Suppose two matrices $A$ and $B$ are unitarily equivalent, that is, $A = UBU^*$ for some unitary matrix $U$. One can interpret this as a change of basis, into say an eigenbasis if all the eigenvectors are orthogonal. You could also list some properties, for example that that $A$ and $B$ must have the same set of eigenvalues.
For the infinite-dimensional case, suppose $A$ and $B$ are self-adjoint operators on a Hilbert space $H$. Then the two are unitarily equivalent if there exists a unitary operator $U$ such that $A = UBU^{-1}$.
The name and the definition suggest that there are some strong parallels between the finite and infinite dimensional cases. How can one interpret the infinite dimensional case, as in is there any nice geometric or algebraic interpretation/motivation?
For example, one can say that the spectral theorem allows us to classify bounded self-adjoint operators on a separable Hilbert space only up to unitary equivalence. Of course one can take this at face value, but what does this really mean?
Best Answer
Given a Hilbert space $H$, let us build another Hilbert space $G$ by taking each vector $\xi $ in $H$ and "painting it green". By this I mean that we are disguising $\xi $ in any way we want, changhing its nature, but keeping track of which vector $\xi $ we started with. If you cannot imagine a green vector, you could alternatively think of the ordered pair $(\xi ,*)$, where $*$ is anything you like.
The set $G$ of all green vectors is then given the structure of a Hilbert space in the most natural way possible. For instance if $x$ and $y$ are a green vectors, we define $x+y$ as follows:
scratch the paint of $x$ and $y$, resulting in vectors $\xi $ and $\eta $ in $H$, respectively,
add $\xi $ and $\eta $, as vectors in $H$, and call the sum $\zeta $,
paint $\zeta $ green and call the resulting vector $z$,
set $x+y:= z$.
We may then define a unitary operator $U:G\to H$ by seting $$ U(\xi ) = \xi \text{ painted green}, \quad \forall \xi \in H. $$
If $T$ is any bounded operator on $H$, one can likewise turn it into an operator $S$ acting on $G$ via:
given $x$ in $G$,
scratch the paint of $x$, resulting in the vector $\xi $ in $H$,
apply $T$ to $\xi $, resulting in the vector $\eta $ in $H$,
paint $\eta $ green, resulting in the vector $y\in G$,
set $S(x)=y$.
Notice that the whole algorithm defining $S$ above may be summarized by $$ S = UTU^{-1}. $$
The moral of the story is that when two operators are conjugate, they should be each viewed as a disguise of the other, and hence they have exactly the same properties, as long as these properties are defined purely in the language of Functional Analysis.