# 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 ${\displaystyle (a_{x},a_{y},a_{z})}$  of a vector ${\displaystyle \mathbf {a} .}$

The sum ${\displaystyle \mathbf {a} +\mathbf {b} }$  of two vectors has the components ${\displaystyle (a_{x}+b_{x},a_{y}+b_{y},a_{z}+b_{z}).}$

• Explain the addition of vectors in terms of arrows.

The dot product of two vectors is the number

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

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

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

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

${\displaystyle {\sqrt {a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}}={\sqrt {(a'_{x})^{2}+(a'_{y})^{2}+(a'_{z})^{2}}}.}$

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

• 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

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

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

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

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

Normal because they are unit vectors:

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

And basis because every vector ${\displaystyle \mathbf {v} }$  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 ${\displaystyle \mathbf {v} }$  (which may be 0):

${\displaystyle \mathbf {v} =v_{x}\mathbf {\hat {x}} +v_{y}\mathbf {\hat {y}} +v_{z}\mathbf {\hat {z}} .}$

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

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

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

${\displaystyle \mathbf {a} \times \mathbf {b} =(a_{y}b_{z}-a_{z}b_{y})\,\mathbf {\hat {x}} +(a_{z}b_{x}-a_{x}b_{z})\,\mathbf {\hat {y}} +(a_{x}b_{y}-a_{y}b_{x})\,\mathbf {\hat {z}} .}$
• Show that the cross product is antisymmetric: ${\displaystyle \mathbf {a} \times \mathbf {b} =-\mathbf {b} \times \mathbf {a} .}$

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

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

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

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

Since ${\displaystyle ab\sin \alpha }$  is also the area ${\displaystyle A}$  of the parallelogram ${\displaystyle P}$  spanned by ${\displaystyle \mathbf {a} }$  and ${\displaystyle \mathbf {b} ,}$  we can think of ${\displaystyle \mathbf {a} \times \mathbf {b} }$  as a vector of magnitude ${\displaystyle A}$  perpendicular to ${\displaystyle P.}$  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 ${\displaystyle \mathbf {a} }$  and ${\displaystyle \mathbf {b} }$  are polar vectors, then ${\displaystyle \mathbf {a} \times \mathbf {b} }$  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:

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