How to eliminate the global phase of a state vector?

Say that I have a qubit that began in the $$|0\rangle$$ state and then the Hadamard gate is applied, resulting in the following state:

$$\begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}$$

Then, apply the Pauli-Y gate to this to get:

$$\begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} \times \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} -i\frac{1}{\sqrt{2}} \\ i\frac{1}{\sqrt{2}} \end{bmatrix}$$

However, this state vector as written does not "cleanly" map on to the Bloch sphere. I know that if we start with a qubit in the $$|0\rangle$$ state and then apply Hadamard and then Pauli-Y, we should end up with:

$$\begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{bmatrix}$$

So somehow:

$$\begin{bmatrix} -i\frac{1}{\sqrt{2}} \\ i\frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{bmatrix}$$

But I am not sure how to explain that equivalence mathematically. I'm guessing this has something to do with the fact that global phase can be ignored. Can someone help me understand this?

• – glS
Sep 13 at 7:54

5 Answers

I think the best way to understand this is via projective geometry. The idea is that in quantum mechanics we always assume that our state vector is normalized and we don't care about global phases, so essentially our states are not defined by vectors, but rather by mathematical objects called rays. Rays are defined by starting out with all nonzero vectors of some fixed vector space, and then identifying all vectors that are scalar multiples of one another. The space of all states is known as projective space.

More concretely, $$n-1$$-dimensional complex projective space is defined as follows: $$\mathbb{CP}^{n-1} = \left( \mathbb{C}^n \setminus \left\{ 0 \right\} \right) / \sim$$ where $$\sim$$ is the equivalence relation on $$\mathbb{C}^n \setminus \left\{ 0 \right\}$$ that identifies two nonzero vectors $$u,v$$ iff there exists some nonzero scalar $$\lambda \in \mathbb{C}^*$$ such that $$u = \lambda v$$. The simplest nontrivial example of a complex projective space is given by $$n=2$$, which is precisely what we need for a qubit. The resulting space, $$\mathbb{CP}^1$$, is topologically a sphere, which can be identified with the Riemann sphere - i.e. the complex plane with an additional point at infinity. To see this, we write the points of $$\mathbb{CP}^1$$ as: $$\left[ a : b \right] , \quad a,b \in \mathbb{C}, \; \left( a, b\ \right) \neq \left( 0, 0 \right) ,$$ where the notation helps us remember that $$\left[ a : b \right]$$ should be thought of representing an equivalence class, in the sense that $$\left[ a : b \right] = \left[ \lambda a : \lambda b \right]$$ for any nonzero $$\lambda$$. For example, now we can see that: $$\left[ 1/\sqrt{2} : -1/\sqrt{2} \right] = \left[ -i/\sqrt{2} : i/\sqrt{2} \right]$$ with $$\lambda = -i$$. This is the mathematical context in which the two vectors you obtained are indeed equal: they are not equal as vectors, but they represent the same ray, so they correspond to the same point of the projective space $$\mathbb{CP}^1$$. Keep in mind that a point of $$\mathbb{CP}^1$$ represents a whole ray (line) of $$\mathbb{C}^2$$; somewhat confusingly, $$\mathbb{CP}^1$$ is often referred to as the complex projective line (since it is one-dimensional as a complex manifold / variety).

Using this language, we note that if $$a \neq 0$$ we can always make it $$1$$ (multiplying both coordinates by $$\lambda = a^{-1}$$); so we get a subspace of $$\mathbb{CP}^1$$ defined by fixing $$a = 1$$. Assuming this "normalization", $$b$$ can obtain any complex value (including zero), and for any two distinct values $$b_1 \neq b_2$$ we have $$\left[ 1 : b_1 \right] \neq \left[ 1 : b_2 \right]$$. So this subspace where $$a=1$$ is just a copy of $$\mathbb{C}$$.

For $$a = 0$$, we only obtain a single point, since in that case $$b$$ is not allowed to be zero, and we have $$\left[ 0 : b_1 \right] = \left[ 0 : b_2 \right]$$ for any two nonzero $$b_1, b_2$$. Thus, we see that $$\mathbb{CP}^1$$ is obtained from the complex plane by adding a single "point at infinity". This added point "closes" the plane on all directions, thus forming a sphere (this process is known as one-point compactification). For example, we can think of the point of infinity as being located at the south pole (an arbitrary choice, of course). There are several ways to see the correspondence with a sphere more concretely. I'll show two of those - one purely mathematical, and the other more "physical".

1. Stereographic projection: start with a vector: $$\begin{bmatrix} a \\ b \end{bmatrix} \in \mathbb{C}^2 ,$$ or equivalently - start with the ray represented by $$\left[ a : b \right]$$. If $$a = 0$$ then this vector maps to the south pole, so we are done. Otherwise, multiply the vector by $$\lambda = a^{-1}$$, to find that $$\left[ a : b \right] = \left[ 1 : a^{-1}b \right]$$, and denote $$z := a^{-1}b$$. If $$z=0$$ we are done again, since we got the north pole. Otherwise, $$z$$ can be written uniquely as $$z = r e^{i \phi}$$ with $$r>0, 0 \leq \phi <2 \pi$$. This $$\phi$$ is our azimuthal angle, and the zenith angle $$\theta$$ is obtained by: $$\theta = 2\cdot \arctan (r) .$$ Note that this strange function maps the inside of the unit circle $$\left\vert z \right\vert^2 < 1$$ to the northern hemisphere, the unit circle itself to the equator, and the rest of the entire complex plane to the southern hemisphere.

Let's see what we get in your example: starting with $$\left[ -\frac{i}{\sqrt{2}} : \frac{i}{\sqrt{2}} \right]$$, we multiply by $$\lambda = \sqrt{2}i$$ to get $$\left[ 1 : -1 \right]$$. So, $$z = -1 = 1 \cdot e^{i\pi}$$. $$\arctan (1) = \pi/4$$, so after multiplying by $$2$$ we can see that we are on the equator ($$\theta = \pi/2$$) with $$\phi=\pi$$. In Cartesian coordinates, that would be: $$x = \sin \theta \cos \phi = -1, \quad y=z=0 .$$ But how can we tell if we chose the "correct" mapping? What does the position on the sphere really mean physically? That leads us to -

1. The Bloch vector as quantum expectation values: starting with any normalized vector $$\vert \psi \rangle \in \mathbb{C}^2$$ (I switched to bra-ket notation since we'll now be doing physics), we can define the following expectation values: $$x := \langle \psi \vert \hat{X} \vert \psi \rangle, \; y := \langle \psi \vert \hat{Y} \vert \psi \rangle, \; z := \langle \psi \vert \hat{Z} \vert \psi \rangle ,$$ where $$\hat{X}, \hat{Y}, \hat{Z}$$ are the Pauli matrices (perhaps up to normalization). The vector $$\mathbb{r} := \left( x, y, z \right)$$ is the Bloch vector, and it turns out that $$\left\vert \mathbb{r} \right\vert^2 = 1$$ (more generally - for a mixed state, $$\left\vert \mathbb{r} \right\vert^2 \leq 1$$). So we think of $$\mathbb{r}$$ as representing position on the Bloch sphere; note that $$\mathbb{r}$$ is independent of the global phase, so we did not need to multiply $$\vert \psi \rangle$$ by a scalar to obtain a "clean" mapping to a sphere (however, this procedure did depend on $$\vert \psi \rangle$$ being normalized).

The neat thing is that the two maps outlined above are in fact the same map. I don't know of any proof that does not require at least some amount of computation, so I will not prove it here. But it is true, and fairly fundamental - in fact, you already have the tools to prove it yourself, which would make good exercise. All this and much more is explained (with helpful pictures!) in the book "Geometry of Quantum States" by Bengtsson and Życzkowski.

• Thank you! One of the key things you said was: "This is the mathematical context in which the two vectors you obtained are indeed equal: they are not equal as vectors, but they represent the same ray, so they correspond to the same point of the projective space ℂℙ1" So basically, these two vectors "point" in the same direction, but have different lengths? So the key to mapping some arbitrary vector onto a point on the surface of the Bloch sphere is to figure out the proper 𝜆 to multiply by? What is the general rule for figuring out the proper 𝜆? Sep 14 at 4:10
• "So basically, these two vectors "point" in the same direction, but have different lengths?" - that is intuitively correct, but not entirely precise. Note that when dealing with projective spaces - even in $\mathbb{R}^3$, with the usual meaning of direction - vectors pointing at opposite directions still correspond to the same ray. So in the real case, a ray identifies a vector up to length and sign; and in the complex case, a ray identifies a vector up to length and global phase. Sep 14 at 12:09
• As to your other questions - I edited my answer to make it clear how to map an arbitrary vector to the Bloch sphere. Sep 14 at 12:12

You're right, in quantum computing it is customary to discard the global phase of the quantum state, since it can not be observed using measurements. (The probabilities of measurement outcomes are defined as squares of absolute values of the amplitudes, so multiplying both amplitudes by the same complex number with absolute value 1, in this case $$-i$$, is not going to impact the measurement outcomes.)

It's not a mathematical equality, since mathematically these vectors are different, but rather an agreement of convenience.

• Or rather, an isomorphism, which is a kind of equivalence. Sep 12 at 19:54
• Right, I was thinking about literal equality rather than the more sophisticated equivalence variants - I've updated the answer to clarify Sep 12 at 21:34

Two-level quantum states are typically mapped onto the Bloch sphere in the following way:

$$\left[ \array{c_0\\ c_1} \right] \rightarrow e^{i\alpha} \left[\array{ \cos\frac{\theta}{2} \\ e^{i\phi} \sin\frac{\theta}{2}} \right]$$

The angles $$\theta$$ and $$\phi$$ are the polar and azimuthal angles identifying the location on the Bloch sphere. Polar angles are typically defined on the interval $$[0,\pi]$$, so we can understand $$\cos\frac{\theta}{2}$$ to always be a positive real number.

This mapping therefore implicitly defines the "global phase" angle $$\alpha$$ as the phase angle of the complex number $$c_0$$. In other words, you should write $$c_0$$ in polar notation as $$c_0 = \cos(\theta/2) \cdot e^{i\alpha}$$.

As you know, this global phase is typically discarded since it has no impact on any physically measurable quantity. That's a physicist's answer - please see @smitke6's thorough answer on projective geometries for a mathematically rigorous formulation of the equivalences. ^_^

In your example $$YH|0⟩$$, our $$c_0 = -i/\sqrt{2}=\frac{1}{\sqrt{2}}⋅e^{-i\pi/2}$$, so our global phase angle is $$\alpha=-\pi/2$$. Factoring that out gives the state you expect from the geometric rotation of the Bloch sphere.

• Thanks. To clarify, aren't polar angles typically on the interval [0,2𝜋] or [-𝜋,𝜋] if you want to use polar coordinate to be able to define any point on a sphere? Also, what exactly is meant by "global phase angle"? Sep 14 at 4:13
• No, the polar angle is the one that defines the latitude. The north pole is typically at 0 and the south pole is at $\pi$. Use the azimuthal angle $\phi\in[0,2π)$ to define the longitude. Some textbooks (especially in mathematics, not so much in physics) use the symbol $\phi$ for the polar angle and $\theta$ for the azimuthal angle, which may account for the confusion. Sep 14 at 14:03
• By "global phase angle," I mean the parameter $\alpha$ in the global phase $\exp(i\alpha)$. I call it that because, if we were to write, say, $c_0$ in polar complex notation as $c_0 = r_0 \exp(i\gamma_0)$, the "global phase angle" $\alpha$ is the same as the angle $\gamma_0$. Some people (including me, normally) may just refer to this as the "phase", but for pedagogical purposes I think it's useful to call $\exp(i\alpha)$ the phase, and $\alpha$ the phase angle. Sep 14 at 14:05

This post is also good for understanding the impact of global phase when it comes to using controlled unitary operations: Do global phases matter when a gate is converted into a controlled gate

Mapping to the Bloch sphere should be global-phase invariant. If you can't cleanly map to the Bloch sphere you essentially skipped a step. For example you can map using $$r_i=\langle\psi|\sigma_i|\psi\rangle$$, where $$r_i$$ is the Bloch sphere position vector and $$\sigma_i$$ are the Pauli matrices.

• What step did I skip? Sep 14 at 4:11