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 $𝐚 .$

The sum $𝐚 + 𝐛$ 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

𝐚 \cdot 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

(𝐚 + 𝐛) \cdot (𝐚 + 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}) = 𝐚 \cdot 𝐚 + 𝐛 \cdot 𝐛 + 2 𝐚 \cdot 𝐛 .

According to Pythagoras, the magnitude of $𝐚$ is $a = \sqrt{a_{x}^{2} + a_{y}^{2} + a_{z}^{2}} .$ If we use a different coordinate system, the components of $𝐚$ will be different: $(a_{x}, a_{y}, a_{z}) \to (a'_{x}, a'_{y}, a'_{z}) .$ But if the new system of axes differs only by a rotation and/or translation of the axes, the magnitude of $𝐚$ 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 $𝐚 \cdot 𝐚,$ $𝐛 \cdot 𝐛,$ and $(𝐚 + 𝐛) \cdot (𝐚 + 𝐛)$ are invariant under rotations, and so, therefore, is the product $𝐚 \cdot 𝐛 .$

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

𝐚 \cdot 𝐛 = a b \cos θ,

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

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

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

Normal because they are unit vectors:

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

And basis because every vector $𝐯$ 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 $𝐯$ (which may be 0):

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

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

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

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

𝐚 \times 𝐛 = (a_{y} b_{z} - a_{z} b_{y}) \hat{𝐱} + (a_{z} b_{x} - a_{x} b_{z}) \hat{𝐲} + (a_{x} b_{y} - a_{y} b_{x}) \hat{𝐳} .

Show that the cross product is antisymmetric: $𝐚 \times 𝐛 = - 𝐛 \times 𝐚 .$

As a consequence, $𝐚 \times 𝐚 = 0 .$

Show that $𝐚 \cdot (𝐚 \times 𝐛) = 𝐛 \cdot (𝐚 \times 𝐛) = 0 .$

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

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

Since $a b \sin α$ is also the area $A$ of the parallelogram $P$ spanned by $𝐚$ and $𝐛,$ we can think of $𝐚 \times 𝐛$ 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 $𝐚$ and $𝐛$ are polar vectors, then $𝐚 \times 𝐛$ 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:

𝐚 \times (𝐛 \times 𝐜) = 𝐛 (𝐜 \cdot 𝐚) - (𝐚 \cdot 𝐛) 𝐜 .

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This Quantum World/Appendix/Vectors

Vectors (spatial)

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