Unitary transformations relating two arbitrary normed vectors in an infinite dimensional Hilbert space?

Question

Let $x$ and $y$ be 2 arbitrary but fixed norm 1 vectors in an infinite dimensional Hilbert space $H$. Is there any unitary operator $U$ on $H$ such that $y = Ux$? How many such unitary operators there are for two given $x$ and $y$?

Answer 1

We don't need to require that our Hilbert space $H$ be infinite-dimensional: it can have any dimension. Here are a few things they explain in any course on Hilbert spaces:

1. Any orthonormal set of vectors in $H$ is contained in some orthonormal basis. (Be careful: in the infinite-dimensional case an orthonormal basis, also called a complete orthonormal set, is an orthonormal set such that every vector in $H$ can be written as a convergent infinite linear combination of vectors in the set. So, it is not a basis in the algebraic sense, known as a Hamel basis.)

2. Any two orthonormal bases of $H$ have the same cardinality, called the Hilbert space dimension of $H$. So, we can index them with the same index set, say $(\psi_\alpha)_{\alpha \in A}$ and $(\phi_\alpha)_{\alpha \in A}$ (Be careful: due to the previous parenthetical remark, the Hilbert space dimension is not the same as the dimension of $H$ as a vector space, except in the finite-dimensional case.)

3. Given any two orthonormal bases $(\psi_\alpha)_{\alpha \in A}$ and $(\phi_\alpha)_{\alpha \in A}$ of $H$ there exists a unique unitary operator $U: H \to H$ with

$$ U(\psi_\alpha) = \phi_\alpha. $$

Using these three remarks we can easily answer your first question. Given two unit vectors $x$ and $y$ we can find an orthonormal basis $(\psi_\alpha)_{\alpha \in A}$ with $\psi_1 = x$ and an orthonormal basis $(\phi_\alpha)_{\alpha \in A}$ with $\phi_1 = y$, by the first two facts. Here I'm writing $1$ for some particular element of $A$. Then by the third fact we get a unitary operator with $U(\psi_\alpha) = \phi_\alpha$, so we get $U(x) = y$.

This also answers your second question if you think about how much free choice you have in picking the orthonormal bases. I'm a bit too tired to explain the details, but unless $H$ is 0-dimensional or 1-dimensional there will be infinitely many different ways to choose the second basis $(\phi_\alpha)_{\alpha \in A}$, each giving a different unitary operator $U$ with $U(x) = y$. If the Hilbert space dimension of $H$ is finite or countably infinite, the exact number of choices will be $2^{\aleph_0}$, the cardinality of the continuum, because you're making countably many of choices, each of which can be made in a continuum of ways.

I recommend reading any decent math book on Hilbert spaces to learn proofs of facts 1-3. You can solve your puzzle without knowing these facts but it's better to know them.

There are lots of books to choose from. When I was an undergrad I adored Reed and Simon's Methods of Modern Mathematical Physics Volume 1: Functional Analysis, which has a great introduction to Hilbert space and operators and their applications to physics. But there may be other, easier ways to learn the results I just mentioned. A lot of people like Halmos' Introduction to Hilbert Space.

Answer 2

Here is an explicit construction:

If $y=\theta x$ for some $|\theta| = 1$, then we can choose $\theta I$.

Otherwise, let $S = \operatorname{sp} \{ x,y\}$ which is a two dimensional subspace. Let $w ={ y-\langle x, y \rangle x \over \| y-\langle x, y \rangle x \|}$.

Let $Ux = y = \langle x, w \rangle x+ \langle y, w \rangle w $ and $ Uw = -\langle y, w \rangle x+ \langle x, w \rangle w$

Define $Uz = z$ for $z \in S^\bot$.

It is straightforward to verify that $U$ is unitary.