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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

Find the (lowest order) relativistic correction to the energy levels of the one-dimensional harmonic oscillator. Hint: Use the technique in Example 2.5 .

Consider a particle of mass m that is free to move in a one-dimensional region of length L that closes on itself (for instance, a bead that slides frictionlessly on a circular wire of circumference L, as in Problem 2.46).

(a) Show that the stationary states can be written in the form $ψ_{n} (x) = \frac{1}{\sqrt{L}} e^{2 π i n x / L}, (- L / 2 < x < L / 2),$

where $n = 0, \pm 1, \pm 2, . . . .$ and the allowed energies are $E_{n} = \frac{2}{m} {(\frac{n π ℏ}{L})}^{2} .$ Notice that with the exception of the ground state (n = 0 ) – are all doubly degenerate.

(b) Now suppose we introduce the perturbation, $H^{'} = - V_{0} e^{- x^{2} / a^{2}}$ where $a ≪ L a$ . (This puts a little “dimple” in the potential at x = 0, as though we bent the wire slightly to make a “trap”.) Find the first-order correction to $E_{n}$ , using Equation 6.27. Hint: To evaluate the integrals, exploit the fact that $a ≪ L a$ to extend the limits from $\pm L / 2 to \pm \infty$ after all, H′ is essentially zero outside $- a < x < a$ .

$E_{\pm}^{1} = \frac{1}{2} [W_{a a} + W_{b b} \pm \sqrt{{(W_{a a} - W_{b b})}^{2} + 4 {|W_{a b}|}^{2}}]$ (6.27).

(c) What are the “good” linear combinations of $ψ_{n}$ and $ψ_{- n}$ , for this problem? Show that with these states you get the first-order correction using Equation 6.9.

$E_{n}^{'} = <ψ_{n}^{0} |H^{'}| ψ_{n}^{0}>$ (6.9).

(d) Find a hermitian operator A that fits the requirements of the theorem, and show that the simultaneous Eigenstates of $H_{0}$ and A are precisely the ones you used in (c).

The Feynman-Hellmann theorem (Problem 6.32) can be used to determine the expectation values of $1 / r$ and $1 / r^{2}$ for hydrogen. ${}^{23}$ The effective Hamiltonian for the radial wave functions is (Equation4.53)

$\frac{ℏ^{2}}{2 m} \frac{d^{2}}{d r^{2}} + \frac{ℏ^{2}}{2 m} \frac{l (l + 1)}{r^{2}} - \frac{e^{2}}{4 π \in_{0}} \frac{1}{r}$

And the eigenvalues (expressed in terms of $l)^{24}$ are (Equation 4.70)

$E_{n} = - \frac{m e^{4}}{32 π^{2} \in_{0}^{2} h^{2} {(j_{\max} + l + 1)}^{2}}$

(a) Use $λ = e$ in the Feynman-Hellmann theorem to obtain $⟨ 1 / r ⟩$ . Check your result against Equation 6.55.

(b) Use $λ = l$ to obtain $<1 / r^{2}>$ . Check your answer with Equation6.56.

If I=0, then j=s, $m_{j} = m_{s}$ , and the "good" states are the same $(n m_{s})$ for weak and strong fields. Determine $E_{z}^{1}$ (from Equation) and the fine structure energies (Equation 6.67), and write down the general result for the I=O Zeeman Effect - regardless of the strength of the field. Show that the strong field formula (Equation 6.82) reproduces this result, provided that we interpret the indeterminate term in square brackets as.

Van der Waals interaction. Consider two atoms a distanceapart. Because they are electrically neutral you might suppose there would be no force between them, but if they are polarizable there is in fact a weak attraction. To model this system, picture each atom as an electron (mass m , charge -e ) attached by a spring (spring constant k ) to the nucleus (charge +e ), as in Figure. We'll assume the nuclei are heavy, and essentially motionless. The Hamiltonian for the unperturbed system is

$H^{0} = \frac{1}{2 m} p_{1}^{2} + \frac{1}{2 k} x_{1}^{2} + \frac{1}{2 m} p_{2}^{2} + \frac{1}{2 k} x_{2}^{2} [6.96]$

The Coulomb interaction between the atoms is

$H^{'} = \frac{1}{4 {πϵ}_{0}} (\frac{e^{2}}{R} - \frac{e^{2}}{R - x_{1}} - \frac{e^{2}}{R + x_{2}} + \frac{e^{2}}{R - x_{1} + x_{2}}$ [6.97]

(a) Explain Equation6.97. Assuming that localid="1658203563220" $| x_{1} |$ and $| x_{2} |$ are both much less than, show that

localid="1658203513972" $H^{'} ≅ - \frac{e^{2} x_{1} x_{2}}{2 π ϵ_{0} R^{3}}$ [6.98]

(b) Show that the total Hamiltonian (Equationplus Equation) separates into two harmonic oscillator Hamiltonians:

$H = [\frac{1}{2 m} p_{+}^{2} + \frac{1}{2} (k - \frac{e^{2}}{2 π ϵ_{0} R^{3}} x_{+}^{2}] + [+ \frac{1}{2 m} p_{-}^{2} + \frac{1}{2} (k + \frac{e^{2}}{2 π ϵ_{0} R^{3}} x_{-}^{2}]$ [6.99]

under the change of variables

$x_{\pm} \equiv \frac{1}{\sqrt{2}} (x_{1} \pm x_{2})$ Which entails $p_{\pm} = \frac{1}{\sqrt{2}} (p_{1} \pm p_{2})$ [6.100]

(c) The ground state energy for this Hamiltonian is evidently

$E = \frac{1}{\sqrt{2}} ℏ (ω_{+} + ω_{-})$ Where $ω_{\pm} = \sqrt{\frac{k \mp (e^{2} / 2 π ϵ_{0} R^{3})}{m}}$ [6.101]

Without the Coulomb interaction it would have been $E_{0} = ħ ω_{0}$ , where $ω_{0} = \sqrt{k / m}$ . Assuming that, show that

$Δ V \equiv E - E_{0} ≅ - \frac{ℏ}{8 m^{2} ω_{0}^{3}} {(\frac{e^{2}}{2 π ϵ_{0}})}^{2} \frac{1}{R^{6}} .$ [6.102]

Conclusion: There is an attractive potential between the atoms, proportional to the inverse sixth power of their separation. This is the van der Waals interaction between two neutral atoms.

(d) Now do the same calculation using second-order perturbation theory. Hint: The unperturbed states are of the form $ψ_{n 1} (x_{1}) ψ_{n 2} (x_{2})$ , where $ψ_{n} (x)$ is a one-particle oscillator wave function with mass mand spring constant $k; Δ V$ is the second-order correction to the ground state energy, for the perturbation in Equation 6.98 (notice that the first-order correction is zero).

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