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Why does the Dirac equation derivation require matrices? Starting from

$$i\hbar \frac{\partial \psi}{\partial t} = \left(\frac{\hbar c}{i}\alpha^k\partial _k + \beta m_0 c^2 \right) \psi =H \psi.$$

My syllabus says that the coefficients ($\alpha _k$ and $\beta$) cannot be numbers because the equation would not be invariant to spatial rotations. Is this obvious? I do not see immediately how they cannot be rotationally invariant.

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  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/163439/2451 $\endgroup$ – Qmechanic Aug 13 '15 at 21:09
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    $\begingroup$ @Qmechanic not really a duplicate. There is indeed an argument from rotational invariance which the OP is explicitly asking about and which is not addressed there. Still related, though. $\endgroup$ – Emilio Pisanty Aug 13 '15 at 21:10
  • $\begingroup$ Not a duplicate. Read the question more carefully next time. $\endgroup$ – Jan M. Aug 21 '15 at 17:00
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To understand:

(1) How the generators of rotation: $-i\sigma^x$, $-i\sigma^y$ and $-i\sigma^z$ are hidden in a spinor wave function and

(2) How the 1st order derivatives of a plane wave can produce the relation $~~p_o^2-p_x^2-p_y^2-p_z^2=m^2$,

you'll need to know the following fundamental identity:

$\exp(-i\phi)~~\xi_s ~~=~~ \exp(-i\phi~~\vec{s}\cdot\vec{\sigma})~~\xi_s$

where $\xi_s$ is a spinor pointing in the direction $\vec{s}=\{s_x, s_y, s_z\}$ and where $\vec{\sigma}=\{\sigma_x, \sigma_y, \sigma_z\}$

This tells us that adding a phase $-i\phi$ to the wave function rotates the spinor field by an angle of $2\phi$ around its own axis. This is a very fundamental relation! If we substitute this in the right way in a plane wave like $\exp(-ip_ot + i\vec{p}\cdot\vec{x})$, we will get the following expression:

$\exp(-ip_ot + i~(\vec{s}\cdot\vec{\sigma})~(\vec{p}\cdot\vec{x})~)$

Using $\vec{s}=\vec{p}/p_o$, because the (light-like transforming) spinor rotates in the direction of its propagation, this gives.

$\exp\left(-i(~p_o^2t - (\vec{p}\cdot\vec{\sigma})~(\vec{p}\cdot\vec{x})~)~/p_o\right)~\xi_s$

The partial derivative in, for instance, the $x$-direction gives us a factor $i~(p_x^2\sigma_x + p_xp_y\sigma_y+p_xp_z\sigma_z)/p_o$ and now you see what the effect is of multiplying the 1st order partial derivatives with the Pauli matrices, since for all squares

$(i\sigma_o)^2=(i\sigma_x)^2=(i\sigma_y)^2=(i\sigma_z)^2~~=~~-I$

and because of the anti-commutation rules cancel the cross terms:

$\sigma_x\sigma_y+\sigma_y\sigma_x=0,~~~~\sigma_y\sigma_z+\sigma_z\sigma_y=0,~~~~\sigma_z\sigma_x+\sigma_x\sigma_z=0$

Therefore the matrix multiplications get rid of the matrices in the partial derivatives and we obtain the simple factors $~~p_o^2/p_o-p_x^2/p_o-p_y^2/p_o-p_z^2/p_o$ . As for the complete two spinor Dirac field it's easiest to use the relativistic representation with two light-like transforming spinor components $\xi_L$ and $\xi_R$ but from the above you get the general idea.

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The term with the derivatives, is really a directional derivative $$ \alpha^k\partial_k=\vec \alpha\cdot \nabla $$ and if applied to a function $f$ it measures its change along that particular direction: $$ \alpha^k\partial_kf=\vec \alpha\cdot \nabla f=|\vec\alpha|\frac{\partial f}{\partial\vec n} =|\vec\alpha|\lim_{\delta x\to0}\frac{f(\vec x+\delta x\:\vec n)-f(\vec x)}{\delta x} $$ where $\vec \alpha=|\vec \alpha|\:\vec n$. If $\alpha$ is fixed - if its entries are simple real numbers that transform as a vector, then this directional derivative will continue to point at that particular direction regardless of the frame you're in. That is, it's not rotationally invariant.

It's not immediately obvious that the problem can be fixed by making them matrices, but it should at least be clear that if they're bigger than 1×1 they no longer need to clearly define a particular direction in space. As it turns out, the Dirac matrices do carry a more complicated representation of the Lorentz group (which includes amongst other things rotations), and this makes them able to retain the rotational invariance of the equation.

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  • $\begingroup$ Is the requirement that the individual components of $\psi$ (since it is an N-component column vector now) stay the same, as opposed to e.g. the scalar product of $\psi$ with itself? $\endgroup$ – Jan M. Aug 13 '15 at 23:43
  • $\begingroup$ The introduction of spinor degrees of freedom forces you to use the appropriate spinor representation of the Lorentz group. This transforms both the states and the operators and should be done with due care. See your favourite QFT textbook for details! $\endgroup$ – Emilio Pisanty Aug 14 '15 at 17:43

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