- #1

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Hello,

I recently saw ##U|v\rangle= e^{ia}|v\rangle, \, a \in \mathbb{R}## and am wondering how to come up with this or how to show this.

My first thought is based on the definition of unitary operators (##UU^\dagger = I##), I would show it something like this:

##(U|v\rangle)^\dagger = \langle v| U^\dagger, \quad (e^{ia}|v\rangle)^\dagger = \langle v|e^{-ia}##

Now ##\langle v|v\rangle = \langle v|I|v\rangle = \langle v| U^\dagger U|v\rangle = \langle v|e^{-ia}e^{ia}|v\rangle = \langle v|1|v\rangle = \langle v|v\rangle##

Is it possible to write it this way? I am not sure if I have really shown ##U|v\rangle= e^{ia}|v\rangle## with this.

I recently saw ##U|v\rangle= e^{ia}|v\rangle, \, a \in \mathbb{R}## and am wondering how to come up with this or how to show this.

My first thought is based on the definition of unitary operators (##UU^\dagger = I##), I would show it something like this:

##(U|v\rangle)^\dagger = \langle v| U^\dagger, \quad (e^{ia}|v\rangle)^\dagger = \langle v|e^{-ia}##

Now ##\langle v|v\rangle = \langle v|I|v\rangle = \langle v| U^\dagger U|v\rangle = \langle v|e^{-ia}e^{ia}|v\rangle = \langle v|1|v\rangle = \langle v|v\rangle##

Is it possible to write it this way? I am not sure if I have really shown ##U|v\rangle= e^{ia}|v\rangle## with this.

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