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The derivation of both Klein-Gordon equation and Dirac equation is due the need of quantum mechanics (or to say more correctly, quantum field theory) to adhere to special relativity. However, excpet that Klein-Gordon has negative probability issue, I do not see difference between these two. What makes Klein-Gordon describe scalar field while Dirac describe spin-1/2 field? Edit: oops. Klein-Gordon does not have non-locality issue. Sorry for writing wrongly.

Edit: Can anyone tell me in detail why $\psi$ field is scalar in Klein-Gordon while $\psi$ in Dirac is spin-1/2? I mean, if solution to Dirac is solution to Klein-Gordon, how does this make sense?

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  • $\begingroup$ Still it has not answered why probability is conserved. Since the solution to Dirac equation also solves Klein-Gordon equation, the problem of non-conservation of probaility is there with Dirac equation too, correct ? $\endgroup$
    – user26288
    Commented Jun 26, 2013 at 2:14
  • $\begingroup$ related physics.stackexchange.com/q/111401/44176 $\endgroup$
    – Nikos M.
    Commented Nov 2, 2014 at 14:12

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Spin is a property of the representation of the rotation group $SO(3)$ that describes how a field transforms under a rotation. This can be worked out for each kind of field or field equation.

The Klein-Gordon field gives a spin 0 representation, while the Dirac equation gives two spin 1/2 representations (which merge to a single representation if one also accounts for discrete symmetries).

The components of every free field satistfy the Klein-Gordon equation, irrespective of their spin. In particular, every component of the Dirac equations solves the Klein-Gordon equation. Indeed, the Klein-Gordon equation only expresses the mass shell constraint and nothing else. Spin comes in when one looks at what happens to the components.

A rotation (and more generally a Lorentz transformation) mixes the components of the Dirac field (or any other field not composed of spin 0 fields only), while on a $k$-component spin 0 field, it will transform each component separately.

In general, a Lorentz transformation given as a $4\times 4$ matrix $\Lambda$ changes a $k$-component field $F(x)$ into $F_\Lambda(\Lambda x)$, where $F_\Lambda=D(\Lambda)F$ with a $k\times k$ matrix $D(\Lambda)$ that depends on the representation. The components are spin 0 fields if and only if $D(\Lambda)$ is always the identity.

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If we say:

"A field has a spin 0, spin 1/2 or spin 1 representation"

then we in fact say something about how the field parameters transform if we go from one reference frame to another.

spin 0: The values of the field do not change if we go from one reference frame to another

spin 1: We have to apply the Lorentz transform matrix $\Lambda$ on the field parameters.

spin 1/2: We have to apply $\Lambda^{1/2}$ on the field parameters.

Remark: The use of an expression like $\Lambda^{1/2}$ should be interpreted in a somewhat symbolic way because vectors and bispinors are different objects. There is an extra factor 1/2 though in the exponent of the $\Lambda^{1/2}$ matrix.

The spin (associated with rotation) gets in here because the transformation matrix $\Lambda$ handles both boosts as well as rotations. The peculiar factor 1/2 however arises also in the 1 dimensional version of the Dirac equation where there is no such thing as spin (or rotation) and the corresponding 1 space + 1 time dimension version of $\Lambda$ only describes boosts.

The deeper reason for the factor 1/2 is that the Dirac equation relates two field components $\psi_R$ and $\psi_L$ which are equal to each other in the rest frame. In the 1 dimensional case these are the right-moving and left-moving components. The ratio of the two transforms as follows

$(\psi_R:\psi_L)\longrightarrow\Lambda~(\psi_R:\psi_L)$

In the normalization of the plane wave eigen functions this then ends up like

$\psi_R\longrightarrow\Lambda^{+1/2}\psi_R$

$\psi_L\longrightarrow\Lambda^{-1/2}\psi_L$

If we now go back to 3 spatial dimensions then $\Lambda$ includes both boosts and rotations and the factor 1/2 as an exponent on the rotation generation matrices leads two what we call spin 1/2 particles.

Hans.

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    $\begingroup$ The spin 1/2 description is wrong. One has to apply $\Lambda$ and not its square root, but as it is applied to a spinor rather than a vector, the action of $\Lambda$ is different. $\endgroup$
    – Arnold Neumaier
    Commented Oct 11, 2012 at 10:47
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    $\begingroup$ I was not talking about currents. - ''Each and every text book uses the expression $Λ^{1/2}$''?? Then please show me where it is used in this form in Weinberg's Volume 1. $\endgroup$
    – Arnold Neumaier
    Commented Oct 11, 2012 at 14:16
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    $\begingroup$ If $\Lambda$ is a Lorentz transform as in $V'^\mu=\Lambda\, V^\nu=\frac{\partial x'^\mu}{\partial x\,^\nu}V^\nu$, then Tensors transform as $T'^{\mu\nu}=\Lambda\!\otimes\!\Lambda\,T^{\alpha\beta}=\frac{\partial x'^\mu}{\partial x\,^\alpha}\!\frac{\partial x'^\beta}{\partial x\,^\nu}\,T^{\alpha\beta}$ or symbolically like $\Lambda^2$ and Spinors transform in the same sense like $\Lambda^{1/2}$ for instance expressed as $\xi'=\sqrt{p^\mu\,\sigma_\mu}~\xi$ where $p^\mu$ transforms like a vector. The absolute scale of the field parameters goes with $\cosh(\vartheta/2)+\Gamma\sinh(\vartheta/2)$ $\endgroup$
    – Hans de Vries
    Commented Oct 11, 2012 at 15:25
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    $\begingroup$ I had asked for a reference in Weinberg as a check on your (most likely false) statement about each and every textbook. - Traditionally, the square root of a 4x4 matrix is another 4x4 matrix, so $\Lambda^{1/2}\psi$ when $\Lambda$ is a Lorentz 4x4 matrix and $\psi$ a 2-dimensional spinor wouldn't make sense. Your different use of the square root is at least very unconventional. $\endgroup$
    – Arnold Neumaier
    Commented Oct 11, 2012 at 15:33
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    $\begingroup$ @HansdeVries: Peskin/Schroeder use $\frac12$ as an index to denote spinor representation, not as an exponent $\endgroup$
    – Christoph
    Commented Oct 11, 2012 at 16:50
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Let's review how the KG equation is recovered from the Dirac: (in natural units where $\hbar=c_0=1)$

$$(i\gamma^\mu \partial_\mu - m)\Psi = 0$$ $$(-i \gamma^\mu \partial_\mu - m)(i \gamma^\mu \partial_\mu - m) = 0$$ $$(\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu + m^2) \Psi = 0$$ $$(\partial^2+m^2)\Psi = 0.$$

In order for us to recover KG, we had to assume $\gamma^\nu \gamma^\mu = \eta^{\mu\nu}$. In other words, you can think of gamma's as doing the dot product of delta's. To take a lousy example, it's like we had an equation describing a velocity "spinor" and then squared it, so now it describes the speed "scalar" which has one less degree of freedom. This doesn't explain much more than how an equation describing a spinor can reduce into an equation describing a scalar.

The reason why the Dirac equation requires spinors and not scalars is because of special relativity. If it weren't for the pesky minus sign in $\eta^{\mu\nu}$, the algebra of the gamma's would be much simpler and we wouldn't need them to be 4x4 matrices. Then $\Psi$ could describe a scalar field.

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Spin is part of what a field IS. The data for two fields of different spins are very different. The KG equation doesn't even make sense for a spin 1/2 field and likewise for the Dirac equation and spin 0 fields.

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Only in the absence of an electromagnetic field do solutions to the Dirac equation also solve the Klein-Gordon equation. The Klein-Gordon equation can be applied to fields of any spin as long as any interaction with the spin can be ignored.

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