Question

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

$\mathbf{H}\mathbf{=}\mathbf{-}\mathbf{\gamma B}$ (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:

$\{\begin{array}{l}x+,withenergy{E}_{+}=-\left(\gamma B_0ħ\right)/2\\ x-,withenergy{E}_{-}=+-\left(\gamma B_0ħ\right)/2\end{array}$ (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′, 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

(a)$\mathbf{<}{\mathbf{x}}_{\mathbf{a}}\left|H^{\text{'}}\right|{\mathbf{x}}_{\mathbf{b}}\mathbf{>}\mathbf{=}\frac{{\mathbf{eB}}_{\mathbf{x}}\mathbf{ħ}}{\mathbf{2}\mathbf{m}}\mathbf{(}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{)}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{1}\\ \mathbf{1}& \mathbf{0}\end{array}\mathbf{\right)}\mathbf{\left(}\begin{array}{c}\mathbf{1}\\ \mathbf{0}\end{array}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{eB}}_{\mathbf{x}}\mathbf{ħ}}{\mathbf{2}\mathbf{m}}\mathbf{(}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{)}\mathbf{\left(}\begin{array}{c}\mathbf{0}\\ \mathbf{1}\end{array}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{eB}}_{\mathbf{x}}\mathbf{ħ}}{\mathbf{2}\mathbf{m}}\mathbf{.}\mathbf{}\mathbf{so}\mathbf{}\mathbf{h}\mathbf{=}\frac{{\mathbf{eB}}_{\mathbf{x}}\mathbf{ħ}}{\mathbf{2}\mathbf{m}}\mathbf{.}$

(b) ${\mathit{E}}_{\mathbf{g}\mathbf{s}}\mathbf{\approx }{\mathit{E}}_{\mathbf{a}}\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{2}}}{(E_b-E_a)}\mathbf{=}\frac{\mathbf{e}\mathbf{ħ}}{\mathbf{2}\mathbf{m}}\mathbf{(}{\mathbf{B}}_{\mathbf{z}}\mathbf{+}\frac{{\mathbf{B}}_{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{B}}_{\mathbf{z}}}\mathbf{)}\mathbf{.}$

(c) ${\mathit{E}}_{\mathbf{g}\mathbf{s}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\sqrt{{\left(\frac{e{B}_zħ}{m}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{4}{\left(\frac{e{B}_zħ}{m}\right)}^{\mathbf{2}}}\mathbf{=}\mathbf{-}\frac{\mathbf{e}\mathbf{ħ}}{\mathbf{2}\mathbf{m}}\sqrt{{{\mathbf{B}}^{\mathbf{2}}}_{\mathbf{z}}\mathbf{+}{{\mathbf{B}}^{\mathbf{2}}}_{\mathbf{z}}}\mathbf{.}$

Step-by-step solution

(a)Finding the matrix elements of H'

For the electron, . (Eq. 4.161).

For consistency with Problem 7.15,

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

$\left\{\begin{array}{l}{x}^{+},\mathrm{with}\mathrm{energy}{\mathrm{E}}_{+}=-\left(\gamma B_0ħ\right)/2\\ x-,\mathrm{with}\mathrm{energy}{\mathrm{E}}_{-}=+\left({\mathrm{\gamma B}}_0\mathrm{ħ}\right)/2\end{array}\right.$

(b)Finding the new ground state energy

From Problem 7.15(b),

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

(c)Finding the variation principle bound

From Problem 7.15(c),

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