9
$\begingroup$

Suppose we have a single qubit with state $| \psi \rangle = \alpha | 0 \rangle + \beta | 1 \rangle$. We know that $|\alpha|^2 + |\beta|^2 = 1$, so we can write $| \alpha | = \cos(\theta)$, $| \beta | = \sin(\theta)$ for some real number $\theta$. Then, since only the relative phase between $\alpha$ and $\beta$ is physical, we can take $\alpha$ to be real. So we can now write

$$| \psi \rangle = \cos(\theta) | 0 \rangle + e^{i \phi} \sin(\theta)| 1 \rangle$$

My Question: Why are points on the Bloch sphere usually associated to vectors written as $$| \psi \rangle = \cos(\theta/2) | 0 \rangle + e^{i \phi} \sin(\theta/2)| 1 \rangle$$ instead of as I have written? Why use $\theta /2$ instead of just $\theta$?

$\endgroup$
7
$\begingroup$

It is a convention, chosen so that $\theta$ is the azimuthal angle of the point representing the state in the Bloch sphere.

To see where this convention comes from, start from a state $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$. Remembering the normalisation constraint $|\alpha|^2+|\beta|^2=1$, and assuming without loss of generality $\alpha\in\mathbb R$, a natural way to parametrise the state is by defining an angle $\gamma$ such that $|\alpha|=\alpha=\cos\gamma$ and $|\beta|=\sin\gamma$: $$|\psi\rangle=\cos\gamma|0\rangle + e^{i\varphi}\sin\gamma|1\rangle.$$ However, let us consider the coordinates of a generic (pure) state $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$ in the Bloch sphere representation: $$\newcommand{\on}[1]{\operatorname{#1}}\newcommand{\bs}[1]{\boldsymbol{#1}} x\equiv\langle\psi|\sigma_x|\psi\rangle,\\ y\equiv\langle\psi|\sigma_y|\psi\rangle,\\ z\equiv\langle\psi|\sigma_z|\psi\rangle. $$ In particular, the last equation above is easily seen to give $z=|\alpha|^2-|\beta|^2$. Using our "natural" parametrisation with $\gamma$ we therefore get that $z=\cos^2\gamma-\sin^2\gamma=\cos(2\gamma)$. But the standard way to parametrise a sphere is by having $z=\cos\theta$.

If we want to be consistent with the regular conventions for spherical coordinates we therefore have to choose $\gamma=\theta/2$.

$\endgroup$
  • $\begingroup$ You mean $z = cos(\theta)$ I think at the end. But very nice! I have never seen the Bloch representation coordinates defined in terms of Pauli matrices. This is exactly what I was looking for. Do you have a reference for where I can read about the Bloch representation the way you defined it? $\endgroup$ – wanderingmathematician Aug 30 '18 at 18:00
  • 1
    $\begingroup$ @user334137 indeed, thanks for the catch. That is how the Bloch representation of a state is defined! More in general, for an arbitrary (non necessarily pure) state $\rho$, the coordinates in the Bloch sphere are defined as $x=\operatorname{Tr}(\rho \sigma_x)$ etc. $\endgroup$ – glS Aug 30 '18 at 18:04
  • $\begingroup$ @gIS Do you have a reference? I would enjoy reading from this more abstract point of view. $\endgroup$ – wanderingmathematician Aug 30 '18 at 18:06
  • $\begingroup$ @user334137 you can find a number of discussions on this topic on physics.SE (see e.g. this and this). I actually can't seem to find now an elementary reference that starts from defining the Bloch sphere in this fashion, but the idea is quite simple. Most texts show you that you can parametrise a state like you show, and then if you compute the expectation values of the Pauli matrices you realise that they correspond exactly to the coordinates in the Bloch sphere. $\endgroup$ – glS Aug 30 '18 at 18:21
  • $\begingroup$ I personally prefer to think about it in the opposite direction: a state is characterised by the expectation values of a complete set of observables (like the Pauli matrices for a single qubit), and it just so happens that the set of these expectation values lies on a sphere (or is at least homeomorphic to one). It follows that we can parametrise states as points in a (generally hyper-)sphere. $\endgroup$ – glS Aug 30 '18 at 18:23
5
$\begingroup$

If we use the convention $$| \psi \rangle = \cos(\theta) | 0 \rangle + e^{i \phi} \sin(\theta)| 1 \rangle$$ then the North ($\theta=0$) and the South ($\theta=\pi)$ are (physically) the same state $|0\rangle$;

If we use the convention$$| \psi \rangle = \cos(\theta/2) | 0 \rangle + e^{i \phi} \sin(\theta/2)| 1 \rangle$$

then North is $|0\rangle$ and South is $|1\rangle$ which is better.

$\endgroup$
5
$\begingroup$

Let $\hat{n}=(\cos\phi\sin\theta,\sin\phi \sin\theta,\cos\theta)$ i.e. the Cartesian coordinate vector for a point on the unit sphere with polar angle $\theta$ and azimuthal angle $\phi$. By sending a spin-1/2 particle through a Stern-Gerlach device with orientation $\hat{n}$, we can measure the observable

\begin{align} S_n:=\vec{S}\cdot \hat{n} &=S_x \cos\phi\sin\theta +S_y \sin\phi \sin\theta+S_z \cos\theta\\ &= \frac{\hbar}{2}\begin{pmatrix} \cos\theta & \cos\phi\sin\theta-i \sin\phi \sin\theta\\ \cos\phi\sin\theta+i \sin\phi \sin\theta & -\cos\theta\end{pmatrix} \\&= \begin{pmatrix} \cos \theta & e^{-i\phi}\sin\theta \\ e^{i\phi}\sin\theta & -\cos\theta\end{pmatrix} \end{align} in the $S_z$ basis. The obvious step is now to determine eigenvalues and eigenvectors. But if we denote the spin-up and spin-down eigenstates of $S_z$ as $|0\rangle$ and $|1\rangle$ respectively, then

$$| \psi \rangle = \cos(\theta/2) |0 \rangle + e^{i \phi} \sin(\theta/2)| 1 \rangle=\begin{pmatrix} \cos(\theta/2)\\ e^{i\phi}\sin(\theta/2)\end{pmatrix}$$ and therefore \begin{align} S_n |\psi\rangle &= \frac{\hbar}{2}\begin{pmatrix} \cos \theta & e^{-i\phi}\sin\theta \\ e^{i\phi}\sin\theta & -\cos\theta\end{pmatrix}\begin{pmatrix} \cos(\theta/2)\\ e^{i\phi}\sin(\theta/2)\end{pmatrix} \\ &= \frac{\hbar}{2}\begin{pmatrix} \cos(\theta)\cos(\theta/2)+\sin(\theta)\sin(\theta/2)\\ e^{i\phi}[\sin(\theta)\cos(\theta/2)-\cos(\theta)\sin(\theta/2)]\end{pmatrix}\\ &= \frac{\hbar}{2}\begin{pmatrix} \cos(\theta/2)\\ e^{i\phi}\sin(\theta/2)\end{pmatrix}=+\frac{\hbar}{2}|\psi\rangle \end{align} where in the second-to-last equality I've used the trigonometric product-to-sum formula. Hence $|\psi\rangle$ is the $S_n=+\hbar/2$ eigenstate. In other words: If a spin-1/2 particle passes through an SG device with orientation $\hat{n}$ and comes out deflected up, then $|\psi\rangle$ is the resulting spin state. (Correspondingly, one can show that $S_{-n}|\psi\rangle=-\hbar/2|\psi\rangle$ i.e. $|\psi\rangle$ will deflect down if the SG device is flipped.) The upshot is that $\theta,\phi$ are not angles in Hilbert space; rather, they're the angles in real space for the SG device for which $|\psi\rangle$ is the spin state of the upward-deflected beam.

Note that the above description is limited to points on the surface of the Bloch sphere i.e pure states. For points on the interior of the Bloch sphere, we need to go to the density matrix formalism as presented by gLs and I'll defer to that answer.

$\endgroup$
1
$\begingroup$

It is just a convention for $ 0 \le \theta \le \pi $. You can write it your way (indeed you can "include" a constant in a variable) but in that case $ 0 \le \theta \le \pi / 2 $.

But we take this convention for unique coordinates. If you refer to the Spherical coordinate system

you can see that if you want a unique set of spherical coordinates for each point of the sphere, you need to restrict their range.

$\endgroup$
  • 1
    $\begingroup$ I agree. Obviously you could simply do a change of variables to see that the two forms are equivalent. But is there a reason for this choice of convention? I would guess so since it is not consistent with standard math or physics convention. Why divide by 2 unless it makes other calculations / viewpoints simpler? $\endgroup$ – wanderingmathematician Aug 30 '18 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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