# Show that a unitary operator maps one ON-basis to another

jhughes

## Homework Statement

Given an inner product space $V$, a unitary operator $U$ and a set $\left\{\epsilon_i\right\}_{i=1,2,\dots}$ which is an orthonormal basis of $V$, show that the image of $\left\{\epsilon_i\right\}$ under $U$ is also an orthonormal basis of $V$

## The Attempt at a Solution

From the definition of a unitary operator we have $U^{\dagger}=U^{-1}$. From this we can pretty easily demonstrate that the inner product is preserved under $U$, i.e.
$\left\langle a|b\right\rangle=\left\langle a^{\prime}|b^{\prime}\right\rangle\quad\mbox{where }\left|a^{\prime}\right\rangle=U\,\left|a\right\rangle\quad\mbox{and }\left|b^{\prime}\right\rangle=U\,\left|b\right\rangle$​

"$\left\{\epsilon_i\right\}_{i=1,2,\dots}$ is orthonormal" can be stated as
$\left\langle \epsilon_i|\epsilon_j\right\rangle=\delta_{i,j}$​
Since $U$ preserves the inner product, if this statement is true for $\left\{\epsilon_i\right\}$ then it must be true for the image under $U$ as well. In other words, if {\epsilon_i\right\} is an orthonormal set, then the image of {\epsilon_i\right\} under $U$ is an orthornomal set as well.

So we have "ortho" and "normal", but where I've hit a mental block is at the "basis" part. For an $N$-dimensional space, it is clear that any set contained in the space which is orthonormal and has the same number of elements as an ON-basis of the space must also be an ON-basis of the space. What I'm stuck on is how to generalize this to an infinite-dimensional space.