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


 * $$\mathbf{a}\cdot\mathbf{b}=a_xb_x+a_yb_y+a_zb_z.$$

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}.$$