Normalization of a wavefunction

[tina21]

Homework Statement

Normalize the wave function
Ѱ(x,0)=A/√a sin(ᴨx/a) +√3/5 sin(3ᴨx/a) + 1/√5a sin(5ᴨx/a)

Relevant Equations

Ѱ(x,0)=A/√a sin(ᴨx/a) +√3/5 sin(3ᴨx/a) + 1/√5a sin(5ᴨx/a)

I tried writing the function as:

Ѱ = c1Φ1 + C2𝚽2 + C3𝚽3

in order to then find mod C1^2...

But ɸ = √2/a sin(ᴨx/a) and not sin(ᴨx/a)

I cannot understand how the factor of "√2/a " comes

 

[PeroK]

I cannot understand how the factor of "√2/a " comes

What do you not understand about $$\phi_n(x) = \sqrt{\frac 2 a}\sin(\frac{n\pi x}{a})$$

 

[tina21]

What do you not understand about $$\phi_n(x) = \sqrt{\frac 2 a}\sin(\frac{n\pi x}{a})$$

According to me the function shouldn't have had the factor √2/a but I now believe the factor arises upon normalising phi (x). Is that correct?

 

[PeroK]

According to me the function shouldn't have had the factor √2/a but I now believe the factor arises upon normalising phi (x). Is that correct?

Yes. $\phi_n(x)$ is, by definition, a normalised wavefunction. For that reason, it is always best to organise things so that you have $$\psi(x) = \sum a_n \phi_n(x)$$ and not $$\psi(x) = \sum a_n \sin (\frac{n \pi x}{a})$$ From that point of view, the question has made things a little difficult for you - but it should be easy enough to take the first step and express your wavefunction in terms of normalised eigenfunctions.

 

[tina21]

Yes. $\phi_n(x)$ is, by definition, a normalised wavefunction. For that reason, it is always best to organise things so that you have $$\psi(x) = \sum a_n \phi_n(x)$$ and not $$\psi(x) = \sum a_n \sin (\frac{n \pi x}{a})$$ From that point of view, the question has made things a little difficult for you - but it should be easy enough to take the first step and express your wavefunction in terms of normalised eigenfunctions.

Thank you. I now understand.

 

[kuruman]

Homework Statement:: Normalize the wave function
Ѱ(x,0)=A/√a sin(ᴨx/a) +√3/5 sin(3ᴨx/a) + 1/√5a sin(5ᴨx/a)
Relevant Equations:: Ѱ(x,0)=A/√a sin(ᴨx/a) +√3/5 sin(3ᴨx/a) + 1/√5a sin(5ᴨx/a)

Shouldn't the coefficient in front of the second term be $\sqrt{\dfrac{3}{5a}}$?
Just asking $\dots$