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Chemistry LibreTexts

Solutions 9A

  • Page ID
    92325
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    Q9.1

    In this problem the perturbation to the Hamiltonian is in the potential term. That's good because we don't need to take any derivatives.

    \[\hat{H}=-\dfrac{(\hbar)^2}{2m}\dfrac{d}{dx^2}+\dfrac{1}{2}kx^2 + ax^3 \tag{1}\]

    \[\hat{H}^1=ax^3 \tag{2}\]

    \[E^1 = \langle \psi_0|\hat{H}^1|\psi_0\rangle \tag{3}\]

    \[E^1 = \langle \psi_0|ax^3|\psi_0\rangle \tag{4}\]

    Looking at the graph of the Harmonic Oscillator wavefunctions we see that \(\psi_0\) is even.

    https://upload.wikimedia.org/Wikipedia/commons/9/9e/HarmOsziFunktionen.png

    from Wikipedia.

    \[E^1 = \langle \psi_0|ax^3|\psi_0\rangle = \langle even|odd|even\rangle = \langle odd\rangle = 0 \tag{5}\]

    Q9.2

    Now we add another term to the potential.

    \[\hat{H}=-\dfrac{(\hbar)^2}{2m}\dfrac{d}{dx^2}+\dfrac{1}{2}kx^2 + ax^3 +ax^4 \tag{6}\]

    \[\hat{H}^1=ax^3 + ax^4 \tag{7}\]

    \[E^1 = \langle \psi_0|\hat{H}^1|\psi_0\rangle \tag{8}\]

    \[E^1 = \langle \psi_0|ax^3 + ax^4|\psi_0\rangle = \langle \psi_0|ax^3|\psi_0\rangle + \langle \psi_0|ax^4|\psi_0\rangle \tag{9}\]

    \[E^1 = 0 + \langle \psi_0|ax^4|\psi_0\rangle \tag{10}\]

    \[E^1 = \langle \psi_0|ax^4|\psi_0\rangle = \int_{-\infty}^{\infty} (\dfrac{\alpha}{\pi})^{1/4}e^{-\alpha x^2 / 2} \times ax^4 \times (\dfrac{\alpha}{\pi})^{1/4}e^{-\alpha x^2 / 2} dx \tag{11}\]

    \[E^1 = a (\dfrac{\alpha}{\pi})^{1/2} \int_{-\infty}^{\infty} x^4 e^{-\alpha x^2} dx = constants\int_{-\infty}^{\infty} even \tag{12}\]

    \[E^1 = 2a (\dfrac{\alpha}{\pi})^{1/2} \int_{0}^{\infty} x^4 e^{-\alpha x^2} dx \tag{13}\]

    \[E^1 = 2a (\dfrac{\alpha}{\pi})^{1/2} \dfrac{3}{8\alpha^2}(\dfrac{\pi}{\alpha})^{1/2} \tag{14}\]

    \[E^1 = \dfrac{3a}{4\alpha^2} \tag{15}\]

    Q9.3

    To find the wavefunction we have to do an infinite number of integrals.... But many of them are equal to \(0\) or so small that we can ignore them.

    \[ | n \rangle = | n^o \rangle + | n^1 \rangle = | n^o \rangle + \sum _{m \neq n} \dfrac{|m^o \rangle \langle m^o | \hat{H}^1| n^o \rangle }{E_n^o - E_m^o} \tag{16}\]

    \[\hat{H}^1=ax^3 \tag{2}\]

    Look at the infinite sum, when do the integrals go to zero?

    \[\langle \psi_{m^o}|ax^3|\psi_{n^o}\rangle = \langle even|odd|even\rangle or \langle odd|odd|odd\rangle = \langle odd\rangle = 0 \tag{17} \]

    Therefore

    \[\langle \psi_0|ax^3|\psi_2\rangle = \langle \psi_0|ax^3|\psi_4\rangle = \langle \psi_2|ax^3|\psi_4\rangle = \langle \psi_1|ax^3|\psi_3\rangle = etc etc = 0 \tag{18} \]

    So many of our integrals are zero.

    Now look at the denominator

    \[E_n^o - E_m^o \tag {19}\]

    is large when there are two very different wavefunctions and will make the sum term small enough to ignore.

    What's left?

    \[ | n \rangle = | \psi_0^0 \rangle + | \psi_0^1 \rangle = | \psi_0 \rangle + \dfrac{|\psi_1 \rangle \langle \psi_1 | \hat{H}^1| \psi_0 \rangle }{E_0^o - E_1^o} + \dfrac{|\psi_ 3\rangle \langle \psi_3 | \hat{H}^1| \psi_0 \rangle }{E_0^o - E_3^o} + \dfrac{|\psi_ 5\rangle \langle \psi_5 | \hat{H}^1| \psi_0 \rangle }{E_0^o - E_5^o} \tag{20}\]

    Fix here with alphas

    One term at a time

    \[\dfrac{|\psi_1 \rangle \langle \psi_1 | \hat{H}^1| \psi_0 \rangle }{E_0^o - E_1^o} = \dfrac{|(\dfrac{4a ^3}{\pi})^{1/4}xe^{-\alpha x^2 /2} \rangle \langle (\dfrac{4a ^3}{\pi})^{1/4}xe^{-a x^2 /2} \times ax^3 \times (\dfrac{a}{\pi})^{1/4}e^{-a x^2 / 2} \rangle }{1/2 h \nu - 3/2 h \nu}\tag{21}\]

    After integration and much simplification which I'm not going to type out the first term is

    \[\dfrac{|\psi_1 \rangle \langle \psi_1 | \hat{H}^1| \psi_0 \rangle }{E_0^o - E_1^o} = \dfrac{-6}{h\nu} (\dfrac{a}{\pi})^{1/4}x e^{ax^2/2} \tag{22}\]

    The second term

    \[\dfrac{|\psi_3 \rangle \langle \psi_3 | \hat{H}^1| \psi_0 \rangle }{E_0^o - E_3^o} = \dfrac{|(\dfrac{a ^3}{9\pi})^{1/4}(2ax^3 -3x) e^{-\alpha x^2 /2} \rangle \langle (\dfrac{a ^3}{9\pi})^{1/4}(2ax^3 -3x) e^{-\alpha x^2 /2} \times ax^3 \times (\dfrac{a}{\pi})^{1/4}e^{-a x^2 / 2} \rangle }{1/2 h \nu - 5/2 h \nu}\tag{23}\]

    Some combination of constants and seperation into 2 integrals gives

    \[\dfrac{|\psi_3 \rangle \langle \psi_3 | \hat{H}^1| \psi_0 \rangle }{E_0^o - E_3^o} = \dfrac{(\dfrac{a ^3}{9\pi})^{1/4}(2ax^3 -3x) e^{-a x^2 /2} \left[ \dfrac{4a^3}{9^{1/4}\pi^{1/2}} \int_0^\infty x^6 e^{-ax^2} dx - \dfrac{6a^2}{9^{1/4}\pi^{1/2}} \int_0^\infty x^4 e^{ax^2} dx \right] }{1/2 h \nu - 5/2 h \nu}\tag{24}\]

    \[\dfrac{|\psi_3 \rangle \langle \psi_3 | \hat{H}^1| \psi_0 \rangle }{E_0^o - E_3^o} =\dfrac{-1}{4h\nu}(\dfrac{a}{\pi})^{1/4}(2ax^3 - 3x) e^{-a x^2 /2}\tag{25}\]

    I'm going to assume that

    \[\dfrac{|\psi_5 \rangle \langle \psi_5 | \hat{H}^1| \psi_0 \rangle }{E_0^o - E_5^o} = 0 \tag{26}\]

    Cause that integral is HARD and th energy difference is bigger.

    Therefore the final answer is:

    \[ | n \rangle = (\dfrac{a}{\pi})^{1/4}e^{-a x^2 / 2} + \dfrac{-6}{h\nu} (\dfrac{a}{\pi})^{1/4} e^{ax^2/2} + \dfrac{-1}{4h\nu}(\dfrac{a}{\pi})^{1/4}(2ax^3 - 3x) e^{-a x^2 /2} \]

    \[= |\psi_0\rangle +\dfrac{-6}{4^{1/4}a^{1/2}h\nu} |\psi_1\rangle + \dfrac{-9^{1/4}}{a^{1/2}4h\nu} |\psi_3\rangle \]

    \[= |\psi_0\rangle +\dfrac{-4.28}{a^{1/2}h\nu} |\psi_1\rangle + \dfrac{-0.433}{a^{1/2}h\nu} |\psi_3\rangle \tag{27}\]

    Q9.4

    An electron moving in a conjugated bond framework of a molecule can be viewed as an electron in a box of length \(L\). If an externally applied electric field of strength \(\epsilon\)

    \[\vec{F}= \epsilon x\]

    that is oriented along the \(x\) axis (the length of the box), it interacts with the negatively charged electron via the following perturbation to the potential energy

    \[V = \epsilon e x\]

    where \(x\) is the position of the electron in the box, \(\epsilon\) is the field strength, and \(e\) is the electron charge. Calculate the first order perturbation to the energy of the ground-state wavefunction.

    The perturbation of the Hamiltonian is \( H^1 = V = \epsilon e x \) and the wavefunctions of the non-perturbed system are particle in a box eigenstates \(\sqrt{\dfrac{2}{L}}\sin{\dfrac{n\pi x}{L}}\) (specifically, for the ground state n=1 - \(\sqrt{\dfrac{2}{L}}\sin{\dfrac{\pi x}{L}}\) )

    The first order correction for the energies in perturbation theory is given by the expression

    \[ E_n^1 = \langle n^o | H^1 | n^o \rangle = \dfrac{2}{L}\int_{0}^{L} \epsilon e x\sin^2\dfrac{\pi x}{L}dx = \epsilon e \dfrac{2}{L}\int_{0}^{L} x\sin^2\dfrac{\pi x}{L}dx = \epsilon e I\]

    Let's integrate \(I\) by parts:

    \[U = x\]

    \[dU=dx\]

    \[dV=\dfrac{2}{L} \sin^2\dfrac{\pi x}{L}dx\]

    \[V = \int \dfrac{2}{L}\sin^2\dfrac{\pi x}{L}dx = \dfrac{2}{L}\dfrac{1}{2}\int (1-cos{\dfrac{2\pi x}{L}})dx= \dfrac{1}{L}x +\dfrac{1}{L}\dfrac{L}{2\pi}(-\sin{\dfrac{2\pi x}{L}}) = \dfrac{1}{L} (x - \dfrac{L}{2\pi}\sin{\dfrac{2\pi x}{L}})\]

    \[I=UV\Biggr\rvert_{0}^{L} -\int_{0}^{L} VdU = x \dfrac{1}{L} (x - \dfrac{L}{2\pi}\sin{\dfrac{2\pi x}{L}}) \Biggr\rvert_{0}^{L} - \int_{0}^{L} \dfrac{1}{L} (x - \dfrac{L}{2\pi}\sin{\dfrac{2\pi x}{L}}) dx = \\ x \dfrac{1}{L} (x - \dfrac{L}{2\pi}\sin{\dfrac{2\pi x}{L}}) \Biggr\rvert_{0}^{L} - \dfrac{1}{L} ( \frac{x^2}{2} +\frac{L^2}{4\pi^2}cos{\dfrac{2\pi x}{L}}) \Biggr\rvert_{0}^{L} = 1 \cdot (L-\dfrac{L}{2\pi}\sin{2\pi}) - 0 - \dfrac{1}{L} ( \frac{L^2}{2} +\frac{L^2}{4\pi^2}cos{2\pi}) + \dfrac{1}{L} ( 0 +\frac{L^2}{4\pi^2}cos{0})= \\ L-\dfrac{L}{2\pi} \cdot 0 - \frac{L}{2} - \dfrac{1}{L} \frac{L^2}{4\pi^2} \cdot 1+ \dfrac{1}{L} \frac{L^2}{4\pi^2} \cdot 1 = L- \frac{L}{2} - \dfrac{1}{L} \frac{L^2}{4\pi^2} + \dfrac{1}{L} \frac{L^2}{4\pi^2} = \dfrac{L}{2} \]

    \[ E_n^1= \epsilon e I = \epsilon e \dfrac{L}{2}\]

    Q9.5

    Calculate the 1st order correction for a 1D potential well for the stationary states with quantum number n = 1 and n=2 with the potential energy:

    \[ V = V_0 \textrm{ for } \frac{1}{4} l \langle x \langle \frac{3}{4} l \]

    \[V = 0 \textrm{ for } 0 \leq x \leq \frac{1}{4} l \]

    \[V = 0 \textrm{ for } \frac{3}{4} l \leq x \leq l \]

    \[V = \infty \textrm{ elsewhere } \]

    where \( V_0 = \frac{\hbar^2}{ml^2} \)

    The perturbation of the Hamiltonian is \( H^1 = V_0 = \frac{\hbar^2}{ml^2} \) in the region \( \frac{1}{4} l \langle x \langle \frac{3}{4} l \)

    The first order correction for the energies in perturbation theory is given by the expression

    \[ E_n^1 = \langle n^o | H^1 | n^o \rangle = \langle n^o | \frac{\hbar^2}{ml^2} | n^o \rangle = \int_{\frac{l}{4}}^{\frac{3l}{4}} \sqrt{\dfrac{2}{l}}\sin{\dfrac{n\pi x}{l}} \frac{\hbar^2}{ml^2} \sqrt{\dfrac{2}{l}}\sin{\dfrac{n\pi x}{l}} = \frac{\hbar^2}{ml^2} \int_{\frac{l}{4}}^{\frac{3l}{4}} \dfrac{2}{l}\sin^2{\dfrac{n\pi x}{l}}dx= \frac{\hbar^2}{ml^2} \dfrac{2}{l}\dfrac{1}{2}\int_{\frac{l}{4}}^{\frac{3l}{4}} (1-cos{\dfrac{2n\pi x}{l}})dx=\\ \frac{\hbar^2}{ml^2} (\dfrac{1}{l}x\Biggr\rvert_{\frac{l}{4}}^{\frac{3l}{4}} +\dfrac{1}{l}\dfrac{l}{2n\pi}(-\sin{\dfrac{2n\pi x}{l}})\Biggr\rvert_{\frac{l}{4}}^{\frac{3l}{4}} )=\frac{\hbar^2}{ml^2} (\dfrac{1}{l}\frac{3l}{4}-\dfrac{1}{l}\frac{l}{4}-\dfrac{1}{2n\pi}\sin{\dfrac{2n\pi \frac{3l}{4}}{l}}+\dfrac{1}{2n\pi}\sin{\dfrac{2n\pi \frac{l}{4}}{l}}) = \\ \frac{\hbar^2}{ml^2} (\frac{1}{2}-\dfrac{1}{2n\pi}\sin{\dfrac{6n\pi}{4}}+\dfrac{1}{2n\pi}\sin{\dfrac{2n\pi}{4}})= \frac{\hbar^2}{ml^2} (\frac{1}{2}-\dfrac{1}{2n\pi}\sin{\dfrac{3n\pi}{2}}+\dfrac{1}{2n\pi}\sin{\dfrac{n\pi}{2}})\]

    For n=1

    \[ E_1^1 = \frac{\hbar^2}{ml^2} (\frac{1}{2}-\dfrac{1}{2\pi}\sin{\dfrac{3\pi}{2}}+\dfrac{1}{2\pi}\sin{\dfrac{\pi}{2}}) = \frac{\hbar^2}{ml^2} (\frac{1}{2}-\dfrac{1}{2\pi}(-1)+\dfrac{1}{2\pi})=\frac{\hbar^2}{ml^2} (\frac{1}{2}+\dfrac{1}{\pi}) = \frac{\hbar^2}{2ml^2} (1+\dfrac{1}{2\pi})\]

    For n=2

    \[ E_2^1 = \frac{\hbar^2}{ml^2} (\frac{1}{2}-\dfrac{1}{4\pi}\sin{\dfrac{6\pi}{2}}+\dfrac{1}{4\pi}\sin{\dfrac{\pi}{2}}) = \frac{\hbar^2}{ml^2} (\frac{1}{2}-\dfrac{1}{4\pi}\sin{3\pi}+\dfrac{1}{4\pi}\sin{\pi}) = \frac{\hbar^2}{ml^2} (\frac{1}{2}-\dfrac{1}{4\pi} \cdot 0+\dfrac{1}{4\pi} \cdot 0) = \frac{\hbar^2}{2ml^2} \]

    The first order correction for the stationary states in perturbation theory is given by the expression

    \[ \underbrace{| n \rangle \approx | n^o \rangle + | n^1 \rangle = | n^o \rangle + \sum _{m \neq n} \dfrac{|m^o \rangle \langle m^o | H^1| n^o \rangle }{E_n^o - E_m^o}}_{\text{First Order Perturbation Theory}} \]

    \[ | n^1 \rangle = \sum _{m \neq n} \dfrac{|m^o \rangle \langle m^o | H^1| n^o \rangle }{E_n^o - E_m^o} = \sum _{m \neq n} \dfrac{|m^o \rangle \langle m^o | \frac{\hbar^2}{ml^2} | n^o \rangle }{E_n^o - E_m^o} = \frac{\hbar^2}{ml^2}\sum _{m \neq n} \dfrac{|m^o \rangle \langle m^o | n^o \rangle }{E_n^o - E_m^o} \]

    Let's calculate \(\langle m^o | n^o \rangle \) first

    \[\langle m^o | n^o \rangle = \int_{\frac{l}{4}}^{\frac{3l}{4}} \dfrac{2}{l}\sin{\dfrac{m\pi x}{l}}\sin{\dfrac{n\pi x}{l}}dx=\dfrac{2}{l}\dfrac{1}{2}\int_{\frac{l}{4}}^{\frac{3l}{4}} \cos{(\dfrac{m\pi x}{l}-\dfrac{n\pi x}{l})}-\cos{(\dfrac{m\pi x}{l}+\dfrac{n\pi x}{l})}dx= \\ \dfrac{1}{l}\int_{\frac{l}{4}}^{\frac{3l}{4}} \cos{\dfrac{(m-n)\pi x}{l}}+\cos{\dfrac{(m+n)\pi x}{l}}dx=\dfrac{1}{l}(\dfrac{l}{(m-n)\pi}\sin{\dfrac{(m-n)\pi x}{l}}+\dfrac{l}{(m+n)\pi}\sin{\dfrac{(m+n)\pi x}{l}})\Biggr\rvert_{\frac{l}{4}}^{\frac{3l}{4}} = \\ \dfrac{1}{l}(\dfrac{l}{(m-n)\pi}\sin{\dfrac{(m-n)\pi \frac{3l}{4}}{l}}+\dfrac{l}{(m+n)\pi}\sin{\dfrac{(m+n)\pi \frac{3l}{4}}{l}}- \dfrac{l}{(m-n)\pi}\sin{\dfrac{(m-n)\pi \frac{l}{4}}{l}}-\dfrac{l}{(m+n)\pi}\sin{\dfrac{(m+n)\pi \frac{l}{4}}{l}}) = \\ \dfrac{1}{(m-n)\pi}\sin{\dfrac{(m-n)3\pi}{4}}+\dfrac{1}{(m+n)\pi}\sin{\dfrac{(m+n)3\pi}{4}}- \dfrac{1}{(m-n)\pi}\sin{\dfrac{(m-n)\pi}{4}}-\dfrac{1}{(m+n)\pi}\sin{\dfrac{(m+n)\pi}{4}}\]

    For n=1 state

    \[ | 1^1 \rangle = \sum _{m \neq n} \dfrac{|m^o \rangle \langle m^o | H^1| 1^o \rangle }{E_1^o - E_m^o} = \sum _{m \neq 1} \dfrac{|m^o \rangle \langle m^o | \frac{\hbar^2}{ml^2} | 1^o \rangle }{E_1^o - E_m^o} = \frac{\hbar^2}{ml^2} \sum _{m \neq 1} \dfrac{|m^o \rangle \langle m^o | 1^o \rangle }{E_1^o - E_m^o} \]

    We can start writing out the terms in the sum

    m = 2

    \[\frac{\hbar^2}{ml^2} \dfrac{|2^o \rangle \langle 2^o | 1^o \rangle }{E_1^o - E_2^o} = \\ \frac{\hbar^2}{ml^2} \dfrac{|2^o \rangle (\dfrac{1}{(m-n)\pi}\sin{\dfrac{(m-n)3\pi}{4}}+\dfrac{1}{(m+n)\pi}\sin{\dfrac{(m+n)3\pi}{4}}- \dfrac{1}{(m-n)\pi}\sin{\dfrac{(m-n)\pi}{4}}-\dfrac{1}{(m+n)\pi}\sin{\dfrac{(m+n)\pi}{4}}) }{E_1^o - E_2^o} = \\ \frac{\hbar^2}{ml^2} \dfrac{|2^o \rangle (\dfrac{1}{(2-1)\pi}\sin{\dfrac{(2-1)3\pi}{4}}+\dfrac{1}{(2+1)\pi}\sin{\dfrac{(2+1)3\pi}{4}} - \dfrac{1}{(2-1)\pi}\sin{\dfrac{(2-1)\pi}{4}}-\dfrac{1}{(2+1)\pi}\sin{\dfrac{(2+1)\pi}{4}}) }{E_1^o - E_2^o} = \\ \frac{\hbar^2}{ml^2} \dfrac{|2^o \rangle (\dfrac{1}{\pi}\sin{\dfrac{3\pi}{4}}+\dfrac{1}{3\pi}\sin{\dfrac{9\pi}{4}}- \dfrac{1}{\pi}\sin{\dfrac{\pi}{4}}-\dfrac{1}{3\pi}\sin{\dfrac{3\pi}{4}}) }{E_1^o - E_2^o} = \\ \frac{\hbar^2}{ml^2} \dfrac{ \dfrac{2}{l}\sin{\dfrac{2\pi x}{l}} (\dfrac{1}{\pi}\sin{\dfrac{3\pi}{4}}+\dfrac{1}{3\pi}\sin{\dfrac{9\pi}{4}}- \dfrac{1}{\pi}\sin{\dfrac{\pi}{4}}-\dfrac{1}{3\pi}\sin{\dfrac{3\pi}{4}}) } {\frac{h^2}{8ml^2} - \frac{4h^2}{8ml^2}} = \\ \frac{\hbar^2}{ml^2} \dfrac{ \dfrac{2}{l}\sin{\dfrac{2\pi x}{l}} (\dfrac{1}{\pi}\frac{\sqrt{2}}{2}+\dfrac{1}{3\pi} \frac{\sqrt{2}}{2}- \dfrac{1}{\pi} \frac{\sqrt{2}}{2} -\dfrac{1}{3\pi} \frac{\sqrt{2}}{2}) } {\frac{h^2}{8ml^2} - \frac{4h^2}{8ml^2}} = 0 \]

    m = 3

    \[ \frac{\hbar^2}{ml^2} \dfrac{|3^o \rangle \langle 3^o | 1^o \rangle }{E_1^o - E_3^o} = \\ \frac{\hbar^2}{ml^2} \dfrac{|3^o \rangle (\dfrac{1}{(m-n)\pi}\sin{\dfrac{(m-n)3\pi}{4}}+\dfrac{1}{(m+n)\pi}\sin{\dfrac{(m+n)3\pi}{4}}- \dfrac{1}{(m-n)\pi}\sin{\dfrac{(m-n)\pi}{4}}-\dfrac{1}{(m+n)\pi}\sin{\dfrac{(m+n)\pi}{4}}) }{E_1^o - E_3^o}= \\ \frac{\hbar^2}{ml^2} \dfrac{|3^o \rangle (\dfrac{1}{(3-1)\pi}\sin{\dfrac{(3-1)3\pi}{4}}+\dfrac{1}{(3+1)\pi}\sin{\dfrac{(3+1)3\pi}{4}}- \dfrac{1}{(3-1)\pi}\sin{\dfrac{(3-1)\pi}{4}}-\dfrac{1}{(3+1)\pi}\sin{\dfrac{(3+1)\pi}{4}}) }{E_1^o - E_3^o} = \\ \frac{\hbar^2}{ml^2} \dfrac{|3^o \rangle (\dfrac{1}{2\pi}\sin{\dfrac{6\pi}{4}}+\dfrac{1}{4\pi}\sin{\dfrac{12\pi}{4}}- \dfrac{1}{2\pi}\sin{\dfrac{2\pi}{4}}-\dfrac{1}{4\pi}\sin{\dfrac{4\pi}{4}}) }{E_1^o - E_3^o} = \frac{\hbar^2}{ml^2} \dfrac{|3^o \rangle (\dfrac{1}{2\pi}\sin{\dfrac{3\pi}{2}}+\dfrac{1}{4\pi}\sin{3\pi}- \dfrac{1}{2\pi}\sin{\dfrac{\pi}{2}}-\dfrac{1}{4\pi}\sin{\pi}) }{E_1^o - E_3^o} = \\ \frac{\hbar^2}{ml^2} \dfrac{|3^o \rangle (\dfrac{1}{2\pi}(-1)+\dfrac{1}{4\pi} \cdot 0 - \dfrac{1}{2\pi} \cdot 1-\dfrac{1}{4\pi} \cdot 1) }{E_1^o - E_3^o} = \\ \frac{\hbar^2}{ml^2} \dfrac{\dfrac{2}{l}\sin{\dfrac{3\pi x}{l}} (-\dfrac{1}{2\pi} - \dfrac{1}{2\pi} -\dfrac{1}{4\pi}) }{\frac{h^2}{8ml^2} - \frac{9h^2}{8ml^2}} = \frac{\hbar^2}{ml^2} \dfrac{\dfrac{2}{l}\sin{\dfrac{3\pi x}{l}} (-\dfrac{5}{4\pi}) }{\frac{h^2}{8ml^2} - \frac{9h^2}{8ml^2}} = \frac{h^2}{4 \pi ml^2} \dfrac{\dfrac{2}{l}\sin{\dfrac{3\pi x}{l}} (-\dfrac{5}{4\pi}) }{\frac{h^2}{8ml^2} - \frac{9h^2}{8ml^2}} = \\ \frac{h^2 2 \sin{\dfrac{3\pi x}{l}} (-5) 8ml^2}{4 \pi ml^2 l (h^2 - 9h^2) 4 \pi}= \frac{ 5 \sin{\dfrac{3\pi x}{l}} }{8\pi^3 l}\]

    m = 4

    \[ \frac{\hbar^2}{ml^2} \dfrac{|4^o \rangle \langle 4^o | 1^o \rangle }{E_1^o - E_4^o} = \\ \frac{\hbar^2}{ml^2} \dfrac{|4^o \rangle (\dfrac{1}{(m-n)\pi}\sin{\dfrac{(m-n)3\pi}{4}}+\dfrac{1}{(m+n)\pi}\sin{\dfrac{(m+n)3\pi}{4}}- \dfrac{1}{(m-n)\pi}\sin{\dfrac{(m-n)\pi}{4}}-\dfrac{1}{(m+n)\pi}\sin{\dfrac{(m+n)\pi}{4}})}{E_1^o - E_4^o}\]

    and so on...

    The same approach would work for n=2 state

    \[ | 2^1 \rangle = \sum _{m \neq 2} \dfrac{|m^o \rangle \langle m^o | H^1| 2^o \rangle }{E_2^o - E_m^o} \]

    Q9.6

    What is the connection between the Pauli Exclusion principle, the Aufbau principle, the indeterminacy of fermions, electron configurations and quantum numbers?

    Both bosons and fermions are indistinguishable, meaning that they cannot be tracked or told apart even in principle. Because of this under exchange the total wavefunction of a multiparticle system can only change by a phase, \(e^{i\phi}\). Thus when the wavefunction is squared the phase term becomes one. In addition the because exchanging the particles twice should return the original state \(\phi\) must be an even or odd multiple of \(\pi\) and thus a multiparticle system can be either symmetric or anti-symmetric. The Pauli exclusion principle states that fermion wavefunctions must be anti-symmetric with respect to the exchange of particles. Because of this requirement fermions cannot inhabit the same state in an atom as this would be a symmetric state. And by the same state we mean have the same quantum numbers in the same atom. This means that atomic states can hold one electron each and after they are filled we have to occupy higher energy states, which is the aufbau principle. This is how we form electron configurations which are the arrangement of electrons in an atom.

    Q9.7

    Why is shielding more effective for electrons in orbitals with lower principal quantum number than for electrons within the same shell?

    Shielding is more effective for lower lever electrons because electrons in lower shells are often between higher shell electrons and the nucleus, because lower shell electrons are physically closer to the nucleus. Electrons in the same shell, however, being equidistant from the nucleus do not provide much shielding for each other.

    Q9.8

    Are the following two-electron wavefunctions symmetric, asymmetric or neither to electron permutation (note: spin and orbitals components are separated)

    1. [1s(1)2s(2)+2s(1)1s(2)][α(1)β(2)+β(1)α(2)]
    2. [1s(1)2s(2)+2s(1)1s(2)][α(1)β(2)β(1)α(2)]
    3. [1s(1)2s(2)2s(1)1s(2)][α(1)β(2)+β(1)α(2)]
    4. [1s(1)2s(2)2s(1)1s(2)][α(1)β(2)β(1)α(2)]
    5. [1s(1)2s(2)2s(1)1s(2)][α(1)α(2)]
    6. 1s(1)2s(2)

    1. [1s(1)2s(2)+2s(1)1s(2)][α(1)β(2)+β(1)α(2)] = [1s(2)2s(1)+2s(2)1s(1)][α(2)β(1)+β(2)α(1)]

    (symmetric)

    2. [1s(1)2s(2)+2s(1)1s(2)][α(1)β(2)β(1)α(2)] = -[1s(2)2s(1)+2s(2)1s(1)][α(2)β(1)β(2)α(1)] (anti-symmetric)

    3. [1s(1)2s(2)2s(1)1s(2)][α(1)β(2)+β(1)α(2)] = [1s(2)2s(1)2s(2)1s(1)][α(2)β(1)+β(2)α(1)] (anti-symmetric)

    4. [1s(1)2s(2)2s(1)1s(2)][α(1)β(2)β(1)α(2)] = [1s(2)2s(1)2s(2)1s(1)][α(2)β(1)β(2)α(1)] (symmetric)

    5. [1s(1)2s(2)2s(1)1s(2)][α(1)α(2)] = -[1s(2)2s(1)2s(2)1s(1)][α(2)α(1)] (anti-symmetric)

    6. 1s(1)2s(2) does not equal 1s(2)2s(1) (neither)

    Q9.9

    Normalize this two-electron wavefunction

    \[ |\Psi(1,2) \rangle = \begin{vmatrix}\alpha (1) & \alpha (2) \\\ \beta(1) & \beta (2) \end{vmatrix} = \alpha (1)\beta (2) - \alpha (2) \beta(1)\]

    Let's include the normalization constant N

    \[ |\Psi(1,2) \rangle = N \begin{vmatrix}\alpha (1) & \alpha (2) \\\ \beta(1) & \beta (2) \end{vmatrix} = N (\alpha (1)\beta (2) - \alpha (2) \beta(1))\]

    Remembering the normalization condition

    \[ \langle \Psi(1,2)| \Psi(1,2) \rangle = N^2 \langle \alpha (1)\beta (2) - \alpha (2) \beta(1)| \alpha (1)\beta (2) - \alpha (2) \beta(1) \rangle =1\]

    \[N^2 (\langle \alpha (1)\beta (2) | \alpha (1)\beta (2) \rangle - \langle \alpha (1)\beta (2) | \alpha (2) \beta(1) \rangle - \langle \alpha (2) \beta(1)| \alpha (1)\beta (2) \rangle +\langle \alpha (2) \beta(1)| \alpha (2) \beta(1) \rangle )=1\]

    Since the integrals above are the double integrals for spin of electron 1 and spin of electron 2 we can break them up into a product of two integrals each of which corresponds only to one electron (either 1 or 2)

    \[N^2 (\langle \alpha (1)| \alpha (1) \rangle \langle \beta (1)| \beta (1) \rangle - \langle \alpha (1)| \beta (1) \rangle \langle \beta (2) | \alpha (2) \rangle - \langle \beta(1)| \alpha (1) \rangle \langle \alpha (2) | \beta (2) \rangle + \langle \beta(1)| \beta(1) \rangle \langle \alpha (2)| \alpha (2) \rangle )=1\]

    Using the orthonormality of spin orbitals \(\langle \alpha (1)| \alpha (1) \rangle =1\), \(\langle \beta (1)| \beta (1) \rangle =1\), \(\langle \alpha (1)| \beta (1) \rangle =0\) and \(\langle \beta (1)| \alpha (1) \rangle =0\), we get \(N^2 (1 \cdot 1 - 0 - 0 + 1 \cdot 1)=2\).

    Therefore, \(2N^2 =1\) and \(N=\dfrac{1}{\sqrt{2}}\).

    Q9.10

    For Be atom the configuration is 1s2 2s2. The spin orbitals are 1sα, 1sβ, 2sα, 2sβ . Therefore, the Slater determinant is

    \[ \Psi(1,2,3,4) = \dfrac{1}{\sqrt{4!}} \begin{vmatrix} 1s(1) \alpha (1) & 1s(2) \alpha (2) & 1s(3) \alpha (3) & 1s(4) \alpha (4) \\\ 1s(1) \beta(1) & 1s(2) \beta (2) & 1s(3) \beta (3) & 1s(4) \beta (4) \\\ 2s(1) \alpha(1) & 2s(2) \alpha (2) & 2s(3) \alpha (3) & 2s(4) \alpha (4) \\\ 2s(1) \beta(1) & 2s(2) \beta (2) & 2s(3) \beta (3) & 2s(4) \beta (4) \end{vmatrix}\]

    Q9.10

    Group IA, configuration ns: H, Li, Na, K, Rb. Group IIA, ns2 : Be, Mg, Ca, Sr. Group VB, ns2np3 : N, P, As, Sb. Group 0: He, Ne, Ar, Kr, Xe. Transition elements: Cr 4s3d5 , Mn 4s23d5 , Mo 5s14d5 , Tc 5s24d5 . Also Cu, Zn, Pd, Ag, Cd, all with d10

    Q9.11

    Find the spectroscopic terms originating from the following configurations (two electrons in two non-equivalent orbitals in the same atom):

    • 1s2s = 3S1 1S0
    • 1s3s = 3S1 1S0
    • 2p3p = 3D3 3D2 3D1 1D2 3P2 3P1 3P0 1P1 3S1 1S0
    • 2s3d = 3D3 3D2 3D1 1D2

    The spectroscopic terms are 2S + 1LJ where S = |Ms1 + Ms2|, |Ms1 + Ms2 - 1|, ...., |Ms1 - Ms2|, L = |l1 + l2|, |l1 + l2 - 1|, ...., |l1 - l2|, and J = |L + S|, |L + S - 1|, ...., |L - S| and since the two electrons in each example are in different orbitals, i.e. 2p3p instead of 2p2p, we do not have to worry about the Pauli exclusion principle rendering certain terms impossible.

    Q9.12

    First excited states:

    • H: 2s which is a 2S state
    • He: 1s12s1 which is a 3S
    • Li: 1s22p1 which is a 2P
    • Be: 1s22s12p1 which is a 3P
    • B: 1s22s23s1 which is a 2S

    For C, N and O, the electron configuration is the same as for the ground state, but the occupation of degenerate p-orbitals is not optimal.

    • C: 2s22p2 which is a 1D
    • N: 2s22p3 which is a 2D
    • O: 2s22p4 which is a 1D
    • F: 2s22p5 which is a 2P
    • Ar: 2s22p53s1 which is a 3P

    Q9.13

    Find the spectroscopic term originating from the ground states configuration of the sodium atom and the boron atom.

    a) Boron: 1s22s22p1 so with one 2p electron not in a filled shell our terms are 2P3/2 and 2P1/2

    b) Sodium: 1s22s22p63s1 so with one 3s electron not in a filled shell our term is 2S1/2

    Q9.14

    Find the spin-orbit energy levels of hydrogen with electron in 2p and 3d.

    a) 2p: 2P3/2 and 2P1/2

    b) 3d: 2D5/2 2D3/2


    Solutions 9A is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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