Chapter 6: Q37P (page 290)

Question: Consider the Stark effect (Problem 6.36) for the states of hydrogen. There are initially nine degenerate states, $ψ_{3 / m}$ (neglecting spin, as before), and we turn on an electric field in the direction.
(a) Construct the matrix representing the perturbing Hamiltonian. Partial answer: $< 300 | z | 310 > = - 3 \sqrt{6} a, < 310 | z | 320 > = - 3 \sqrt{3} a, < 31 \pm 1 | z | 32 \pm 1 > = - (9 / 2) a$ ,,
(b) Find the eigenvalues, and their degeneracies.

Short Answer

Expert verified

Answer

(b)The value Eigenvalues and corresponding degeneracies are

Step by step solution

Definition of the Stark effect.

Due to the existence of an external electric field, the Stark effect causes the spectral lines of atoms and molecules to move and split.

It's the electric-field equivalent of the Zeeman phenomenon, in which the presence of a magnetic field splits a spectral line into many components

Step 2: (a) Creation of the matrix

$\frac{e E_{ext}}{3^{8} \sqrt{3} a^{6}} \cdot {(\frac{3 a}{2})}^{7} \cdot \frac{8}{15} \int_{0}^{\infty} x^{7} (6 - \frac{3}{2} x) e^{- x} d x = \frac{e a E_{ext}}{360 \sqrt{3}} (6 \cdot 6! - \frac{3}{2} \cdot 7!) <310 |H'| 320> = - 3 \sqrt{3} e a E_{ext}$

Calculate perturbations in energy for . In order to do that, the wave functions of hydrogen must be known.

$ψ_{300} = \frac{1}{81 \sqrt{3 π a^{3}}} (27 - \frac{18 r}{a} + \frac{2 r^{2}}{a^{2}}) e^{- r / (3 a)} ψ_{31 \pm 1} = \frac{1}{81 \sqrt{π a^{3}}} (6 - \frac{r}{a}) \frac{r}{a} e^{- r / (3 a)} \sin θ e^{\pm i φ} ψ_{310} = \frac{\sqrt{2}}{81 \sqrt{π a^{3}}} (6 - \frac{r}{a}) \frac{r}{a} e^{- r / (3 a)} \cos θ ψ_{32 \pm 2} = \frac{1}{162 \sqrt{π a^{3}}} \frac{r^{2}}{a^{2}} e^{- r / (3 a)} \sin^{2} θ e^{\pm 2 i φ} ψ_{32 \pm 1} = \frac{1}{81 \sqrt{π a^{3}}} \frac{r^{2}}{a^{2}} e^{- r / (3 a)} \sin θ \cos θ e^{\pm i φ} ψ_{320} = \frac{1}{81 \sqrt{6 π a^{3}}} \frac{r^{2}}{a^{2}} e^{- r / (3 a)} (3 \cos^{3} θ - 1)$

Perturbation matrix has 81 elements and it is a 9x9 matrix. The value of maximum number of elements is zero, except for 8 elements. So, determine the value of 3 of them.

Matrix elements which have $\int_{0}^{2 π} e^{\pm m i φ} d φ$ vanish. Those elements are as follows,

$<300 |H'| 31 \pm 1>, <300 |H'| 32 \pm 1>, <300 |H'| 32 \pm 2> <310 |H'| 31 \pm 1>, <310 |H'| 32 \pm 1>, <310 |H'| 32 \pm 2> <320 |H'| 31 \pm 1>, <320 |H'| 32 \pm 1>, <320 |H'| 32 \pm 2> <31 \pm 1 |H'| 32 \pm 2>, <31 \pm 1 |H'| 32 \pm 1>, <32 \pm 1 |H'| 32 \pm 2>$

Even the diagonal elements also vanish, as integral with vanish. As $\int_{0}^{π} \cos^{2 k + 1} ϑ \sin ϑ d ϑ = 0$ for k = 0, 1, 2 , below matrix elements are zero:

$<311 |H'| 31 - 1>, <311 |H'| 32 - 1>, <300 |H'| 322>, <300 |H'| 320>, <322 |H'| 32 - 2>$

Thus, that leaves us with the matrix elements as follows,

$<300 |H'| 310>, <31 \pm 1 |H'| 32 \pm 1>, <310 |H'| 320> <300 |H'| 310> = \frac{e E_{ext}}{81 \sqrt{3 π a^{3}}} \frac{\sqrt{2}}{81 \sqrt{π a^{3}}} \int (27 - \frac{18 r}{a} + \frac{2 r^{2}}{a^{2}}) e^{- r / (3 a)} r \cos θ (6 - \frac{r}{a}) \frac{r}{a} e^{- r / (3 a)} \cos θ r^{2} d r \sin θ d θ d φ$

Simplify the above expression.

$\frac{e E_{ext} 2 \sqrt{2}}{3^{8} \sqrt{3} a^{4}} \int_{0}^{\infty} (27 - \frac{18 r}{a} + \frac{2 r^{2}}{a^{2}}) (6 - \frac{r}{a}) e^{- 2 r / (3 a)} r^{4} d r \int_{0}^{π} \cos^{2} θ \sin θ d θ$

Find the value of dr .

localid="1659008811088" $x = \frac{2 r}{3 a} r = \frac{3 a}{2} x d r = \frac{3 a}{2} d x$

Substitute the above value in the expression.

$\frac{e E_{ext} 2 \sqrt{2}}{3^{8} \sqrt{3} a^{4}} \cdot \frac{2}{3} {(\frac{3 a}{2})}^{5} \int_{0}^{\infty} x^{4} (27 - 27 x + \frac{9}{2} x^{2}) (6 - \frac{3}{2} x) e^{- x} d x = \frac{e a E_{ext}}{324 \sqrt{6}} \int_{0}^{\infty} x^{4} (162 - \frac{405}{2} x + \frac{135}{2} x^{2} - \frac{27}{4} x^{3}) e^{- x} d x = \frac{e a E_{ext}}{324 \sqrt{6}} (162 \cdot 4! - \frac{405}{2} \cdot 5! + \frac{135}{2} \cdot 6! - \frac{27}{4} \cdot 7!) = \frac{e a E_{ext}}{324 \sqrt{6}} \cdot (- 5832) <300 |H'| 310> = - 3 \sqrt{6} e a E_{ext}$

Calculate for the next function.

$<310 |H'| 320> = \frac{e E_{ext}}{81 \sqrt{6 π a^{3}}} \frac{\sqrt{2}}{81 \sqrt{π a^{3}}} \int \frac{r^{2}}{a^{2}} (3 \cos^{2} θ - 1) e^{- r / (3 a)} r \cos θ (6 - \frac{r}{a}) \frac{r}{a} e^{- r / (3 a)} \cos θ r^{2} d r \sin θ d θ d φ = \frac{e E_{ext}}{3^{8} \sqrt{3} a^{6}} \int_{0}^{\infty} (6 - \frac{r}{a}) r^{6} e^{- 2 r / (3 a)} d r \int_{0}^{π} (3 \cos^{4} θ \sin θ - \cos^{2} θ \sin θ) d θ$ Substitute the value of in the above expression.

Evaluate the next function.

$<31 \pm 1 |H'| 32 \pm 1> = \frac{e E_{ext}}{81 \sqrt{π a^{3}}} \frac{1}{81 \sqrt{π a^{3}}} \int (6 - \frac{r}{a}) \frac{r}{a} e^{- r / (3 a)} e^{\pm i φ} \sin θ r \cos θ \frac{r^{2}}{a^{2}} e^{- r / (3 a)} e^{\mp i φ} \cos θ \sin θ r^{2} d r \sin θ d θ d φ = \frac{2 e E_{ext}}{3^{8} a^{6}} \int_{0}^{\infty} (6 - \frac{r}{a}) r^{6} e^{- 2 r / (3 a)} d r \int_{0}^{π} \cos^{2} θ \sin^{3} θ d θ$

Substitute the value of in the above expression.

$\frac{2 e E_{ext}}{3^{8} a^{6}} \cdot {(\frac{3 a}{2})}^{7} \cdot \frac{4}{15} \int_{0}^{\infty} x^{6} (6 - \frac{3}{2} x) e^{- x} d x = \frac{e a E_{ext}}{720} \cdot (6 \cdot 6! - \frac{3}{2} \cdot 7!) <31 \pm 1 |H'| 32 \pm 1> = - \frac{9 e a E_{ext}}{2}$

Write the perturbation matrix.

Thus, reversed the rows and columns to create attractive blocks that may be diagonalized separately. Rows and columns can be rearranged, but carefully: W must be Hermitian in the final matrix.: $W_{i j} = W_{j i}^{*}$ .

(b) Determination of the value of eigenvalues and degenderizes

There are one 3X3 block, two 2X2 blocks and one 1X1 block.

Find the determinant.

$|\begin{matrix} - E & 3 \sqrt{6} λ & 0 \\ 3 \sqrt{6} λ & - E & 3 \sqrt{3} λ \\ 0 & 3 \sqrt{3} λ & - E \end{matrix}| = 0$

Simplify the above.

$- E (E^{2} - 27 λ^{2}) - 3 \sqrt{6} λ (- E \cdot 3 \sqrt{6} λ) = 0 E (54 λ^{2} - E^{2} + 27 λ^{2}) = 0 E^{2} = 81 λ^{2} E = \mp 9 e a E_{ext}$

Solve further.

$|\begin{matrix} - E & 9 λ / 2 \\ 9 λ / 2 & - E \end{matrix}| = 0 E^{2} - \frac{81 λ^{2}}{4} = 0 |\begin{matrix} - E & 0 \\ 0 & - E \end{matrix}| = 0 E^{2} = 0 E = 0$