Why does superposition principle and Copenhagen interpretation not contradict with themselves?

In quantum mechanics, when we say that a particle in a state $$|x_1\rangle$$, physically the states $$|x_1 \rangle$$ and $$c |x_1\rangle$$ (for some $$c\not = 0\in \mathbb{C}$$) are the same, i.e they correspond to the same physical states.

However, when we talk about a superposition of states,say (ket notation is omitted ) $$\alpha = N\cdot (a_1 x_1 + a_2 x_2), \quad a_1,a_2 \in \mathbb{C}\quad N \in \mathbb{R},$$ those coefficients $$a_i$$s physically means something, i.e they correspond to the probability of measuring the particle (in state $$\alpha$$ priori to measurement ) in the state $$x_i$$.

Now, if $$x_1$$ and $$c\cdot x_1$$ are physically the same states, why should changing $$a_1 x_1$$ to $$x_1$$ in the expression $$\alpha = N\cdot (a_1 x_1 + a_2 x_2)$$ result in physically different state $$\alpha$$ ?

• What do you mean by " if $x_1$ and $x_2$ are physically the same states"? If $x_1$ and $x_2$ are the same states, then why do they have different lables? – Hugo V Jan 18 '19 at 10:33
• @HugoV There is a typo, see me edit please. – onurcanbektas Jan 18 '19 at 10:40

Changing the overall multiplacation factor of a state has no effect, but changing the relative "amount" of each state in it surely affects it. So in your example it means that when you have a pure state like $$|\psi \rangle = |x_1 \rangle$$, it doesn't matter wheter you multiply it by any $$c \in\mathbb{C}$$, because what you care about, the probability of finding it in state $$|x_1 \rangle$$, will always be:

$$P = \frac {|\langle x_1|\psi \rangle|^2}{|\langle \psi|\psi \rangle|^2} = \frac {|c|^2}{|c|^2}=1$$

In your next example, the same is true for your state $$|\alpha \rangle$$, you can multiply it by any $$N \in\mathbb{C}$$, and you will get the same physical significance, that is, the relative probabilities will be the same. Explicitly:

$$P_{|x_1 \rangle} = \frac {|\langle x_1|\alpha \rangle|^2}{|\langle \alpha|\alpha \rangle|^2} = \frac {|N \cdot a_1|^2}{|N|^2}=|a_1|^2$$

$$P_{|x_2 \rangle} = \frac {|\langle x_2|\alpha \rangle|^2}{|\langle \alpha|\alpha \rangle|^2} = \frac {|N \cdot a_2|^2}{|N|^2}=|a_2|^2$$

So, as wanted, independent of $$N$$. But its not true that you can multiply each individual contribution to $$|\alpha \rangle$$, because that will change it to a different state. To make it simple, you can relate this to vectors in $$\mathbb{R}^3$$, and you can think that a state is a vector, but you only care about its direction, not its length. So, in this situation, for any vector $$\vec{v} = a \cdot \vec{x}+b \cdot \vec{y}$$, it is true that multiplying $$\vec{v}$$ by any constant wouldn't change it, but multiplying any of its components independently would for sure change your state.

• Thanks for the answer @HugoV; the analogy is a nice one. – onurcanbektas Jan 18 '19 at 11:43
• @onurcanbektas No problem! – Hugo V Jan 18 '19 at 11:56