# Extra Credit 0

• simons.hec.utah.edu/TheoryPag...&Solutions.pdf
• simons.hec.utah.edu/TheoryPag...tionsPart2.pdf

Section ? problems and solutions (pg 103 in part 2 pdf)

• E1, E2, E3, E4, E5, E6, E7, E8 (and solutions on page 105)

Section 5 problems and solutions (pg 141 in part 2 pdf)

• E1, E2, E3, P1, P2, (solutions on page 145 and up)

Section 5 problems and solutions (pg 157 in part 2 pdf)

• E1, E2, (solutions on page 160 and up)

## Q1

Show that the configuration (determinant) corresponding to the Li$$^+$$ 1s($$\alpha$$)1s($$\alpha$$) state vanishes.

## Q2

Construct the 3 triplet and 1 singlet wavefunctions for the Li$$^+$$ 1s$$^1$$2s$$^1$$ configuration. Show that each state is a proper eigenfunction of S$$^2$$ and S$$_z$$ (use raising and lowering operators for S$$^2$$).

## Q3

Construct wavefunctions for each of the following states of CH$$_2$$:

1. $$^1B_1 (1a_1 ^22a_1 ^21b_2 ^23a_1 ^11b_1^1)$$
2. $$^3B_1 (1a_1 ^22a_1 ^21b_2 ^23a_1 ^11b_1^1)$$
3. $$^1A_1 (1a_1 ^22a_1 ^21b_2 ^23a_1 ^2)$$

## Q4

Construct wavefunctions for each state of the $$1\sigma^22\sigma^23\sigma^21\pi^2$$ configuration of NH.

5. Construct wavefunctions for each state of the $$1s^12s^213s^1$$ configuration of Li.

6. Determine all term symbols that arise from the $$1s^22s^22p^23d^1$$ configuration of the excited N atom.

7. Calculate the energy (using Slater Condon rules) associated with the 2p valence electrons for the following states of the C atom.

• $$^3P(M_L=1,M_S=1)$$,
• $$^3P(M_L=0,M_S=0)$$,
• $$^1S(M_L=0,M_S=0)$$,
• $$^1D(M_L=0,M_S=0)$$.

8. Calculate the energy (using Slater Condon rules) associated with the p valence electrons for the following states of the NH molecule.

• $$^1\Delta (M_L=2, M_S=0)$$,
• $$^1\Sigma (M_L=0, M_S=0)$$,
• $$^3\Sigma (M_L=0, M_S=0)$$.

Solutions

## S1

1. $$\lvert1s\alpha 1s\alpha\rvert =\dfrac{1}{\sqrt{2!}} \begin{vmatrix} 1\alpha(1) & 1s\alpha(2)\\1s\alpha(1) & 1s\alpha(2) \end{vmatrix}$$

$$=\dfrac{1}{\sqrt{2}}(1s\alpha(1)1s\alpha(2)-1s\alpha(1)1s\alpha(2)$$

$$=0$$

2. Starting with the $$M_S=1 ^3S$$ state (which in a "box" for this $$M_L=0, M_S=1$$ case would contain only one product function; $$\lvert1s\alpha2s\alpha\rvert)$$ and applying S. gives:

S.$$^3S(S=1,M_S=1) \quad = \sqrt{1(1+1) - 1(1-1)}\hbar ^S(S=1, M_S=0)$$

$$\qquad \qquad \qquad \qquad = \hbar\sqrt{2} ^3S(S=1,M_S=0)$$

$$\qquad \qquad \qquad \qquad = (S.(1)+S.(2)) \lvert1 1s\alpha2s\alpha \rvert$$

$$\qquad \qquad \qquad \qquad = S.(1)\lvert1s\alpha2s\alpha\rvert + S.(2)\lvert1s\alpha2s\alpha\rvert$$

$$\qquad \qquad \qquad \qquad = \hbar\sqrt{\dfrac{1}{2}\left(\dfrac{1}{2}+1\right) -\dfrac{1}{2}\left(\dfrac{1}{2}-1\right)} \lvert1s\beta2s\alpha\rvert + \hbar\sqrt{\dfrac{1}{2}\left(\dfrac{1}{2}+1\right) -\dfrac{1}{2}\left(\dfrac{1}{2}-1\right)} \lvert1s\alpha2s\beta\rvert$$

$$\qquad \qquad \qquad \qquad = \hbar(\lvert1s\beta2s\alpha\rvert + \lvert1s\alpha2s\beta\rvert)$$

$$\qquad So, \quad \hbar\sqrt{2} \quad ^3S(S=1,M_S=0)=\hbar(\lvert1s\beta2s\alpha\rvert + \lvert1s\alpha2s\beta\rvert)$$

$$\qquad \qquad \quad \quad ^3S(S=1,M_S=0) = \dfrac{1}{\sqrt{2}}(\lvert1s\beta2s\alpha\rvert + \lvert1s\alpha2s\beta\rvert)$$

$$\qquad \text{The three triplet states are then:}$$

$$\qquad \qquad \qquad \qquad ^3S(S=1, M_S=1) = \lvert1s\alpha2s\alpha\rvert,$$

$$\qquad \qquad \qquad \qquad ^3S(S=1, M_S=0) = \dfrac{1}{\sqrt{2}}(\lvert1s\beta2s\alpha\rvert + \lvert1s\alpha2s\beta\rvert), \text{and}$$

$$\qquad \qquad \qquad \qquad ^3S(S=1, M_S=-1) = \lvert1s\beta2s\beta\rvert$$.

$$\qquad$$ The singlet state which must be constructed orthogonal to the three single states (and in particular to the $$^3S(S=1,M_S=-1)$$ state) can be seen to be:

$$\qquad \qquad \qquad \qquad ^1S(S=0,M_S=0) = \dfrac{1}{\sqrt{2}}(\lvert1s\beta2s\alpha\rvert - \lvert1s\alpha2s\beta\rvert).$$

$$\qquad \text{Applying} S^2 \text{and} S_Z \text{to each of these states gives:}$$

$$\qquad \qquad \qquad \qquad S_Z\lvert1s\alpha2s\alpha\rvert = (S_Z(1) + S_Z(2) \lvert1s\alpha2s\alpha\rvert$$

$$\qquad \qquad \qquad \qquad \qquad \qquad= S_Z(1)\lvert1s\alpha2s\alpha + S_Z(2))\lvert1s\alpha2s\alpha\rvert$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = \hbar\left(\dfrac{1}{2}\right) \lvert1s\alpha2s\alpha\rvert + \hbar\left(\dfrac{1}{2}\right)\lvert1s\alpha2s\alpha\rvert$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = \hbar\lvert1s\alpha2s\alpha\rvert$$

$$\qquad \qquad \qquad \qquad S^2\lvert1s\alpha2s\alpha\rvert = (S.S_+ + S_Z^2 +\hbar S_Z)\lvert1s\alpha2s\alpha\rvert$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = S.S_+\lvert1s\alpha2s\alpha\rvert + S_Z^2\lvert1s\alpha2s\alpha\rvert + \hbar S_Z\lvert1s\alpha2s\alpha\rvert$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = 0 + \hbar^2\lvert1s\alpha2s\alpha + \hbar^2\lvert1s\alpha2s\alpha\rvert$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = 2\hbar^2\lvert1s\alpha2s\alpha\rvert$$

$$\qquad S_Z\dfrac{1}{\sqrt{2}}(\lvert1s\beta2s\alpha\rvert + \lvert1s\alpha2s\beta\rvert) = (S_Z(1) + S_Z(2))\dfrac{1}{\sqrt{2}}(\lvert1s\beta2s\alpha\rvert + \lvert1s\alpha2s\beta\rvert)$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = \dfrac{1}{\sqrt{2}}(S_Z(1)+S_Z(2)) \lvert1s\beta2s\alpha\rvert + \dfrac{1}{\sqrt{2}} (S_Z(1)+S_Z(2)) \lvert1s\alpha2s\beta\rvert$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = \dfrac{1}{\sqrt{2}}\left( \hbar \left( \dfrac{-1}{2}\right) +\hbar\left(\dfrac{1}{2}\right)\right)\lvert1s\beta2s\alpha\rvert + \dfrac{1}{\sqrt{2}}\left( \hbar \left( \dfrac{1}{2}\right) +\hbar\left(\dfrac{-1}{2}\right)\right)\lvert1s\alpha2s\beta\rvert$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = 0\hbar\dfrac{1}{\sqrt{2}}(\lvert1s\beta2s\alpha\rvert + \lvert1s\alpha2s\beta\rvert)$$

$$\qquad S^2\dfrac{1}{\sqrt{2}}(\lvert1s\beta2s\alpha\rvert + \lvert1s\alpha2s\beta\rvert) = (S.S_+ + S_Z^2 +\hbar S_Z)\dfrac{1}{\sqrt{2}}(\lvert1s\beta2s\alpha\rvert + \lvert1s\alpha2s\beta\rvert)$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = S.S_+\dfrac{1}{\sqrt{2}}(\lvert1s\beta2s\alpha\rvert + \lvert1s\alpha2s\beta\rvert)$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = \dfrac{1}{\sqrt{2}}(S.(S_+(1) + S_+(2))\lvert1s\beta2s\alpha\rvert + S.(S_+(1) + S_+(2))\lvert1s\alpha2s\beta\rvert)$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = \dfrac{1}{\sqrt{2}}\left(S.\hbar\lvert1s\alpha2s\alpha\rvert + S.\hbar\lvert1s\alpha2s\alpha\rvert\right)$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = 2 \hbar\dfrac{1}{\sqrt{2}}((S.(1)+S.(2))\lvert1s\alpha2s\alpha\rvert)$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = 2 \hbar \dfrac{1}{\sqrt{2}}(\hbar\lvert1s\beta2s\alpha\rvert + \hbar\lvert1s\alpha2s\beta\rvert)$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = 2 \hbar^2 \dfrac{1}{\sqrt{2}}(\lvert1s\beta2s\alpha\rvert + \lvert1s\alpha2s\beta\rvert)$$

$$\qquad \qquad \qquad \qquad S_Z\lvert1s\beta2s\beta\rvert = (S_Z(1) + S_Z(2)\lvert1s\beta2s\beta\rvert$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = S_Z(1)\lvert1s\beta2s\beta\rvert + S_Z(2))\rvert1s\beta2s\beta\rvert$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = \hbar\left(\dfrac{-1}{2}\right)\lvert1s\beta2s\beta\rvert + \hbar\left(\dfrac{-1}{2}\right)\lvert1s\beta2s\beta\rvert$$

$$\qquad \qquad \qquad \qquad \qquad \qquad =-\hbar\lvert1s\beta2s\beta\rvert$$

$$\qquad \qquad \qquad \qquad S^2\lvert1s\beta2s\beta\rvert = (S_+S.+S_Z^2 -\hbar S_Z)\lvert1s\beta2s\beta\rvert$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = S_+S.\lvert1s\beta2s\beta\rvert + S_Z^2\lvert1s\beta2s\beta\rvert - \hbar S_Z\lvert1s\beta2s\beta2\rvert$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = 0 + \hbar^2\lvert1s\beta2s\beta\rvert +\hbar^2\rvert1s\beta2s\beta\rvert$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = 2\hbar^2\lvert1s\beta2s\beta\rvert$$

$$\qquad \qquad \qquad \qquad S_Z\lvert1s\beta2s\beta\rvert = (S.S_+ + S_Z^2 +\hbar S_Z)\dfrac{1}{\sqrt{2}}(\lvert1s\beta2s\alpha\rvert - \lvert1s\alpha2s\beta\rvert)$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = S.S_+\dfrac{1}{\sqrt{2}}(\rvert1s\beta2s\alpha - \lvert1s\alpha2s\beta\rvert)$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = \dfrac{1}{\sqrt{2}}\rvert 1s\beta2s\alpha - \lvert1s\alpha2s\beta\rvert)$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = \dfrac{1}{\sqrt{2}}(S.(S_+(1) + S_+(2))\lvert1s\beta2s\alpha\rvert - S.(S_+(1) + S_+(2))\lvert1s\alpha2s\beta\rvert)$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = \dfrac{1}{\sqrt{2}}(S.(S_+(1) + S_+(2))\lvert1s\alpha2s\alpha\rvert - S.\hbar \lvert1s\alpha2s\alpha\rvert)$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = 0\hbar\dfrac{1}{\sqrt{2}}((S.(1) + S.(2))\lvert1s\alpha2s\alpha\rvert)$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = 0\hbar\dfrac{1}{\sqrt{2}}(\hbar\lvert1s\beta2s\alpha\rvert - \hbar\lvert1s\alpha2s\beta\rvert)$$

$$\qquad \qquad \qquad \qquad \qquad \qquad = 0\hbar^2\dfrac{1}{\sqrt{2}}(\lvert1s\beta2s\alpha\rvert - \hbar\lvert1s\alpha2s\beta\rvert)$$

3.

A.) Once the spatial symmetry has been determined by multiplication of the irreducible representations, the spin coupling is identical to exercise 2 and gives the result:

$\dfrac{1}{\sqrt{2}}(\lvert3a_1\alpha1b_1\beta\rvert - \lvert3a_1\beta1b_1\alpha\rvert )$

B.) There are three states here (again analogous to exercise 2):

1. ) $$\lvert3a_1\alpha1b_1\alpha\rvert,$$
2. ) $$\dfrac{1}{\sqrt{2}}(\lvert3a_1\alpha1b_1\beta\rvert + \lvert3a_1\beta1b_1\alpha\rvert), \text{and}$$
3. ) $$\lvert 3a_1\beta1b_1\beta\rvert$$

C.) $$\lvert3a_1\alpha3a_1\beta\rvert$$

4. As shown in review exercise 1c, for two equivalent $$\pi$$ electrons one obtains six states:

$$\qquad \qquad ^1\Delta (M_L=2); \text{one state} (M_S=0),$$

$$\qquad \qquad ^1\Delta (M_L=-2); \text{one state} (M_S=0),$$

$$\qquad \qquad ^1\Sigma (M_L=0); \text{one state} (M_S=0), \text{and}$$

$$\qquad \qquad ^3\Sigma(M_L=0); \text{three states} (M_S=1,0, \text{and} -1)$$.

$$\qquad \text{By inspecting the "box" in review exercise 1c, it should be fairly straightforward to write down the wavefunctions for each of these:}$$

$$\qquad \qquad ^1\Delta (M_L=2); \lvert\pi_1\alpha\pi_1\beta\rvert$$

$$\qquad \qquad ^1\Delta (M_L=-2); \lvert\pi_{-1}\alpha\pi_{-1}\beta\rvert$$

$$\qquad \qquad ^1\Sigma(M_L=0); \dfrac{1}{\sqrt{2}}(\lvert\pi_1\beta\pi_{-1}\alpha\rvert - \lvert\pi_1\alpha\pi_{-1}\beta\rvert)$$

$$\qquad \qquad ^3\Sigma (M_L=0, M_S=1); \lvert\pi_1\alpha\pi_1\alpha\rvert$$

$$\qquad \qquad ^3\Sigma (M_L=0, M_S=0); \dfrac{1}{\sqrt{2}}(\lvert\pi_1\beta\pi_{-1}\alpha\rvert + \lvert\pi_1\alpha\pi_{-1}\beta\rvert)$$

$$\qquad \qquad ^3\Sigma (M_L=0, M_S=-1); \lvert\pi_1\beta\pi_{-1}\beta\rvert$$

5. We can conveniently couple another s electron to the states generated from the $$1s^12s^1$$ configuration in exercise 2:

$$\qquad \qquad ^3S(L=0,S=1) \text{with} 3s^1(L=0, S=\dfrac{1}{2}) \text{giving:}$$

$$\qquad \qquad L=0, S=\dfrac{3}{2}, \dfrac{1}{2}; ^4S (\text{4 states}) and ^2S (\text{2 states}).$$

$$\qquad \qquad ^1S(L=0, S=0) \text{with} 3s^1(L=0, S= \dfrac{1}{2}) giving:$$

$$\qquad \qquad L=0, S=\dfrac{1}{2}; ^2S (\text{2 states}).$$

Constructing a "box" for this case would yield:

 $$M_L$$ 0 $$M_S$$ $$\dfrac{3}{2}$$ $$\lvert1s\alpha2s\alpha3s\alpha\rvert$$ $$\dfrac{1}{2}$$ $$\lvert1s\alpha2s\alpha3s\beta\rvert \lvert1s\alpha2s\beta3s\alpha\rvert \lvert1s\beta2s\alpha3s\alpha\rvert$$

One can immediately identify the wavefunctions for two of the quartets (they are single entries):

$$\qquad \qquad ^4S(S=\dfrac{3}{2}, M_S=\dfrac{3}{2}): \lvert1s\alpha2s\alpha3s\alpha\rvert$$

$$\qquad \qquad 4^S(S=\dfrac{3}{2}, M_S=\dfrac{-3}{2}): \lvert1s\beta3s\beta\lvert$$

$$\qquad \qquad \text{Applying S. to } ^4S(S=\dfrac{3}{2}, m_S=\dfrac{3}{2}) \text{yields:}$$

$$\qquad S. ^4S(S=\dfrac{3}{2}, M_S=\dfrac{3}{2}) = \hbar \sqrt{\dfrac{3}{2}\left( \dfrac{3}{2}+1\right) - \dfrac{3}{2}\left( \dfrac{3}{2}-1\right)} \quad ^4S(S=\dfrac{3}{2}, M_S=\dfrac{-1}{2})$$

$$\qquad \qquad \qquad \qquad \qquad = \hbar \sqrt{3} ^4S(S=\dfrac{3}{2}, M_S=\dfrac{1}{2})$$

$$\qquad \qquad S.\lvert1s\alpha2s\alpha3s\alpha\rvert = \hbar(\lvert1s\beta2s\alpha3s\alpha\rvert + \lvert1s\alpha2s\beta3s\alpha\rvert + \lvert1s\alpha2s\alpha3s\beta\rvert)$$

So, $$^4S(S=\dfrac{3}{2}, M_S=\dfrac{1}{2}) = \dfrac{1}{\sqrt{3}}(\lvert1s\beta2s\alpha3s\alpha\rvert + \lvert1s\alpha2s\beta3s\alpha\rvert + \lvert1s\alpha2s\alpha3s\beta\rvert)$$

$$\qquad$$ Applying $$S_+ \text{to} ^4S(S=\dfrac{3}{2}, M_S=\dfrac{-3}{2}) \text{yields}:$$

$$\qquad \qquad S_+^4S(S=\dfrac{3}{2}, M_S=\dfrac{-3}{2}) = \hbar \sqrt{ \dfrac{3}{2}\left(\frac{3}{2}+1\right) - - \dfrac{3}{2}\left(\dfrac{-3}{2}+1\right)} \quad ^4S(S=\dfrac{3}{2}, M_S=\dfrac{-1}{2})$$

$$\qquad \qquad \qquad \qquad \qquad = \hbar \sqrt{3} \quad ^4S(S=\dfrac{3}{2}, M_S=\dfrac{-1}{2})$$

$$\qquad \qquad S_+\lvert1s\beta2s\beta3s\beta\rvert = \hbar(\lvert1s\alpha2s\beta3s\beta\rvert + \lvert1s\beta2s\alpha3s\beta\rvert + \lvert1s\beta2s\beta3s\alpha\rvert)$$

$$\qquad \text{So}, ^4S(S=\dfrac{3}{2}, M_S=\dfrac{-1}{2}) = \dfrac{1}{\sqrt{3}}(\lvert1s\alpha2s\beta3s\beta\rvert + \lvert1s\beta2s\alpha3s\beta\rvert + \lvert1s\beta2s\beta3s\alpha\rvert)$$

It only remains to construct the doublet states which are orthogonal to these quartet states. Recall that the orthogonal combinations for systems having three equal components (for example when symmetry adapting the 3 sp$$^2$$ hybrids in $$C_{2v}$$ or $$D_{3h}$$ symmetry) give results of + + +, +2 - -, and 0 + -. Notice that the quartets are the + + + combinations and therefore the doublets can be recognized as:

$$^2S(S=\dfrac{1}{2} ,M_S=\dfrac{1}{2} ) = \dfrac{1}{\sqrt{6}}( \lvert1s\beta2s\alpha3s\alpha\rvert + \lvert1s\alpha2s\beta3s\alpha\rvert - 2\lvert1s\alpha2s\alpha3s\beta\rvert)$$

$$^2S(S=\dfrac{1}{2} ,M_S=\dfrac{1}{2}) = \dfrac{1}{\sqrt{2}}( \lvert1s\beta2s\alpha3s\alpha\rvert - \lvert1s\alpha2s\beta3s\alpha\rvert + 0\lvert1s\alpha2s\alpha3s\beta\rvert)$$

$$^2S(S=\dfrac{1}{2} ,M_S=\dfrac{-1}{2} ) = \dfrac{1}{\sqrt{6}}(\lvert1s\alpha2s\beta3s\beta\rvert + \lvert1s\beta2s\alpha3s\beta\rvert - 2\lvert1s\beta2s\beta3s\alpha\rvert)$$

$$^2S(S=\dfrac{1}{2} ,M_S=\dfrac{-1}{2} ) = \dfrac{1}{\sqrt{3}}(\lvert1s\alpha2s\beta3s\beta\rvert - \lvert1s\beta2s\alpha3s\beta\rvert + 0\lvert1s\beta2s\beta3s\alpha\rvert)$$

6. As illustrated in this chapter a p2 configuration (two equivalent p electrons) gives rise to the term symbols: $$^3$$P, $$^1$$D, and $$^1$$S. Coupling an additional electron (3d$$^1$$) to this p$$^2$$ configuration will give the desired $$1s^22s^22p^23d^1$$ term symbols:

$$\qquad ^3P(L=1,S=1) \text{with} ^2D(L=2,S=1 2 ) \text{generates};$$

$$\qquad L=3,2,1, and S=\dfrac{3}{2}, \dfrac{1}{2} \text{with term symbols} ^4F, ^2F, ^4D, ^2D, ^4P, \text{and} ^2P,$$

$$\qquad ^1D(L=2,S=0) \text{with} ^2D(L=2,S=1 2 ) \text{generates};$$

$$\qquad L=4,3,2,1,0, \text{and} S=\dfrac{1}{2} \text{with term symbols} ^2G, ^2F, ^2D, ^2P, \text{and} ^2S,$$

$$\qquad ^1S(L=0,S=0) \text{with} ^2D(L=2,S=1 2 ) \text{generates};$$

$$\qquad L=2 and S=1 2 \text{with term symbol} ^2D.$$

7. The notation used for the Slater Condon rules will be the same as used in the text:

(a.) zero (spin orbital) difference;

$$<\lvert F + G \rvert> = \sum\limits_{i}<\phi_{i}\lvert f \rvert\phi_{i}> + \sum\limits_{i>j}\left( <\phi_{i}\phi_{j}\lvert g \rvert\phi_{i}\phi_{j}> - < \phi_{i}\phi_{j}\lvert g \rvert\phi_{j}\phi_{i} >\right)$$

$$= \sum\limits_{i} f_{ii} + \sum\limits_{i>j} (g_{ijij} - g_{ijji})$$

(b.) one (spin orbital) difference $$(\phi_p \ne \phi_{p'});$$

$$<\lvert F + G \rvert> = <\phi_{p}\lvert f \rvert\phi_{p'}> + \sum\limits_{j\ne p;p'j}\left( <\phi_{p}\phi_{j}\lvert g \rvert\phi_{p'}\phi_{j}> - < \phi_{p}\phi_{j}\lvert g \rvert\phi_{j}\phi_{p'} >\right)$$

$$f_{pp'} + \sum\limits_{j\ne p;p'}(g_{pjp'j} - g_{pjjp'})$$

(c.) two (spin orbital) differences $$( \phi_p \ne \phi_{p'} \text{and} \phi_q \ne \phi_{q'});$$

$$<\lvert F + G \rvert> = 0$$

7. i. $$^3P(M_L=1, M_S=1) = \lvert p_1\alpha p_0\alpha\rvert$$

$$<\lvert p_1\alpha p_0\alpha\lvert H\rvert p_1\alpha p_0\alpha> =$$

Error!. Using the Slater Condon rule (a.) above (SCa):

$$<\lvert 10\lvert H \rvert 10\rvert> = f_{11} + f_{00} +g_{1010} - g_{1001}$$

7. ii. $$^3P(M_L=0, M_S=0) = \dfrac{1}{\sqrt{2}}(\lvert p_1\alpha p_{-1}\beta\rvert + \lvert p_1\beta p_{-1}\alpha\rvert)$$

$$< ^3P(M_L=0, M_S=0) \lvert H \rvert ^3P(M_L=0, M_S=0)>$$

$$= \dfrac{1}{2}(< \lvert p_{1}\alpha p_{-1} \beta \lvert H \rvert p_1 \alpha p_{-1} \beta \rvert> + < \lvert p_{1}\alpha p_{-1} \beta \lvert H \rvert p_1 \beta p_{-1} \alpha \rvert>$$

$$+ < \lvert p_{1}\beta p_{-1} \alpha \lvert H \rvert p_1 \alpha p_{-1} \beta \rvert> + < \lvert p_{1}\beta p_{-1} \alpha \lvert H \rvert p_1 \beta p_{-1} \alpha \rvert>$$

Evaluating each matrix element gives:

$$<\lvert p_1 \alpha p_{-1} \beta \lvert H \rvert p_1 \alpha p_{-1}\beta \rvert> = f_{1\alpha 1\alpha} + f_{-1\beta -1\beta} + g_{1\alpha -1\beta 1\alpha -1\beta} -g_{1 \alpha -1 \beta -1 \beta 1\alpha} (SCa)$$

$$= f_{11} + f_{-1-1} + g_{1-11-1} -0$$

$$<\lvert p_1 \alpha p_{-1} \beta \lvert H \rvert p_1 \alpha p_{-1}\alpha \rvert> = g_{1\alpha -1\beta 1\beta -1\alpha} - g_{1\alpha -1\beta -1\alpha 1\beta} \text{(SCc)}$$

$$= 0 - g_{1 -1 -1 1}$$

$$<\lvert p_1 \beta p_{-1} \alpha \lvert H \rvert p_1 \alpha p_{-1}\beta \rvert> = g_{1\beta -1\alpha 1\alpha -1\beta} - g_{1\beta -1\alpha -1\beta 1\alpha} \text{(SCc)}$$

$$= 0 - g_{1 -1 -1 1}$$

$$<\lvert p_1 \beta p_{-1} \alpha \lvert H \rvert p_1 \beta p_{-1}\alpha \rvert> = f_{1\beta 1\beta} + f_{-1\alpha -1\alpha} + g_{1\beta -1\alpha 1\beta -1\alpha} - g_{1\beta - 1\alpha -1\alpha 1\beta} \text{(SCa)}$$

$$=f_{11} + f_{-1-1} + g_{1 -1 1 -1} -0$$

Substitution of these expressions give:

$$< ^3P(M_L=0,M_S=0) \lvert H \rvert ^3P(M_L=0,M_S=0)>$$

$$= \dfrac{1}{2}(f_{11} + f_{-1 -1} +g_{1 -1 1 -1} -g_{1 -1 -1 1} - g_{1 -1 -1 1} + f_{11} + f_{-1 -1} + g_{1 -1 1 -1} )$$

$$= f_{11} + f_{-1 -1} + g_{1 -1 1 -1} -g_{1 -1 -1 1}$$

7. iii. $$^1S(M_L=0,M_S); \dfrac{1}{\sqrt{3}}(\lvert p_0\alpha p_0\beta \rvert - \lvert p_1\alpha p_{-1}\beta\rvert - \lvert p_{-1}\alpha p_1\beta \rvert)$$

$$<^1S(M_L=0,M_S=0)\lvert H \rvert ^1S(M_L=0,M_S=0)>$$

$$=\dfrac{1}{3}(<\lvert p_0\alpha p_0\beta \lvert H \rvert p_0\alpha p_0\beta \rvert> - <\lvert p_0\alpha p_0\beta \lvert H \rvert p_1\alpha p_{-1}\beta \rvert>$$

$$- <\lvert p_0\alpha p_0\beta \lvert H \rvert p_{-1}\alpha p_1\beta \rvert> - <\lvert p_1\alpha p_{-1}\beta \lvert H \rvert p_0\alpha p_{0}\beta \rvert>$$

$$+ <\lvert p_1\alpha p_{-1}\beta \lvert H \rvert p_1\alpha p_{-1}\beta \rvert> + <\lvert p_1\alpha p_{-1}\beta \lvert H \rvert p_{-1}\alpha p_1\beta \rvert>$$

$$- <\lvert p_{-1}\alpha p_1\beta \lvert H \rvert p_0\alpha p_0\beta \rvert> + <\lvert p_{-1}\alpha p_1\beta \lvert H \rvert p_1\alpha p_{-1}\beta \rvert>$$

$$+ <\lvert p_{-1}\alpha p_1\beta \lvert H \rvert p_{-1}\alpha p_1\beta \rvert>$$

Evaluating each matrix element gives:

$$<\lvert p_0\alpha p_0\beta \lvert H \rvert p_0\alpha p_0\beta \rvert> = f_{0\alpha 0\alpha} + f_{0\beta 0\beta} + g_{0\alpha0\beta 0\alpha 0\beta} - g_{0\alpha 0\beta 0 \beta 0\alpha} \text{(SCa)}$$

$$= f_{00} + f_{00} + g_{0000} -0$$

$$<\lvert p_0\alpha p_0\beta \lvert H \rvert p_1\alpha p_{-1}\beta \rvert> = <\lvert p_1\alpha p_{-1}\beta \lvert H \rvert p_0\alpha p_{0}\beta \rvert>$$

$$= g_{0\alpha 0\beta 1\alpha -1\beta} - g_{0\alpha 0\beta -1\beta 1\alpha} \text{(SCc)}$$

$$= g_{001-1} -0$$

$$<\lvert p_0\alpha p_0\beta \lvert H \rvert p_{-1}\alpha p_{1}\beta \rvert> = <\lvert p_{-1}\alpha p_{1}\beta \lvert H \rvert p_0\alpha p_{0}\beta \rvert>$$

$$= g_{0\alpha 0\beta -1\alpha 1\beta} - g_{0\alpha 0\beta 1\beta -1\alpha} \text{(SCc)}$$

$$= g_{00-11} -0$$

$$<\lvert p_1\alpha p_{-1}\beta \lvert H \rvert p_{1}\alpha p_{-1}\beta \rvert> = f_{1\alpha 1\alpha} + f_{-1\beta -1\beta} + g_{1\alpha -1\beta 1\alpha -1\beta} - g_{1\alpha -1\beta -1\beta 1\alpha} \text{(SCa)}$$

$$f_{11} + f_{-1-1} + g_{1-11-1} -0$$

$$<\lvert p_1\alpha p_{-1}\beta \lvert H \rvert p_{-1}\alpha p_{1}\beta \rvert> = <\lvert p_{-1}\alpha p_{1}\beta \lvert H \rvert p_{1}\alpha p_{-1}\beta \rvert>$$

$$g_{1\alpha -1\beta -1\alpha 1\beta} - g_{1\alpha -1\beta 1\beta -1\alpha1} \text{(SCa)}$$

$$= g_{1-1-11} -0$$

$$<\lvert p_{-1}\alpha p_{1}\beta \lvert H \rvert p_{-1}\alpha p_{1}\beta \rvert> = f_{-1\alpha -1\alpha} + f_{1\beta 1\beta} + g_{-1\alpha 1\beta -1\alpha 1\beta} -g_{-1\alpha 1\beta 1\beta -1\alpha} \text{(SCa)}$$

$$= f_{-1-1} + f_{11} +g_{-11-11} -0$$

Substitution of these expressions give:

$$< ^1S(M_L=0,M_S=0) \lvert H \rvert ^1S(M_L=0,M_S=0)>$$

$$\dfrac{1}{3}(f_{00} + f_{00} +g_{0000} - g_{001-1} - g_{00-11} - g_{001-1} + f_{11} + f_{-1-1}$$

$$+ g_{1-11-1} + g_{1-1-11} -g_{00-11} + g_{1-1-11} + f_{-1-1} + f_{11} + g_{-11-11})$$

$$\dfrac{1}{3}(2f_{00} + 2f_{11} + 2f_{-1-1} + g_{0000} - 4g_{001-1} + 2g_{1-11-1} + 2g_{1-1-11})$$

7. iv. $$^1D(M_L=0,M_S=0) = \dfrac{1}{\sqrt{6}}(2\lvert p_0\alpha p_0\beta \rvert + \lvert p_1\alpha p_{-1}\beta \rvert + \lvert p_{-1}\alpha p_1\beta \rvert)$$

Evaluating $$<^1D(M_L=0,M_S=0)\lvert H \rvert ^1D(M_L=0,M_S=0)>$$ we note that all the Slater Condon matrix elements generated are the same as those evaluated in part iii. (the signs for the wavefunction components and the multiplicative factor of two for one of the components, however, are different).

$$<^1D(M_L=0,M_S=0) \lvert H \rvert ^1D(M_L=0,M_S=0)>$$

$$= \dfrac{1}{6}(4f_{00} + 4f_{00} + 4g_{0000} + 2g_{001-1} + 2_{00-11} + 2g_{001-1} + f_{11}$$

$$+f_{-1-1} + g_{1-11-1} + g_{1-1-11} + 2g_{00-11} + g_{1-1-11} + f_{-1-1} + f_{11} + g_{-11-11}$$

$$= \dfrac{1}{6}(8f_{00} + 2f_{11} + 2f_{-1-1} + 4g_{0000} + 8_g{001-1} + 2g_{1-11-1} + 2g_{1-1-11})$$

8. i. $$^1\Delta(M_L=2,M_S=0) = \lvert \pi_1\alpha \pi_1\beta \rvert$$

$$<^1\Delta(M_L=2,M_S=0) \lvert H \rvert ^1\Delta(M_L=2,M_S=0)>$$

$$= <\rvert \pi_1 \alpha \pi_1\beta \lvert H \rvert \pi_1 \alpha \pi_1 \beta \rvert>$$

$$= f_{1\alpha 1\alpha} + f_{1\beta 1\beta} + g_{1\alpha 1\beta 1\alpha 1\beta} - g_{1\alpha 1\beta 1\beta 1\alpha} \text{(SCa)}$$

$$= f_{11} + f_{11} + g_{1111} - 0$$

$$= 2f_{11} + g_{1111}$$

ii. $$^1\sum(M_L=0,M_S=0) = \dfrac{1}{\sqrt{2}}( \lvert \pi_1 \alpha \pi_{-1}\beta \rvert - \lvert \pi_1 \beta \pi_{-1} \alpha \rvert)$$

$$<^3\sum(M_L=0,M_S=0) \lvert H \rvert ^3\sum(M_L=0,M_S=0)>$$

$$= \dfrac{1}{2}( \vert\pi_1 \alpha\beta \lvert H \rvert \pi_{1}\alpha\pi_{-1}\beta \rvert> - <\lvert \pi_1 \alpha \pi_{-1}\beta \lvert H \rvert \pi_1\beta \pi_{-1}\alpha \rvert >$$

$$- <\lvert \pi_1 \beta\pi_{-1}\alpha \lvert H \rvert \pi_1 \alpha \pi_{-1} \beta \rvert> + <\lvert \pi_{1}\beta\pi_{-1}\alpha \lvert H \rvert \pi_{1}\beta\pi_{-1}\alpha\rvert>)$$

Evaluating each matrix element gives:

$$\vert\pi_1 \alpha\beta \lvert H \rvert \pi_{1}\alpha\pi_{-1}\beta \rvert> = f_{1\alpha 1\alpha} + f_{-1\beta -1\beta} + g_{1\alpha -1\beta 1\alpha -1\beta} - g_{1\alpha -1\beta -1\beta 1\alpha} \text{(SCa)}$$

$$= f_{11} + f_{-1-1} + g_{1-11-1} - 0$$

$$<\lvert \pi_1 \alpha \pi_{-1}\beta \lvert H \rvert \pi_1\beta \pi_{-1}\alpha \rvert > = g_{1\alpha -1\beta 1\beta -1\alpha} - g_{1\alpha -1\beta -1\alpha 1\beta} \text{(SCc)}$$

$$= 0 -g_{1 -1-11}$$

$$<\lvert \pi_1 \beta\pi_{-1}\alpha \lvert H \rvert \pi_1 \alpha \pi_{-1} \beta \rvert> = g_{1\beta -1\alpha 1\alpha -1\beta} - g_{1\beta -1\alpha -1\beta 1\alpha} \text{(SCc)}$$

$$= 0 -g_{1-1-11}$$

$$<\lvert \pi_{1}\beta\pi_{-1}\alpha \lvert H \rvert \pi_{1}\beta\pi_{-1}\alpha\rvert> = f_{1\beta 1\beta} + f_{-1\alpha -1\alpha} + g_{1\beta -1\alpha 1\beta -1\alpha} - g_{1\beta -1\alpha -1\alpha 1\beta} \text{(SCa)}$$

$$= f_{11} + f_{-1-1} + g_{1-11-1} - 0$$

Substitution of these expressions give:

$$< ^3\sum(M_L=0,M_S=0)\rvert H \lvert ^3\sum(M_L=0,M_S=0)>$$

$$= \dfrac{1}{2}(f_{11} + f_{-1-1} + g_{1-11-1} + g_{1-1-11} + g_{1-1-11} + f_{11} + f_{-1-1} + g_{1-11-1})$$

$$= f_{11} + f_{-1-1} + g_{1-11-1} + g_{1-1-11}$$

iii. $$^3\sum(M_L=0,M_S=) = \dfrac{1}{\sqrt{2}}(\lvert\pi_{1}\alpha\pi_{-1}\beta\lvert + \rvert \pi_{1}\beta\pi_{-1}\alpha\rvert)$$

$$<^3\sum(M_L=0,M_S=0)\lvert H \rvert ^3\sum(M_L=0,M_S=0)>$$

$$=\dfrac{1}{2}(<\lvert\pi_{1}\alpha\pi_{-1}\beta\lvert H \rvert\pi_{1}\alpha\pi_{-1}\beta\rvert > + <\lvert\pi_{1}\alpha\pi_{-1}\beta\lvert H \rvert\pi_{1}\beta\pi_{-1}\alpha\rvert >$$

$$+ <\lvert \pi_{1}\beta\pi_{-1}\alpha\lvert H \rvert\pi_{1}\alpha\pi_{-1}\beta\rvert> + <\lvert\pi_{1}\beta\pi_{-1}\alpha\lvert H \rvert\pi_{1}\beta\pi_{-1}\alpha\rvert >)$$

Evaluating each matrix element gives:

$$<\lvert\pi_{1}\alpha\pi_{-1}\beta\lvert H \rvert\pi_{1}\alpha\pi_{-1}\beta\rvert > = f_{1\alpha 1\alpha} + f_{-1\beta -1\beta} + g_{1\alpha -1\beta 1\alpha -1\beta} - g_{1\alpha -1\beta -1\beta 1\alpha} \text{(SCa)}$$

$$=f_{11} + f_{-1-1} + g_{1-11-1} - 0$$

$$<\lvert\pi_{1}\alpha\pi_{-1}\beta\lvert H \rvert\pi_{1}\beta\pi_{-1}\alpha\rvert > = g_{1\alpha -1\beta 1\beta -1\alpha} - g_{1\alpha -1\beta -1\alpha 1\beta} \text{(SCc)}$$

$$= 0 - g_{1-1-11}$$

$$<\lvert \pi_{1}\beta\pi_{-1}\alpha\lvert H \rvert\pi_{1}\alpha\pi_{-1}\beta\rvert> = g_{1\beta -1\alpha 1\alpha -1\beta} - g_{1\beta -1\alpha -1\beta 1\alpha} \text{(SCc)}$$

$$= 0 - g_{1-1-11}$$

$$<\lvert\pi_{1}\beta\pi_{-1}\alpha\lvert H \rvert\pi_{1}\beta\pi_{-1}\alpha\rvert > = f_{1\beta 1\beta} + f_{-1\alpha -1\alpha} + g_{1\beta -1\alpha 1\beta -1\alpha} - g_{1\beta -1\alpha -1\alpha 1\beta} \text{(SCa)}$$

$$=f_{11} + f_{-1-1} + g_{1-11-1} - 0$$

Substitution of these expressions give:

$$<^3\sum(M_L=0,M_S=0) \lvert H \rvert ^3\sum(M_L=0,M_S=0)>$$

$$= \dfrac{1}{2}(f_{11} + f_{-1-1} + g_{1-11-1} - g_{1-1-11} - g_{1-1-11} + f_{11} + f_{-1-1} + g_{1-11-1})$$

$$= f_{11} + f_{-1-1} + g_{1-11-1} - g_{1-1-11}$$

Section 5 Exercises, Problems and Solutions

1. Time dependent perturbation theory provides an expression for the radiative lifetime of an excited electronic state, given by $$\tau_{R}$$:

$\tau_{R} = \frac{3\hbar^4c^3}{4(E_{i} - E_{f})^3|\mu_{fi}|^2 },$

where i refers to the excited state, f refers to the lower state, and $$\mu_{fi}$$ is the transition dipole.

a. Evaluate the z-component of the transition dipole for the $$2P_{z} \rightarrow 1s$$ transition in a hydrogenic atom of nuclear charge Z, given:

$\Psi_{1s} = \frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_{0}}\right)^{\frac{3}{2}}e^{\frac{-Zr}{a_0}}, and \Psi_{2p_{z}} = \frac{1}{4\sqrt{2\pi}}\left( \frac{Z}{a_0}\right)^{\frac{5}{2}}rCos\theta e^{\frac{-Zr}{2a_0}}.$

Express your answer in units of $$ea_{0}.$$

b. Use symmetry to demonstrate that the x- and y- components of $$\mu_{fi}$$ are zero, i.e.

$$<2p_z| e x |1s>$$ = $$<2p_z| e y |1s>$$= 0.

c. Calculate the radiative lifetime $$\tau_{R}$$ of a hydrogenlike atom in its 2p$$_z$$ state. Use the relation $$e^2 = \frac{\hbar^2}{m_ca_0}$$ to simplify your results.

2. Consider a case in which the complete set of states {$$\phi_k$$} for a Hamiltonian is known.

a. If the system is initially in the state m at time t=0 when a constant pertubation V is suddenly turned on, find the probability amplitudes C$$_k^{(2)}(t)$$ and $$C_m^{(2)}(t)$$, to second order in V, that describe the system being in a different state k or the same state m at time t.

b. If the pertubation is turned on adiabatically, what are C$$_k^{(2)}(t)$$ and $$C_m^{(2)}(t)$$?

Here, consider that the initial time is $$t_0 \rightarrow - \inf$$, and the potential is V $$e^{\eta t}$$, where the positive parameter $$\eta$$ is allowed to approach zero $$\eta \rightarrow 0$$ in order to describe the adiabatically (i.e., slowly) turned on pertubation.

c. Compare the results of parts a. and b. and explain any differences.

d. Ignore first order contributions (assume they vanish) and evaluate the transition rates $$\frac{d}{dt}{|C_k^{(2)}(t)|}^2$$ for the results of part b. by taking the limit $$\eta \rightarrow 0^{+}$$, to obtain the adiabatic results.

3. If a system is initially in a state m, conservation of probability requires that the total probabulity of transitions out of state m be obtainable from the decrease in the probability of being in state m. Prove this to the lowest order by using the results of exercise 2, i.e. show that: $$|C_m|^2 = 1 - \sum\limits_{k \neq m} |C_k|^2.$$

Problems:

1. Consider an interaction or perturbation which is carried out suddenly (instantaneously,

e.g., within an interval of time $$\Delta$$t which is small compared to the natural period $$\omega_{nm}^{-1}$$ corresponding to the transition from state m to state n), and after that is turned off

adiabatically (i.e., extremely slowly as V $$e^{\eta t}$$). The transition probability in this case is given as:

$T_{nm} \approx \frac{|<n|V|m>|^2}{\hbar^2 \omega_{nm}^2}$

where V corresponds to the maximum value of the interaction when it is turned on. This

formula allows one to calculate the transition probabilities under the action of sudden

perturbations which are small in absolute value whenever perturbation theory is applicable.

Let's use this "sudden approximation" to calculate the probability of excitation of an

electron under a sudden change of the charge of the nucleus. Consider the reaction:

$^3_1H \rightarrow ^3_2 He^+ + e^-,$

and assume the tritium atom has its electron initially in a 1s orbital.

a. Calculate the transition probability for the transition 1s $$\rightarrow$$ 2s for this reaction using the above formula for the transition probability.

b. Suppose that at time t = 0 the system is in a state which corresponds to the

wavefunction $$\psi_m$$, which is an eigenfunction of the operator $$H_0$$. At t = 0, the sudden change of the Hamiltonian occurs (now denoted as H and remains unchanged). Calculate the same 1s $$\rightarrow$$ 2s transition probability as in part a., only this time as the square of the magnitude of the coefficient, A$$_{1s,2s}$$ using the expansion: $\Psi(r,0) = \psi_m(r) = \sum\limits_n A_{nm}\psi_n (r), where A_{nm} = \int \psi_m(r)\psi_n(r)d^3r$

Note, that the eigenfunctions of H are $$\psi_n$$ with eigenvalues E$$_n.$$ Compare this "exact" value with that obtained by pertubation theory in part a.

2. The methyl iodide molecule is studied using microwave (pure rotational) spectroscopy.

The following integral governs the rotational selection rules for transitions labeled J, M, K

$$\rightarrow$$ J', M', K': $I = <D^{J'}_{M'K'}|\vec{\varepsilon}\cdot \vec{\mu}| D^{J}_{MK}>.$

The dipole moment $$\vec{\mu}$$ lies along the molecule's C$$_3$$ symmetry axis. Let the electric field of the light $$\vec{\varepsilon}$$ define the lab-fixed Z-direction.

a. Using the fact that Cos($$\beta) = D^1_{00},$$ show that

$I = 8 \pi^2\mu \varepsilon (-1)^{(M+K)} \begin{pmatrix}J' & 1 & J \\ M & 0 & M \\ \end{pmatrix} \begin{pmatrix} J' & 1 & J \\ K & 0 & K \\ \end{pmatrix} \delta_{M'M}\delta_{K'K}$

b. What restrictions does this result place on $$\Delta$$ J = J' -J? Explain physically why the K quantum number can not change.

Solutions

Exercises:

1. a. Evaluate the z-component of $$\mu_{fi}:$$

$$\mu_{fi} = <2p_{z}|e rCos \theta |1s>,$$ where $$\psi_{1s} = \frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_{0}}\right)^{\frac{3}{2}} e^{\frac{-Zr}{a_{0}}},$$ and $$\psi 2p_z = \frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a_0}\right)^{\frac{5}{2}} r Cos \theta e^{\frac{-Zr}{2a_{0}}}.$$

$\mu_{fi} = \frac{1}{4\sqrt{2 \pi}}\left(\frac{Z}{a_0}\right)^{\frac{5}{2}}\frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{\frac{3}{2}}<rCos \theta e^{\frac{-Zr}{2a_0}}|e rCos\theta |e^{\frac{-Zr}{a_0}}>$

$=\frac{1}{4\pi \sqrt{2}}\left(\frac{Z}{a_0}\right)^4 <rCos\theta e^{\frac{-Zr}{2a_0}} |e rCos\theta |e^{\frac{-Zr}{a_0}}>$

$= \frac{e}{4\pi \sqrt{2}}\left(\frac{Z}{a_0}\right)^4 \int\limits_0^\infty r^2dr \int\limits^\pi_0 Sin\theta d\theta \int\limits_0^{2\pi}d\varphi \left(r^2 e^{\frac{-Zr}{2a_0}}e^{\frac{-Zr}{a_0}}\right) Cos^2\theta$

$=\frac{e}{4\pi\sqrt{2}}2 \pi\left(\frac{Z}{a_0}\right)^4 \int\limits_0^\infty \left(r^4 e^{\frac{-3Zr}{2a_0}}\right)dr \int\limits _0^{\pi} Sin\theta Cos^2\theta d\theta$

Using integral equation 4 to integrate over r and equation 17 to integrate over $$\theta$$ we obtain:

$= \frac{e}{4\pi\sqrt{2}}2\pi \left(\frac{Z}{a_0}\right)^4 \frac{4\!}{\left(\frac{3Z}{2a_0}\right)^5}\left(\frac{-1}{3}\right) Cos^3\theta \bigg|_0^\pi$

$= \frac{e}{4\pi\sqrt{2}}2\pi\left(\frac{Z}{a_0}\right)^4 \frac{2^5a_0^5 4\!}{3^5Z^5}\left(\frac{-1}{3}\right)((-1)^3-(1)^3)$

$=\frac{3}{\sqrt{2}}\frac{2^8a_0}{3^5Z}= \frac{ea_0}{Z}\frac{2^8}{\sqrt{2}3^5}= 0.7449\frac{ea_0}{Z}$

b. Examine the symmetry of the integrands for $$<2p_z| e x | 1s>$$ and $$<2p_z| e y | 1s>.$$

Consider reflection in the xy plane:

Function Symmetry
2p$$_z$$ -1
x +1
1S +1
y +1

Under this operation the integrand of $$<2p_z| e x |1s>$$is$$(-1)(1)(1) = -1$$(it is antisymmetric) and hence $$<2p_z| e x |1s> =0.$$

Similarly, under this operation the integrand of $$<2p_z| e y |1s>$$ is (-1)(1)(1)= -1 (it is also antisymmetric) and hence $$<2p_z| e y |1s> =0.$$

c. $$\tau_{R} = \frac{3\hbar^4 c^3}{4(E_i-E_f )^3|\mu_{fi}|^2},$$ \\

$$E_i = E_{2p_z} = -\frac{1}{4}Z^2\left(\frac{e^2}{2a_0}\right)$$ \\

$$E_f = E_{1s} = -Z^2 \left(\frac{e^2}{2a_0}\right)$$

$$E_i - E_f = \frac{3}{8}\left(\frac{e^2}{a_0}\right)Z^2$$

Making the substitutions for $$E_i - E_f$$and $$|\mu_{fi}|$$ in the expression for $$\tau_{R}$$we obtain: \\

$\tau_{R} = \frac{3 \hbar^4 c^3}{4\left(\frac{3}{8}\left(\frac{e^2}{a_0}\right)Z^2\right)^3\left(\left(\frac{ea_0}{Z}\right)\frac{2^8}{\sqrt{2}3^5}\right)^2}$

$= \frac{3\hbar^4c^3}{4\frac{3^3}{8^3}\left(\frac{e^6}{a_0^3}\right)Z^6\left(\frac{e^2a_0^2}{Z^2}\right)\frac{2^{16}}{(2)3^{10}}}$

$= \frac{\hbar^4c^33^8a_0}{e^8Z^42^8},$

Inserting $$e^2 = \frac{\hbar^2}{m_ea_0}$$ we obtain:

$\tau=\frac{\hbar^4c^3 3^8 a_0m_{e}^4a_0^4}{\hbar^8Z^42^8}= \frac{3^8}{2^8}\frac{c^3 a_0^5 m_e^4}{\hbar^4Z^4}$

$= 25.6289\frac{c^3a_0^5 m_e^4}{\hbar^4Z^4}$

$= 25.6289\left(\frac{1}{Z^4}\right)\frac{(2.998x10^{10}cm \cdot sec^{-1})^3(0.529177x10^{-8}cm)^5(9.109x10^{-28}g)^4}{(1.0546x10^{-27}g\cdot cm^2 \cdot sec^{-1})^4}$

$= 1.595x10^{-9}\left(\frac{1}{Z^4}\right)$

So, for example:

Atom $$\tau_R$$
H 1.595 ns
He$$^+$$ 99.7 ps
Li$$^{+2}$$ 19.7 ps
Be$$^{+3}$$ 6.23 ps
Ne$$^{+9}$$ 159 fs

2. a. H = $$H_0 + \lambda H'(t), H'(t) = V\theta (t), H_{0\varphi k} = E_{k\varphi k}, \omega_k=\frac{E_{k}}{\hbar}$$

$$i\hbar\frac{\partial \psi}{\partial t}= H\psi$$

let $$\psi(r,t) = i \hbar \sum\limits_j c_j(t)\varphi_j e^{-i\omega_j t}$$ and insert into the above expression:

$$i\hbar \sum\limits_j \left[ \dot{c}_j -i\omega_jc_j\right] e^{-i\omega_jt}\varphi_j = i\hbar \sum\limits_jc_c(t) e^{-i\omega_jt}(H_0+\lambda H'(t))\varphi_j$$

$$\sum\limits_j \left[ i\hbar \dot{c}_j + E_jc_j - c_jE_j - c_j\lambda H' \right]e^{-i\omega_jt} = 0$$

$$\sum\limits_j \left[ i\hbar\dot{c}_j<m|j> - c_j\lambda<m|H'|j> \right]e^{-i\omega_jt}=0$$

$$i\hbar\dot{c}_m e^{-i\omega_mt} = \sum\limits_j c_j\lambda H'_{mj} e^{-i\omega_jt}$$

So,

$$\dot{c}_m = \frac{1}{i\hbar}\sum\limits_j c_j\lambda H'_{mj}e^{-i(\omega_{jm})t}$$

Going back a few equations and multiplying the left by $$\varphi_k$$ instead of $$\varphi_m$$ we obtain:

$$\sum\limits_j \left[ i\hbar\dot{c}_j<k|j> - c_j\lambda <k|H'|j> \right]e^{-i\omega_jt} = 0$$

$$i\hbar \dot{c}_k e^{-i\omega_kt} = \sum\limits_j c_j\lambda H'_{kj} e^{-i\omega_jt}$$

So,

$$\dot{c}_k = \frac{1}{i\hbar}\sum\limits_j c_j\lambda H'_{kj} e^{-(\omega_{jk})t}$$

Now, let:

$$c_m=c_m^{(0)} + c_m^{(1)}\lambda + c_m^{(2)}\lambda^2 + ...$$

$$c_k = c_k^{(0)} + c_k^{(1)}\lambda + c_k^{(2)}\lambda^2 + ...$$

and substituting into above we obtain:

$$\dot{c}_m^{(0)} + \dot{c}_m^{(1)}\lambda + \dot{c}_m^{(2)}\lambda^2 + ... = \frac{1}{i\hbar}\sum\limits_j \left[ c_j^{(0)} + c_j^{(1)}\lambda + c_j^{(2)}\lambda^2 + ... \right]$$

$\lambda H'_{mj}e^{-i(\omega_{jm})t}$

first order:

$$\dot{c}_m^{(0)} = 0 \Rightarrow c_m^{(0)} = 1$$

second order:

$$\dot{c}_m^{(1)} = \frac{1}{i\hbar} \sum\limits_j c_j^{(0)} H'_{mj} e^{-i(\omega_{jm})t}$$

(n+1)$$^st$$ order:

$$\dot{c}_m^{(n)} = \frac{1}{i \hbar}\sum\limits_j c_j^{(n-1)}H'_{mj} e^{-i(\omega_{jm})t}$$

Similarly:

first order:

$$\dot{c}_k^{(0)} = 0 \Rightarrow c_{k\neq m}^{(0)} = 0$$

second order:

$$\dot{c}_k^{(1)} = \frac{1}{i\hbar}\sum\limits_j c_j^{(n-1)} H'_{kj}e^{-i(\omega_{jk})t}$$

(n+1)$$^{st}$$ order:

$$\dot{c}_k(n) = \frac{1}{i\hbar}\sum\limits_j c_j^{(n-1)}H'_{kj}e^{-i(\omega_{jk})t}$$

So,

$$\dot{c}_m^{(1)} = \frac{1}{i\hbar}c_m^{(0)} H'_{mm} e^{-i(\omega_{mm})t} = \frac{1}{i\hbar}H'_{mm}$$

$$c_m^{(1)} = \frac{1}{i\hbar} \int\limits_0^t dt' V_{mm} = \frac{V_{mm}t}{i\hbar}$$

and similarly,

$$\dot{c}_k^{(1)} = \frac{1}{i\hbar}c_m^{(0)}H'_{km}e^{-i(\omega_{mk})t} = \frac{1}{i\hbar}H'_{km}e^{-i(\omega_{mk})t}$$

$$c_k^{(1)}(t) = \frac{1}{i\hbar}V_{km}\int\limits_0^t dt' e^{-i(\omega_{mk})t'} = \frac{V_{km}}{\hbar\omega_{mk}}\left[ e^{-i(\omega_{mk})t}-1\right]$$

$$\dot{c}_m^{(2)} = \frac{1}{i\hbar}\sum\limits_j c_j^{(1)} H'_{mj}e^{-i(\omega_{jm})t}$$

$$\dot{c}_m^{(2)}= \frac{1}{i\hbar} \sum\limits_{j \neq m} \frac{1}{i\hbar}\frac{V_{jm}}{\hbar \omega_{mj}} \left[ e^{-i(\omega_{mj})t} - 1 \right] H'_{mj} e^{-i(\omega_{mj})t} + \frac{1}{i\hbar}\frac{V_{mm}t}{i \hbar}H'_{mm}$$

$$\dot{c}_m^{(2)} = \sum\limits_{j \neq m} \frac{1}{i\hbar}\frac{V_{jm}V_{mj}}{\hbar \omega_{mj}} \int\limits_0^t dt' e^{-i(\omega_{jm})t'}\left[ e^{-i(\omega_{mj})t'} -1 \right] - \frac{V_{mm}V_{mm}}{\hbar^2}\int\limits_0^t t'dt'$$

$$=\sum\limits_{j \neq m} \frac{V_{jm}V_{mj}}{i\hbar^2 \omega_{mj}}\int\limits_0^tdt'[1 - e^{-i(\omega_{jm})t'}] - \frac{{|V_{mm}|}^2}{\hbar^2}\frac{t^2}{2}$$

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