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During the course of going to the covariant formulation of the Dirac equation, we have the following:

$$ i\hbar \partial_{t} \psi = [ c(\mathbf{\alpha}\cdot\mathbf{\hat{p}}) + \beta m c^{2}]\psi $$

At this point, it seems that $\psi$ is most likely a state "ket" in the Hilbert space, since everything is basis independent. In order to move to the covariant formulation of this equation, the first step is to replace $\mathbf{\hat{p}}$ by $-i\hbar\nabla$ and multiply the above by $\beta$ from the left. Then, $$ i\hbar[\gamma^{0}\partial_{0} + \gamma^{i}\partial_{i}]\psi = i \hbar \gamma^{\mu}\partial_{\mu}\psi = m c\psi $$ And, that is it.

However, as far as I am aware (for example, in Sakurai, chapter 1), when replacing $\mathbf{\hat{p}}$ by $-i\hbar\nabla$ what we actually do is this:

$$ \langle \mathbf{x} | \mathbf{\hat{p}} | \Psi \rangle = - i\hbar \mathbf{\nabla}\langle\mathbf{x}|\Psi\rangle = - i\hbar \mathbf{\nabla}\psi(\mathbf{x})$$ That is, we explicitly choose the coordinate basis and the $\psi$ here is a wavefunction and not a state vector (which is coordinate/basis independent as a vector should be) anymore. This would imply that in the Dirac equation (covariant form), we no longer have a "vector", but this is not true either since taking the conjugate of the equation gives us $\psi^\dagger$ and not $\psi^{*}$ - in fact, it is a $4 \times 1$ column vector.

So what exactly is happening here? Are we choosing the coordinate representation or is it something else? If we are choosing the representation, how are we still left with a column vector?

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  • $\begingroup$ For reference, see here, section 6. The author uses the coordinate representation of $\mathbf{\hat{p}}$ but is talking about a 4x1 column vector, $\psi$. $\endgroup$
    – ShKol
    May 12, 2023 at 4:41
  • $\begingroup$ Also, the notation $\partial_{t}$ might be unorthodox, but what I mean is just $\partial / \partial t$. Thanks. $\endgroup$
    – ShKol
    May 12, 2023 at 4:46

3 Answers 3

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Yes, you are correct. What it really is, is the covariant derivative form.

The moment you make the replacement of $\vec{\hat p}\to-i\hslash\vec{\hat\nabla}$, that is already a choice of the position representation. We are indeed using the amplitudes $\left<\vec x\mid\psi\right>$ in place of Hilbert space ket vectors $\left|\psi\right>$

In fact, you should also look at $\hat{\mathcal H}\to i\hslash\hat{\partial_t}$ as part of the choice of position representation too. The truly covariant and basis independent version of the equation is $$ \left(mc\mathbb 1-\gamma^t\,\hat{\mathcal H}+\vec\gamma\cdot\vec{\hat p}\right)\left|\psi\right>=0 $$ where you can then decide to choose the position representation or momentum representation, really the choice of the bra vector to multiply on the left.


The column vector is still there because each single Dirac particle is a superposition of 4 wavefunctions, that you should not view as separate. There are quite a bit of choices involved, but one popular separation is into spin up and spin down of electrons and positrons, i.e. $$ \left|\psi\right>=\left|-\uparrow\right>\otimes\left|-\downarrow\right>\otimes\left|+\uparrow\right>\otimes\left|+\downarrow\right>= \begin{pmatrix}\left|-\uparrow\right>\\\left|-\downarrow\right>\\\left|+\uparrow\right>\\\left|+\downarrow\right> \end {pmatrix}$$ where $-$ refers to the electron's negative charge. When you act on the left of this with the position representation bra, you are doing it on each of the 4 components.
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  • $\begingroup$ So, if I understand this correctly, in the equation I had, the "operator" ($i \hbar \gamma^{\mu} \partial_{\mu}$) was acting on each component (i.e., the $\langle x | \Psi \rangle$) of the total state vector, $| \Psi \rangle$? $\endgroup$
    – ShKol
    May 12, 2023 at 5:21
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    $\begingroup$ Correct, that is precisely what they are doing, and the $\gamma$ matrices mixes the components together, hence why it is not ok to separate them. It also means that the position bra vector has to have four copies, same values at the same point, but different components to match the ket vectors. $\endgroup$ May 12, 2023 at 5:24
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Dirac equation is a equation of classical fermion field $\psi$, not a equation of quantum mechanical wavefuntion $| \psi \rangle$.

Any attempt of interpreting Dirac equation as quantum mechanical equation, such as the ket notation $| \psi \rangle$ in the Hilbert space, is misguided and will run into serious problems. One obvious example is that since $\psi$ is a Grassmann-valued fermion field, it would be absurd to interpret $|\psi|^2$ as a probability function as in quantum mechanics.

Historically, Dirac equation has be derived as a relativistic generalization of Schrodinger equation, ergo the quantum mechanic interpretation. But later on, it has been realized that Dirac equation is just a classical equation of the fermion field, which needs to be second-quantized. The "second-quantization" is also a misnomer, since there is only one quantization, which is quantization of the classical Dirac spinor.

Another myth about the "quantum"-association is originated from the Planck constant showing up in the Dirac equation: $$ i \hbar \gamma^{\mu}\partial_{\mu}\psi = m c\psi $$ However, let's take a look at the "second-quantized" path integral formulation of massless Dirac spinor: $$ \int D\psi \exp\left[\frac{i}{\hbar}S_\text{Dirac}\right] = \int D\psi\exp\left[\frac{i}{\hbar}\int dx^4 i\hbar \bar{\psi}\not{\partial}\psi \right]= \int D\psi\exp\left[i\int dx^4 i\bar{\psi}\not{\partial}\psi\right]. $$ As we can see above, the Planck constant $\hbar$ in the Dirac spinor Lagrangian $i\hbar\bar{\psi}\not{\partial}\psi$ is canceled out by the $\frac{1}{\hbar}$ factor in the path integrand. In other words, the Planck constant $\hbar$ is irrelevant for a massless Dirac fermion, at least in the quantum field theory context. See more details here.

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The Dirac equation is a classical field equation as is the Schrödinger equation, one in the real Clifford algebra over $(1,i)$ over a 3-manifold, the other in the real Clifford algebra $(1,e_{i,j,k})^2$ over a 3-manifold, too.

In physics, both obey a first order time development equation for the spinor. The time development equation for the spinar is given by the Hamiltonian in space cordinates only, derivatives as actinge as operators in Hilbert spaces over $\mathbb R^3$.

The main difference is a positive spectrum of Schrödinger real component equation $$\partial_{tt}(\Re \psi, \Im \psi) = ((-\Delta)^2 (\Re \psi,\Im \psi)$$ compared to the Klein-Gordon equation obtained by squaring the Dirac operator with its supersymmetric spectrum of pairs of both signs.

The constant $\hbar$, needed to map the canonical Poisson-algebra of of position differentials and momentum differentials $dx, p=\partial_{\dot x^i}\ \mathit L$ to the Heisenberg algebra $dx,dp = -i \partial_x) $ is indeed misplaced in first quantization (the so called mystery) as a factor multiplying the derivative operators in the Dirac equation. (The same is true for the i, that converts the antisymmetric derivative in Stokes theorem of exterioer calculus to a selfadjoint oparator in the Hilbert space's bilinear form)

By a division by $c \ \hbar$ the combination goes to the coupling constant and the mass term, the Lorentz norm of the 4-momentum, and thereby exhibits second quantization of elementary charges and masses.

$$\gamma^i\ \bigl(\partial_{x^i}\ \psi + i \frac{e^2}{4 \pi \ c \ \hbar \epsilon_0}\ A_i(x)\ \psi\bigr) \ = \ \frac{i m c}{\hbar} \ \psi$$

The equation is formulated in the lenght scale of the inverse Compton wavelength that occurs in experiment as the equivalent photon wavelength by in a complete decay of the mass.

The unsecondquantized Dirac equation in external fields is a perfect object to study, see Thallers book.

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