Is there anyone could tell me why if $$\sum_{k \geq 0} e^{it \sqrt{-\lambda_k}}=\int_{-\infty}^{\infty} (\sum_{k \geq 0} e^{it \sqrt{-\lambda_k}} \mu_k^2(x))dx= \sum_{k \geq 0} e^{it \sqrt{-\lambda_k}} \int_{-\infty}^{\infty}\mu_k^2(x)dx,$$

then necessarily, $\int_{-\infty}^{\infty}\mu_k^2(x)dx=1$? Here $\mu_k(x)$ is the eigenfunction of the eigenvalue $\lambda_k$, and I suppose that the eigenfunctions are taken on the real line.

I know this is probably a silly question, but I don't know the fundamental reason.


To get a better comprehension of the problem, you could look at page $25$ of pdf

  • $\begingroup$ You can scale an eigenfunction, but it is convenient if the eigenfunctions are orthonormal. $\endgroup$ – grdgfgr May 24 '16 at 22:25

Any eigenvalue of any operator has an entire vector space of eigenfunctions. So $au_k$ is also a eigenfunction of $\lambda_k$ for any $a\ne0$. This gives him the freedom to choose his eigenfunctions to be normalized. If $v_k$ is an arbitrary eigenfunction of $\lambda_k$, then he can define $$ a = \left(\int_{-\infty}^{\infty}v_k^2(x)dx\right)^{-1/2}$$

and then choose $u_k = av_k$. Then $$\int_{-\infty}^{\infty}u_k^2(x)dx = \int_{-\infty}^{\infty}(av_k)^2(x)dx\\=a^2\int_{-\infty}^{\infty}v_k^2(x)dx = 1$$

This is such a very common thing to do, he didn't feel the need to explain it, though he would have been wiser to say

where $u_k$ is the normalized eigenfunction associated to the eigenvalue $\lambda_k.$


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