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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 6: Q38P (page 290)

Calculate the wavelength, in centimeters, of the photon emitted under a hyperfine transition in the ground state (n=1) of deuterium. Deuterium is "heavy" hydrogen, with an extra neutron in the nucleus; the proton and neutron bind together to form a deuteron, with spin 1 and magnetic moment
$渭_{d l} = \frac{g_{d} e}{2 m_{d}} S_{d}$
he deuteron g-factor is 1.71.

Short Answer

Expert verified

The difference in energy is $位_{d} 鈮� 92 cm$

Step by step solution

01

Definehyperfine transition.

The interaction of electrons with the magnetic moments of the nucleus splits a spectral line of an atom or molecule into two or more closely spaced components.

02

Step 2: Draw the difference in energy of deuterium.

Given

$S_{e}^{2} = \frac{1}{2} (\frac{3}{2}) {鈩�}^{2} = \frac{3}{4} {鈩�}^{2}$

渭_{d} = \frac{g_{d} e}{2 m_{d}} S_{d}, E_{h f}^{1} = \frac{渭_{0} g_{d} e^{2}}{3 蟺 m_{d} m_{e} a^{3}} <S_{d} 路 S_{e}>

Spin $S_{d}^{2} = 1 (2) {鈩�}^{2} = 2 {鈩�}^{2}$

$<S_{d} 路 S_{e}> = \{\begin{cases} \frac{1}{2} (\frac{3}{4} {鈩�}^{2} - \frac{3}{4} {鈩�}^{2} - 2 {鈩�}^{2}) = - {鈩�}^{2} \\ \frac{1}{2} (\frac{15}{4} {鈩�}^{2} - \frac{3}{4} {鈩�}^{2} - 2 {鈩�}^{2}) = \frac{1}{2} {鈩�}^{2} \end{cases}$

Difference of two state

$\frac{1}{2} {鈩�}^{2} - (- {鈩�}^{2}) = \frac{3}{2} {鈩�}^{2}$

$= \frac{g_{d} e^{2} {鈩�}^{2}}{2 蟺蔚_{0} c^{2} m_{d} m_{e} a^{3}}$

$= \frac{2 g_{d} e^{2} {鈩�}^{2}}{4 蟺蔚_{0} c^{2} m_{d} m_{e} a^{3}}$

$= \frac{2 g_{d} e^{2} {鈩�}^{4}}{c^{2} m_{d} m^{2} a^{4}}$

$= \frac{3}{2} \frac{g_{d}}{g_{p}} \frac{m_{p}}{m_{d}} 螖 E_{hydrogen}$

$位 = \frac{c}{谓} = \frac{c h}{螖 E}$

localid="1657135928063" $鈬� 位_{d} = \frac{2 g_{d}}{3 g_{p}} \frac{m_{d}}{m_{p}} 位_{h}, m_{d} = 2 m_{p}$

$鈬� 位_{d} = \frac{4}{3} (\frac{5.59}{1.71}) 路 21 cm 鈮� 92 cm$

Thus, the difference in energy is $位_{d} 鈮� 92 cm$

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Most popular questions from this chapter

Work out the matrix elements of $H_{Z}^{'} {and H_{fs}}^{'}$ construct the W matrix given in the text, for n = 2.

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 $蠄_{n} (x) = \frac{1}{\sqrt{L}} e^{2 蟺 i n x / L}, (- L / 2 < x < L / 2),$

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

(b) Now suppose we introduce the perturbation, $H^{'} = - V_{0} e^{- x^{2} / a^{2}}$ where $a 鈮� L a$ . (This puts a little 鈥渄imple鈥� in the potential at x = 0, as though we bent the wire slightly to make a 鈥渢rap鈥�.) Find the first-order correction to $E_{n}$ , using Equation 6.27. Hint: To evaluate the integrals, exploit the fact that $a 鈮� L a$ to extend the limits from $卤 L / 2 to 卤鈭�$ after all, H鈥� is essentially zero outside $- a < x < a$ .

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

(c) What are the 鈥済ood鈥� linear combinations of $蠄_{n}$ and $蠄_{- n}$ , for this problem? Show that with these states you get the first-order correction using Equation 6.9.

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

When an atom is placed in a uniform external electric field ,the energy levels are shifted-a phenomenon known as the Stark effect (it is the electrical analog to the Zeeman effect). In this problem we analyse the Stark effect for the n=1 and n=2 states of hydrogen. Let the field point in the z direction, so the potential energy of the electron is

$H'_{s} = e E_{e x t} z = e E_{e x t} r c o s 胃$

Treat this as a perturbation on the Bohr Hamiltonian (Equation 6.42). (Spin is irrelevant to this problem, so ignore it, and neglect the fine structure.)

(a) Show that the ground state energy is not affected by this perturbation, in first order.

(b) The first excited state is 4-fold degenerate: $Y_{200}, Y_{211}, Y_{210}, Y_{200}, Y_{21 - 1}$ Using degenerate perturbation theory, determine the first order corrections to the energy. Into how many levels does $E_{2}$ split?

(c) What are the "good" wave functions for part (b)? Find the expectation value of the electric dipole moment $(p_{e} = - er)$ in each of these "good" states.Notice that the results are independent of the applied field-evidently hydrogen in its first excited state can carry a permanent electric dipole moment.

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

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

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

localid="1658203513972" $H^{'} 鈮� - \frac{e^{2} x_{1} x_{2}}{2 蟺系_{0} R^{3}}$ [6.98]

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

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

under the change of variables

$x_{卤} 鈮� \frac{1}{\sqrt{2}} (x_{1} 卤 x_{2})$ Which entails $p_{卤} = \frac{1}{\sqrt{2}} (p_{1} 卤 p_{2})$ [6.100]

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

$E = \frac{1}{\sqrt{2}} 鈩� (蝇_{+} + 蝇_{-})$ Where $蝇_{卤} = \sqrt{\frac{k 鈭� (e^{2} / 2 蟺系_{0} R^{3})}{m}}$ [6.101]

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

$螖 V 鈮� E - E_{0} 鈮� - \frac{鈩�}{8 m^{2} 蝇_{0}^{3}} {(\frac{e^{2}}{2 蟺系_{0}})}^{2} \frac{1}{R^{6}} .$ [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 $蠄_{n 1} (x_{1}) 蠄_{n 2} (x_{2})$ , where $蠄_{n} (x)$ is a one-particle oscillator wave function with mass mand spring constant $k; 螖 V$ 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).

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.

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