Q39P Question: In a crystal, the elec... [FREE SOLUTION]

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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 6: Q39P (page 291)

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 $r 鈮� d_{1}, r 鈮� d_{2}, r 鈮� d_{3}$ show that
$H' = V_{0} + 3 (尾_{1} x^{2} + 尾_{2} y^{2} + 尾_{3} z^{2}) - (尾_{1} + 尾_{2} + 尾_{3}) r^{2}$
where
$尾_{i} 鈮� - \frac{e}{4 {蟺蔚}_{0}} \frac{q_{i}}{d_{i}^{3}}, a n d V_{0} = 2 (尾_{1} d_{1}^{2} + 尾_{2} d_{2}^{2} + 尾_{3} d_{3}^{2})$
(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 $尾_{1} = 尾_{2} = 尾_{3};$ , (ii) in the case of tetragonal symmetry $尾_{1} = 尾_{2} 鈮� 尾_{3};$ , (iii) in the general case of orthorhombic symmetry (all three different)?

Short Answer

Expert verified

Answer

(a) The given expression is verified.

(b) The lowest-order correction to the ground state energy is $V_{0}$ .

(ii) If $尾_{1} = 尾_{2} 鈮� 尾_{3},$ three level

(iii). if $尾_{1} 鈮� 尾_{2} 鈮� 尾_{3},$ no level and then all states have a different energy, and there is no degeneracy.

Step by step solution

Step 1: Definition of the ground state energy.

A quantum mechanical system's ground state is its stationary, lowest energy state; this energy is often referred to as the system's zero-point energy.

(a) Verification of the expression

Consider that interaction between electron and charges at $x = 卤 d$ .

$V = - \frac{e q}{4 蟺蔚_{0}} (\frac{1}{\sqrt{{(x + d)}^{2} + y^{2} + z^{2}}} + \frac{1}{\sqrt{{(x - d)}^{2} + y^{2} + z^{2}}})$

Simplify the expression, $\sqrt{{(x + d)}^{2} + y^{2} + z^{2}}$ .

${[{(x 卤 d)}^{2} + y^{2} + z^{2}]}^{- 1 / 2} = {[x^{2} 卤 2 x d + d^{2} + y^{2} + z^{2}]}^{- 1 / 2} {[{(x 卤 d)}^{2} + y^{2} + z^{2}]}^{- 1 / 2} = {[d^{2} 卤 2 x d + r^{2}]}^{- 1 / 2} {[{(x 卤 d)}^{2} + y^{2} + z^{2}]}^{- 1 / 2} = \frac{1}{d} {[1 卤 \frac{2 x}{d} + \frac{r^{2}}{d^{2}}]}^{- 1 / 2} 鈮� \frac{1}{d} (1 鈭� \frac{x}{d} - \frac{r^{2}}{2 d^{2}} + \frac{3 x^{2}}{2 d^{2}}) = \frac{1}{d} (1 鈭� \frac{x}{d} + \frac{3 x^{2} - r^{2}}{2 d^{2}})$

Substitute the above value in .

$V = - \frac{e q}{4 蟺蔚_{0} d} (1 - \frac{x}{d} + \frac{3 x^{2} - r^{2}}{2 d^{2}} + 1 + \frac{x}{d} + \frac{3 x^{2} - r^{2}}{2 d^{2}}) = - \frac{e q}{4 蟺蔚_{0} d} (2 + \frac{3 x^{2} - r^{2}}{d^{2}})$

Substitute $尾$ for $- \frac{e q}{4 蟺蔚_{0} d^{3}}$ in the above expression.

For all six charges,

$H' = 2 (尾_{1} d_{1}^{2} + 尾_{2} d_{2}^{2} + 尾_{3} d_{3}^{2}) + 3 (尾_{1} x^{2} + 尾_{2} y^{2} + 尾_{3} z^{2}) - r^{2} (尾_{1} + 尾_{2} + 尾_{3}) = V_{0} + 3 (尾_{1} x^{2} + 尾_{2} y^{2} + 尾_{3} z^{2}) - r^{2} (尾_{1} + 尾_{2} + 尾_{3})$

Thus, the given expression is verified.

(b) Determination of the lowest order correction

Perform the correction on ground state energy.

$蠄_{100} = \frac{e^{- r / a}}{\sqrt{蟺 a^{3}}} E_{1}^{1} = <蠄_{100} |H'| 蠄_{100}> = <蠄_{100} |V_{0}| 蠄_{100}> + 3 (尾_{1} <x^{2}> + 尾_{2} <y^{2}> + 尾_{3} <z^{2}>)$

The first term is equal to $V_{0}$ as the wave function is normalized. Write the value of $r^{2}$ .

$r^{2} = x^{2} + y^{2} + z^{2} <r^{2}> = <x^{2}> + <y^{2}> + <z^{2}>$

The function is spherically symmetric. Write the value of .

$<x^{2}> = <y^{2}> = <z^{2}> <x^{2}> = \frac{<r^{2}>}{3}$

Write the value of the lowest order correction to the ground state.

$E_{1}^{1} = V_{0} + 3 (尾_{1} + 尾_{2} + 尾_{3}) \frac{<r^{2}>}{3} - 3 <r^{2}> (尾_{1} + 尾_{2} + 尾_{3}) = V_{0}$

Thus, the lowest-order correction to the ground state energy is v₀ .

(c) Determination of the first order corrections of energy when n=2

Write the expression for the wave function.

$蠄_{200} = \frac{1}{\sqrt{2 蟺 a}} \frac{1}{2 a} (1 - \frac{r}{2 a}) e^{- r / (2 a)} 蠄_{21 卤 1} = 鈭� \frac{1}{\sqrt{蟺 a}} \frac{1}{8 a^{2}} r e^{- r / (2 a)} \sin 胃 e^{卤 i 蠁} 蠄_{210} = \frac{1}{\sqrt{2 蟺 a}} \frac{1}{4 a^{2}} r e^{- r / (2 a)} \cos 胃$

Construct the perturbation matrix.

$<蠄_{200} |H'| 蠄_{200}> = V_{0} <蠄_{21 卤 1} |H'| 蠄_{21 卤 1}> = V_{0} + 3 (尾_{1} <x^{2}> + 尾_{2} <y^{2}> + 尾_{3} <z^{2}>) - <r^{2}> (尾_{1} + 尾_{2} + 尾_{3}) <r^{2}> = \frac{n^{2} a^{2}}{2} [5 n^{2} - 3 l (l + 1) + 1]$

When n = 2, I = 1 then $<r^{2}> = 30 a^{2}$ .

$<y^{2}> ~ {鈭�}_{0}^{2 蟺} \sin^{2} 蠁 d 蠁 <r^{2}> = 2 <x^{2}> + <z^{2}> 鈫� <x^{2}> = \frac{1}{2} (<r^{2}> - <z^{2}>)$

Calculate the value of $x^{2}, y^{2}, a n d z^{2}$

role="math" localid="1659007692072" $<z^{2}> = \frac{1}{2 蟺 a} \frac{1}{16 a^{4}} 鈭� r^{2} e^{- r / a} \cos^{2} 胃 r^{2} \cos^{2} 胃 r^{2} d r \sin 胃 d 胃 d 蠁 = \frac{1}{16 a^{5}} {鈭�}_{0}^{鈭�} r^{6} e^{- r / a} d r {鈭�}_{0}^{蟺} \cos^{4} 胃 \sin 胃 d 胃 = \frac{1}{16 a^{5}} 路 a^{7} 6! 路 \frac{2}{5} <x^{2}> = <y^{2}> = 6 a^{2} <蠄_{210} |H^{'}| 蠄_{210}> = V_{0} - 12 a^{2} (尾_{1} + 尾_{2}) + 24 a^{2} 尾_{3}$

Determinethe off-diagonal element.

$<蠄_{200} |H^{'}| 蠄_{21 卤 1}> = 3 (尾_{1} <x^{2}> + 尾_{2} <y^{2}> + 尾_{3} <z^{2}>) - <r^{2}> (尾_{1} + 尾_{2} + 尾_{3}) <x^{2}> ~ {鈭�}_{0}^{2 蟺} \cos^{2} 蠁 e^{卤 i 蠁} d 蠁 = 0 <z^{2}> ~ {鈭�}_{0}^{蟺} \cos^{3} 胃 \sin 胃 d 胃 = 0 <r^{2}> ~ {鈭�}_{0}^{蟺} \cos 胃 \sin 胃 d 胃 = 0 <蠄_{211} |H^{'}| 蠄_{21 - 1}> = 3 (尾_{1} <x^{2}> + 尾_{2} <y^{2}> + 尾_{3} <z^{2}>) - <r^{2}> (尾_{1} + 尾_{2} + 尾_{3}) <x^{2}> = - \frac{1}{蟺 a} \frac{1}{64 a^{4}} 鈭� r^{2} e^{- 蟿 / a} r^{2} \sin^{2} 胃 e^{- 2 i 蠁} r^{2} \sin^{2} 胃 \cos^{2} 蠁 r^{2} d r \sin 胃 d 胃 d 蠁 = - \frac{1}{64 蟺 a^{5}} {鈭�}_{0}^{鈭�} r^{6} e^{- r / a} d r {鈭�}_{0}^{蟺} \sin^{5} 胃 d 胃 {鈭�}_{0}^{2 蟺} \cos^{2} 蠁 e^{- 2 i 蠁} d 蠁 = - \frac{1}{64 蟺 a^{5}} 路 6! a^{7} 路 \frac{16}{15} 路 \frac{蟺}{2}$

$<x^{2}> = - 6 a^{2} A p p l y {鈭�}_{0}^{2 蟺} \sin^{2} 蠁 e^{- 2 i 蠁} d 蠁 = - 蟺 / 2 <y^{2}> = - <x^{2}> = 6 a^{2} <蠄_{211} |H^{'}| 蠄_{21 - 1}> = 18 a^{2} (- 尾_{1} + 尾_{2})$

Construct the perturbation matrix.

$W = (\begin{matrix} V_{0} & 0 & 0 & 0 \\ 0 & V_{0} - 12 a^{2} (尾_{1} + 尾_{2} - 2 尾_{3}) & 0 & 0 \\ 0 & 0 & V_{0} + 6 a^{2} (尾_{1} + 尾_{2} - 2 尾_{3}) & 18 a^{2} (- 尾_{1} + 尾_{2}) \\ 0 & 0 & 18 a^{2} (- 尾_{1} + 尾_{2}) & V_{0} + 6 a^{2} (尾_{1} + 尾_{2} - 2 尾_{3}) \end{matrix})$

Determine the eigenvalues andseparately diagonalize.

$(V_{0} - E) [V_{0} - 12 a^{2} (尾_{1} + 尾_{2} - 2 尾_{3}) - E] = 0 E_{1} = V_{0}, E_{2} = V_{0} - 12 a^{2} (尾_{1} + 尾_{2} - 2 尾_{3}) V_{0} + 6 a^{2} (尾_{1} + 尾_{2} - 2 尾_{3}) - E 18 a^{2} (尾_{2} - 尾_{1}) 18 a^{2} (尾_{2} - 尾_{1}) V_{0} + 6 a^{2} (尾_{1} + 尾_{2} - 2 尾_{3}) - E {[V_{0} + 6 a^{2} (尾_{1} + 尾_{2} - 2 尾_{3}) - E]}^{2} - {[18 a^{2} (尾_{2} - 尾_{1})]}^{2} = 0$

Thus,

i) If $尾_{1} = 尾_{2} = 尾_{3},$ } then: $E_{1} = E_{2} = E_{3} = E_{4} = V_{0}$

one level of degeneracy =4

ii) If $尾_{1} = 尾_{2} 鈮� 尾_{3},$

iii). if $尾_{1} 鈮� 尾_{2} 鈮� 尾_{3},$ then all states have different energy, and there is no degeneracy

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Short Answer

Step by step solution

Step 1: Definition of the ground state energy.

(a) Verification of the expression

(b) Determination of the lowest order correction

(c) Determination of the first order corrections of energy when n=2

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