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Qmechanic ♦
Qmechanic ♦
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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$c\in \mathbb{C}$ and you won't be changing the physics in any way. I understand this mThematicallymathematically, 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$c$ is not of the form $e^{i\phi}$. Thanks!

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 and you won't be changing the physics in any way. I understand this mThematically, 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 $e^{i\phi}$. Thanks!

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 $e^{i\phi}$.

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user34801
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Quantum states as rays as opposed to vectors

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 and you won't be changing the physics in any way. I understand this mThematically, 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 $e^{i\phi}$. Thanks!