- #1

jhughes

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## Homework Statement

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

## Homework Equations

## The Attempt at a Solution

From the definition of a unitary operator we have [itex]U^{\dagger}=U^{-1}[/itex]. From this we can pretty easily demonstrate that the inner product is preserved under [itex]U[/itex], i.e.

[itex]\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[/itex]

"[itex]\left\{\epsilon_i\right\}_{i=1,2,\dots}[/itex] is orthonormal" can be stated as

[itex]\left\langle \epsilon_i|\epsilon_j\right\rangle=\delta_{i,j}[/itex]

Since [itex]U[/itex] preserves the inner product, if this statement is true for [itex]\left\{\epsilon_i\right\}[/itex] then it must be true for the image under [itex]U[/itex] as well. In other words, if {\epsilon_i\right\} is an orthonormal set, then the image of {\epsilon_i\right\} under [itex]U[/itex] 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 [itex]N[/itex]-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.

...or am I going about this the wrong way?