Orthogonality of 2 complex exponentials

[Nana113]

For 2 complex functions, to find the orthogonality, one of the function has to be in complex conjugate? Because in the lecture note, the first formula is without complex conjugate, so I’m a bit confused

 

[PeroK]

The inner product of two complex things always involves the complex conjugate of one of the things. In this case:$$\langle u, v \rangle = \int_a^b u(x)^*v(x) \ dx$$There are loads of hits on Goggle if you search for "complex inner product".

 

[Nana113]

The inner product of two complex things always involves the complex conjugate of one of the things. In this case:$$\langle u, v \rangle = \int_a^b u(x)^*v(x) \ dx$$There are loads of hits on Goggle if you search for "complex inner product".

so this condition always applies and the integral always equates to 0 when asking for orthogonality?

 

[PeroK]

so this condition always applies and the integral always equates to 0 when asking for orthogonality?

That's the definition of orthogonality. It's a generalization of the concept of orthogonality for 2D or 3D vectors. Vectors (or functions) are orthogonal if $$\langle u, v \rangle = 0$$In this case that means that two functions are orthogonal if$$\int_a^b u(x)^*v(x) \ dx = 0$$

 

[Orodruin]

Just to add a couple of things.

The inner product of two complex things always involves the complex conjugate of one of the things. In this case:$$\langle u, v \rangle = \int_a^b u(x)^*v(x) \ dx$$There are loads of hits on Goggle if you search for "complex inner product".

The reason for this being that the inner product is supposed to satisfy $\langle v,u\rangle = \langle u,v\rangle^*$, resulting in linearity in one argument and anti-linearity in the other.

In this case that means that two functions are orthogonal if$$\int_a^b u(x)^*v(x) \ dx = 0$$

It should however be noted that an inner product on a function space will often come with an additional weight function. In this case it does not, but it is good to be aware.

 

[martinbn]

Just to add a couple of things.The reason for this being that the inner product is supposed to satisfy $\langle v,u\rangle = \langle u,v\rangle^*$, resulting in linearity in one argument and anti-linearity in the other.

It also, i'd say the main point, makes it positive definite. $\langle u, u \rangle > 0$ for non zero vectors.

 

[pasmith]

The inner product of two complex things always involves the complex conjugate of one of the things. In this case:$$\langle u, v \rangle = \int_a^b u(x)^*v(x) \ dx$$There are loads of hits on Goggle if you search for "complex inner product".

An inner product on a complex vector space is by definition linear in its first argument and Hermitian (see eg. here), so that $$\langle \alpha u, v \rangle = \alpha \langle u, v \rangle$$ but $$\langle u, \alpha v \rangle = (\langle \alpha v, u \rangle)^{*} = \alpha^{*}\langle u, v \rangle.$$ Hence we must have $$\langle u, v \rangle = \int_a^b u(x)v^{*}(x)\,dx.$$

 

[PeroK]

An inner product on a complex vector space is by definition linear in its first argument and Hermitian (see eg. here), so that $$\langle \alpha u, v \rangle = \alpha \langle u, v \rangle$$ but $$\langle u, \alpha v \rangle = (\langle \alpha v, u \rangle)^{*} = \alpha^{*}\langle u, v \rangle.$$ Hence we must have $$\langle u, v \rangle = \int_a^b u(x)v^{*}(x)\,dx.$$

Yeah, but that messes up Dirac notation!

 

[Orodruin]

An inner product on a complex vector space is by definition linear in its first argument

Note: This is the typical definition among mathematicians. Among phycisists, the typical convention is that the linearity is in the second argument. One needs to be careful to ensure oneself which convention a particular source uses.

 

[Orodruin]

Yeah, but that messes up Dirac notation!

Because Dirac notation is typically used by physicists. See above.

 

[Nana113]

so my working out of the question attached is correct?

 

[PeroK]

so my working out of the question attached is correct?

Looks fine to me.