**Problem Detail:**

If we take angle names as shown on the Bloch sphere below and a typical qubit representation written as $\cos(\theta/2)|0\rangle + \mathrm{e}^{-\mathrm{i}\phi}\sin(\theta/2)|1\rangle$, why do we divide $\theta$ by $2$ in the cosine and sine parameters?

###### Asked By : 3yakuya

#### Answered By : Niel de Beaudrap

**Short diagrammatic answer.** Because states which are orthogonal as vectors, are represented as antipodal points on the Bloch sphere — orthogonal states of qubits are as far from each other as possible and represent a sort of 'opposite' of each other, and the Bloch sphere represents this idea.

**Longer mathematical answer.** Consider the vector-valued function $\def\ket#1{|#1\rangle}$ $$ \ket{\psi_{\theta,\phi}} = \cos(\theta/2) \ket0 + \mathrm e^{i \phi} \sin(\theta/2) \ket{1}. $$ This function has a period of $4\pi$ in the argument $\theta$, as a result of that division by 2; this might be part of what you are finding strange. However, when we consider *quantum state vectors*, we are not actually interested in the vector itself, as we are in the equivalence class of all vectors *which are the same up to scalar factors* (*i.e.* global phases).

Because $\ket{\psi_{\theta,\phi}} \ne \mathbf 0$, we always have $\ket{\psi_{\theta,\phi}} \ne \ket{\psi_{\theta+2\pi,\phi}} = -\ket{\psi_{\theta,\phi}}$; but we **do** have $\ket{\psi_{\theta,\phi}} \propto \ket{\psi_{\theta+2\pi,\phi}}$ instead. That is, adding $2\pi$ to the angle $\theta$ gives you a vector $\ket{\psi_{\theta+2\pi,\phi}}$ which represents the same state. Because we only care about the state-vector inasmuch as it represents a state, this vector-valued function with peridod $4\pi$ is used to stand for a *state-valued* function with period $2\pi$.

This becomes clearer if we use density operators to represent states, because global phases of state vectors do not affect the density operator. Define the operator $$ \rho_{\theta,\phi} = |\psi_{\theta,\phi}\rangle\!\langle\psi_{\theta,\phi}| = \begin{bmatrix} \cos^2(\theta/2) & \mathrm e^{-i\phi}\!\cos(\theta/2)\sin(\theta/2) \\ \mathrm e^{i\phi} \!\cos(\theta/2)\sin(\theta/2) & \sin^2(\theta/2) \end{bmatrix};$$ using the double-angle formulae $$ \begin{gather*} \cos(\theta) = 2\cos(\theta/2)^2 - 1= 1 - 2\sin(\theta/2)^2\\ \sin(\theta) = 2\cos(\theta/2)\sin(\theta/2) \end{gather*} $$ we may re-write $\rho_{\theta,\phi}$ as $$ \rho_{\theta,\phi} = \frac{1}{2}\begin{bmatrix} \cos(\theta) + 1 \;&\; \mathrm e^{-i\phi} \sin(\theta) \\ \mathrm e^{i\phi} \sin(\theta) \;&\; 1 - \cos(\theta) \end{bmatrix} ,$$ which is not the best representation for thinking about qubits, but certainly makes it very clear that the density operator corresponding to $\ket{\psi_{\theta,\phi}}$ is periodic in $\theta$ with period $2\pi$.

As icing on the cake, note that $\rho_{\theta,\phi}$ is Hermitian by construction. So are the Pauli spin operators $\mathbf 1, X, Y, Z$, and those four are linearly independent. That means that we can decompose $\rho_{\theta,\phi}$ as a linear combination of those four operators. We may easily show, from the last expression for $\rho_{\theta,\phi}$, that $$ \rho_{\theta,\phi} = \tfrac{1}{2}\Bigl(\mathbf 1 + \cos(\phi)\sin(\theta) X + \sin(\theta)\sin(\phi) Y + \cos(\theta) Z \Bigr).$$ If you take the coefficients for $X$, $Y$, and $Z$, and map them to coordinates in $\mathbb R^3$, what you obtain is the Bloch sphere representation: that is, the Bloch sphere not only represents qubit states, but it is specifically representing the decomposition of the density operator into spin matrices.

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Question Source : http://cs.stackexchange.com/questions/34065

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