that

$H' = V_{0} + 3 (β_{1} x^{2} + β_{2} y^{2} + β_{3} z^{2}) - (β_{1} + β_{2} + β_{3}) r^{2}$

where

$β_{i} \equiv - \frac{e}{4 {πε}_{0}} \frac{q_{i}}{d_{i}^{3}}, a n d V_{0} = 2 (β_{1} d_{1}^{2} + β_{2} d_{2}^{2} + β_{3} d_{3}^{2})$

(b) Find the lowest-order correction to the ground state energy.

(c) Calculate the first-order corrections to the energy of the first excited states Into how many levels does this four-fold degenerate system split,

(i) in the case of cubic symmetry $β_{1} = β_{2} = β_{3};$ , (ii) in the case of tetragonal symmetry $β_{1} = β_{2} \neq β_{3};$ , (iii) in the general case of orthorhombic symmetry (all three different)?

Analyze the Zeeman effect for the $n = 3$ states of hydrogen, in the weak, strong, and intermediate field regimes. Construct a table of energies (analogous to Table 6.2), plot them as functions of the external field (as in Figure 6.12), and check that the intermediate-field results reduce properly in the two limiting cases.

Question: Evaluate the following commutators :

a) $[L \cdot S, L]$

b) $[L \cdot S, S]$

c)role="math" localid="1658226147021" $[L \cdot S, J]$

d) $[L \cdot S, L^{2}]$

e) $[L \cdot S, S^{2}]$

f) $[L \cdot S, J^{2}]$

Hint: L and S satisfy the fundamental commutation relations for angular momentum (Equations 4.99 and 4.134 ), but they commute with each other.

$[L_{X}, L_{Y}] = {ihL}_{z}; [L_{y}, L_{z}] = {ihL}_{x}; [L_{z}, L_{x}] = {ihL}_{y} . . . . . . . 4.99 [S_{X}, S_{Y}] = {ihS}_{z}; [S_{y}, S_{z}] = {ihS}_{x}; [S_{z}, S_{x}] = {ihS}_{y} . . . . . . . . 4.134$

Consider a charged particle in the one-dimensional harmonic oscillator potential. Suppose we turn on a weak electric field (E), so that the potential energy is shifted by an amount $H^{'} = - qEx$ .(a) Show that there is no first-order change in the energy levels, and calculate the second-order correction. Hint: See Problem 3.33.

(b) The Schrödinger equation can be solved directly in this case, by a change of variables $x^{'} \equiv x - (qE / {mω}^{2})$ . Find the exact energies, and show that they are consistent with the perturbation theory approximation.

Question: In Problem 4.43you calculated the expectation value of $r^{s}$ in the state $ψ_{321}$ . Check your answer for the special cases s = 0(trivial), s = -1(Equation 6.55), s = -2(Equation 6.56), and s = -3(Equation 6.64). Comment on the case s = -7.

See all solutions

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