# This Quantum World/Appendix/Vectors

### Vectors (spatial)

A vector is a quantity that has both a magnitude and a direction. Vectors can be visualized as arrows. The following figure shows what we mean by the components $(a_x,a_y,a_z)$ of a vector $\mathbf{a}.$

The sum $\mathbf{a}+\mathbf{b}$ of two vectors has the components $(a_x+b_x,a_y+b_y,a_z+b_z).$

• Explain the addition of vectors in terms of arrows.

The dot product of two vectors is the number

$\mathbf{a}\cdot\mathbf{b}=a_xb_x+a_yb_y+a_zb_z.$

Its importance arises from the fact that it is invariant under rotations. To see this, we calculate

$(\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})= (a_x+b_x)^2+(a_y+b_y)^2+(a_z+b_z)^2=$
$a_x^2+a_y^2+a_z^2+b_x^2+b_y^2+b_z^2+2\,(a_xb_x+a_yb_y+a_zb_z)= \mathbf{a}\cdot\mathbf{a}+\mathbf{b}\cdot\mathbf{b}+2\,\mathbf{a}\cdot\mathbf{b}.$

According to Pythagoras, the magnitude of $\mathbf{a}$ is $a=\sqrt{a_x^2+a_y^2+a_z^2}.$ If we use a different coordinate system, the components of $\mathbf{a}$ will be different: $(a_x,a_y,a_z)\rightarrow(a'_x,a'_y,a'_z).$ But if the new system of axes differs only by a rotation and/or translation of the axes, the magnitude of $\mathbf{a}$ will remain the same:

$\sqrt{a_x^2+a_y^2+a_z^2}=\sqrt{(a'_x)^2+(a'_y)^2+(a'_z)^2}.$

The squared magnitudes $\mathbf{a}\cdot\mathbf{a},$ $\mathbf{b}\cdot\mathbf{b},$ and $(\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})$ are invariant under rotations, and so, therefore, is the product $\mathbf{a}\cdot\mathbf{b}.$

• Show that the dot product is also invariant under translations.

Since by a scalar we mean a number that is invariant under certain transformations (in this case rotations and/or translations of the coordinate axes), the dot product is also known as (a) scalar product. Let us prove that

$\mathbf{a}\cdot\mathbf{b}=ab\cos\theta,$

where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}.$ To do so, we pick a coordinate system $\mathcal{F}$ in which $\mathbf{a}=(a,0,0).$ In this coordinate system $\mathbf{a}\cdot\mathbf{b}=ab_x$ with $b_x=b\cos\theta.$ Since $\mathbf{a}\cdot\mathbf{b}$ is a scalar, and since scalars are invariant under rotations and translations, the result $\mathbf{a}\cdot\mathbf{b}=ab\cos\theta$ (which makes no reference to any particular frame) holds in all frames that are rotated and/or translated relative to $\mathcal{F}.$

We now introduce the unit vectors $\mathbf{\hat x},\mathbf{\hat y},\mathbf{\hat z},$ whose directions are defined by the coordinate axes. They are said to form an orthonormal basis. Ortho because they are mutually orthogonal:

$\mathbf{\hat x}\cdot\mathbf{\hat y}=\mathbf{\hat x}\cdot\mathbf{\hat z}=\mathbf{\hat y}\cdot\mathbf{\hat z}=0.$

Normal because they are unit vectors:

$\mathbf{\hat x}\cdot\mathbf{\hat x}=\mathbf{\hat y}\cdot\mathbf{\hat y}= \mathbf{\hat z}\cdot\mathbf{\hat z}=1.$

And basis because every vector $\mathbf{v}$ can be written as a linear combination of these three vectors — that is, a sum in which each basis vector appears once, multiplied by the corresponding component of $\mathbf{v}$ (which may be 0):

$\mathbf{v}=v_x\mathbf{\hat x}+v_y\mathbf{\hat y}+v_z\mathbf{\hat z}.$

It is readily seen that $v_x=\mathbf{\hat x}\cdot\mathbf{v},$ $v_y=\mathbf{\hat y}\cdot\mathbf{v},$ $v_z=\mathbf{\hat z}\cdot\mathbf{v},$ which is why we have that

$\mathbf{v}=\mathbf{\hat x}\,(\mathbf{\hat x}\cdot\mathbf{v})+\mathbf{\hat y}\,(\mathbf{\hat y}\cdot\mathbf{v})+\mathbf{\hat z}\,(\mathbf{\hat z}\cdot\mathbf{v}).$

Another definition that is useful (albeit only in a 3-dimensional space) is the cross product of two vectors:

$\mathbf{a}\times\mathbf{b}=(a_yb_z-a_zb_y)\,\mathbf{\hat x}+(a_zb_x-a_xb_z)\,\mathbf{\hat y}+(a_xb_y-a_yb_x)\,\mathbf{\hat z}.$
• Show that the cross product is antisymmetric: $\mathbf{a}\times\mathbf{b}=-\mathbf{b}\times\mathbf{a}.$

As a consequence, $\mathbf{a}\times\mathbf{a}=0.$

• Show that $\mathbf{a}\cdot(\mathbf{a}\times\mathbf{b})=\mathbf{b}\cdot(\mathbf{a}\times\mathbf{b})=0.$

Thus $\mathbf{a}\times\mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}.$

• Show that the magnitude of $\mathbf{a}\times\mathbf{b}$ equals $ab\sin\alpha,$ where $\alpha$ is the angle between $\mathbf{a}$ and $\mathbf{b}.$ Hint: use a coordinate system in which $\mathbf{a}=(a,0,0)$ and $\mathbf{b}= (b\cos\alpha,b\sin\alpha,0).$

Since $ab\sin\alpha$ is also the area $A$ of the parallelogram $P$ spanned by $\mathbf{a}$ and $\mathbf{b},$ we can think of $\mathbf{a}\times\mathbf{b}$ as a vector of magnitude $A$ perpendicular to $P.$ Since the cross product yields a vector, it is also known as vector product.

(We save ourselves the trouble of showing that the cross product is invariant under translations and rotations of the coordinate axes, as is required of a vector. Let us however note in passing that if $\mathbf{a}$ and $\mathbf{b}$ are polar vectors, then $\mathbf{a}\times\mathbf{b}$ is an axial vector. Under a reflection (for instance, the inversion of a coordinate axis) an ordinary (or polar) vector is invariant, whereas an axial vector changes its sign.)

Here is a useful relation involving both scalar and vector products:

$\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=\mathbf{b}(\mathbf{c}\cdot\mathbf{a})-(\mathbf{a}\cdot\mathbf{b})\mathbf{c}.$