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What is the meaning of writing a state $|\psi\rangle$ in its Bloch representation. Would I be correct in saying it's just writing out its Bloch vector?

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    $\begingroup$ The short answer is "yes." Read Sanchayan's answer for a complete understanding :) $\endgroup$ – Will Apr 27 at 22:14
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Yes. The Bloch sphere formalism is used for geometrically representing pure and mixed states of two-dimensional quantum systems a.k.a qubits. Any pure state $|\Psi\rangle$ of a qubit can be written in the form:

$$|\Psi\rangle = \cos\frac{\theta}{2}|0\rangle + e^{i\phi}\sin\frac{\theta}{2}|1\rangle$$ where $0\leq \theta\leq \pi$ and $0\leq \phi\leq 2\pi$. This $|\Psi\rangle$ can be represented on the Bloch sphere as:

Bloch Sphere

The Bloch vector $\vec{a}\in \Bbb R^3$ is basically $(\sin\theta \cos\phi, \sin\theta\sin\phi, \cos \theta) = (a_1,a_2,a_3)$.

To represent mixed states you need to consider the corresponding density operator $\rho$. the set of states of a single qubit can be described in terms of $2\times 2$ density matrices and as $\{I,X,Y,Z\}$ forms a basis for the vector space of $2\times 2$ Hermitian matrices, you can write the density operator as $$\rho = a_0I+a_1X+a_2Y+a_3Z = \frac{1}{2}\begin{pmatrix}1+a_3 & a_1-ia_2 \\ a_1+ia_2 & 1-a_3\end{pmatrix}.$$ As density matrices always have trace $1$, and here $\mathrm{tr}(\rho)=2a_0$, so $a_0$ is necessarily $\frac{1}{2}$. So from here you can find out the Bloch coordinates of the any mixed state i.e. $(a_1,a_2,a_3)$ after performing the Pauli decompostion of the density matrix. If you're wondering what ensures that $|\vec{a}|\leq 1$, it's the positive semidefiniteness! The two eigenvalues of $\rho$ are $\frac{1}{2}(1+|\vec{a}|)$ and $\frac{1}{2}(1-|\vec{a}|)$. Thus, to ensure that the second eigenvalue is non-negative, $|\vec{a}|\leq 1$. The three properties of density matrices which you should drill into your brain are: self-adjointness, positive-semidefiniteness and unit trace; prove them as an exercise.

Once you determine the values $a_1,a_2$ and $a_3$ from the density operator, you can easily find the location of the qubit state $(\sin\theta \cos\phi, \sin\theta\sin\phi, \cos \theta)$ inside the Bloch sphere. Let me emphasize on this point: pure states lie on the Bloch sphere (i.e. $|\vec{a}|=1$) whereas mixed states lie inside the Bloch sphere (i.e. $|\vec{a}|<1$). If you're mathematically inclined, you'll also love to think about the Bloch sphere in terms of stereographic projections; it's excellently summarized in this Physics SE answer. I'll attach the image from there, which is originally from this blogpost (the article is in French, sorry :).

enter image description here

Here are a few "further readings" for you:

Essentially, go through the tag; you'll find several interesting questions and answers, which should clarify most of your beginner confusions about the Bloch sphere formalism.

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