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Nana113
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Thread moved from the technical forums to the schoolwork forums
so this condition always applies and the integral always equates to 0 when asking for orthogonality?PeroK said: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".
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$$\int_a^b u(x)^*v(x) \ dx = 0$$Nana113 said:so this condition always applies and the integral always equates to 0 when asking for orthogonality?
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.PeroK said: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".
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.PeroK said:In this case that means that two functions are orthogonal if$$\int_a^b u(x)^*v(x) \ dx = 0$$
It also, i'd say the main point, makes it positive definite. ##\langle u, u \rangle > 0## for non zero vectors.Orodruin said: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.
PeroK said: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".
Yeah, but that messes up Dirac notation!pasmith said:An inner product on a complex vector space is by definition linear in its first argument and Hermitian (see eg. here), so that [tex]\langle \alpha u, v \rangle = \alpha \langle u, v \rangle[/tex] but [tex]\langle u, \alpha v \rangle = (\langle \alpha v, u \rangle)^{*} = \alpha^{*}\langle u, v \rangle.[/tex] Hence we must have [tex]\langle u, v \rangle = \int_a^b u(x)v^{*}(x)\,dx.[/tex]
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.pasmith said:An inner product on a complex vector space is by definition linear in its first argument
Because Dirac notation is typically used by physicists. See above.PeroK said:Yeah, but that messes up Dirac notation!
Looks fine to me.Nana113 said:so my working out of the question attached is correct?