How do you normalize the wavefunction ψ = e^(iψ) in quantum mechanics?

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To normalize the wavefunction ψ = e^(iψ) over the interval 0 to 2π, the normalization condition requires calculating N^2 ∫ (ψ)^2 dψ. The integral simplifies to N^2 ∫ e^(-iψ) * e^(iψ) dψ, which equals N^2 ∫ 1 dψ from 0 to 2π. This integral evaluates to 2π, leading to the equation N^2 * 2π = 1. Solving for N gives N = sqrt(1/2π), confirming the normalization process.
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Please I need help with wavefunctions!

Okay so I have attempted to understand wavefunctions in my pchem class but I am a little lost... Here is the problem.
unnormalized wavefunction (ψ) = e^(iψ) with 0≤ψ≤2∏, normalize this equation

Homework Equations



N^2∫ (ψ)^2 dψ

**I uploaded a pic of relevant equations bc it would be too messy to write out on here



The Attempt at a Solution



Okay so I know the answer is N = sqrt (1/2pi)

but I get to this step...

N^2 ∫ e^(-iψ)* e(iψ) dψ

and I am not sure where to go from here... I know ψ^2 = 1 but I am still stuck
 

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tbair200 said:
Okay so I have attempted to understand wavefunctions in my pchem class but I am a little lost... Here is the problem.
unnormalized wavefunction (ψ) = e^(iψ) with 0≤ψ≤2∏, normalize this equation

Homework Equations



N^2∫ (ψ)^2 dψ

**I uploaded a pic of relevant equations bc it would be too messy to write out on here



The Attempt at a Solution



Okay so I know the answer is N = sqrt (1/2pi)

but I get to this step...

N^2 ∫ e^(-iψ)* e(iψ) dψ

and I am not sure where to go from here... I know ψ^2 = 1 but I am still stuck

Not sure why you are stuck. The integral of 1 from 0 to 2pi is 2pi. So N^2*(2pi)=1. What's N?
 
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