Q16P As an explicit example of the me... [FREE SOLUTION]

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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 8: Q16P (page 338)

As an explicit example of the method developed inProblem 7.15, consider an electron at rest in a uniform magnetic field $B = B_{2} \overset{铃�}{K}$ for which the Hamiltonian is (Equation 4.158):
$H = - 纬 B$ (4.158).
$H^{0} = \frac{e B_{z}}{m} S_{z}$ (7.57).
The eigenspinors localid="1656062306189" $x_{a} and x_{b} and the corresponding energies, E_{a} and E_{b}$ , are given in Equation 4.161. Now we turn on a perturbation, in the form of a uniform field in the x direction:
${\begin{cases} x +, with energy E_{+} = - ({纬叠}_{0} 魔) / 2 \\ x -, with energy E_{-} = - ({纬叠}_{0} 魔) / 2 \end{cases}$ (4.161).
$H^{'} = \frac{e b_{x}}{m} S_{x}$ (7.58).
(a) Find the matrix elements of H鈥�, and confirm that they have the structure of Equation 7.55. What is h?
(b) Using your result inProblem 7.15(b), find the new ground state energy, in second-order perturbation theory.
(c) Using your result inProblem 7.15(c), find the variation principle bound on the ground state energy.

Short Answer

Expert verified

$(a) < x_{a} |H^{'}| x_{b} > = \frac{e B_{x} 魔}{2 m} (\begin{matrix} 0 & 1 \end{matrix}) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} 1 \\ 0 \end{matrix}) = \frac{e B_{x} 魔}{2 m} (\begin{matrix} 0 & 1 \end{matrix}) (\begin{matrix} 1 \\ 0 \end{matrix}) = \frac{e B_{x} 魔}{2 m} . s o h = \frac{e B_{x} 魔}{2 m} . (b) E_{g s} 鈮� E_{a} - \frac{h^{2}}{(E_{b} - E_{a})} = - \frac{e 魔}{2 m} (B_{z} + \frac{B_{x}^{2}}{2 B_{z}}) . (c) E_{g s} = \frac{1}{2} \sqrt{{(\frac{e B_{x} 魔}{2 m})}^{2} + 4 {(\frac{e B_{x} 魔}{2 m})}^{2}} = \frac{e 魔}{2 m} \sqrt{B_{z}^{2} + B_{x}^{2}} .$

Step by step solution

(a)Finding the matrix elements of H'

For the electron, $纬 = - e / m, s o E_{卤} = 卤 e B_{z} 魔 .$ (Eq. 4.161).

For consistency with Problem 7.15, Eb>Ea,

$s o X_{B} = x_{+} = (\begin{matrix} 1 \\ 0 \end{matrix}), x_{a} = x_{-} = (\begin{matrix} 0 \\ 1 \end{matrix}), E_{b} = E_{+} = \frac{e B_{z} 魔}{2 m}, E_{a} = E_{-} = - \frac{e B_{z} 魔}{2 m} . \{\begin{cases} x^{+}, w i t h e n e r g y E_{+} = - (纬 B_{0} 魔) / 2 \\ x^{-}, w i t h e n e r g y E_{+} = + (纬 B_{0} 魔) / 2 \end{cases} (4.161) <x_{a} |H^{'}| x_{a}> = \frac{e B_{x} 魔}{m} (\begin{matrix} 0 & 1 \end{matrix}) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} 0 \\ 1 \end{matrix}) = \frac{e B_{x} 魔}{2 m} (\begin{matrix} 0 & 1 \end{matrix}) (\begin{matrix} 1 \\ 0 \end{matrix}) . <x_{b} |H^{'}| x_{b}> = \frac{e B_{x} 魔}{2 m} (\begin{matrix} 1 & 0 \end{matrix}) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} 1 \\ 0 \end{matrix}) = 0; <x_{b} |H^{'}| x_{a}> = \frac{e B_{x} 魔}{2 m} (\begin{matrix} 1 & 0 \end{matrix}) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} 0 \\ 1 \end{matrix}) = \frac{e B_{x} 魔}{2 m} (10) (0110) (10) . <x_{a} |H^{'}| x_{b}> = \frac{e B_{x} 魔}{2 m} (\begin{matrix} 0 & 1 \end{matrix}) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} 1 \\ 0 \end{matrix}) = \frac{e B_{x} 魔}{2 m} (\begin{matrix} 0 & 1 \end{matrix}) (\begin{matrix} 0 \\ 1 \end{matrix}) = \frac{e B_{x} 魔}{2 m} . s o h = \frac{e B_{x} 魔}{2 m} . and the conditions of Problem 7.15 are met .$

(b)Finding the new ground state energy

From Problem 7.15(b),

$E_{g s} 鈮� E_{4} - \frac{h^{2}}{(E_{b} - E_{a})} = - \frac{e B_{z} 魔}{2 m} - \frac{{(e B_{z} 魔 / 2 m)}^{2}}{(e B_{z} 魔 / m)} = - \frac{e 魔}{2 m} (B_{z} + \frac{B_{x}^{2}}{2 B_{z}}) .$

(c)Finding the variation principle bound

From Problem 7.15(c),

$E_{g s} = \frac{1}{2} [E_{a} + E_{b} - \sqrt{{(E_{b} - E_{a})}^{2} + 4 h^{2}}], (it 鈥� s actually the exact ground state) . E_{gs} = \frac{1}{2} \sqrt{{(\frac{{eB}_{2} 魔}{m})}^{2} + 4 {(\frac{{eB}_{2} 魔}{2 m})}^{2}} = \frac{别魔}{2 m} \sqrt{B_{z}^{2} + B_{x}^{2}} . which was obvious from the start, since the square root is simply the magnitude of the total field) .$

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Short Answer

Step by step solution

(a)Finding the matrix elements of H'

(b)Finding the new ground state energy

(c)Finding the variation principle bound

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