# What is the meaning of writing a state in its Bloch representation?

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?

• The short answer is "yes." Read Sanchayan's answer for a complete understanding :) – Will Apr 27 '19 at 22:14

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:

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 :).

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.