Normalizing Orthogonal Functions

Classical Mechanics Level 4

Let the wavefunction of a particle be

Ψ ( x , 0 ) = A ( 2 ψ 1 ( x ) + 5 ψ 2 ( x ) ) , \Psi(x,0)=A(\sqrt{2}{\psi}_{1}(x)+\sqrt{5}{\psi}_{2}(x)),

where ψ 1 ( x ) {\psi}_{1}(x) and ψ 1 ( x ) {\psi}_{1}(x) are orthogonal. Calculate the normalization constant A A to three decimal places.


The answer is 0.378.

Steven Zheng by Steven Zheng , 26, China

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1 solution

Deepanshu Gupta
Oct 12, 2015

Use following Facts : \text{Use following Facts :}

∙ \bullet Any two wave functions \text{Any two wave functions} ψ 1 ( x ) , ψ 2 ( x ) \displaystyle{{ \psi }_{ 1 }(x),{ { \psi }_{ 2 }(x) }} are said to be orthogonal if : \text{are said to be orthogonal if :}

∫ ψ 1 ( x ) . ψ 2 ( x ) ∗ d x = 0 = ∫ ψ 1 ( x ) ∗ . ψ 2 ( x ) d x \displaystyle{\int { { \psi }_{ 1 }(x).{ { \psi }_{ 2 }(x) }^{ * }dx } =0=\int { { { \psi }_{ 1 }(x) }^{ * }.{ \psi }_{ 2 }(x)dx } }

∙ \bullet Any wave function \text{Any wave function} , ψ ( x ) { \psi }(x) is said to be normalized if, probability of finding the particle in total space is equal to 1 :: \text{is said to be normalized if, probability of finding the particle in total space is equal to 1 ::}

∫ ψ ( x ) . ψ ( x ) ∗ d x = 1 \displaystyle{{\int { { \psi }(x).{ { \psi }(x) }^{ * }dx } =1}}

Hence: \text{Hence:} A = 1 7 = 0.377 \displaystyle{\boxed { A=\cfrac { 1 }{ \sqrt { 7 } } =0.377 } }

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