Solutions 1

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.