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?

Bloch Sphere


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.