Square of the sum of two orthonormal functions?

[Boltzman Oscillation]

Homework Statement

Given:
Ψ and Φ are orthonormal find
(Ψ + Φ)^2

Homework Equations

None

The Attempt at a Solution

Since they are orthonormal functions then can i do this?

(Ψ + Φ) = (Ψ + Φ)(Ψ* + Φ*)?

 

[fresh_42]

Homework Statement

Given:
Ψ and Φ are orthonormal find
(Ψ + Φ)^2

Homework Equations

None

The Attempt at a Solution

Since they are orthonormal functions then can i do this?

(Ψ + Φ) = (Ψ + Φ)(Ψ* + Φ*)?

Yes. Now go ahead. Multiply it and apply what you know.

 

[Boltzman Oscillation]

Yes. Now go ahead. Multiply it and apply what you know.

Oh goodie!
ΨΨ* + ΦΦ* + ΨΦ* + Ψ*Ψ
The dot product of two orthonormal functions is 0. The dot product of two equal functions is 1.

1 + 1 + 0 + 0 = 2

 

[Delta2]

There are some typos in this thread, most important one is that we want to find
$|\Psi+\Phi|^2$ not just $(\Psi+\Phi)^2$ , right?

 

[bjnartowt]

Does the squaring operation indicate scalar/vector/etc. multiplication, or does it indicate the inner product under which you have orthonormality? After all, $\sin n\pi x$ and $\sin m\pi x$ for $m\ne n$ are orthogonal (and also normal with the appropriate coefficient), and:
${(\sin n\pi x + \sin m\pi x)^2} = {(\sin n\pi x + \sin m\pi x)^*}(\sin n\pi x + \sin m\pi x) = {\left| {\sin n\pi x} \right|^2} + {\left| {\sin m\pi x} \right|^2} + {(\sin m\pi x)^*}\sin n\pi x + {(\sin n\pi x)^*}\sin m\pi x$

I apologize if my pedantry glosses over some context, but I'm just trying to be careful.

 

[Boltzman Oscillation]

There are some typos in this thread, most important one is that we want to find
$|\Psi+\Phi|^2$ not just $(\Psi+\Phi)^2$ , right?

Yes. I was asked to normalize and equation and it contained two orthonormal functions. I forgot how important the absolute value is.

 

[fresh_42]

I forgot how important the absolute value is.

Technically, it is not the absolute value which counts! It is only a common indication, that $\psi \chi = \langle \psi , \chi \rangle$ is meant to be the multiplication. One could as well simply define the multiplication given by the inner product and then no absolute value would be necessary. So it's the way the square has to be interpreted which counts, not the shape of the brackets.

 

[Boltzman Oscillation]

Technically, it is not the absolute value which counts! It is only a common indication, that $\psi \chi = \langle \psi , \chi \rangle$ is meant to be the multiplication. One could as well simply define the multiplication given by the inner product and then no absolute value would be necessary. So it's the way the square has to be interpreted which counts, not the shape of the brackets.

I thought the absolute value's function was to make every section of the function positive so that the integral over that function doesn't cancel out. Kinda like taking the energy of a wave. Right?

 

[fresh_42]

I thought the absolute value's function was to make every section of the function positive so that the integral over that function doesn't cancel out. Kinda like taking the energy of a wave. Right?

No, this is not right. It only makes the resulting values positive, not the individual function values. The square does the same.

The point is, that - normally - $|\psi + \chi|^2=\langle \psi + \chi, \psi + \chi \rangle$ and $(\psi + \chi)^2= \psi^2 + \psi \chi + \chi \psi + \chi^2$ where the former leads to a scalar as result and the latter to another function. In cases where the latter doesn't exist or isn't defined, resp. given, then only the former is left as possibility. By the way you stated the question, it is automatically implied that we are talking about the inner product, simply because no other product was available.