# 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:

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