2

Is a ray in Hilbert space the same thing as a vector? Or is a ray a special kind of vector?

Qmechanic
  • 201,751
Lory
  • 1,035
  • 4
    A ray in Hilbert space is a set of vectors, where all vectors of the set are multiples of each other. To specify a ray you pick a vector $\psi$ then the ray $[\psi]$ is the set $$[\psi]={\lambda \psi \in {\cal H} : \lambda \in \mathbb{C}}.$$ – Gold Oct 07 '21 at 01:02
  • 1
    $\lambda=0$ is not permitted… – Valter Moretti Sep 21 '23 at 01:15

2 Answers2

12

They are not actually the same.

A ray is a set of vectors. For example, $\psi$ and $\chi$ belong to the same ray in the Hilbert space if $\psi = \lambda\chi$ where {$\lambda \in \mathbb C:\lambda\ne 0$} is an arbitrary phase factor.

In other words, we can say that any set of vectors that differ only by a complex phase factor, form a ray in the Hilbert space.

joseph h
  • 29,356
  • 1
    Complex phase doesn't sound as the same thing as multiplication by a non-zero complex number. If \lambda is 2, the phase is the same, isn't it? I suppose that the answer might want to incorporate unitarity somehow? – Jirka Hanika Sep 20 '23 at 06:23
1

If ${\cal H}$ is a complex vector space, the rays of it are the equivalence classes of the equivalence relation

if $\psi,\phi\in {\cal H}$, then $\psi \sim \phi$ if and only if $\psi = a \phi$ for some $a\in \mathbb C$, $a\neq 0$.

So, $[\psi] :=\{a\psi \:|\: a\in \mathbb{C}\setminus\{0\}\}$ is a generic ray.

Usually, the equivalence class made of the zero vector $[0]$ is not considered a ray. I adopt this convention henceforth.

If ${\cal H}$ is equipped with a Hermitian inner product, in particular if the space is Hilbert, we can extract a unit vector to every equivalence class.

Evidently, the unit vectors of every given equivalence class can only differ for a factor of absolute value $1$, i.e. a phase.

In this sense, without loss of information, we can define the rays from scratch by starting from the unit sphere $S^1\subset {\cal H}$ instead of the space itself.

if $\psi,\phi\in S^1$, then $\psi \sim \phi$ if and only if $\psi = a \phi$ for some $a\in \mathbb C$, $|a|=1$.

So, $[\psi] :=\{a\psi \:|\: a\in \mathbb{C}, |a|=1\}$, when $\psi \in S^1$, is a generic ray of the second definition.

It is clear that there is a one-to-one correspondence between the equivalence classes of the former relation (with the said convention on the zero ray) and the equivalence classes of the latter relation.

The second definition of rays is technically convenient in QM because it automatically takes the normalisation of the states into account.

Rays are not vectors, but sets of vectors. There is no natural vector space structure on the set of rays. However it is possible to define a notion of distance, with a physical interpretation in terms of transition probability between corresponding states.

Valter Moretti
  • 71,578