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Intuitive understanding of quaternions

  1. Jul 2, 2014 #1
    Complex numbers were fine. I understand that [tex] \imath=\sqrt{-1} [/tex].

    But what I am struggling with is the idea that j and k are somehow different in the context of quaternions. I don't know if j and k are also equal to[tex] \sqrt{-1} [/tex] or something different since they do not commute.

    Also the idea that an algebra can be non-commutative can be very confusing.

    If anyone could help me get a better understanding of these concepts that would be greatly appreciated.

  2. jcsd
  3. Jul 2, 2014 #2


    Staff: Mentor

    I think of it in terms of dimensions the i, j and k components are in separate dimensions which are orthogonal to each other.

    Hamilton was trying to extend the notion of complex numbers into something that could be used more broadly in physics. He succeeded for a time but because it was too cumbersome and not intuitive enough, physicists most notably Heaviside and Gibbs borrowed from it to create vectors and vector calculus (the i,j,k unit vector notation).


    Recently, however physicists have been revisiting quaternions as it brings rotation into clearer focus over 3x3 matrices and euler angles.


    In particular, see the section "Comparison with other representations of rotations" in the above article.
  4. Jul 2, 2014 #3
    Much more helpful and accurate to say that i2 = -1. Because i is not the only complex number with that property. -i has the same property, and can't be distinguished from i in the same way that you can distinguish the square roots of 1.
  5. Jul 2, 2014 #4

    D H

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    A quaternion is a sum of a scalar (the "real" part) and three quaternion parts: [itex]q = a + bi + cj + dk[/itex]. Multiplying a quaternion by a real is commutative: [itex]qs=sq[/itex]. Multiplying a quaternion by another quaternion? That's not necessarily the case. The above multiplication rules readily yield [itex]ij=k[/itex], [itex]jk=i[/itex], and [itex]ki=j[/itex]. Switching the order yields a different picture: [itex]ji=-k[/itex], [itex]kj=-i[/itex], and [itex]ik=ij[/itex].

    One view of quaternions that might help you gain that "intuitive understanding" is to look at a quaternion [itex]q[/itex] as comprising a scalar real part [itex]a[/itex] and a vectorial "imaginary" (better: "pure quaternion") part [itex]\vec b[/itex]: [itex]q = \begin{pmatrix} a \\ \vec b \end{pmatrix}[/itex]. Note the similarity to a complex number, which can be viewed as comprising a scalar real part and a scalar imaginary part: [itex]z= \begin{pmatrix} a \\ b \end{pmatrix}[/itex]. The difference is that the imaginary part is a scalar with the complex numbers but is a 3-vector with the quaternions: [itex]b = b_1 \hat i + b_2 \hat j + b_3 \hat k[/itex].

    Aside: You probably have seen [itex]\hat i[/itex], [itex]\hat j[/itex], and [itex]\hat k[/itex] used to represent the unit vectors along the [itex]x[/itex], [itex]y[/itex], and [itex]z[/itex] axes. Where did that notation come from? The answer is the quaternions.

    What happens when we multiply two quaternions [itex]q_1[/itex] and [itex]q_2[/itex] represented in this form, [itex]q_1 = \begin{pmatrix} a \\ \vec b \end{pmatrix}[/itex] and [itex]q_2 = \begin{pmatrix} c \\ \vec d \end{pmatrix}[/itex] ? It's similar to how complex numbers are multiplied, but with a twist. There are two ways to "multiply" 3-vectors, the dot product and the cross product. Both of these products show up in the scalar+vector representation of a quaternion:
    [tex] q_1 q_2 = \begin{pmatrix} a \\ \vec b \end{pmatrix} \begin{pmatrix} c \\ \vec d \end{pmatrix} =
    \begin{pmatrix} a - \vec b \cdot \vec d \\ a\vec d + c\vec b + \vec b \times \vec d \end{pmatrix} [/tex]

    This is a very useful concept, one that goes well beyond quaternions. For example, consider the set of NxN matrices. Two such matrices A and B add, element by element. They also multiply, and since A and B are square, both AB and BA are well defined. However, AB in general is not equal to BA. The difference between AB and BA turns out to be a very useful concept, such a useful concept that it has its own special notation: [itex][A,B]=AB-BA[/itex]. This "Lie bracket" pops up in many, many different places.
    Last edited: Jul 2, 2014
  6. Jul 3, 2014 #5
    Get a Rubik's cube - it's a really good way of getting a good understanding of the properties of quaternions, among other things.

    First solve your cube. This is fun to learn to do and is highly instructive in its own right in the art and science of groups, rotations and symmetry.

    Next, name each axis of the cube i,-i; j,-j; k,-k.

    There are only two things you can do with a Rubik's cube: rotate the entire cube to change its orientation relative to yourself, or rotate one face by 90 degrees. In quaternion terms this basic operation is a multiplication.

    If you rotate the "i" or "j" or "k" face twice, in other words if you "square" the basic operation (a multiplication performed twice is a squaring) then you find that you have multiplied the position of the face by -1. Thus, i^2,=j^2=k^2=-1

    Similarly, if you choose a cube element to follow and rotate it about each axis in turn you will find that its final resting place confirms that i*j*k=-1 in other words, it will be on the other side of the cube vertically, horizontally and sideways from where it began its journey.

    Then try i*j and see where you end up (k), j*i, (-k) and so on.

    So you can see the cube as a quaternion abacus! It is also a highly compact information store, being permutable in 4.3*10^19 ways so a single 3x3x3 array can store twice as much information as a 64 bit binary number. And it does quaternion maths automatically by virtue of its structure. Nice.
    Last edited: Jul 3, 2014
  7. Jul 3, 2014 #6
    This is something I've given quite a bit of thought, but unfortunately, at the moment, I don't know if I will be able to give as coherent a description as I could have several months ago when I was thinking about it and seemed to reach reasonably clear conclusions.

    For starters, you should try to understand complex numbers really--particularly, the way that complex numbers are good at dealing with rotations or dilations in 2 dimensions. Hamilton was trying to go up one dimension and deal with 3-dimensional rotations. There are some snags that he ran into with doing that, so he was stumped for 10 years, until he had the epiphany that he needed these gadgets to have 4 components, rather than 3.

    The first thing to notice is the requirement of non-commutativity. Actually, this is a special case of non-commutavity for matrices, so it shouldn't be so strange if you've done any linear algebra. Anyway, the non-commutavity can be seen to arise from the fact that 3-D rotations don't commute. Try to think of an example of rotations that don't commute and you will understand the significance of non-commutativity. Try fiddling around experimenting with a book with 180 degree rotations around different axes and then switch the order of the rotations.

    I could say more, but I think maybe I'll have to post again later when I have collected my old thoughts again.
  8. Jul 4, 2014 #7


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    This is a nice explanation, Blibbler, how would you interpret in this setup, rotations about skew axes?
  9. Jul 4, 2014 #8

    I've never given skew axes much thought in this context, but seeing as a Rubik's cube has three orthogonal axes I would suggest that it can't give much insight.

    However, wire frame models, a torch (flashlight) and a tabletop, a ruler and a protractor could all help nicely...

    I suppose that what I am trying to say is that maths doesn't have to be seen as an intellectual abstraction - it's not about manipulating typographical strings according to rules. It's also about how things in the "real" world work, and there is much in consensus reality that can give insight into the abstract world of maths just as is the case vice versa.
  10. Jul 4, 2014 #9


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    Thanks, Blibbler; I agree with your thesis here, if Math was just a collection of empty abstractions , I believe it would have imploded long ago. Not a rigorous claim, but one I think is defensible.
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