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}.


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


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


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:


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


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: