Show this integral defines a scalar product.

[PhysStudent12]

Hi,

I'm stuck on a problem from my quantum homework. I have to show <p1|p2> = ∫(from -1 to 1) dx (p1*)(p2)
is a scalar product (p1 and p2 are single variable complex polynomials). I've figured out how to show that they satisfy linearity and positive definiteness, but I'm completely stuck on showing that they have conjugate symmetry. Anyone know how to show this integral has conjugate symmetry? Thanks!

 

[PeterDonis]

I have moved this thread to the homework forum. @PhysStudent12 please provide the information requested in the homework template--basically, what equations are applicable, and what attempts at solution you have made.

 

[PhysStudent12]

I have moved this thread to the homework forum. @PhysStudent12 please provide the information requested in the homework template--basically, what equations are applicable, and what attempts at solution you have made.

Sure, sorry about that, I'm new here.

The problem:

I have to show <p1|p2> = ∫_-1¹ dx (p1*)(p2)
is a scalar product (p1 and p2 are single variable complex polynomials).

Relevant Equations:

A scalar product has three criteria:
Linearity: a<x|y> = <ax|y> and <x+y|z> = <x|z> + <y|z>
Positive Definiteness: <x|x> ≥ 0 and <x|x> = 0 iff x = 0
Conjugate symmetry: <x|y> = the conjugate of <y|x>

Solution Attempt:

Linearity is pretty easy, <ap1|p2> = ∫_-1¹ dx a(p1*)(p2) = a∫_-1¹dx (p1*)(p2) = a<p1|p2> and so forth. For positive definiteness, <0|0> = 0 trivially, and <p1|p1> should be positive since the negative parts should multiply by each other in the integral (correct me if I'm wrong here).

The part I need help on is showing conjugate symmetry. I'm not sure how to approach that part.

Thanks!

 

[S.G. Janssens]

For positive definiteness, <0|0> = 0 trivially, and <p1|p1> should be positive since the negative parts should multiply by each other in the integral (correct me if I'm wrong here).

You need to do more here. The non-negativity actually follows from the non-negativity of the modulus of a complex number. Perhaps more importantly, remember the "iff" in

Positive Definiteness: <x|x> ≥ 0 and <x|x> = 0 iff x = 0

So, indeed $\langle 0 | 0 \rangle = 0$. But can you also deduce that $\langle p | p \rangle = 0$ implies $p = 0$ identically?

 

[Demystifier]

@PhysStudent12 I would like to see your contribution on this thread:
https://www.physicsforums.com/threads/explain-your-avatar.805265/