How do $|1⟩$ and $-|1⟩$ represent the same state? I am a newbie to quantum mechanics (I come from more of an engineering/CS background), and today my quantum computing professor stated that $\def\ket#1{\left|#1\right\rangle}\ket{1}$ and $-\ket{1}$ represent the same state. I don't quite understand how this is possible, if someone might be able to help me out!
Also, I am not sure I understand how this implies that $\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ and $\frac{1}{\sqrt{2}} (\ket{0} − \ket{1})$ are also the same. Do you have any tips?
 A: $\vert 1\rangle$ and $-\vert 1\rangle $ are obviously different as vectors of a vector space. However, one has to keep in mind that mathematical objects are related to physical entities via some postulated correspondence. The claim (that your professor is correctly making) is that they represent the same physical states. In other words, there is no observable difference between physical systems that one might describe as $\vert 1\rangle$ or as $-\vert 1\rangle$. Why?! Because the job of a mathematical object standing in correspondence with a physical state is to tell us everything there is to tell us about the object. In quantum mechanics, this means encoding the probabilities pertaining to the outcome of all the possible measurements that one can perform on the physical state. If $\vert b\rangle$ is a possible outcome of a measurement being performed on a given physical state $\vert a\rangle$, then the probability of that outcome coming about is given by the squared norm of the inner--product of the two vectors, i.e., $\vert \langle b  \vert a\rangle\vert ^2$. As you can see, this probability will be completely unchanged if I replace $\vert a\rangle\to e^{i\phi}\vert a\rangle$ where $\phi$ is a real number. Thus, an overall phase of a vector is completely irrelevant to the physically observable aspects of the physical state that is being described. Since $-1$ is of the form $e^{i\phi}$, it is true that $\vert a\rangle$ and $-\vert a\rangle$ describe the same physical state.
However, notice that $\vert c\rangle + \vert a\rangle$ and $\vert c\rangle -\vert a\rangle$ do not represent the same physical state! Because the minus is not an overall minus -- and thus, it does not play the role of an overall phase. For example, you can calculate the probability of the kind that we calculated earlier and see that unlike $\vert a\rangle$ and $e^{i\phi}\vert a\rangle$, $\vert c\rangle + \vert a\rangle$ and $\vert c\rangle -\vert a\rangle$ do not keep the probabilities invariant.
A: To add a very simplistic statement to the other very complete and well thought out answers:
Absolute phase is not physical, relative phase is physical.
That's why the two states in your first example are physically the same, whereas the two states in your latter example are not.
A: $\def\ket#1{\left|#1\right\rangle}\def\bra#1{\left\langle#1\right|}\def\braket#1#2{\left\langle#1|#2\right\rangle}$In quantum mechanics, the overall phase of the wavefunction is not physical. So  $e^{i\phi}\ket{\psi}$ represents one physical state, regardless of the value of $\phi$.
This does not imply that ${1\over\sqrt{2}} (\ket{0} + \ket{1})$ and ${1\over\sqrt{2}} (\ket{0} − \ket{1})$ are the same. Those represent two different physical states. One way to show this is to take the inner product of the two:
$$\begin{align}&{1\over\sqrt{2}}\left(\bra{0}-\bra{1}\right){1\over\sqrt{2}}\left(\ket{0}+\ket{1}\right)\\=&\ {1\over2}\left(\braket{0}{0}-\braket{1}{0}+\braket{0}{1}-\braket{1}{1}\right)\\=&\ {1\over2}(1-0+0-1)\\=&\ 0\end{align}$$
So the two states are not only different, they are orthogonal to one another.
