At t = 0 the hydrogen atom is in the superposition state Ψ(𝑟⃗, 0) =1√3𝜓100 + 𝐴𝜓210 where A is a real positive constant. Find A by normalization
Given wave function is
ψ(r,0)=13ψ100+Aψ210\psi(r,0) = \frac{1}{\sqrt {3}} \psi _{100}+ A \psi_{210}ψ(r,0)=31ψ100+Aψ210
According to Normalization condition,
∫∣ψ(r,0)∣2dr=1\int |\psi(r,0)|^2 dr=1∫∣ψ(r,0)∣2dr=1
∫∣13ψ100+Aψ210∣2dr=1\int |\frac{1}{\sqrt {3}} \psi _{100}+ A \psi_{210}|^2 dr=1∫∣31ψ100+Aψ210∣2dr=1
13∫∣ψ100∣2dr+A2∫∣ψ210∣2dr=1\frac{1}{3}\int |\psi _{100}|^2 dr + A^2 \int |\psi _{210}|^2 dr=131∫∣ψ100∣2dr+A2∫∣ψ210∣2dr=1
13+A2=1\frac{1}{3}+A^2=131+A2=1
Since ∫∣ψ100∣2dr=1and\int |\psi _{100}|^2 dr =1 and∫∣ψ100∣2dr=1and ∫∣ψ210∣2dr=1\int |\psi _{210}|^2 dr=1∫∣ψ210∣2dr=1
On simplifying above equation, we get
A=23A=\sqrt\frac{2}{3}A=32
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