Quaternions in spherical coordinates A $\mathbb{R^2}$ unit vector can be expressed as a complex number, using a $\textit{spiral phase quadrature}$ ( Larkin 2001 ):
$$(x,y)\in\mathbb{R^2} \longrightarrow z \in \mathbb{C} =x+iy = e^{i\theta}$$
where $\theta$ has the same meaning than the $\theta'$ from polar coordinates.
Would it possible to express a $\mathbb{R^3}$ unit vector as a quaternion?
$$(x,y,z)\in\mathbb{R^3} \longrightarrow q \in \mathbb{H} =ix + jy +kz = e^{j\theta}e^{k\phi}$$
where $\theta$ and $\phi$ are equivalent to $\theta'$ and $\phi'$ from spherical coordinates.
I know that an unit quaternion can be expressed as:
$$q = e^{i\alpha/2}e^{k\beta/2}e^{j\gamma/2} $$
where $ \alpha$, $\beta$ and $\gamma$ are the Euler angles.
but I would prefer to work in spherical coordinates. Is there an analogy to the spiral phase in 3D?
 A: Let's work out your proposed mapping:
\begin{align}
e^{j\theta}e^{k\phi}
 &= (\cos\theta + j\sin\theta)(\cos\phi + k\sin\phi) \\
 &= \cos\theta\cos\phi + (j\sin\theta)(k\sin\phi)
     + (j\sin\theta)\cos\phi + (\cos\theta)(k\sin\phi) \\
 &= \cos\theta\cos\phi + i\sin\theta\sin\phi
     + j\sin\theta\cos\phi + k\cos\theta\sin\phi \\
\end{align}
Now, since we know $q = e^{i\alpha/2}e^{k\beta/2}e^{j\gamma/2}$
is a unit quaternion, 
it follows that $e^{j\theta}e^{k\phi}$ also is a unit quaternion
(just set $\alpha = 0, \beta = 2\phi, \gamma = 2\theta)$;
but since it has a non-zero real part (except when $\cos\theta=0$
or $\cos\phi=0$),
the imaginary part $ix + jy + kz$ generally will not correspond to a
unit vector in $\mathbb R^3.$
In fact, an even more obvious problem is that if $\theta = 0,$ as $\phi$ varies freely we get a set of vectors whose imaginary parts are all in one direction, while if $\phi = 0,$ as $\theta$ varies we get a set of vectors whose imaginary parts are all in a different direction.
In spherical coordinates we can have only one coordinate such that setting that coordinate to zero forces the vector to be in a particular direction.

But let's consider the interpretation of a quaternion as a rotation.
It is well-known
that a rotation by $\theta$ radians around the axis through the origin
in the direction of the unit vector $\hat{\mathbf u}$ can be represented
by a unit quaternion $q = q_0 + \mathbf q,$ 
where $q_0 = \cos(\theta/2)$ is real and 
$\mathbf q = (\sin(\theta/2))\hat{\mathbf u}$ is the imaginary part of the quaternion interpreted as a vector.
The rotated image $v'$ of a three-dimensional vector $v$ (interpreted as the imaginary part of a quaternion) is computed by $v' = qvq^*,$
where $q^* = q_0 - \mathbf q$ is the conjugate of $q.$
A rotation by $\alpha$ radians around the $x$-axis (or respectively around the $y$- or $z$-axis) is therefore represented by the quaternion
$\cos\frac\alpha2 + i\sin\frac\alpha2 = e^{i\alpha/2}$
(or respectively by $\cos\frac\alpha2 + j\sin\frac\alpha2 = e^{j\alpha/2}$
or $\cos\frac\alpha2 + k\sin\frac\alpha2 = e^{k\alpha/2}$).
Moreover, the conjugate of a quaternion $q$ for the rotation $R$ is a quaternion $q^*$ for the inverse rotation $R^{-1},$ since $q^*q = qq^* = 1,$ which is a rotation by a zero angle (or if you would prefer a different explanation, because reversing the sign of the imaginary part, which is conjugation, is equivalent to reversing the sign of the angle $\theta$ in 
$q = \cos(\theta/2) + (\sin(\theta/2))\hat{\mathbf u}$).
Due to the way rotation by means of a quaternion is defined,
the first rotation in a sequence is the rightmost factor in the product of rotation quaternions.
So if we want to rotate the vector $k$ (the unit vector pointing upward along the $z$ axis) by an angle $\phi$ around the $y$-axis and then by an angle $\theta$ around the $z$-axis, the quaternion for this rotation is
$p_{\theta,\phi} = e^{k\theta/2} e^{j\phi/2},$ its conjugate is
$p_{\theta,\phi}^* = e^{-j\phi/2} e^{-k\theta/2},$
and the purely imaginary unit quaternion equivalent to the vector with spherical coordinates $(1, \theta, \phi)$ is
$$
q_{\theta,\phi} = e^{k\theta/2} e^{j\phi/2} k e^{-j\phi/2} e^{-k\theta/2}.
$$
It turns out the answer looks something like your initial guess, with some important distinctions: we use $i\theta/2$ and $i\phi/2$ as exponents instead of $i\theta$ and $i\phi,$ and in order to get the coordinates of the actual desired vector we must use these exponents to rotate the unit vector $k.$

For completeness, here's another way to derive the same answer.
I decided to describe this method later because it requires much more calculation.
Any unit quaternion $ix + jy + kz,$ where $i,$ $j,$ and $k$ are the three imaginary units of the quaternion and $(x,y,z) \in \mathbb R^3,$ represents a $180$-degree rotation around the unit vector $(x,y,z).$
And if
\begin{align}
x &= \sin\phi \cos\theta, \\
y &= \sin\phi \sin\theta, \\
z &= \cos\phi,
\end{align}
then $(1, \theta, \phi)$ are spherical coordinates of the axis of rotation,
and hence the quaternion corresponds almost uniquely to those coordinates.
There is a complication, namely, the $180$ degree rotation around the axis is the same for the axis given by $(x,y,z)$ as for $(-x,-y,-z),$
and likewise the quaternions $q$ and $-q$ represent the same rotation;
hence there is some ambiguity about whether a particular quaternion corresponds to particular spherical coordinates $(1, \theta, \phi)$ or to the coordinates $(1, \theta + \pi, \pi - \phi),$ 
that is, the antipodes of $(1, \theta, \phi).$
Setting aside that ambiguity for now, the question is just how to construct such a rotation.
One way is to rotate the point at spherical coordinates $(1, \theta, \phi)$
to the positive $z$-axis, rotate by $\pi$ radians around the $z$-axis,
then rotate the positive $z$-axis back to the spherical coordinates 
$(1, \theta, \phi).$
The rotations to do this are a rotation by $-\theta$ radians around the $z$-axis, which brings the desired rotation axis into the $x,z$ plane,
followed by a rotation by $-\phi$ radians around the $y$-axis to bring the axis onto the $z$-axis, followed by a rotation by $\pi$ radians around the $z$-axis, then $\phi$ radians around the $y$-axis and finally
$\theta$ radians round the $z$-axis to bring the rotation axis back to its original location.
This rotation described is represented by the quaternion
$$
q_{\theta,\phi} = e^{k\theta/2} e^{j\phi/2} e^{k\pi/2} e^{-j\phi/2} e^{-k\theta/2}.
$$
Working this out in detail, first note that 
$e^{k\pi/2} = \cos\frac\pi2 + k\sin\frac\pi2 = k$; then
\begin{align}
e^{j\phi/2} e^{k\pi/2} e^{-j\phi/2} 
&= \left(\cos\frac\phi2 + j\sin\frac\phi2\right) k
    \left(\cos\frac\phi2 - j\sin\frac\phi2\right) \\
&= k \cos^2\frac\phi2 + jk \sin\frac\phi2\cos\frac\phi2
   - kj \cos\frac\phi2\sin\frac\phi2 - jkj \sin^2\frac\phi2 \\
&= k \left(\cos^2\frac\phi2 - \sin^2\frac\phi2\right)
     + 2i \sin\frac\phi2\cos\frac\phi2 \\
&= k \cos\phi + i \sin\phi \\
\end{align}
and so
\begin{align}
q_{\theta,\phi} &= e^{k\theta/2} (k \cos\phi + i \sin\phi) e^{-k\theta/2} \\
&= \left(\cos\frac\theta2 + k\sin\frac\theta2\right)
   (k \cos\phi + i \sin\phi)
   \left(\cos\frac\theta2 - k\sin\frac\theta2\right) \\
&= \left(k \cos^2\frac\theta2 + k^2\sin\frac\theta2\cos\frac\theta2
 - k^2\sin\frac\theta2\cos\frac\theta2
 - k^3 \sin^2\frac\theta2\right)\cos\phi \\
&\qquad + \left(i \cos^2\frac\theta2 + ki \sin\frac\theta2\cos\frac\theta2
 - ik \cos\frac\theta2\sin\frac\theta2
 - kik \sin^2\frac\theta2\right)\sin\phi \\
&= k \left(\cos^2\frac\theta2 + \sin^2\frac\theta2\right)\cos\phi
 + i \left(\cos^2\frac\theta2 - \sin^2\frac\theta2\right)\sin\phi
 + 2j \sin\frac\theta2\cos\frac\theta2\sin\phi \\
&= i \cos\theta\sin\phi + j \sin\theta\sin\phi + k \cos\phi,\\
\end{align}
which is exactly what we want for the unit vector with
spherical coordinates $(1, \theta, \phi).$
A: The traditional vectors are in quaternions represented in the following way for $(x,y,z)\in\Bbb R^3$, $xi+yj+zk\in\Bbb H$. However we have for all that
$$e^{ix}=\cos x + i\sin x$$
with respective $i,j,k$. So if we try to get a product we run into the issue of that any product of $3$ such will still give a real number value, and if you try to only use $2$ of those basis then the last one will still appear except for a small set of fixed angles. As such you cannot do it as a mere product.
A: Hamilton coined the term Vector for the "pure Quaternion" part, Gibbs "stole" the term and gave it the Cartesian Vector meaning used today
Quaternions in historical practice used the implicit duality of the currently taught Gibbs Vectors to the "pure Quaternion" elements and Cayley clarified the rotation properties of the full 4 element Quaternion product that includes the scalar term
There was a war of ideas between Gibbs system and Hamilton's, giving more good search terms https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=vector%20quaternion%20war
I especailly like https://arxiv.org/pdf/1509.00501.pdf
Today I would look at Hestenes work popularizing Grassman/Clifford "Geometric Algebra" for understanding the relation of Gibbs Vector Algebra and Quaternions - both are subalgebras of the 3D Euclidean "Geometric Algebra"/real valued Clifford Algebra: https://en.wikipedia.org/wiki/Geometric_algebra
"Geometric Algebra" is becoming better known in general, in computer graphics in particular, and the relation with Quaternions are pretty well explained by many:
: https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=%22geometric+algebra%22+quaternion&pws=0
Again search shows multiple explainations, uses of Spherical Coordinates in GA: https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=geometric+algebra+spherical+coordinates
