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As an explicit example of the method developed inProblem 7.15, consider an electron at rest in a uniform magnetic field $\mathbf{B}\mathbf{=}{\mathbf{B}}_{\mathbf{2}}\stackrel{\mathbf{铃�}}{\mathbf{k}}$for which the Hamiltonian is (Equation 4.158):

$\mathbf{H}\mathbf{=}\mathbf{-}\mathbf{纬叠}$ (4.158).

${\mathit{H}}^{\mathbf{0}}\mathbf{=}\frac{\mathbf{e}{\mathbf{B}}_{\mathbf{z}}}{\mathbf{m}}{\mathit{S}}_{\mathbf{z}}$ (7.57).

The eigenspinors, ${\mathit{x}}_{\mathbf{a}}$arelocalid="1655969802629" ${\mathbf{x}}_{\mathbf{b}}\mathbf{,}\mathbf{and}\mathbf{}\mathbf{the}\mathbf{}\mathbf{corresponding}\mathbf{}\mathbf{energies}\mathbf{,}{\mathbf{E}}_{\mathbf{a}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{E}}_{\mathbf{b}}\mathbf{,}\mathbf{}\mathbf{are}$given in Equation 4.161. Now we turn on a perturbation, in the form of a uniform field in the x direction:

$\{\begin{array}{l}x+,withenergy{E}_{+}=-\left(纬B_0魔\right)/2\\ x-,withenergy{E}_{-}=+-\left(纬B_0魔\right)/2\end{array}$ (4.161).

${\mathit{H}}^{\mathbf{\text{'}}}\mathbf{=}\frac{\mathbf{e}{\mathbf{B}}_{\mathbf{x}}}{\mathbf{m}}{\mathit{S}}_{\mathbf{x}}$ (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.

Step by step solution

01

(a)Finding the matrix elements of H'

For the electron, . (Eq. 4.161).

For consistency with Problem 7.15,

$$\begin{gathered} so{x}_b{x}_{+}=\left(\begin{array}{c}1\\ 0\end{array}\right),x_a=x_{-}=\left(\begin{array}{c}0\\ 1\end{array}\right), \\ {E}_b=E_x=\frac{{\mathrm{eB}}_{\mathrm{z}}\mathrm{魔}}{2\mathrm{m}},E_a=E_{-}=-\frac{{\mathrm{eB}}_{\mathrm{z}}\mathrm{魔}}{2\mathrm{m}}. \end{gathered}$$

$\left\{\begin{array}{l}{x}^{+},\mathrm{with}\mathrm{energy}{\mathrm{E}}_{+}=-\left(纬B_0魔\right)/2\\ x-,\mathrm{with}\mathrm{energy}{\mathrm{E}}_{-}=+\left({\mathrm{纬叠}}_0\mathrm{魔}\right)/2\end{array}\right.$

02

(b)Finding the new ground state energy

From Problem 7.15(b),

$E_{gs}鈮�E_a-\frac{h^2}{\left(E_b{E}_a\right)}=-\frac{e{B}_x魔}{2m}-\frac{{\left(e{B}_x魔/2m\right)}^2}{(e{B}_x魔/m)}=-\frac{e魔}{2m}\left(B_z+\frac{B_x^2}{2B_z}\right).$

03

(c)Finding the variation principle bound

From Problem 7.15(c),

$$\begin{gathered} E_{gs}=\frac{1}{2}\left[E_a+E_a-\sqrt{{\left(E_b-E_a\right)}^2+4h^2}\right],(\mathrm{it}鈥�\mathrm{s}\mathrm{actually}\mathrm{the}\mathrm{exact}\mathrm{ground}\mathrm{state}). \\ {E}_{gs}=\frac{1}{2}\sqrt{{\left(\frac{e{B}_z魔}{m}\right)}^2+4{\left(\frac{e{B}_z魔}{m}\right)}^2}=-\frac{e魔}{2m}\sqrt{B_z^2+B_x^2}.\mathrm{which}\mathrm{was}\mathrm{obvious}\mathrm{from}\mathrm{the}\mathrm{start}, \\ \mathrm{since}\mathrm{the}\mathrm{square}\mathrm{root}\mathrm{is}\mathrm{simply}\mathrm{the}\mathrm{magnitude}\mathrm{of}\mathrm{the}\mathrm{total}\mathrm{field}). \end{gathered}$$

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