On the Non-uniqueness Problem of the Covariant Dirac Theory and the Spin-Rotation Coupling

Abstract

Gorbatenko and Neznamov [arXiv:1301.7599, 2013] recently claimed the absence of the title problem. In this paper, the reason for that problem is reexplained by using the notions of a unitary transformation and of the mean value of an operator, invoked by them. Their arguments actually aim at proving the uniqueness of a particular prescription for solving this problem. But that prescription is again shown non-unique. Two Hamiltonians in the same reference frame in a Minkowski spacetime, only one of them including the spin-rotation coupling term, are proved to be physically non-equivalent. This confirms that the reality of that coupling should be checked experimentally.

Appendix: Can the Energy Be Shifted by a Constant?

It is generally considered that the energy of a quantum-mechanical system is defined only up to a real constant. Indeed, if we replace the wave function Ψ by

$$ \widetilde{\varPsi}(t,x)\equiv e^{-iCt}\, \varPsi(t,x), $$

(33)

with C a real constant, then the starting Schrödinger equation (12) is equivalent to the following one:

$$ i \frac{\partial\widetilde{\varPsi} }{\partial t}= \widetilde{\mathrm {H}} \widetilde{\varPsi}, $$

(34)

with the new Hamiltonian

$$ \widetilde{\mathrm{H}}\equiv\mathrm{H}+C. $$

(35)

Then, also the energy operator E [Eq. (17)] is replaced by $\widetilde{\mathrm{E}}\equiv\mathrm{E}+C$. Thus, all energy eigenvalues and mean values are just shifted by one and the same constant C. Note that the foregoing applies to any quantum-mechanical Hamiltonian. Coming back to the covariant Dirac equation, the transformation (33) can be seen as a local similarity transformation, for which the matrix S in Eqs. (1) and (4) has the special form

$$ T=e^{iCt}\,{\bf1}_4. $$

(36)

[This leaves the γ ^μ matrices unchanged by Eq. (1).] Note that indeed the transformation law (15) of the Hamiltonian after a similarity transformation gives again (35) when it is applied to the particular transformation (36). ⁵

Recall that it is the energy operator E which is relevant to the energy, and that for the covariant Dirac equation it is in general different from the Hamiltonian H. Therefore, we might make the definition of equivalent energy operators less restrictive by adopting the following one:

Definition 1

The energy operators E and $\widetilde{\mathrm{E}}$, before and after the unitary transformation $\mathcal{U}$ associated with a similarity transformation S [Eq. (4)], are physically equivalent iff there is a real constant C such that the energy mean values are just shifted by C:

$$ \forall\varPsi\in\mathcal{D} \quad \mathrm{with}\ (\varPsi\mid\varPsi )=1,\quad\bigl(\mathcal{U}\varPsi\ \widetilde{\mid}\ \, \widetilde {\mathrm {E}}(\mathcal{U}\varPsi) \bigr) - (\varPsi\mid\mathrm{E} \, \varPsi)=C. $$

(37)

By extending and formalizing the line of reasoning used around Eqs. (8)–(10), we shall first prove the following:

Lemma 2

In order that we have (37), it is necessary and sufficient that

$$ \widetilde{\mathrm{E}}=S^{-1}\,\mathrm{E}\,S +C{\bf1}_4 . $$

(38)

Proof

Consider the sesquilinear forms defined on $\mathcal {D}\times\mathcal{D}$:

$$ \mathcal{S}(\varPsi,\varPhi)\equiv\bigl(\mathcal{U}\varPsi\ \widetilde{\mid}\ \, \widetilde{\mathrm{E}}(\mathcal{U}\varPhi) \bigr) $$

(39)

and

$$ \mathcal{S}'(\varPsi,\varPhi)\equiv(\varPsi\mid\mathrm{E}\varPhi) +C (\varPsi \mid\varPhi). $$

(40)

Denoting $Q(\varPsi)\equiv\mathcal{S}(\varPsi,\varPsi)$ and $Q'(\varPsi )\equiv\mathcal{S}'(\varPsi,\varPsi)$ the associated quadratic forms defined on $\mathcal{D}$, (37) is equivalent to:

$$ \forall\varPsi\in\mathcal{D} \quad \mathrm{with} \ (\varPsi\mid\varPsi )=1,\quad Q(\varPsi)=Q'(\varPsi). $$

(41)

Due to the homogeneity of degree 2, this of course means that Q=Q′. Hence, by (9), this is equivalent to $\mathcal {S}=\mathcal {S}'$. Thus, (37) is equivalent to:

$$ \forall\varPsi,\varPhi\in\mathcal{D},\quad\bigl( \mathcal{U}\varPsi\ \widetilde {\mid}\ \, \widetilde{\mathrm{E}}(\mathcal{U} \varPhi) \bigr) = (\varPsi\mid \mathrm{E}\varPhi) +C (\varPsi\mid\varPhi). $$

(42)

As with (8), the pushforward energy operator $\breve{ \mathrm{E}}=S^{-1}\mathrm{E}S$ verifies:

$$ \forall\varPsi,\varPhi\in\mathcal{D}\quad\bigl(\mathcal{U} \varPsi\ \widetilde {\mid}\ \breve{ \mathrm{E}} \, (\mathcal{U}\varPhi) \bigr) = ( \varPsi\mid \mathrm {E} \, \varPhi). $$

(43)

Hence, using also the unitarity (5) of $\mathcal {U}$, (42) may be rewritten as:

$$ \forall\varPsi,\varPhi\in\mathcal{D}\quad\bigl( \mathcal{U}\varPsi\ \widetilde {\mid}\ \, \widetilde{\mathrm{E}}(\mathcal{U} \varPhi) \bigr) = \bigl(\mathcal {U}\varPsi \ \widetilde{\mid}\ \breve{ \mathrm{E}}(\mathcal{U}\varPhi)\bigr) +C (\mathcal{U}\varPsi\ \widetilde{\mid}\ \mathcal{U}\varPhi). $$

(44)

Because $\breve{ \mathcal{D}}\equiv\mathcal{U}(\mathcal{D})$, which is the domain of each among the two operators $\widetilde{\mathrm{E}}$ and $\breve{ \mathrm{E}}$, is in one-to-one correspondence with $\mathcal {D}$ under $\mathcal{U}$, this is still equivalent to

$$ \forall\varXi,\varOmega\in\breve{ \mathcal{D}} \quad(\varXi\ \widetilde {\mid}\ \, \widetilde{\mathrm{E}}\varOmega) = (\varXi\ \widetilde{\mid }\ \breve{ \mathrm{E}}\varOmega) +C (\varXi\ \widetilde{\mid}\ \varOmega )\equiv \bigl(\varXi\ \widetilde{\mid}\ (\breve{ \mathrm{E}}+C{ \bf1}_4)\varOmega\bigr). $$

(45)

But since $\breve{ \mathcal{D}}$ is dense in the Hilbert space $\widetilde{ \mathcal{H}}$, the latter means that we have $\widetilde {\mathrm{E}}=\breve{ \mathrm{E}}+C{\bf1}_{4}$, that is, precisely (38). The Lemma is proved. □

However, it turns out that (38) can happen after an admissible local similarity transformation only if C=0:

Theorem 3

Suppose that the local similarity transformation S is admissible, i.e., $S(X) \in{\sf Spin(1,3)}$ for any X∈V. If the energy operators E and $\widetilde{\mathrm{E}}$ before and after the application of S are physically equivalent in the sense of (37), then C=0.

Proof

From the Lemma 2, we know that the condition (37) is equivalent to (38). Setting $\delta\mathrm{E}\equiv S\widetilde {\mathrm {E}}S^{-1} -\mathrm{E}$, Eq. (38) is equivalent to

$$ \delta\mathrm{E} = C {\bf1}_4. $$

(46)

In Ref. [2], Eq. (73), the following expression has been derived most generally for δE:

$$ \delta\mathrm{E} =-iB^{-1} \bigl[B( \partial_0S)S^{-1} \bigr]^a, \qquad Q^a\equiv\frac{1}{2}\bigl(Q-Q^\dagger\bigr), $$

(47)

where B≡Aγ ⁰. Note that S is a smooth function of X, as is by definition a local similarity transformation. Hence, the assumption that $S(X) \in{\sf Spin(1,3)}$ for all X implies that S is deduced from the smooth local Lorentz transformation L≡Λ∘S. (Here $\varLambda:{\sf Spin(1,3)}\rightarrow{\sf SO(1,3)}$ is the two-to-one covering map of the special Lorentz group by the spin group.) In other words, S is got from a smooth change of the tetrad field by using the spinor representation. Using the fact that the explicit form of the spinor representation (e.g. Ref. [17]) is generated by the commutators of the “flat” Dirac matrices γ ^♮α:

$$ s^{\alpha\beta} \equiv\bigl[\gamma^{\natural\alpha} , \gamma^{\natural \beta} \bigr], $$

(48)

it has been shown in Ref. [2, Eq. (81) and below there] that we have then

$$ \frac{\partial S}{\partial t}S^{-1}= \omega_{\alpha\beta} s^{\alpha \beta}, $$

(49)

in which the six coefficients ω _αβ =−ω _βα are real and depend on the spacetime point X as does S. Hence, we may rewrite (47) more explicitly as:

$$ 2\delta\mathrm{E} =-iB^{-1} \bigl[B\bigl( \omega_{\alpha\beta}s^{\alpha \beta}\bigr)-\bigl(B \omega_{\alpha\beta}s^{\alpha\beta} \bigr)^\dagger \bigr]=-i\omega_{\alpha\beta} \bigl[s^{\alpha\beta}-B^{-1} \bigl(s^{\alpha \beta} \bigr)^\dagger\,B^\dagger \bigr]. $$

(50)

Now, from the definition (48), and since tr MN=tr NM for two square matrices M and N having the same dimension, we have:

$$ \mathrm{tr}\,s^{\alpha\beta}=0, $$

(51)

hence also tr (s ^αβ)^†=0. We have also B ^†=B from the definition of the hermitizing matrix A {Ref. [2], Eq. (7)}. Therefore, it follows from Eq. (50) that

$$ \mathrm{tr}\, \delta\mathrm{E}=0. $$

(52)

However, if Eq. (46) is verified, we have tr δE=4C, hence C=0 from (52). This proves the Theorem 3. □

That Theorem 3 says that, for the standard version of the covariant Dirac equation (DFW), the energy mean values can not be shifted by a constant number. If a local similarity transformation is admissible for DFW and leaves the energy mean values unchanged up to a constant shift as in Eq. (37), then the new energy operator $\widetilde {\mathrm {E}}$ is just the pushforward operator $\breve{ \mathrm{E}}$ that leaves the energy mean values exactly unchanged. Said differently: if we have $\widetilde{\mathrm{E}} \ne\breve{ \mathrm{E}}$, then the mean values of the energy operators E and $\widetilde{\mathrm {E}}$ before and after the similarity transformation differ by a number which depends on the state Ψ. This result is true also for the Hamiltonian operator. (The proof is almost the same but a bit simpler.)