# 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

$\mathbf {a} \cdot \mathbf {b} =a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z}.$

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

$(\mathbf {a} +\mathbf {b} )\cdot (\mathbf {a} +\mathbf {b} )=(a_{x}+b_{x})^{2}+(a_{y}+b_{y})^{2}+(a_{z}+b_{z})^{2}=$
$a_{x}^{2}+a_{y}^{2}+a_{z}^{2}+b_{x}^{2}+b_{y}^{2}+b_{z}^{2}+2\,(a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z})=\mathbf {a} \cdot \mathbf {a} +\mathbf {b} \cdot \mathbf {b} +2\,\mathbf {a} \cdot \mathbf {b} .$

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:

$\mathbf {a} \times \mathbf {b} =(a_{y}b_{z}-a_{z}b_{y})\,\mathbf {\hat {x}} +(a_{z}b_{x}-a_{x}b_{z})\,\mathbf {\hat {y}} +(a_{x}b_{y}-a_{y}b_{x})\,\mathbf {\hat {z}} .$
• 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:

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