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Difference between revisions of "Dual basis"

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Given a basis of ''n'' vectors '''e<sub>i</sub>''' spanning the [[direct space]] ''E<sup>n</sup>'', and a vector
 
Given a basis of ''n'' vectors '''e<sub>i</sub>''' spanning the [[direct space]] ''E<sup>n</sup>'', and a vector
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'''x''' = ''x<sup>i</sup>'' '''e<sub>i</sub>''', let us consider the ''n'' quantities defined by the
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'''x''' = ''x<sup> i</sup>'' '''e<sub>i</sub>''', let us consider the ''n'' quantities defined by the
 
scalar products of '''x''' with the basis vectors, '''e<sub>i</sub>''':
 
scalar products of '''x''' with the basis vectors, '''e<sub>i</sub>''':
  
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''x<sub>i</sub>'' = '''x''' . '''e<sub>i</sub>''' = ''x<sup>j</sup>'' '''e<sub>j</sub>''' . '''e<sub>i</sub>''' = ''x<sup>j</sup> g<sub>ji</sub>'',
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''x<sub>i</sub>'' = '''x''' . '''e<sub>i</sub>''' = ''x<sup> j</sup>'' '''e<sub>j</sub>''' . '''e<sub>i</sub>''' = ''x<sup> j</sup> g<sub>ji</sub>'',
 
   
 
   
 
where the ''g<sub>ji</sub>'' 's are the doubly covariant components of the [[metric tensor]].
 
where the ''g<sub>ji</sub>'' 's are the doubly covariant components of the [[metric tensor]].
  
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By solving these equations in terms of ''x<sup>j</sup>'', one gets:
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By solving these equations in terms of ''x<sup> j</sup>'', one gets:
  
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''x^j^'' = ''x,,i,, g^ij^''
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''x<sup> j</sup>'' = ''x<sub>i</sub> g<sup>ij</sup>''
  
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where the matrix of the  ''g^ij^'' 's is inverse of that of the ''g,,ij,,'' 's (''g^ik^g,,jk,,'' = &#948;^i^,,j,,). The development of vector '''x''' with respect to basis vectors '''e,,i,,''' can now also be written:
+
where the matrix of the  ''g<sup>ij</sup>'' 's is inverse of that of the ''g<sub>ij</sub>'' 's (''g<sup>ik</sup>g<sub>jk</sub>'' = &#948;<sup>i</sup><sub>j</sub>). The development of vector '''x''' with respect to basis vectors '''e<sub>i</sub>''' can now also be written:
  
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'''x''' = ''x^i^'' '''e,,i,,''' = ''x,,i,, g^ij^'' '''e,,j,,'''
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'''x''' = ''x<sup> i</sup>'' '''e<sub>i</sub>''' = ''x<sub>i</sub> g<sup>ij</sup>'' '''e<sub>j</sub>'''
  
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The set of ''n'' vectors '''e^i^''' = ''g^ij^'' '''e,,j,,''' that span the space ''E^n^'' forms a basis since vector '''x''' can be written:
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The set of ''n'' vectors '''e<sup>i</sup>''' = ''g<sup>ij</sup>'' '''e<sub>j</sub>''' that span the space ''E<sup>n</sup>'' forms a basis since vector '''x''' can be written:
  
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'''x''' = ''x,,i,,'' '''e^i^'''
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'''x''' = ''x<sub>i</sub>'' '''e<sup>i</sup>'''
  
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This basis is the ''dual basis'' and the ''n'' quantities ''x,,i,,'' defined above are the
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This basis is the ''dual basis'' and the ''n'' quantities ''x<sub>i</sub>'' defined above are the
 
coordinates of '''x''' with respect to the dual basis. In a similar way one can express the direct basis vectors in terms of the dual basis vectors:
 
coordinates of '''x''' with respect to the dual basis. In a similar way one can express the direct basis vectors in terms of the dual basis vectors:
  
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'''e,,i,,''' = ''g,,ij,,'' '''e^j^'''
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'''e<sub>i</sub>''' = ''g<sub>ij</sub>'' '''e<sup>j</sup>'''
  
 
The scalar products of the basis vectors of the dual and direct bases are:
 
The scalar products of the basis vectors of the dual and direct bases are:
  
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''g^i^,,j,,'' = '''e^i^''' . '''e,,j,,''' = ''g^ik^'' '''e,,k,,''' . '''e,,j,,''' = ''g^ik^g,,jk,,'' = &#948;^i^,,j,,.
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''g<sup>i</sup><sub>j</sub>'' = '''e<sup>i</sup>''' . '''e<sub>j</sub>''' = ''g<sup>ik</sup>'' '''e<sub>k</sub>''' . '''e<sub>j</sub>''' = ''g<sup>ik</sup>g<sub>jk</sub>'' = &#948;<sup>i</sup><sub>j</sub>.
  
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One has therefore, since the matrices ''g^ik^'' and ''g,,ij,,'' are inverse:
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One has therefore, since the matrices ''g<sup>ik</sup>'' and ''g<sub>ij</sub>'' are inverse:
  
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''g^i^,,j,,'' = '''e^i^''' . '''e,,j,,''' = &#948;^i^,,j,,.
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''g<sup>i</sup><sub>j</sub>'' = '''e<sup>i</sup>''' . '''e<sub>j</sub>''' = &#948;<sup>i</sup><sub>j</sub>.
  
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These relations show that the dual basis vectors satisfy the definition conditions of the reciprocal vectors. In a three-dimensional space the dual basis and the basis of ["reciprocal space"] are identical.
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These relations show that the dual basis vectors satisfy the definition conditions of the reciprocal vectors. In a three-dimensional space the dual basis and the basis of [[reciprocal space]] are identical.
  
 
== Change of basis ==
 
== Change of basis ==
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In a change of basis where the direct basis vectors and coordinates transform like:
 
In a change of basis where the direct basis vectors and coordinates transform like:
  
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'''e'<sub>j</sub>''' = ''A<sub>j</sub><sup>i</sup>'' '''e<sub>i</sub>'''; ''x'<sup>j</sup>'' = ''B<sub>i</sub?<sup>j</sup>'' ''x^i^'',
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'''e'<sub>j</sub>''' = ''A<sub>j</sub><sup>i</sup>'' '''e<sub>i</sub>'''; ''x'<sup>j</sup>'' = ''B<sub>i</sub><sup> j</sup>'' ''x<sup>i</sup>'',
  
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where ''A<sub>j</sub><sup>i</sup>'' and ''B<sub>i</sub><sup>j</sup>'' are transformation matrices, transpose of one another,
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where ''A<sub>j</sub><sup>i</sup>'' and ''B<sub>i</sub><sup> j</sup>'' are transformation matrices, transpose of one another,
 
the dual basis vectors '''e<sup>i</sup>''' and the coordinates ''x<sub>i</sub>'' transform according to:
 
the dual basis vectors '''e<sup>i</sup>''' and the coordinates ''x<sub>i</sub>'' transform according to:
  
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'''e'<sup>j</sup> ''' = ''B<sub>i</sub>^j^'' '''e<sup>i</sup>'''; ''x'<sub>j</sub>'' = ''A<sub>j</sub><sup>i</sup>x<sub>i</sub>''.
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'''e'<sup>j</sup> ''' = ''B<sub>i</sub><sup> j</sup>'' '''e<sup>i</sup>'''; ''x'<sub>j</sub>'' = ''A<sub>j</sub><sup>i</sup>x<sub>i</sub>''.
  
 
The coordinates of a vector in reciprocal space are therefore ''covariant'' and the dual basis vectors (or reciprocal vectors) ''contravariant''.
 
The coordinates of a vector in reciprocal space are therefore ''covariant'' and the dual basis vectors (or reciprocal vectors) ''contravariant''.

Revision as of 13:08, 25 January 2006

Dual basis

Other languages

Base duale (Fr).

Definition

The dual basis is a basis associated to the basis of a vector space. In three-dimensional space, it is isomorphous to the basis of the reciprocal lattice. It is mathematically defined as follows:

Given a basis of n vectors ei spanning the direct space En, and a vector x = x i ei, let us consider the n quantities defined by the scalar products of x with the basis vectors, ei:

xi = x . ei = x j ej . ei = x j gji,

where the gji 's are the doubly covariant components of the metric tensor.

By solving these equations in terms of x j, one gets:

x j = xi gij

where the matrix of the gij 's is inverse of that of the gij 's (gikgjk = δij). The development of vector x with respect to basis vectors ei can now also be written:

x = x i ei = xi gij ej

The set of n vectors ei = gij ej that span the space En forms a basis since vector x can be written:

x = xi ei

This basis is the dual basis and the n quantities xi defined above are the coordinates of x with respect to the dual basis. In a similar way one can express the direct basis vectors in terms of the dual basis vectors:

ei = gij ej

The scalar products of the basis vectors of the dual and direct bases are:

gij = ei . ej = gik ek . ej = gikgjk = δij.

One has therefore, since the matrices gik and gij are inverse:

gij = ei . ej = δij.

These relations show that the dual basis vectors satisfy the definition conditions of the reciprocal vectors. In a three-dimensional space the dual basis and the basis of reciprocal space are identical.

Change of basis

In a change of basis where the direct basis vectors and coordinates transform like:

e'j = Aji ei; x'j = Bi j xi,

where Aji and Bi j are transformation matrices, transpose of one another, the dual basis vectors ei and the coordinates xi transform according to:

e'j = Bi j ei; x'j = Ajixi.

The coordinates of a vector in reciprocal space are therefore covariant and the dual basis vectors (or reciprocal vectors) contravariant.

See also

metric tensor
reciprocal space

The Reciprocal Lattice (Teaching Pamphlet of the International Union of Crystallography)

Section 1.1.2 of International Tables of Crystallography, Volume D




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