Normalization of Quantum State \(\Phi_1\) in Hydrogen Atom

(29 marks) Consider the electron of a hydrogen atom. Let the states Ψ𝑛,𝑙,𝑚 with 𝑛, 𝑙, 𝑚 integers, 𝑛 = 1, 2, 3, ...; 𝑙 = 0, 1, ..., 𝑛 − 1; −𝑙 ≤ 𝑚 ≤ 𝑙, be an orthonormal basis made of common eigenstates of the Hamiltonian, total angular momentum and angular momentum around the 𝑧 axis. That is, 𝐻ˆΨ𝑛,𝑙,𝑚 = 𝐸𝑛Ψ𝑛,𝑙,𝑚 where 𝐸𝑛 = − 𝐸0 𝑛 2 , and 𝐸0 = 13.6𝑒𝑉, 𝐿ˆ 2Ψ𝑛,𝑙,𝑚 = 𝑙(𝑙 + 1)ℏ 2Ψ𝑛,𝑙,𝑚, 𝐿ˆ 𝑧Ψ𝑛,𝑙,𝑚 = 𝑚ℏΨ𝑛,𝑙,𝑚. The corresponding orthonormality conditions are ⟨Ψ𝑛,𝑙,𝑚|Ψ𝑛 ′ ,𝑙′ ,𝑚′⟩ = 𝛿𝑛,𝑛′𝛿𝑙,𝑙′𝛿𝑚,𝑚′. a) (2 marks) Consider the state Φ1 = √ 2Ψ1,0,0 − 2Ψ2,1,1 + (2 + 𝑖)Ψ2,1,−1. Normalise Φ1

To normalize the state (\Phi_1), we need to ensure that the inner product (\langle \Phi_1 | \Phi_1 \rangle = 1). The state (\Phi_1) is given by:

[ \Phi_1 = \sqrt{2} \Psi_{1,0,0} - 2 \Psi_{2,1,1} + (2 + i) \Psi_{2,1,-1} ]

First, we calculate the inner product (\langle \Phi_1 | \Phi_1 \rangle):

[ \langle \Phi_1 | \Phi_1 \rangle = \left( \sqrt{2} \Psi_{1,0,0} - 2 \Psi_{2,1,1} + (2 + i) \Psi_{2,1,-1} \right)^\dagger \left( \sqrt{2} \Psi_{1,0,0} - 2 \Psi_{2,1,1} + (2 + i) \Psi_{2,1,-1} \right) ]

Using the orthonormality conditions (\langle \Psi_{n,l,m} | \Psi_{n',l',m'} \rangle = \delta_{n,n'} \delta_{l,l'} \delta_{m,m'}), we can simplify this expression:

[ \langle \Phi_1 | \Phi_1 \rangle = (\sqrt{2})^2 \langle \Psi_{1,0,0} | \Psi_{1,0,0} \rangle + (-2)^2 \langle \Psi_{2,1