Orthogonality of Momentum Eigenstates: Why is ##\int e^{-ix}e^{2ix}dx=0##?

[Isaac0427]

Consider two momentum eigenstates $\phi_1$ and $\phi_2$ representing momenta $p_1$ and $p_2$. For the sake of easy numbers, $p_1=1*\hbar$ (with $k=1$) and $p_2=2*\hbar$ (with $k=2$). Thus, $\phi_1=e^{ix}$ and $\phi_2=e^{2ix}$. Orthogonality states that
$\int \phi_1^*\phi_2dx=\int e^{-ix}e^{2ix}dx=0$
Why is this?

I understand how orthogonality would work with dirac deltas (i.e. I know why position eigenfunctions are orthogonal in position space and why momentum eigenfunctions are orthogonal in momentum space, etc.) but I am unclear of how it works with plane waves.

Also, I am specifically asking why $\int e^{-ix}e^{2ix}dx=0$, not why orthogonality works in general (I understand its derivation using the definition of hermitian operators and the inner product).

 

[hilbert2]

You could first try to plot some graphs of real functions $\sin k_1 x \sin k_2 x$ where $k_2$ is much larger than $k_1$ and convince yourself of the fact that there's practically as much "positive" as "negative" surface area between the graph and the x-axis.

 

[Isaac0427]

You could first try to plot some graphs of real functions $\sin k_1 x \sin k_2 x$ where $k_2$ is much larger than $k_1$ and convince yourself of the fact that there's practically as much "positive" as "negative" surface area between the graph and the x-axis.

I guess then that my question is how do you numerically evaluate that improper integral?

 

[hilbert2]

When talking about rigorous mathematics, the improper integral does not converge, but I guess you can show that if you calculate an integral $\int_{-\infty}^{\infty} e^{-a|x|}e^{-ix}e^{2ix}dx$ (which does converge) and take the limit $a \rightarrow 0+$, then you get value 0.