Lecture 6: Change of Basis ( BK 1.5 RHB 7.14)

6. 1. Linear Transformation of Basis

Suppose ei and ei0 are two different orthonormal bases, how do we relate them?

Clearly e10 can be written as a linear combination of the vectors e1, e2, e3. Let us write the linear combination as

e10 = λ11e1 + λ12e2 + λ13e3 Similarly we may write

ei0 = λij ej ,

(assuming summation convention) where λij (i = 1, 2, 3 and j = 1, 2, 3) are the 9 numbers relating the basis vectors e10, e20 and e30 to the basis vectors e1, e2 and e3. Notes

(i) λij are nine numbers defining the change of basis (abbreviated to ‘c.o.b’) or ‘linear transformation’. They are sometimes known as ‘direction cosines’. [ Here a linear transformation is ‘passive’ and only the basis changes. In your maths courses you may have also met ‘active transformations’ which are mappings between vector spaces]

(ii) Since ei0 are orthonormal e 0 e 0 = δ . i · j ij Now the l.h.s. of this equation may be written as

(λ e ) (λ e ) = λ λ δ = λ λ ik k · jl l ik jl kl ik jk (in the final step we have used the sifting property of δkl) and we deduce

λikλjk = δij

Since there are 6 distinct relations, only 3 of the 9 numbers λij are independent.

(iii) In order to determine λij from the two bases consider

e 0 e = (λ e ) e = λ δ = λ . i · j ik k · j ik kj ij Thus

e 0 e = λ . i · j ij

6. 2. Inverse Relations

Consider expressing the unprimed basis in terms of the primed basis and suppose that

21 ei = µij ej 0.

Then λsi = es0 ei = µij (es0 ej 0) = µij δsj = µis . Therefore · · µ = λ = λT ij ji
ij

Note that e e = δ = λ (e 0 e ) = λ λ and so i · j ij si s · j si sj

λsiλsj = δij .

6. 3. The Transformation Matrix

The numbers λij may be arranged in a square matrix, denoted by λ .

λ11 λ12 λ13   λ = λ21 λ22 λ23 Transformation Matrix. ⇐  λ31 λ32 λ33 

Recall that in matrix form δij is the identity matrix 11. The relations λkiλkj = λikλjk = δij can now be written as:-

T T 1 T λλ = λ λ = 11 , i.e. λ− = λ .

This is the condition for an orthogonal matrix and the transformation (from the ei basis to the ei0 basis) is called an orthogonal transformation. Now since λλT = 11 = 1 = λ λT and λT = λ , we have that λ 2 = 1 hence | | | | | | | | | | | | | | λ = 1 . | | ±

If λ = +1 the orthogonal transformation is said to be ‘proper’ | | If λ = 1 the orthogonal transformation is said to be ‘improper’ | | −

6. 4. Examples of Orthogonal Transformations

(a) Rotation about the e3 axis. For a rotation of θ, we have:–

e1 6 ¢ e10 e30 = e3 e30 e1 = e30 e2 = 0 ¢ ⇒ · · ¢ e10 e1 = cos θ θ¢ · ¢ e10 e2 = cos (π/2 θ) = sin θ e · − P¢ - 2 e 0 e = cos θ PθP 2 2 PP · Pq e 0 e = cos (π/2 + θ) = sin θ e20 2 · 1 −

22 Thus cos θ sin θ 0 λ =  sin θ cos θ 0  . −  0 0 1  It is easy to check that λλT = 11. Since λ = cos2 θ + sin2 θ = 1, this is a proper transformation. Note that rotations cannot| change| the handedness of the basis vectors.

(b) Inversion or Parity transformation. This is defined such that e 0 = e . i − i 1 0 0 i.e. λ = δ or λ =  −0 1 0  = 11 . ij − ij − −  0 0 1  − Clearly λλT = 11. Since λ = 1, this is an improper transformation. Note that the handedness of the basis is| reversed.| −

6e3 ¨¨*e10 e20 ¨ e λ ¨ - 2 ¨ = ¨e1 ⇒ R.H. ?e30 L.H.

(c) Reflection. Consider reflection of the axes in e –e plane so that e 0 = e , e 0 = e 2 3 1 − 1 2 2 and e30 = e3. The transformation matrix is:– 1 0 0 λ =  −0 1 0  .  0 0 1  Since λ = 1, this is an improper transformation. Again the handedness of the basis changes.| | −

e30 6e3 6 e10 ¨¨* - e λ ¨ - e 0 ¨ 2 = 2 ¨e¨ 1 R.H. ⇒ L.H.

6. 5. Products of Transformations

Consider a transformation λ to the ei0 basis followed by a transformation µ to another basis the e 00 basis i λ µ e = e 0 = e 00 i ⇒ i ⇒ i Clearly there must be an orthogonal transformation ξ e = e 00 i ⇒ i Now ei00 = µijej 0 = µijλjkek = (µλ)ikek so

ξ = µλ Note order of the product!

23 Notes

(i) In general transformations do not commute e.g. rotation of θ about e3 then reflection in e e 2 − 3 1 0 0 cos θ sin θ 0 cos θ sin θ 0  −0 1 0   sin θ cos θ 0  =  − sin θ −cos θ 0  − −  0 0 1   0 0 1   0 0 1  whereas cos θ sin θ 0 1 0 0 cos θ sin θ 0  sin θ cos θ 0   −0 1 0  =  − sin θ cos θ 0  −  0 0 1   0 0 1   0 0 1 

(ii) The inversion and the identity transformations commute with all transformations.

6. 6. Improper Transformations

We may write any improper transformation λ (for which λ = 1) as | | − λ = ( 11) µ where µ = λ and µ = +1 − − | | Thus an improper transformation can always be expressed as a proper transformation fol- lowed by an inversion. e.g. consider λ for a reflection in the 1 3 plane which may be written as − 1 0 0 1 0 0 1 0 0 λ =  0 1 0  =  −0 1 0   −0 1 0  − −  0 0 1   0 0 1   0 0 1  − − Identifying µ from λ = ( 11) µ we see µ is a rotation of π about e . − 2 e 63 6e 00 e2 ¨¨*e20 3 ¨* e10 ¨ ¨ e µ 11 e 00 ¨ - 1 ¨ - 1 = −= ¨ ⇒ ⇒ ¨e200 ?e30

6. 7. Summary

If λ = +1 we have a proper orthogonal transformation which is equivalent to rotation of| axes.| It can be proven that any rotation is a proper orthogonal transformation and vice-versa. If λ = 1 we have an improper orthogonal transformation which is equivalent to rotation of| axes| then− inversion. This is known as an improper rotation since it changes the handedness of the basis.

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