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Are two quantum states defined by two linearly dependent vectors the same state or is the norm of the state vector essential part of the definition of an specific quantum state?

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  • $\begingroup$ See e.g en.wikipedia.org/wiki/… $\endgroup$ – Nephente Sep 23 '16 at 9:28
  • $\begingroup$ Two quantum states can only be linearly dependent if they are proportional to each other. Since quantum states are normalized the proportionality constant must therefore be a pure phase factor. Such a phase factor has no physical significance. Hence these to states would therefore represent the same state. $\endgroup$ – flippiefanus Sep 23 '16 at 10:37
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In Quantum Mechanics, a physical state is always normalized by definition. So, in fact, you have to normalize your linearly dependent vectors, obtaining the same normalized vector, and, in fact, both represent the same physical state. Norm is not important.

Note however that scalars in QM are complex numbers, so your two vectors could be $|{\psi_1} \rangle$ and $|{\psi_2} \rangle = i|{\psi_1}\rangle$. Note that if the first vector is normalized, the second is also normalized: so now you have two linearly dependent vectors, correctly normalized. But they still represent the same state! This is because if you take any linear operator $A$, which is an observable (like position or momentum), then the expected value is the same. Let's check it:

$$\langle\psi _2|A|\psi _2 \rangle=|\psi_2\rangle ^\dagger A |\psi_2\rangle= (i|\psi _1\rangle)^\dagger A (i |\psi_1\rangle)=(-i)i\langle\psi_1|A|\psi_1\rangle = \langle\psi_1 |A| \psi_1\rangle$$

In general, a complex phase $e^{i \alpha }$ doesn't affect the expected values we measure in QM, so all these states represent the same physical state, even if they have the same norm and are linearly dependent.

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