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 $B = B_{2} \overset{⏜}{K}$ for which the Hamiltonian is (Equation 4.158):
$H = - γ B$ (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:
${\begin{cases} x +, with energy E_{+} = - ({γB}_{0} ħ) / 2 \\ x -, with energy E_{-} = - ({γB}_{0} ħ) / 2 \end{cases}$ (4.161).
$H^{'} = \frac{e b_{x}}{m} S_{x}$ (7.58).
(a) Find the matrix elements of H′, 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

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

Step by step solution

(a)Finding the matrix elements of H'

For the electron, $γ = - e / m, s o E_{\pm} = \pm e B_{z} ħ .$ (Eq. 4.161).

For consistency with Problem 7.15, Eb>Ea,

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

(b)Finding the new ground state energy

From Problem 7.15(b),

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

(c)Finding the variation principle bound

From Problem 7.15(c),

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

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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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