What does a negative ket state represent?

Question

I'm studying - well trudging through - Quantum Computing for Everyone by Bernhardt. Towards the bottom of the book's physical page 41 he states, "spins are given by either," and then shows an up state ket $|\uparrow\rangle$ and then a negative up state ket $-| \uparrow \rangle$. Nowhere did he mention a negative state ket and what it means. He then goes on to use the negative ket on page 42, but I'm lost.

I've looked on the Internet, but to no avail.

What does a negative ket $-| \uparrow \rangle$, or negative bra (if exists) for that matter, mean/represent within the introductory scope of Bernhardt's book?

Much appreciated.

Answer 1

At the beginning of $40$ page of the same book an arbitrary spin state is written as:

$$c_1| \uparrow \rangle + c_2|\downarrow \rangle$$

where $|c_1|^2 + |c_2|^2 = 1$, $c_1$ and $c_2$ are in general complex numbers, but how I understand from the parts of the book the author uses only real $c_1$ and $c_2$ here for some (maybe more pedagogical) reasons. There are infinite combinations of $c_1$ and $c_2$ that will produce different valid quantum states. Nevertheless, there exist some states that are not distinguishable experimentally because of the global phase ambiguity that is nicely explained in this answer.

At the $41$ and $42$ pages, the author shows an example of states that are not distinguishable. Both $| \uparrow \rangle$ ($c_1 = 1$ and $c_2 = 0$) and $-| \uparrow \rangle$ ($c_1 = -1$ and $c_2 = 0$) are valid quantum states that cannot be distinguished with some measurement procedure. There is no measurement procidure that will say if we have $| \uparrow \rangle$ or $-| \uparrow \rangle$. Moreover this two (equivalently valid) states are also not distinguishable (similar and more general example is shown in this answer (replace $|0\rangle \rightarrow |\uparrow\rangle$ and $|1\rangle \rightarrow |\downarrow\rangle$ in order to have the same notations):

$$|\psi_1 \rangle = c_1| \uparrow \rangle + c_2|\downarrow \rangle \qquad |\psi_2 \rangle = -(c_1| \uparrow \rangle + c_2|\downarrow \rangle)$$

A general statement: there is no experiment that will indicate what is the global phase of the state. Let me rephrase this a little bit for this special case (restricted to real numbers $c_1$ and $c_2$ discussed in the book): there is no experiment that will indicate if there is a minus sign before the state or not.

Also: ket vectors by definition of allowed values of $c_1$ and $c_2$ can have a minus sign before them and hence their coresponding bra vectors also will have a minus sign: $(\pm |\uparrow \rangle)^\dagger = \pm\langle \uparrow |$. In the vector representation:

$|\uparrow \rangle = \begin{pmatrix} 1 \\0 \end{pmatrix} \qquad -|\uparrow \rangle = \begin{pmatrix} -1 \\0 \end{pmatrix} \qquad \langle\uparrow | = \begin{pmatrix} 1 & 0 \end{pmatrix} \qquad -\langle \uparrow | = \begin{pmatrix} -1 & 0 \end{pmatrix}$