2
$\begingroup$

The Bloch Sphere is regarded as the most "intuitive" way of explaining a 2-level quantum system in computation and rotations of states described on Bloch sphere provides a really easy picture. Despite that, I've been having some problems with understanding this representation of quantum states. I read quite a few articles and answers on StackExchange too, but most of them try to explain things using Pauli Matrices and Density Matrices. But the real problem I've been struggling to solve and understand would be better stated as follows-

Prove that a 2-level Quantum State (qubits) is mathematically equivalent to a point on the Unit Sphere in $\mathbb{R}^3$.

Most of the explanations I've seen are based on hand-wavy arguments and lack proper formalism. I've been trying to prove this without involving Pauli matrices or Density matrices. By using the arguments of normalization and global-phase invariance, I've been able to deduce that we can write an arbitrary state as follows- $$|\psi \rangle=r|0\rangle + (a+ib)|1\rangle $$ where $r\in \mathbb{R}^{\geq0}$ and $ a,b\in \mathbb{R}$ with the constraint $r^2+a^2+b^2=1.$ If $a,b,r$ are the co-ordinates in the Cartesian system, then since we have $r\geq0$, the equation results in a hemisphere.

If anyone has an explanation or a proof of how qubits have an equivalent description on the Bloch sphere, or some suggestion, I would be grateful.

$\endgroup$
7
3
$\begingroup$

The idea is to find a nice "real" representation for pure qubit states, that is, vectors $(\alpha,\beta)\in\mathbb C^2$ defined up to global phase and norm. Remember you can always write such a unit complex vector as $$\cos(\theta)\lvert0\rangle+\sin(\theta) e^{i\phi}\lvert 1\rangle$$ for some $\theta,\phi\in\mathbb R$. This comes from observing that if $x,y\in\mathbb R$ satisfy $x^2+y^2=1$, then there is some angle $\theta$ such that $x=\cos\theta$ and $y=\sin\theta$, and that $\lvert\alpha\rvert^2,\lvert\beta\rvert^2$ are an example of such real numbers. You then observe that rescaling by a global phase you can always reduce to a case in which $\alpha\in\mathbb R$.

Keeping in mind the physical interpretation of quantum states, an obvious starting point for such parametrisation it to use the probability $p_0\equiv \lvert\alpha\rvert^2=\cos(\theta)^2$. We don't need $p_1=\lvert\beta\rvert^2$ as the norm constraint tells us that $p_1=1-p_0$.

Fixed $p_0$, what freedom is left? Well there is the phase difference to take into account. This is encoded into the phase $\phi$ in the formula above. A natural way to represent geometrically such a phase is via a corresponding unit vector in $\mathbb R^2$, that is, through the mapping $\phi\to(\cos\phi,\sin\phi)$. However, this phase is attached to the chosen value of $p_0$. In particular, for $p_0=0$ and $p_0=1$ all phases $\phi$ should represent the same state, which makes our looking for a nice representation a bit trickier.

So we just found that we can represent a state using (1) the parameter $p_0$, and (2) the phase $\phi$ represented as $(\cos\phi,\sin\phi)$, with the added requirement that (3) the phase $\phi$ shouldn't matter for $p_0=0$ and $p_0=1$. An easy solution is then to use a three-dimensional representation in which $p_0\in[0,1]$ is on one axis (say the $z$-axis), and at each value of $p_0$ we have a circle around it. The radius of such circles depends on $p_0$, and is vanishing for $p_0=0,1$.

The most natural choice is then obviously to use a sphere of radius $1/2$ centred around $p_0=1/2$. By simply translating the $p_0$ axis and rescaling, you get your unit sphere centred around the origin, i.e. the Bloch sphere.

It's worth stressing that you cannot really derive the Bloch sphere representation from the sole definition of quantum state. You can prove that the Bloch sphere representation works to faithfully represent states, and show that it is convenient for all sorts of reasons, but if the only requirement is to represent states as points in $\mathbb R^3$, it is not uniquely defined: other equivalent (though probably highly inconvenient) representations are possible.

$\endgroup$
4
  • $\begingroup$ Yes. I'm exactly along these lines of reasoning. But the point where it gets tricky is that the usual representation of a general state is $\cos(\theta /2)\lvert0\rangle+\sin(\theta /2) e^{i\phi}\lvert 1\rangle$. If you instead just put $\theta$ (as you did) you'll realize that $ 0 < \theta < \pi/2 $ and that gives a hemisphere. I do understand your argument about vanishing circles but is there a way to formally prove it using algebra? Thanks. $\endgroup$ – Tachyon209 Jul 26 '20 at 21:25
  • 1
    $\begingroup$ @Tachyon209 I didn't use the half-angle in my definition, which is why I end up with a sphere not centred around the origin, but that doesn't matter. We translate it afterward only because representing states on a unit sphere is more convenient from a mathematical point of view. There is no completely formal way to "derive" the Bloch sphere representation, because it is not unique. As per my argument, you need to have certain properties, but you could use different representations than a sphere if you really wanted to $\endgroup$ – glS Jul 27 '20 at 1:05
  • 1
    $\begingroup$ in other words, you don't "prove" the Bloch sphere from the definition of ket states. You prove that the Bloch sphere representation works to represent states. You see this hidden in other answers as well. For example, it is true that $\mathbb C^2/\mathbb C^*\simeq S^2$. This is another way to say that the set of vectors in $\mathbb C^2$ defined up to a complex scalar is homeomorphic to $S^2$. But that's it: it is equivalent to $S^2$ in regards to the properties we are interesting in, it doesn't mean that it is equal to it $\endgroup$ – glS Jul 27 '20 at 1:06
  • $\begingroup$ Yess. Now I get it. Can you add this part to the answer so that others too can be benefited by this and don't have to go through the comments. :) $\endgroup$ – Tachyon209 Jul 27 '20 at 9:32
5
$\begingroup$

The general state is a superposition of $| 0 \rangle$ and $| 1 \rangle$ state: $$ \psi = \alpha | 0 \rangle + \beta | 1 \rangle \qquad \alpha, \beta \in \mathbb{C} $$ However, states, which are related by multiplication on arbitrary complex number $\lambda \neq 0$, are identified : $\psi \sim \lambda \psi$. So the space of quantum states is: $$ \mathbb{C}^2 / \mathbb{C}^{*} = \mathbb{CP}^{1} \simeq S^2 $$

$\endgroup$
7
  • $\begingroup$ Sorry, but I didn't quite follow it. Can you elaborate a bit on the last line since I don't understand much of the operators that you are using. $\endgroup$ – Tachyon209 Jul 25 '20 at 7:19
  • $\begingroup$ @Tachyon209, ok , this is simply the illustration of the above sentence in mathematical symbols : the $\mathbb{C}^2$ refers to $\alpha, \beta$ - and the $/$ - means that we factorize (regard as the same elements, obtained by multiplication on $\lambda$), and the space obtained by such identificiation is called complex projective space $\mathbb{CP}^{1}$ - en.wikipedia.org/wiki/Riemann_sphere. $\endgroup$ – spiridon_the_sun_rotator Jul 25 '20 at 8:42
  • 1
    $\begingroup$ It’s unlikely that $\mathbb{C}^2/\mathbb{C}^*~\sim \mathbb{CP}^1$ will be very useful if the OP want something simpler than density matrices. $\endgroup$ – ZeroTheHero Jul 27 '20 at 1:42
  • 1
    $\begingroup$ @Tachyon209 The way bloch sphere is described here, is a correct way, it is a projection (shadow) of a 4-d sphere onto a 3D sphere in a vector space due to the constraint of $|\alpha|^2 + |\beta|^2 =1$. You can read the book geometry of quantum mechanics or read this article. arxiv.org/abs/quant-ph/9906086 $\endgroup$ – Chetan Waghela Jul 27 '20 at 5:49
  • 1
    $\begingroup$ 4d sphere because, there are 4 variables, real parts of alpha and beta, and imaginary parts of alpha and beta. $\endgroup$ – Chetan Waghela Jul 27 '20 at 5:56
1
$\begingroup$

Parametrizing a general state as \begin{align} \cos(\vartheta/2)\vert +\rangle + e^{i\varphi}\sin(\vartheta/2)\vert -\rangle \tag{1} \end{align} (any state up to a global phase can be written that way) and computing \begin{align} \langle \sigma_x\rangle =\sin(\varphi)\sin(\vartheta)\, ,\qquad \langle \sigma_y \rangle =\cos(\varphi)\sin(\vartheta)\, ,\qquad \langle \sigma_z\rangle =\cos(\vartheta) \end{align} gives a vector \begin{align} \hat n=\left(\langle \sigma_x\rangle ,\langle \sigma_y\rangle,\langle\sigma_z\rangle \right) \end{align} which points to a point on the Bloch sphere.

$\endgroup$

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