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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition · 469 pages

ISBN: 9780131118928

Chapter 6: Q16P (page 276)

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

Short Answer

Expert verified

a) $i h (L \times S)$

b) $i h (S \times L)$

c) 0

d) 0

e) 0

f) 0

Step by step solution

01

Step 1:(a)

$[L \cdot S, L_{x}] = [L_{x} S_{x} + L_{y} S_{y} + L_{z} S_{z}, L_{X}] = S_{x} [L_{x}, L_{x}] + S_{y} [L_{y}, L_{x}] + S_{z} [L_{z}, L_{x}] = S_{x} (0) + S_{y} (- i h L_{z}) + S_{z} (- i h L_{y}) = i h (L_{y} S_{z} - L_{z} S_{y}) = i h {(L \times S)}_{x}$

Same goes for the other two components, so

$[L \cdot S, L] = i h (L \times S)$

02

Step 2:(b)

$[L \cdot S, S]$ is identical, only with $L \leftrightarrow S$ :

$[L \cdot S, S] = i h (S \times L)$

03

:(c)

$[L \cdot S, J] = [L \cdot S, L] + [L \cdot S, S] = i h (L \times S + S \times L) = 0$

04

Step 4:(d)

$L^{2}$ Commutes with all components of L (and S),

So $[L . S, L^{2}] = 0$

05

Step 5:(e)

Likewise $[L . S, S^{2}] = 0$

06

Step 6:(f)

$[L . S, J^{2}] = [L . S, L^{2}] + [L . S, S^{2}] + 2 [L . S, L . S] = 0 + 0 + 0 = [L . S, J^{2}] = 0$

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Most popular questions from this chapter

Let aand bbe two constant vectors. Show that

$\int (a . \overset{⏜}{r}) (b . \overset{⏜}{r}) sinθdθdϕ = \frac{4 π}{3} (a . b)$

(the integration is over the usual range: $0 < θ < π, 0 < ϕ < 2 π$ ). Use this result to demonstrate that

$(\frac{3 (S_{p} . \overset{⏜}{r}) (S_{e} . \overset{⏜}{r}) - S_{p} . S_{e}}{r^{3}}) = 0$

For states with I=0. Hint: $\overset{⏜}{r} = sinθcosϕ \overset{⏜}{i} + sinθsinϕ \overset{⏜}{Φ} + cosθ \overset{⏜}{k}$ .

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

$μ_{d l} = \frac{g_{d} e}{2 m_{d}} S_{d}$

he deuteron g-factor is 1.71.

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).

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)?

Problem 6.6 Let the two "good" unperturbed states be

$ψ_{\pm}^{0} = α \pm ψ_{a}^{0} + β_{\pm} ψ_{b}^{0}$

where $α_{\pm}$ and $β_{\pm}$ are determined (up to normalization) by Equation 6.22(orEquation6.24). Show explicitly that

(a)are orthogonal;role="math" localid="1655966589608" $(⟨ ψ_{+}^{0} ∣ ψ_{-}^{0} ⟩ = 0);$

(b) $⟨ ψ_{+}^{0} | H^{'} | ψ_{-}^{0} ⟩ = 0;$

(c) $⟨ ψ_{\pm}^{0} | H^{'} | ψ_{\pm}^{0} ⟩ = E_{\pm}^{1},$ with $E_{\pm}^{1}$ given by Equation 6.27.

See all solutions

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