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 $\alpha$ ?

  • $\begingroup$ 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? $\endgroup$ – Hugo V Jan 18 '19 at 10:33
  • $\begingroup$ @HugoV There is a typo, see me edit please. $\endgroup$ – 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.

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

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