<p>If we have $a|0\rangle + b|1\rangle$, there are two quantities we need to know to specify the qubit: </p>

<ul>
<li>the ratio $a/b$, which is encoded by the latitude: $\left|0\right\rangle$ is the north pole and $\left|1\right\rangle$ the south,</li>
<li>the relative phase of $a$ and $b$, which gives the longitude. </li>
</ul>

<p>That is, you can treat either $a$ or $b$ as a real number, but not both. </p>

<p>A circle can only specify one quantity.</p>

CC BY-SA 3.0
2014-11-29T16:17:32.937Z
I am also a bit confused about |0> and |1> being the poles: isn't |0> a vector orthogonal to |1>? I'd also like to ask you about some intuition, as I considered latitude to somehow represent probability of measuring 1 or 0. Could you provide some intuition behind longitude (the relative phase of a and b)? Am I right to relate it to the complex part of the "probability"? Is there some intuition, or am I messing it up? Thank you
3yakuya
CC BY-SA 3.0
2014-11-29T17:27:28.513Z
@Byakuya: any two antipodal points on the Bloch sphere are orthogonal. The latitude is not actually the ratio; it can be deduced from it.
Peter Shor 
CC BY-SA 3.0
2014-11-29T17:28:35.893Z
@Byakuya: you're right, it may be confusing, because you're used to the picture of the unit circle in $R^2$ (I hated that myself when I first read about the Bloch sphere!). Here $a$ and $b$ don't correspond directly to the coordinates. The [wikipedia page](https://en.wikipedia.org/wiki/Bloch_sphere#Definition) does a good job with explaining how to map $a$ and $b$ to the coordinates in the Bloch sphere.
Alessandro Cosentino
CC BY-SA 3.0
2014-11-29T18:36:14.190Z
Thank you both for further explanation (and yes, mind stuck in Cartesian coordinates is a problem sometimes), but could you point me a direction for intuition about that relative phase? From what I understand: in standard notation a|0> + b|1> a and b represent amplitudes, which somehow correspond to probabilities of measuring one value or another. As they are complex values they can be rewritten as cos(theta/2)|0>+e^(i*fi)sin(theta/2)|1> where theta is the angle from our vector to the axis z (with basis states 0 and 1) and fi is the phase. Am I correct? How should I interpret the phase?
3yakuya
CC BY-SA 3.0
2014-11-29T19:04:26.087Z
in the expression $\cos(\theta/2) \left| 0 \right\rangle + \sin(\theta/2) e^{i \phi} \left| 1 \right\rangle$, $\theta$ is the latitude and $\phi$ is the longitude.
Peter Shor 
CC BY-SA 3.0
2014-11-29T15:38:21.897Z
2014-11-29T15:38:21.897Z
Peter Shor 
<p>Bloch sphere is a common visualisation of possible qubit values, but I fail to understand why do we need a sphere to represent them. Assuming we want to represent the single qubit a|0> + b|1> theoretically we know both a and b coefficients may be complex. However, we also know that the global phase factor is insignificant (we treat two unit vectors related by a global phase as equivalent) so we may treat a or b to be a real number. Wouldn't a circle be enogh to represent a qubit then (like taking axis y away from the original Bloch sphere)?</p>

<p>I feel like I am missing something simple (or some basic information I should know), any suggested read or explanation would be appreciated. Thank you in advance.</p>

CC BY-SA 3.0
2014-11-29T17:45:43.707Z
The 'long' answer is what Peter Shor has already responded. The 'short' answer is: remember that the amplitudes are in $\mathbb C$! A circle is enough to represent unit vectors in $\mathbb R^2$, but to represent $\frac1{\sqrt 2}|0\rangle + \frac i{\sqrt2}|1\rangle$, you need at least one other axis.
Niel de Beaudrap
CC BY-SA 3.0
2014-11-29T18:28:22.327Z
I was not sure about my intuition here, but I considered the axis y from the typical drawing (like https://upload.wikimedia.org/wikipedia/commons/f/f4/Bloch_Sphere.svg) to be related to imaginary unit i. It possibly depends on how you visualise it, but is my intuition right here?
3yakuya
CC BY-SA 3.0
2014-11-30T17:58:21.807Z
That's about right: on the Bloch sphere, the y axis essentially governs how large the imaginary component is in the relative phase, while the x and z axes represent the qubit well enough if you consider only states with real components. But in my opinion, what really motivates the Bloch sphere is density operators (which also do not contain global phases), and in particular the decomposition of density operators as a real-valued linear combination of the Pauli operators $\{I, X, Y, Z\}$ (which is useful because of the relationship of that decomposition to measurement of Hermitian observables).
Niel de Beaudrap
<quantum-information>
Why do we need a sphere to graphically represent a qubit?
3yakuya
# Why do we need a sphere to graphically represent a qubit?

(Asked by: username_0 on 2014-11-29)

Bloch sphere is a common visualisation of possible qubit values, but I fail to understand why do we need a sphere to represent them. Assuming we want to represent the single qubit a|0> + b|1> theoretically we know both a and b coefficients may be complex. However, we also know that the global phase factor is insignificant (we treat two unit vectors related by a global phase as equivalent) so we may treat a or b to be a real number. Wouldn't a circle be enogh to represent a qubit then (like taking axis y away from the original Bloch sphere)?

I feel like I am missing something simple (or some basic information I should know), any suggested read or explanation would be appreciated. Thank you in advance.

**Tags:** quantum-information
> **Comments:**
> **username_1**: The 'long' answer is what Peter Shor has already responded. The 'short' answer is: remember that the amplitudes are in $\mathbb C$! A circle is enough to represent unit vectors in $\mathbb R^2$, but to represent $\frac1{\sqrt 2}|0\rangle + \frac i{\sqrt2}|1\rangle$, you need at least one other axis.  
> **username_0**: I was not sure about my intuition here, but I considered the axis y from the typical drawing (like https://upload.wikimedia.org/wikipedia/commons/f/f4/Bloch_Sphere.svg) to be related to imaginary unit i. It possibly depends on how you visualise it, but is my intuition right here?  
> **username_1**: That's about right: on the Bloch sphere, the y axis essentially governs how large the imaginary component is in the relative phase, while the x and z axes represent the qubit well enough if you consider only states with real components. But in my opinion, what really motivates the Bloch sphere is density operators (which also do not contain global phases), and in particular the decomposition of density operators as a real-valued linear combination of the Pauli operators $\{I, X, Y, Z\}$ (which is useful because of the relationship of that decomposition to measurement of Hermitian observables).  

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## 1 answers

### Answer by username_2 on 2014-11-29. Score: 3 | Accepted answer

If we have $a|0\rangle + b|1\rangle$, there are two quantities we need to know to specify the qubit:

* the ratio $a/b$, which is encoded by the latitude: $\left|0\right\rangle$ is the north pole and $\left|1\right\rangle$ the south,
* the relative phase of $a$ and $b$, which gives the longitude.

That is, you can treat either $a$ or $b$ as a real number, but not both.

A circle can only specify one quantity.

> **Comments:**
> **username_0**: I am also a bit confused about |0> and |1> being the poles: isn't |0> a vector orthogonal to |1>? I'd also like to ask you about some intuition, as I considered latitude to somehow represent probability of measuring 1 or 0. Could you provide some intuition behind longitude (the relative phase of a and b)? Am I right to relate it to the complex part of the "probability"? Is there some intuition, or am I messing it up? Thank you  
> **username_2**: @Byakuya: any two antipodal points on the Bloch sphere are orthogonal. The latitude is not actually the ratio; it can be deduced from it.  
> **username_3**: @Byakuya: you're right, it may be confusing, because you're used to the picture of the unit circle in $R^2$ (I hated that myself when I first read about the Bloch sphere!). Here $a$ and $b$ don't correspond directly to the coordinates. The [wikipedia page](https://en.wikipedia.org/wiki/Bloch_sphere#Definition) does a good job with explaining how to map $a$ and $b$ to the coordinates in the Bloch sphere.  
> **username_0**: Thank you both for further explanation (and yes, mind stuck in Cartesian coordinates is a problem sometimes), but could you point me a direction for intuition about that relative phase? From what I understand: in standard notation a|0> + b|1> a and b represent amplitudes, which somehow correspond to probabilities of measuring one value or another. As they are complex values they can be rewritten as cos(theta/2)|0>+e^(i*fi)sin(theta/2)|1> where theta is the angle from our vector to the axis z (with basis states 0 and 1) and fi is the phase. Am I correct? How should I interpret the phase?  
> **username_2**: in the expression $\cos(\theta/2) \left| 0 \right\rangle + \sin(\theta/2) e^{i \phi} \left| 1 \right\rangle$, $\theta$ is the latitude and $\phi$ is the longitude.  

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