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According to the Wikipedia article on the Bloch sphere, a pure state of a qubit can always be represented as $$| \psi \rangle = \cos \left( \frac{\theta}{2} \right)| 0 \rangle + e^{i \phi} \sin\left(\frac{\theta}{2}\right)|1 \rangle$$ The parameters $\theta$ and $\phi$ specify a point $$\vec{a} = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$$ on the unit sphere in $\mathbb{R}^3$, as depicted below

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Image source

Question:

If $\hat{z} = |0 \rangle$ and $-\hat{z} = |1 \rangle$ then wouldn't any linear combination of $|0 \rangle$ and $|1 \rangle$ be along the $z$ axis again with complex coefficients? Also, is $| \psi \rangle = \vec{a}$?

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The Bloch sphere is not a vector space. In particular, the "vectors"/arrows you draw on it cannot be added like usual vectors.

The Bloch sphere is what you get when you take the two-dimensional complex vector space $\mathbb{C}^2$ and ask what you get when you make no distinction between vectors that only differ by multiplication with a complex scalar. This is called a projective Hilbert space.

Note that $\hat{z}$ and $-\hat{z}$ differ only by multiplication by the scalar $-1$, so they would be identified if the points on the Bloch sphere were still vectors in the original Hilbert space. They are not and the only points corresponding to pure states are really on the sphere, neither in its interior nor its exterior.

Where a linear combination of $\lvert 0\rangle$ and $\lvert 1 \rangle$ sits on the Bloch sphere is exactly what your formula for $\lvert \psi\rangle$ tells you - the linear combination $\lvert \psi\rangle$ sits at the coordinates $(\phi,\theta)$ on the sphere, where those are the usual polar coordinates on a sphere.

Writing $\lvert 0\rangle = \hat{z}$ and $\lvert 1\rangle = -\hat{z}$ is very confusing - the equality is not an equality as vectors, it is just an equality as points on the sphere. You get to writing this by looking at your general form for $\lvert \psi\rangle$ and defining the map $\mathbb{C}P^2\to S^2, \lvert \psi\rangle\mapsto (\phi,\theta)$ using polar coordinates for the $S^2$, then by embedding the $S^2$ into an $\mathbb{R}^3$ and then observing that the two poles get mapped to the $\hat{z}$ and $-\hat{z}$ in the $\mathbb{R}^3$.

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    $\begingroup$ Your second statement is not quite true. The Bloch sphere (ball, really) is not a vector space but it is a convex space, i.e. one can take convex linear combinations of the form $p_1 \vec a_1+p_2\vec a_2$ where $p_1,p_2\geq 0$ and $p_1+p_2=1$; this corresponds to forming a mixed state with probabilities $p_i$ of the pure states that correspond to the $a_i$, and there is a clear correspondence between the entries of the resulting density matrix in the Pauli-matrix basis and the coordinates inside the Bloch ball. That's a bit over-the-top for the OP's formulation, though. $\endgroup$ – Emilio Pisanty Feb 20 '17 at 18:11
  • $\begingroup$ @EmilioPisanty Added purity. $\endgroup$ – ACuriousMind Feb 20 '17 at 18:15
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I think that the problem in the wikipedia image is stating $\hat z=\left|0\right>$ and $-\hat z=\left|1\right>$. That's mathematically incorrect. The Bloch sphere is used to represent the spin state of a spin $\frac12$ particle. Thus, the associated Hilbert space is two-dimensional. You denote $\left|0\right>$ and $\left|1\right>$ the basis vectors. Now, you can write any state vector in that Hilbert space as \begin{equation} \left|\psi\right>=\alpha_0\left|0\right>+\alpha_1\left|1\right>,\tag{1}\end{equation} such that $|\alpha_0|^2+|\alpha_1|^2=1$. A good way to treat quantities such that their squares add to some constant is parametrising them in spheres (or circles, or hyperspheres, depending on the dimension).

Now, if you had real parameters $\alpha_k$, you'd use a circle, defining probably $\alpha_0=\cos\theta$ and $\alpha_1=\sin\theta$, which would automatically agree with (1). But in a complex Hilbert space you have one more degree of freedom, which is adding a relative phase between $\alpha_0$ and $\alpha_1$ (a global phase is useless). You need then a two-parametrical representation for $\alpha_k$ satisfying (1). The most natural one is taking the two angles of a unit sphere, and that's what Bloch sphere is.

Then, the unit vector $\hat z$ represents the Hilbert space vector $\left|0\right>$, and $-\hat z$ represents the Hilbert space vector $\left|1\right>$, but they are not equal as that wiki image says.

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To complement previous answers: part of the continuing confusion with the use of the Bloch sphere is that vectors orthogonal in the Hilbert space, like $\vert 0\rangle$ and $\vert 1\rangle$, get mapped to vectors on the Bloch sphere which are no longer orthogonal but in opposite direction. Thus, scalar products and linear combinations taken in the Hilbert space must be handled with care when "exported" to the Bloch sphere.

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