N
Geometrical Aspects of Gauge Fields
N.1 Covariant derivatives and coordinate
transformations
Let us go back to the U(1) case, equations (13.4)–(13.7). There, the introduction of the (gauge) covariant derivative D
μ produced an object, D
μ ψ(x),
which transformed like ψ(x) under local U(1) phase transformations, unlike
the ordinary derivative ∂
μ ψ(x) which acquired an ‘extra’ piece when transformed. This followed from simple calculus, of course – but there is a slightly
different way of thinking about it. The derivative involves not only ψ(x) at
the point x, but also ψ at the infinitesimally close, but different, point x + dx;
and the transformation law of ψ(x) involves α(x), while that of ψ(x + dx)
would involve the different function α(x + dx). Thus we may perhaps expect
something to ‘go wrong’ with the transformation law for the gradient.
To bring out the geometrical analogy we are seeking, let us write ψ =
ψ R + iψ I and α(x) = qχ(x) so that (13.3) becomes (cf (2.64))
′
ψ R (x) = cosα(x)ψ R (x) − sin α(x)ψ I (x)
(N.1)
′
ψ I (x) = sinα(x)ψ R (x) + cos α(x)ψ I (x).
If we think of ψ R (x) and ψ I (x) as being the components of a ‘vector’ ψ → (x) along
the → e R and → e I axes, respectively, then (N.1) would represent the components
′
′
of ψ → (x) as referred to new axes → e and → e I , which have been rotated by −α(x)
R
about an axis in the direction → e R ×→ e I (i.e. normal to the → e R −→ e I plane), as shown
in figure N.1. Other such ‘vectors’ φ → 1 (x), φ → 2 (x), . . . (i.e. other wavefunctions
for particles of the same charge q) when evaluated at the same point x will
have ‘components’ transforming the same as (N.1) under the axis rotation
′
′
→ e R , → e I → → e , → e . But the components of the vector ψ → (x + dx) will behave
R I
differently. The transformation law (N.1) when written at x + dx will involve
′
α(x+dx), which (to first order in dx) is α(x)+∂ μ α(x)dx
μ . Thus for ψ (x+dx)
R
′
and ψ (x + dx) the rotation angle is α(x) + ∂ μ α(x)dx
μ rather than α(x).
I
Now comes the key step in the analogy: we may think of the additional
μ
angle ∂ μ α(x)dx as coming about because, in going from x to x + dx, the
μ
coordinate basis vectors → e R and → e I have been rotated through +∂ μ α(x)dx
453
Geometrical Aspects of Gauge Fields
N.1 Covariant derivatives and coordinate
transformations
Let us go back to the U(1) case, equations (13.4)–(13.7). There, the introduction of the (gauge) covariant derivative D
μ produced an object, D
μ ψ(x),
which transformed like ψ(x) under local U(1) phase transformations, unlike
the ordinary derivative ∂
μ ψ(x) which acquired an ‘extra’ piece when transformed. This followed from simple calculus, of course – but there is a slightly
different way of thinking about it. The derivative involves not only ψ(x) at
the point x, but also ψ at the infinitesimally close, but different, point x + dx;
and the transformation law of ψ(x) involves α(x), while that of ψ(x + dx)
would involve the different function α(x + dx). Thus we may perhaps expect
something to ‘go wrong’ with the transformation law for the gradient.
To bring out the geometrical analogy we are seeking, let us write ψ =
ψ R + iψ I and α(x) = qχ(x) so that (13.3) becomes (cf (2.64))
′
ψ R (x) = cosα(x)ψ R (x) − sin α(x)ψ I (x)
(N.1)
′
ψ I (x) = sinα(x)ψ R (x) + cos α(x)ψ I (x).
If we think of ψ R (x) and ψ I (x) as being the components of a ‘vector’ ψ → (x) along
the → e R and → e I axes, respectively, then (N.1) would represent the components
′
′
of ψ → (x) as referred to new axes → e and → e I , which have been rotated by −α(x)
R
about an axis in the direction → e R ×→ e I (i.e. normal to the → e R −→ e I plane), as shown
in figure N.1. Other such ‘vectors’ φ → 1 (x), φ → 2 (x), . . . (i.e. other wavefunctions
for particles of the same charge q) when evaluated at the same point x will
have ‘components’ transforming the same as (N.1) under the axis rotation
′
′
→ e R , → e I → → e , → e . But the components of the vector ψ → (x + dx) will behave
R I
differently. The transformation law (N.1) when written at x + dx will involve
′
α(x+dx), which (to first order in dx) is α(x)+∂ μ α(x)dx
μ . Thus for ψ (x+dx)
R
′
and ψ (x + dx) the rotation angle is α(x) + ∂ μ α(x)dx
μ rather than α(x).
I
Now comes the key step in the analogy: we may think of the additional
μ
angle ∂ μ α(x)dx as coming about because, in going from x to x + dx, the
μ
coordinate basis vectors → e R and → e I have been rotated through +∂ μ α(x)dx
453