This Quantum World/Appendix/Vectors

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Vectors (spatial)[edit]

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}.

Vector1.png

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}.