# Dirac Equation in RQM (as opposed to QFT) is written in which representation?

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 just:

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

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"?

• this is another one of the reasons very few people care for RQM. The real theory, the one that is useful, is QFT. May 24, 2016 at 20:31
• 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. May 24, 2016 at 20:57

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 $$\langle\boldsymbol r|H=-i\alpha^i\partial_i+m\beta$$