Peskin and Schroeder, equation 3.28, states that the Klein-Gordon equation $$(\partial^2+m^2)\psi=0 \tag{3.28}$$ is a valid choice of equation for a Dirac spinor field. Their explanation makes sense (the spinor transformation matrices obviously commute with differentiation and scalars and it was verified earlier that the scalar part of the transformation law works with that equation), but on a more foundational level I'm having trouble. In particular, how exactly do we assign a meaning to the object $\partial_\mu\psi$? In theory it should be a four-vector, but since $\psi$ is already a spinor it seems odd to just differentiate each component of $\psi$ w.r.t. $x_\mu$ (since $\psi$ carries a spinor index, not a vector index).

In short, what exactly does $\partial_\mu\psi$ mean when $\psi$ is not a scalar field?

  • $\begingroup$ Why would differentiating the components be "odd"? $\endgroup$
    – ACuriousMind
    Aug 13, 2015 at 19:36

2 Answers 2


Formally, the meaning you assign is just the usual meaning of the derivative.

$$\partial_\mu \psi(x^\nu) = \lim_{h \to 0} \frac{\psi(x^\nu + h\delta^\nu_\mu) - \psi(x^\nu)}{h}$$

You can indeed compute it componentwise, because you can subtract two spinors, as in the equation above, just by subtracting their components. The object you get has sixteen components and two indices.

It carries a four-vector index from the differentiation and a spinor index from the spinor field. Under a Lorentz transformation, the four-vector index contracts with a normal (four-vector) Lorentz transformation matrix and the spinor index contracts with a spinor Lorentz transformation matrix. More precisely you could say that the representation of the Lorentz group it lives in is formed as the tensor product of the four-vector representation and the Dirac spinor representation.

Your intuition is correct that if you take a single component of the spinor and you differentiate it w.r.t. space and time, the resulting object, with four indices, will not be a four-vector, because the Lorentz transformation not only mixes up the space and time components but also mixes up this spinor component with the other spinor components, whereas an actual four-vector would transform in a way that only requires you to know its own four components. It is the same as how a single row or column of, say, the stress-energy tensor fails to transform as a four-vector.

  • $\begingroup$ Of course! All I was missing is that it carries a spinor index and a four-vector index. Thanks! $\endgroup$
    – theage
    Aug 14, 2015 at 1:07

Basically your confusion is caused by a bias about what a vector is.

Everyone agrees that you can add two vector and get another.

Everyone agrees you can scale a vector and get another vector.

Sometimes we square a vector and get a scalar, but some people say that is "merely" an abuse of notation. But that is just a special case of $frac{\gamma_\mu\gamma_\nu+\gamma_\mu\gamma_\mu}{2}$ equalling $\pm 1$ or zero depending on whether the indexes are equal and timelike, equal and spacelike, or the indexes are different.

So here is a perspective. Assume that some day in some physics application you are going to represent vectors in spacetime and multiple them together in a way where the square of a vector will be equal to the square length.

It happens in intro physics, but you could always take the length and then multiply that It happens in relativistic quantum mechanics and there is no avoiding it.

So start out assuming you are going to do that someday. Then accept that the result of multiplying vectors is just as normal as the result of adding vectors. So the vectors live in a space of multivectors of which the vectors just make up a portion.

Now vectors are just particular kinds if multivectors and multivectors are things you can scale, add and also multiply and you get other multivectors. And what do multivectors look like? They look like linear combinations of products of vectors. That's it. Nothing outrageous.

Now. Is a vector a column matrix? Is it a row matrix? Is it a square matrix? Is it just an element of an algebra that has a scaling an addition and a multiplication? The later is the real story. But the square matrix can be nice in that we already know how to scale, add and multiply matrices. But note the algebra only has sixteen linearly independent elements.

So a complex 4x1 matrix can have enough information, a complex 1x4 can have enough information. And a complex 4x4 can be read to use the scaling, addition and multiplication of matrices you already know. But if they are just your surrogates for the elements of the algebra then which surrogates you use doesn't matter.

And here is the key about how silly your bias is. You could pick any four elements of your algebra that anticommute and square to giving the right number of $\pm 1$ times the identity element. Then you could call them vectors and call them $\gamma_0,$ $\gamma_1,$ $\gamma_2,$ and $\gamma_3.$ And you can associate them with any orthonormal frame of vectors in space time.

There is no reason to associate a particular element with a particular direction in spacetime, its just like choosing a basis.

Choosing a matrix to stand in for your vector is just like picking a basis. It is irrelevant but if that means you can use the matrix scaling and addition (like you do in intro physics) or even the matrix multiplication too (like many people do when studying relativistic quantum mechanics) that is just you trying to remind yourself how to scale, add, and multiply.

The multivectors already lived in an algebra they already had a scaling an addition and a multiplication. Instead of learning it (which is simple, just multiply one linear combination of products of vectors by another by first distributing and then you have a linear combination of products of vectors then if necessary you can break into orthogonal basis vectors and distribute then if you see a basis vector repeated anticommute it past the other vectors until it is by its twin then replace with $\pm 1$ until there are no repeats left for any product, now you can combine like terms and you have a canonical form for that basis) which you can just do a few times for a smaller set of generators until you get the idea.

Instead of learning the algebra, you fixate on a basis and then a representation and then worry about when to represent something by a 4x4 versus by a 4x1 (use the former if you are going to multiply it to the left of something or else stop using matrices for everything and just use the algebra because who cares which orthonormal frame in spacetime you like best, nature doesn't care).

So vectors are just special multivectors. And multivector can be scaled, added, and multiplied. You Xan wrote them as matrices if that helps you to scale, add and multiply them. If you are never going to multiply something on the left of something else you can write it as a column matrix. If you want to.

But here is the thing. The laws of physics just see the equation they can't tell which orthonormal frame you like, they can't tell if you are using 4x4 complex matrices to represent everything. They don't care. And you don't have to care either.

You can have spacetime and you can have multivector valued fields defined on them. If you want to use matrices to help you remember which elements you get when you multiply have that right, but don't expect nature to care.

Its like how nature doesn't care whether you wrote the formula for a curl on the front side or the back side of tour equation sheet. Nature doesn't care.

So we know what a vector is. When you write the laplacian you either just need to take some derivatives and combine them with the right signs and the four vector I devices are trying to remind you or else it is something like a repeated derivative.

And there is one derivative, $\gamma^\mu\partial_\mu$ and since these live in an algebra you know how to scale and add them so you can take the partial and then you also know how to multiply by $\gamma^\mu=\pm \gamma_\mu$ so there is no mystery about how to take a derivative.

If the partials don't exist it could still have a derivative as the limit of $\int_S f x dS/\int_V dV$ where again this is just a scaling by $1/\int_VdV$ of a sum (integral) of $f x$ a product. All things that exist in an algebra.

And spinors are, like vectors, just a subset of all the multivectors.


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