Solutions 1

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Q1

Write out the complete time-independent Hamiltonians for (each term is explicitly given):

1. the helium atom, electrons 1,2: $-\dfrac{\hbar^2}{2m_e}(\nabla^2_1 + \nabla^2_2) - \dfrac{Ze^2}{4\pi\epsilon}(\dfrac{1}{r_1}+\dfrac{1}{r_2}) + \dfrac{e^2}{4\pi\epsilon r_{12}}$
2. the $$H_2^+$$ ion, nuclei A,B: $-\dfrac{\hbar^2}{2m_p}(\nabla^2_A + \nabla^2_B) - \dfrac{Ze^2}{4\pi\epsilon}(\dfrac{1}{r_A}+\dfrac{1}{r_B}) + \dfrac{e^2}{4\pi\epsilon r_{AB}}$
3. the $$H_2$$ molecule, nuclei A,B electrons 1,2: $-\dfrac{\hbar^2}{2m_p}(\nabla^2_A + \nabla^2_B) -\dfrac{\hbar^2}{2m_e}(\nabla^2_1 + \nabla^2_2) - \dfrac{e^2}{4\pi\epsilon} (\dfrac{1}{r_{1A}}+\dfrac{1}{r_{1B}} + \dfrac{1}{r_{2A}}+\dfrac{1}{r_{2B}} -\dfrac{1}{r_{AB}} - \dfrac{1}{r_{12} })$

Q2

What is the Born-Oppenheimer approximation and what do we use it for? When would it fail?

The Born-Oppenheimer approximation assumes that the nuclei in a system are stationary relative to the faster moving electrons. This allows separation of the wavefunction into a nuclear contribution and an independent electronic contribution. The approximation fails when nuclear motion is non-neglible (i.e., if the nuclei are moving faster than expected such as if very vibrational excited).

Q3

Confirm two s-p orbitals are orthonormal. The two $$|sp \rangle$$ orbitals are:

$| sp_1 \rangle = \dfrac{1}{\sqrt {2}}( |2s\rangle + |2p \rangle)$

$| sp_2 \rangle = \dfrac{1}{\sqrt {2}}( |2s\rangle - |2p \rangle)$

Orthonormal means fulfilling both orthogonality and normality.

Normality means showing:

$\langle sp_1 | sp_1 \rangle = \langle sp_2 | sp_2 \rangle = 1$

Orthogonality means showing:

$\langle sp_1 | sp_2 \rangle = \langle sp_2 | sp_1 \rangle = 0$

We must keep in mind that s and p orbitals are orthonormal:

$\langle s|s \rangle = \langle p|p \rangle = 1$

and

$\langle s|p \rangle = \langle p|s \rangle = 0$

Therefore using the decomposition of the hybrid orbitals....

$\langle sp_1 | sp_1 \rangle = \dfrac{1}{\sqrt {2}}( \langle2s| + \langle2p |) \dfrac{1}{\sqrt {2}}( |2s\rangle + |2p \rangle) = \dfrac{1}{2}( \langle s|s \rangle + \langle s|p \rangle + \langle p|s \rangle + \langle p|p \rangle ) = \dfrac{1}{2}( 1 + 0 + 0 + 1) = 1$

$\langle sp_2 | sp_2 \rangle = \dfrac{1}{\sqrt {2}}( \langle2s| - \langle2p |) \dfrac{1}{\sqrt {2}}( |2s\rangle - |2p \rangle) = \dfrac{1}{2}( \langle s|s \rangle - \langle s|p \rangle - \langle p|s \rangle + \langle p|p \rangle ) = \dfrac{1}{2}( 1 - 0 - 0 + 1) = 1$

$\langle sp_1 | sp_2 \rangle = \dfrac{1}{\sqrt {2}}( \langle2s| + \langle2p |) \dfrac{1}{\sqrt {2}}( |2s\rangle - |2p \rangle) = \dfrac{1}{2}( \langle s|s \rangle - \langle s|p \rangle - \langle p|s \rangle - \langle p|p \rangle ) = \dfrac{1}{2}( 1 - 0 + 0 - 1) = 0$

$\langle sp_2 | sp_1 \rangle = \dfrac{1}{\sqrt {2}}( \langle2s| - \langle2p |) \dfrac{1}{\sqrt {2}}( |2s\rangle + |2p \rangle) = \dfrac{1}{2}( \langle s|s \rangle - \langle s|p \rangle - \langle p|s \rangle - \langle p|p \rangle ) = \dfrac{1}{2}( 1 + 0 - 0 - 1) = 0$

Q4

Show $$|sp^2\rangle = \dfrac{|s\rangle + \sqrt{2} |p \rangle}{\sqrt{3}}$$ is normalized if $$|s \rangle$$ and $$|p \rangle$$ are normalized. This means showing that $$\langle sp^2|sp^2\rangle = 1$$:

$\langle sp^2|sp^2\rangle = (\dfrac{\langle s | + \sqrt{2} \langle p |}{\sqrt{3}} )( \dfrac{|s\rangle + \sqrt{2} |p \rangle}{\sqrt{3}}) = \dfrac{1}{3}(\langle s | s \rangle + \sqrt{2} \langle s|p \rangle = \sqrt{2} \langle p | s \rangle + 2 \langle p|p \rangle) = \dfrac{1}{3}(1 + 0 + 0 + 2) = 1$

Q5

What is the average energy of a Hydrogen atom $$|sp^2 \rangle$$ hybrid orbital if energy of $$|s \rangle$$ is $$E_s$$ and energy of $$|p \rangle$$ is $$E_p$$?

We are given that $$\langle s| \hat{H} | s \rangle = E_s$$ and $$\langle p| \hat{H} | p \rangle = E_p$$.

It is also useful to realize that

$\langle s| \hat{H} | p \rangle = \langle p| \hat{H} | s \rangle = 0$

$E = \dfrac{ \langle sp^2| \hat{H} | sp^2 \rangle}{ \langle sp^2|sp^2 \rangle} = \left(\dfrac{\langle s | + \sqrt{2} \langle p |}{\sqrt{3}} \right) \hat{H} \left( \dfrac{|s\rangle + \sqrt{2} |p \rangle}{\sqrt{3}} \right)$

$= \dfrac{1}{3}( \langle s| \hat{H} | s \rangle + \sqrt{2}\langle s| \hat{H} | p \rangle + \sqrt{2}\langle p| \hat{H} | s \rangle + 2 \langle p| \hat{H} | p \rangle = \dfrac{1}{3}(E_s + 0 + 0 + 2 E_p)$

= $\dfrac{1}{3}E_s + \dfrac{2}{3}E_p$

Q6

When one s orbital and two p orbitals are used to generate hybrid orbitals, a total of three orbitals are generated.

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