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I'm unclear why the Bloch sphere representation of a maximally entangled qubit shows the state of the bit as being at the origin of the sphere.

For example, this illustration

enter image description here

shows the effect of the simple circuit

enter image description here

over time, with $q_0$ on the left and $q_1$ on the right. Both qubits end up at the origin of their respective spheres following application of $CNOT$ ($q_1$ "waits" at its initial value until after $H$ moves $q_1$ to $x$).

Why is a maximally entangled qubit shown at the origin of a Bloch sphere?

An explanation of sorts is provided here, but I'm too much of a beginner to follow it.

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  • $\begingroup$ This is a good question with good answers. Partial trace and density matrix formalism is necessary to understand the answers. Without these tools, we can only provide the most shallow description of what's going on. $\endgroup$ – psitae Jan 9 at 21:13
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Let $(x,y,z)$ be a point in the unit sphere with $x^2+y^2+z^2 \leq 1$.

The state associated with this point is

\begin{eqnarray*} \rho &=& \frac{1}{2} (I_2 + x \sigma_x + y \sigma_y + z \sigma_z)\\ &=& \frac{1}{2} \begin{pmatrix} 1+z&x-iy\\ x+iy&1-z\\ \end{pmatrix} \end{eqnarray*}

This is just a convenient way to parameterize all $2\times 2$ density matrices. This doesn't work as nicely for qudits with $d \neq 2$. But since we are talking $d=2$, we might as well use this nice parameterization.

In particular let $(x,y,z)=(0,0,0)$, the associated $\rho$ is

\begin{eqnarray*} \rho &=& \frac{1}{2} \begin{pmatrix} 1+0&0-i0\\ 0+i0&1-0\\ \end{pmatrix}\\ &=& \begin{pmatrix} \frac{1}{2}&0\\ 0&\frac{1}{2} \end{pmatrix} \end{eqnarray*}

This is the maximally mixed state.

What is being shown is the state for only 1 qubit. This is the result after taking a partial trace over the other qubit.

So if looking at the first one $q_0$. It starts off in the state

\begin{eqnarray*} \rho &=& | 0 \rangle \langle 0 |\\ \end{eqnarray*}

which corresponds to $(x,y,z)=(0,0,1)$

Then it goes to

\begin{eqnarray*} \rho &=& H | 0 \rangle \langle 0 | H\\ \end{eqnarray*}

But after the CNOT it is

\begin{eqnarray*} \rho &=& \text{Tr}_2 (\operatorname{CNOT}_{12} H | 0 0 \rangle \langle 00 | H \operatorname{CNOT}_{12}) \\ \end{eqnarray*}

which ends up being the maximally mixed state corresponding to $(x,y,z)=(0,0,0)$

Edit: As stated above "This is just a convenient way to parameterize all $2\times 2$ density matrices. This doesn't work as nicely for qudits with $d \neq 2$. But since we are talking $d=2$, we might as well use this nice parameterization." So even if density matrices still make you queasy, don't think of the center of a sphere as being something particularly meaningful. It is just a convenient way to draw all states and in this case the center happens to line up with the maximally mixed state. So no it is not something fundamental. It doesn't generalize to other $d$ or more qubits. Don't take this particular parameterization too seriously, it just allows us to plot the state in a way to quickly convey the information visually.

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  • $\begingroup$ See my comment to DaftWullie's answer. $\endgroup$ – orome Jan 9 at 19:10
  • $\begingroup$ Edited to say how it is not fundamental. $\endgroup$ – AHusain Jan 9 at 21:13
  • $\begingroup$ You say this doesn't work as nicely for d ≠ 2 but the visualization seems to still be commonly used for larger dimensions. $\endgroup$ – orome Jan 9 at 21:48
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    $\begingroup$ What they're doing is like this circuit, they're showing each qubit after tracing out the others. Just like with this circuit showing 2 spheres. What I was saying is about trying to visualize d by d density matrices for the entire system. They get too big and complicated to draw. $\endgroup$ – AHusain Jan 9 at 22:15
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The Bloch sphere only represents the state of a single qubit. What you’re talking about is taking a multi-qubit state, and representing the state of just one of those qubits on the Bloch sphere.

If the multi-qubit state is a product state (pure and separable), then the state of the single qubit is a pure state, and is represented as a point on the surface of the Bloch sphere. If the overall state is entangled, then the individual qubit is not pure, and is represented by a point that is on the interior of the Bloch sphere. The shorter the distance to the centre, the more mixed the individual qubit is, and hence the more entangled the global state is. The maximally entangled state yields the shortest possible distance, i.e. the point right at the centre of the sphere. AHussain’s answer gives you the mathematics of how to formally calculate that.

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  • $\begingroup$ These answers are helpful, but not quite at the level I'm looking for, which is both very basic (density operators still make me a bit queasy) and higher level, namely: why represent entanglement as a distance from the center of the sphere? Is there some natural or compelling reason for doing so; does it follow from something else that is well established or fundamental? $\endgroup$ – orome Jan 9 at 19:09
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    $\begingroup$ Let me reiterate, the Bloch Sphere is not representing entanglement. It is representing the state of one qubit. If that one qubit is part of a two-qubit pure state, then the extent to which one qubit is not in a product state is the extent to which it is entangled. But, fundamentally, this is a property of density operators for single qubits. You cannot hide from that. $\endgroup$ – DaftWullie Jan 10 at 8:04
  • $\begingroup$ Here's the crucial bit, I think: "then the extent to which one qubit is not in a product state is the extent to which it is entangled". That provides the rational I was looking for. $\endgroup$ – orome Jan 10 at 22:47

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