# Quantum Mechanics – Quantum States as Rays vs. Vectors

hilbert-spacequantum mechanicswavefunction

I recently read that a quantum state is actually defined by a ray and not a vector. That is it is possible to multiply a state $\psi$ by any complex number $c\in \mathbb{C}$ and you won't be changing the physics in any way. I understand this mathematically, but I don't understand what the physical meaning of such an "equivalent state" would be since the new state need not be normalised if $c$ is not of the form $e^{i\phi}$.

There is no particularly interesting new physical significance to such a state vector. As you already stated, it represents exactly the same physical state. The only difference is that, on taking the modulus squared, the new state gives an unnormalised probability distribution over possible measurement outcomes. You can easily extract the probability of obtaining a measurement outcome corresponding to the (possibly unnormalised) state $|\phi\rangle$ from an unnormalised state $|\psi\rangle$ by using the Born rule: $$\mathrm{Pr}(\phi) = \frac{|\langle \phi | \psi \rangle |^2}{\langle \phi | \phi \rangle \langle \psi | \psi \rangle }.$$ Clearly, using normalised states is just a handy convention that avoids any worries about calculating the denominator above. There is nothing wrong with formulating quantum mechanics without normalising state vectors to unity. Indeed, most people avoid bothering with the overall normalisation factors until they are needed at the very end of the calculation, since they just add unsightly clutter to the mathematics and have no physical significance.