I would like to know why we have $\phi'(x')=\phi(x)$ for a field satisfying Klein-Gordon equation.

Is it an assumption or can it be proved?

The $'$ means a Lorentz transformation: $\phi'$ is the field in $R'$ deduced from $R$ by a Lorentz transformation

  • $\begingroup$ What do your primes represent? $\endgroup$ – gautampk Apr 18 '17 at 17:02
  • $\begingroup$ yes sorry : it is a Lorentz transformation : $\phi'$ represent the field in $R'$ where $R'$ is deduced from $R$ by Lorentz transformation $\endgroup$ – StarBucK Apr 18 '17 at 17:09
  • $\begingroup$ Okay, well if $\phi$ follows the Klein Gordon then it's a scalar field, from which it follows that the transformed field is equal to the untransformed field. It's the definition of a scalar field. $\endgroup$ – gautampk Apr 18 '17 at 17:17
  • $\begingroup$ You can show this by looking at the 1D representation of the Lorentz group. You find that all transformations are identically 1, meaning that any object transforming under this representation is a scalar. $\phi$ is then defined to be an object transforming under this representation $\endgroup$ – gautampk Apr 18 '17 at 17:18

The point is that at the transformed point $x'$, evaluating the transformed field $\phi'$ is the same as the original field $\phi$ evaluated at the original point.

Suppose we have the transformed field $\phi'$ and wish to evaluate it at an original point $x$, then this is equivalent to,

$$\phi'(x) = \phi(\Lambda^{-1}x).$$

If we think of purely rotations, I can rotate the scalar field $\phi$ by say $\frac{\pi}{2}$ counter-clockwise and let us say that at $(1,0)$ the value is $a$ and the value at $(0,1)$ is $b$ for the original field.

The rotated field $\phi'$ would then have the value $a$ at the point $(0,1)$ instead of $b$. This would be the same as $\phi$ evaluated at $(1,0)$ which is $x$ rotated by $\frac{\pi}{2}$ clockwise, the inverse transformation. This explains $\phi'(x) = \phi(\Lambda^{-1}x)$. Now substitute in $x \to \Lambda x$ and we have that,

$$\phi'(x') = \phi(x)$$

since $\Lambda^{-1}\Lambda x = x$, where $x' = \Lambda x$.


The Lorentz group is the group of rotations in spacetime, and any quantity that remains identical under a Lorentz transformation is defined to be a Lorentz scalar.

The group is usually formed by the exponentiation of the six generators, $J_i$ and $K_i$, for $i$ from 1 to 3. These correspond to the three Euclidean rotations and three hyperbolic rotations (aka boosts) of Minkowski spacetime. The generators form the basis of the Lorentz algebra and follow the commutation relations:

$$ [J_i,J_j] = i\varepsilon_{ijk}J_k \\ [K_i,K_j] = -i\varepsilon_{ijk}J_k \\ [J_i,K_j] = i\varepsilon_{ijk}K_k $$

You can represent these by a set of six $n \times n$ dimensional matrices. In the case of $n=1$, you'll notice that only $0$ fulfills the requirements. To construct a general transformation $\Lambda$ you 'exponentiate' the generator, $\Lambda\propto e^{\theta J_i}$ for some parameter $\theta$. Since all the generators are $0$, all the transformations are $1$.

The $n$ dimensional representation of the Lorentz group acts on $n \times 1$ dimensional matrices ('column vectors'). So the 1D rep is just the set of real numbers. Therefore all reals are Lorentz scalars.

You can construct further 'reducible' representations of the group by taking direct sums of other representations. For example, the direct sum of two $1$s, $1 \oplus 1$, is the 2D identity matrix. Hence the direct sum of two 1D scalar representations is the 2D scalar representation. This acts on complex numbers (represented as an $2\times1$ matrix) so complex numbers are also scalars.

You then define your field, $\phi$, to transform under either the basic (irreducible) scalar representation, or one of the reducible scalar representations. Hence:

$$ \phi(x^{\mu}) \to \Lambda\phi(\Lambda x^{\mu}) = \phi({x^{\mu}}') $$

holds by definition.

From this premise you can work out the form of a Lagrangian, and then derive the Klein-Gordon equation.


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