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In introductory Quantum Mechanics treatments it is common to see the Schrödinger's equation being written, simply as:

$$-\dfrac{\hbar^2}{2m}\nabla^2\Psi(\mathbf{r},t)+V(\mathbf{r})\Psi(\mathbf{r},t)=i\hbar \dfrac{\partial \Psi}{\partial t}(\mathbf{r},t).$$

When I first encountered it I got the wrong impression that $\Psi$ was a function defined on spacetime.

Later, studying Quantum Mechanics in a little bit more advanced level than this one, I've learned the postulates. What we have, in truth is one abstract state space (the space of kets) $\mathcal{E}$, we have a position observable $\mathbf{R} = (X,Y,Z)$ and this observable gives rise to a basis $|\mathbf{r}\rangle$ of eigenstates.

In that sense, the evolution equation is in truth jus:

$$H|\psi(t)\rangle = i\hbar \dfrac{d|\psi(t)\rangle}{dt},$$

and the Shcrödinger's equation which appear on introductory treatments is just the projection of that equation onto the basis $|\mathbf{r}\rangle$ as long as we write $\Psi(\mathbf{r},t)=\langle \mathbf{r}|\psi(t)\rangle$.

In almost all the treatments I've seem up to now of the Dirac Equation, the equation is directly written as:

$$(i\gamma^\mu \partial _\mu -m)\psi=0.$$

It is then said that $\gamma^\mu$ must be matrices and this implies that $\psi$ must be a column vector with four lines. Indeed, we have $\psi : \mathcal{M}\to \mathbb{C}^4$, where $\mathcal{M}$ is spacetime.

Now we ask ourselves: why it makes sense, in the Schrödinger's equation to write it in terms of a function $\Psi(\mathbf{r},t)$? And the answer is: because we have a position basis and time is a parameter of evolution.

Now, as I've found out, time is not an observable! Hence, there is no basis of eigenvectors associated to time. In that case, it makes no sense in talking about one "spacetime basis" $|\mathbf{r}\rangle \otimes |t\rangle$. This, again, doesn't exist, because time and space are treated differently in QM: time is a parameter, position is an observable.

In that case, the Dirac equation is written in which representation? I mean, the Dirac equation is what equation in the abstract state space $\mathcal{E}$ and what is the representation we project it into to get the "spacetime" equation?

How does the Dirac equation fit into the formalism of Quantum Mechanics of the abstract state space if there is no "spacetime basis"?

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    $\begingroup$ this is another one of the reasons very few people care for RQM. The real theory, the one that is useful, is QFT. $\endgroup$ – AccidentalFourierTransform May 24 '16 at 20:31
  • $\begingroup$ Absolutely nothing stops you from making time an observable, but it won't be any more useful than making spatial coordinates observables. Quantum theory, at least at this level, doesn't describe spacetime itself. It describes motion of matter on the classical background of spacetime. $\endgroup$ – CuriousOne May 24 '16 at 20:57
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The basis is still $\{|\boldsymbol r\rangle\}$. The abstract Schrödinger equation is $$ i\frac{\mathrm d}{\mathrm dt}|\psi\rangle=H|\psi\rangle $$ where $|\psi\rangle$ is a set of four kets, (with a slight abuse of notation) $$ |\psi\rangle=\begin{pmatrix}|\psi_1\rangle\\|\psi_2\rangle\\|\psi_3\rangle\\|\psi_4\rangle\end{pmatrix} $$

Time is still a parameter, $|\psi\rangle=|\psi\rangle(t)$; to get the equation of motion in the position basis, you just have to project it into the "ket" $\langle \boldsymbol r|$: $$ i\langle\boldsymbol r|\frac{\mathrm d}{\mathrm dt}|\psi\rangle=\langle\boldsymbol r|H|\psi\rangle $$ which is just the Dirac equation if you identify $\langle\boldsymbol r|\psi\rangle=\psi(\boldsymbol r,t)$ and $$ \langle\boldsymbol r|H=-i\alpha^i\partial_i+m\beta $$

As you can see, time and position are treated differently. RQM is better understood without reference to abstract spaces. When written in an abstract manner, the covariance of the theory is not explicit. But the final equation, Dirac's equation, is covariant, so everything works out just fine. Anyway, I'd like to stress that RQM is not really useful. Dirac's equation makes no sense as a relativistic wave equation. It is only useful because it also used in QFT. You may enjoy the first few chapters of Srednicki's book on QFT (there is an open access copy in his webpage), where he discusses the subtleties of constructing relativistic quantum theories.

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  • $\begingroup$ Actually, OP, I really encourage you to read the first few chapters of Srednicki's book. I think it contains exactly what you're looking for. As usual, if you have any question feel free to ask. $\endgroup$ – AccidentalFourierTransform May 24 '16 at 20:53

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