Dirac spin operator in quantum field theory

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4.

The total spin of a multiparticle Dirac state is a bit trickier to calculate than the total momentum. For the momentum, the result turned out to be

$\displaystyle \mathbf{P}=\sum_{r,\mathbf{p}}\mathbf{p}\left(N_{r}\left(\mathbf{p}\right)+\bar{N}_{r}\left(\mathbf{p}\right)\right) \ \ \ \ \ (1)$

which is just the sum of the momenta of the particles and antiparticles. This works because a multiparticle state is an eigenstate of all the number operators, with the eigenvalues being just the number of particles in each state with spin ${r}$ and momentum ${\mathbf{p}}$, and ${\mathbf{p}}$ itself is just a 3-vector which multiplies the result. For example, we can operate on a multiparticle state to get the total momentum like this:

$\displaystyle \mathbf{P}\left|\psi_{r_{1}\mathbf{p}_{1}}\psi_{r_{2}\mathbf{p}_{2}}\psi_{r_{1}\mathbf{p}_{3}}\bar{\psi}_{r_{1}\mathbf{p}_{1}}\right\rangle =2\mathbf{p}_{1}+\mathbf{p}_{2}+\mathbf{p}_{3} \ \ \ \ \ (2)$

If we tried something analogous for the spin component ${\Sigma_{j}}$ we would get

$\displaystyle \Sigma_{j,tot}=\sum_{r,\mathbf{p}}\Sigma_{j}\left(N_{r}\left(\mathbf{p}\right)+\bar{N}_{r}\left(\mathbf{p}\right)\right) \ \ \ \ \ (3)$

As with the momentum, a multiparticle state is still an eigenstate of the number operators, but ${\Sigma_{j}}$ is a matrix operator which can operate only on single particle states, that is, states containing a single 4-component spinor. That is, the operation

$\displaystyle \Sigma_{j}\left|\psi_{r_{1}\mathbf{p}_{1}}\psi_{r_{2}\mathbf{p}_{2}}\psi_{r_{1}\mathbf{p}_{3}}\bar{\psi}_{r_{1}\mathbf{p}_{1}}\right\rangle \ \ \ \ \ (4)$

is not defined, so 3 is not a well-defined quantity.

The solution turns out to be defining the total spin operator as

$\displaystyle _{QFT}\Sigma_{j}=\int_{V}\psi^{\dagger}\Sigma_{j}\psi\;d^{3}x \ \ \ \ \ (5)$

where the ${\Sigma_{j}}$ in the integrand is the usual Dirac spin operator, and ${\psi}$ and ${\psi^{\dagger}}$ are the general solutions to the Dirac equation

 $\displaystyle \psi$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[c_{r}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)e^{-ipx}+d_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \psi^{+}+\psi^{-}\ \ \ \ \ (7)$ $\displaystyle \psi^{\dagger}$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)v_{r}^{\dagger}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)u_{r}^{\dagger}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \psi^{\dagger+}+\psi^{\dagger-} \ \ \ \ \ (9)$

Klauber evaluates the integral in his section 4.9.1 for the case of ${\psi}$ and ${\psi^{\dagger}}$ containing only ${c}$ and ${c^{\dagger}}$ operators. The integration uses the usual property of such integrals that any term in the integrand containing an exponential ${e^{ipx}}$ goes to zero because of the boundary conditions. We are left with

$\displaystyle _{QFT}\Sigma_{j}=\sum_{r,s,\mathbf{p}}\frac{m}{E_{\mathbf{p}}}u_{r}^{\dagger}\left(\mathbf{p}\right)\Sigma_{j}u_{s}\left(\mathbf{p}\right)c_{r}^{\dagger}\left(\mathbf{p}\right)c_{s}\left(\mathbf{p}\right) \ \ \ \ \ (10)$

If we apply this operator to some state ${\left|\psi_{s^{\prime}\mathbf{p}^{\prime}}\right\rangle }$ then because the ${c_{s}\left(\mathbf{p}\right)}$ operator (an annihiliation operator) is the first one to operate on the state, only operators with ${s=s^{\prime}}$ and ${\mathbf{p}=\mathbf{\mathbf{p}^{\prime}}}$ will produce a non-zero result. In those cases, the ${s^{\prime}\mathbf{\mathbf{p}^{\prime}}}$ particle is annihilated and then replaced with a ${r\mathbf{p}}$ particle because of the creation operator ${c_{r}^{\dagger}\left(\mathbf{p}\right)}$. That is, the sum over ${\mathbf{p}}$ collapses to a single term where ${\mathbf{p}=\mathbf{\mathbf{p}^{\prime}}}$ and the sum over ${s}$ is eliminated:

$\displaystyle _{QFT}\Sigma_{j}\left|\psi_{s^{\prime}\mathbf{\mathbf{p}^{\prime}}}\right\rangle =\sum_{r}\frac{m}{E_{\mathbf{\mathbf{p}^{\prime}}}}u_{r}^{\dagger}\left(\mathbf{\mathbf{p}^{\prime}}\right)\Sigma_{j}u_{r}\left(\mathbf{p}^{\prime}\right)\left|\psi_{s^{\prime}\mathbf{\mathbf{p}^{\prime}}}\right\rangle \ \ \ \ \ (11)$

Now suppose we apply this operator to a 2-particle state ${\left|\psi_{s^{\prime}\mathbf{\mathbf{p}^{\prime}}}\psi_{s^{\prime\prime}\mathbf{p}^{\prime\prime}}\right\rangle }$. We’ll get a term like 11 for each particle, with the result

$\displaystyle _{QFT}\Sigma_{j}\left|\psi_{s^{\prime}\mathbf{\mathbf{p}^{\prime}}}\psi_{s^{\prime\prime}\mathbf{p}^{\prime\prime}}\right\rangle =\left[\sum_{r}\frac{m}{E_{\mathbf{\mathbf{p}^{\prime}}}}u_{r}^{\dagger}\left(\mathbf{\mathbf{p}^{\prime}}\right)\Sigma_{j}u_{r}\left(\mathbf{p}^{\prime}\right)+\sum_{r}\frac{m}{E_{\mathbf{\mathbf{p}^{\prime\prime}}}}u_{r}^{\dagger}\left(\mathbf{p}^{\prime\prime}\right)\Sigma_{j}u_{r}\left(\mathbf{p}^{\prime\prime}\right)\right]\left|\psi_{s^{\prime}\mathbf{\mathbf{p}^{\prime}}}\psi_{s^{\prime\prime}\mathbf{p}^{\prime\prime}}\right\rangle \ \ \ \ \ (12)$

Thus in general, because only the terms in the sum over ${\mathbf{p}}$ in 10 that correspond to the momenta of the particles in the many particle state will be non-zero when this operator is applied to that many particle state, and the sum over ${s}$ also collapses, we can write 10 as

 $\displaystyle _{QFT}\Sigma_{j}$ $\displaystyle =$ $\displaystyle \sum_{r,\mathbf{p}}\frac{m}{E_{\mathbf{p}}}u_{r}^{\dagger}\left(\mathbf{p}\right)\Sigma_{j}u_{r}\left(\mathbf{p}\right)c_{r}^{\dagger}\left(\mathbf{p}\right)c_{r}\left(\mathbf{p}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{r,\mathbf{p}}\frac{m}{E_{\mathbf{p}}}u_{r}^{\dagger}\left(\mathbf{p}\right)\Sigma_{j}u_{r}\left(\mathbf{p}\right)N_{r}\left(\mathbf{p}\right) \ \ \ \ \ (14)$

[Klauber’s equation 4-113 is a bit sloppy since he applies 10 to a state which he calls ${\left|\psi_{s\mathbf{p}}\right\rangle }$ which contains the two summation variables ${s}$ and ${\mathbf{p}}$, when in fact this state should refer to a specific spin and momentum and not be part of the sum over ${s}$ and ${\mathbf{p}}$. Likewise, he retains the sum over ${\mathbf{p}}$ in 4-114 even though the annihilation operator removes all but one momentum. The final result 4-115 does appear to be correct however.]

The expectation value of the operator ${_{QFT}\Sigma_{j}}$ between two multiparticle states ${\left|\psi_{1}\right\rangle }$ and ${\left|\psi_{2}\right\rangle }$ is therefore

$\displaystyle \left\langle \psi_{1}\left|_{QFT}\Sigma_{j}\right|\psi_{2}\right\rangle \ \ \ \ \ (15)$

Because the term ${u_{r}^{\dagger}\left(\mathbf{p}\right)\Sigma_{j}u_{r}\left(\mathbf{p}\right)}$ is the product of a ${1\times4}$ row vector (${u_{r}^{\dagger}\left(\mathbf{p}\right)}$), a ${4\times4}$ matrix (${\Sigma_{j}}$) and a ${4\times1}$ column vector (${u_{r}\left(\mathbf{p}\right)}$), the result is just a scalar, that is, a number. Also, because the ${u_{r}}$ spinors are eigenspinors of the ${\Sigma_{3}}$ operator, we have ${\Sigma_{3}u_{up}=+\frac{1}{2}u_{up}}$ and ${\Sigma_{3}u_{down}=-\frac{1}{2}u_{down}}$. Finally, the inner product ${u_{r}^{\dagger}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)=E_{\mathbf{p}}/m}$. So applying 14 to the state ${\left|\psi_{up,\mathbf{\mathbf{p}^{\prime}}}\psi_{up,\mathbf{p}^{\prime\prime}}\right\rangle }$, for example, and calculating the expectation value in that state, we get

 $\displaystyle \left\langle \psi_{up,\mathbf{\mathbf{p}^{\prime}}}\psi_{up,\mathbf{p}^{\prime\prime}}\left|_{QFT}\Sigma_{3}\right|\psi_{up,\mathbf{\mathbf{p}^{\prime}}}\psi_{up,\mathbf{p}^{\prime\prime}}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi_{up,\mathbf{\mathbf{p}^{\prime}}}\psi_{up,\mathbf{p}^{\prime\prime}}\left|\left[+\frac{1}{2}+\frac{1}{2}\right]\right|\psi_{up,\mathbf{\mathbf{p}^{\prime}}}\psi_{up,\mathbf{p}^{\prime\prime}}\right\rangle \ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle +1\left|\psi_{up,\mathbf{\mathbf{p}^{\prime}}}\psi_{up,\mathbf{p}^{\prime\prime}}\right\rangle \ \ \ \ \ (17)$

The expectation value of ${_{QFT}\Sigma_{j}}$ between any two different states produces zero because different multiparticle states are orthogonal.

Momentum of particles in a Dirac field

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problems 4.26 – 4.27.

To work out the total 3-momentum of the particles (and antiparticles) in a Dirac field, we can start with the 3-momentum density

$\displaystyle p^{i}=-\pi_{r}\frac{\partial\phi^{r}}{\partial x^{i}} \ \ \ \ \ (1)$

where the ${r}$ index is summed, and for the Dirac field, ${\phi^{1}=\psi}$ and ${\phi^{2}=\bar{\psi}}$. The conjugate momenta for the Dirac field are

 $\displaystyle \pi_{1}$ $\displaystyle =$ $\displaystyle \pi^{1/2}=i\psi^{\dagger}\ \ \ \ \ (2)$ $\displaystyle \pi_{2}$ $\displaystyle =$ $\displaystyle \bar{\pi}^{1/2}=0 \ \ \ \ \ (3)$

The two field solutions are

 $\displaystyle \psi$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[c_{r}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)e^{-ipx}+d_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (4)$ $\displaystyle \bar{\psi}$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)\bar{v}_{r}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)e^{ipx}\right] \ \ \ \ \ (5)$

The latter can be converted into the Hermitian conjugate by post-multiplying the equation by ${\gamma^{0}}$, which gives

$\displaystyle \psi^{\dagger}=\sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)v_{r}^{\dagger}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)u_{r}^{\dagger}\left(\mathbf{p}\right)e^{ipx}\right] \ \ \ \ \ (6)$

The 3-momentum density is therefore

 $\displaystyle p^{i}$ $\displaystyle =$ $\displaystyle -i\sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)v_{r}^{\dagger}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)u_{r}^{\dagger}\left(\mathbf{p}\right)e^{ipx}\right]\times\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle$ $\displaystyle i\sum_{s=1}^{2}\sum_{\mathbf{q}}\sqrt{\frac{m}{VE_{\mathbf{q}}}}\left[\left(-q_{i}\right)c_{s}\left(\mathbf{q}\right)u_{s}\left(\mathbf{q}\right)e^{-iqx}+q_{i}d_{s}^{\dagger}\left(\mathbf{q}\right)v_{s}\left(\mathbf{q}\right)e^{iqx}\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)v_{r}^{\dagger}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)u_{r}^{\dagger}\left(\mathbf{p}\right)e^{ipx}\right]\times\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle$ $\displaystyle \sum_{s=1}^{2}\sum_{\mathbf{q}}\sqrt{\frac{m}{VE_{\mathbf{q}}}}\left[q^{i}c_{s}\left(\mathbf{q}\right)u_{s}\left(\mathbf{q}\right)e^{-iqx}-q^{i}d_{s}^{\dagger}\left(\mathbf{q}\right)v_{s}\left(\mathbf{q}\right)e^{iqx}\right] \ \ \ \ \ (10)$

The total momentum is the integral of ${p^{i}}$ over the volume ${V}$:

$\displaystyle P^{i}=\int p^{i}d^{3}x \ \ \ \ \ (11)$

and, as usual, any terms with an exponential term will integrate to zero due to the requirement that an integral number of wavelengths must fit into the volume. This means that in the product of the first term of 9 with the first term of 10, and in the product of the last term of 9 with the last term of 10, we must have ${\mathbf{q}=-\mathbf{p}}$. However, due to the orthogonality of the spinors we have

$\displaystyle v_{r}^{\dagger}\left(\mathbf{p}\right)u_{s}\left(-\mathbf{p}\right)=u_{r}^{\dagger}\left(\mathbf{p}\right)v_{s}\left(-\mathbf{p}\right)=0 \ \ \ \ \ (12)$

so these terms all equal zero. The remaining terms result from the product of the first term of 9 with the last term of 10, and the product of the last term of 9 with the first term of 10, where in both cases we must have ${\mathbf{p}=\mathbf{q}}$. Further, because

$\displaystyle u_{r}^{\dagger}\left(\mathbf{p}\right)u_{s}\left(\mathbf{p}\right)=v_{r}^{\dagger}\left(\mathbf{p}\right)v_{s}\left(\mathbf{p}\right)=\frac{E_{\mathbf{p}}}{m}\delta_{rs} \ \ \ \ \ (13)$

the double sum over spins ${r}$ and ${s}$ collapses to a single sum over ${r}$. Because ${\mathbf{p}=\mathbf{q}}$, the exponential terms cancel out and using ${\int d^{3}x=V}$ we get

$\displaystyle P^{i}=\sum_{r,\mathbf{p}}p^{i}\left[c_{r}^{\dagger}\left(\mathbf{p}\right)c_{r}\left(\mathbf{p}\right)-d_{r}\left(\mathbf{p}\right)d_{r}^{\dagger}\left(\mathbf{p}\right)\right] \ \ \ \ \ (14)$

Using the anticommutators

$\displaystyle \left[d_{r}\left(\mathbf{p}\right),d_{s}^{\dagger}\left(\mathbf{p}^{\prime}\right)\right]_{+}=\delta_{rs}\delta_{\mathbf{pp}^{\prime}} \ \ \ \ \ (15)$

we get

$\displaystyle P^{i}=\sum_{r,\mathbf{p}}p^{i}\left[c_{r}^{\dagger}\left(\mathbf{p}\right)c_{r}\left(\mathbf{p}\right)+d_{r}^{\dagger}\left(\mathbf{p}\right)d_{r}\left(\mathbf{p}\right)-1\right] \ \ \ \ \ (16)$

Because the sum over ${\mathbf{p}}$ includes both negative and positive momenta, the ${-1}$ in the sum cancels out and we’re left with

 $\displaystyle P^{i}$ $\displaystyle =$ $\displaystyle \sum_{r,\mathbf{p}}p^{i}\left[c_{r}^{\dagger}\left(\mathbf{p}\right)c_{r}\left(\mathbf{p}\right)+d_{r}^{\dagger}\left(\mathbf{p}\right)d_{r}\left(\mathbf{p}\right)\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{r,\mathbf{p}}p^{i}\left(N_{r}\left(\mathbf{p}\right)+\bar{N}_{r}\left(\mathbf{p}\right)\right) \ \ \ \ \ (18)$

For the full vector 3-momentum, we thus have

$\displaystyle \mathbf{P}=\sum_{r,\mathbf{p}}\mathbf{p}\left(N_{r}\left(\mathbf{p}\right)+\bar{N}_{r}\left(\mathbf{p}\right)\right) \ \ \ \ \ (19)$

That is, the total momentum is just the sum of the momenta over all the particles. For a state ${\left|\psi_{r_{1}\mathbf{p}_{1}}\psi_{r_{2}\mathbf{p}_{2}}\psi_{r_{1}\mathbf{p}_{3}}\bar{\psi}_{r_{1}\mathbf{p}_{1}}\right\rangle }$ the momentum is

$\displaystyle \mathbf{P}\left|\psi_{r_{1}\mathbf{p}_{1}}\psi_{r_{2}\mathbf{p}_{2}}\psi_{r_{1}\mathbf{p}_{3}}\bar{\psi}_{r_{1}\mathbf{p}_{1}}\right\rangle =2\mathbf{p}_{1}+\mathbf{p}_{2}+\mathbf{p}_{3} \ \ \ \ \ (20)$

Using a similar derivation, we can find the total charge in a state to be (see Klauber’s section 4.7.2)

$\displaystyle Q=-e\sum_{r,\mathbf{p}}\left(N_{r}\left(\mathbf{p}\right)-\bar{N}_{r}\left(\mathbf{p}\right)\right) \ \ \ \ \ (21)$

where ${-e}$ is the electron charge. Note that the antiparticle number operator ${\bar{N}_{r}\left(\mathbf{p}\right)}$ contributes a charge of ${+e}$ for each antiparticle. Klauber’s derivation uses the conserved probability current and integrates it using a method similar to the above.

Anticommutators, creation and annihilation operators in the Dirac equation

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.25.

In the case of the scalar field, the second quantization postulate is that the Poisson brackets of classical field theory translate into commutators in quantum field theory. For the Dirac equation, however, the four solution vectors contain 4-d spinors, for which there are no analogies in classical theory. As a result, there are no Poisson brackets for spinor fields, so we can’t apply second quantization in the same way as for the Klein-Gordon field. In fact, it turns out that the coefficients in the general solution of the Dirac equation in quantum field theory obey anticommutation relations. The general solutions are

 $\displaystyle \psi$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[c_{r}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)e^{-ipx}+d_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \psi^{+}+\psi^{-}\ \ \ \ \ (2)$ $\displaystyle \bar{\psi}$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)\bar{v}_{r}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \bar{\psi}^{+}+\bar{\psi}^{-} \ \ \ \ \ (4)$

where the coefficients ${c_{r}\left(\mathbf{p}\right)}$ and ${d_{r}\left(\mathbf{p}\right)}$ (and their Hermitian conjugates) are operators rather than numbers. It is these operators that obey the anticommutation relations, which as far as I can tell, are just postulates of the theory. The anticommutation relations are

$\displaystyle \left[c_{r}\left(\mathbf{p}\right),c_{s}^{\dagger}\left(\mathbf{p}^{\prime}\right)\right]_{+}=\left[d_{r}\left(\mathbf{p}\right),d_{s}^{\dagger}\left(\mathbf{p}^{\prime}\right)\right]_{+}=\delta_{rs}\delta_{\mathbf{pp}^{\prime}} \ \ \ \ \ (5)$

[Note that some books use braces ${\left\{ \right\} }$ to indicate anticommutators rather than the ${\left[\right]_{+}}$ notation used by Klauber.] All other anticommutators are taken to be zero. One consequence of this latter property is that all the operators anticommute with themselves, that is

 $\displaystyle \left[c_{r}\left(\mathbf{p}\right),c_{r}\left(\mathbf{p}\right)\right]_{+}$ $\displaystyle =$ $\displaystyle \left[c_{r}^{\dagger}\left(\mathbf{p}\right),c_{r}^{\dagger}\left(\mathbf{p}\right)\right]_{+}=0\ \ \ \ \ (6)$ $\displaystyle \left[d_{r}\left(\mathbf{p}\right),d_{r}\left(\mathbf{p}\right)\right]_{+}$ $\displaystyle =$ $\displaystyle \left[d_{r}^{\dagger}\left(\mathbf{p}\right),d_{r}^{\dagger}\left(\mathbf{p}\right)\right]_{+}=0 \ \ \ \ \ (7)$

Or, in other words

$\displaystyle \left[c_{r}\left(\mathbf{p}\right)\right]^{2}=\left[c_{r}^{\dagger}\left(\mathbf{p}\right)\right]^{2}=\left[d_{r}\left(\mathbf{p}\right)\right]^{2}=\left[d_{r}^{\dagger}\left(\mathbf{p}\right)\right]^{2}=0 \ \ \ \ \ (8)$

Just as in the case of the scalar field, these operators prove to be creation and annihilation operators. As shown by Klauber in his section 4.6.1, if we operate on a state containing a single particle in the state ${\left|\psi_{r,\mathbf{p}}\right\rangle }$ with the operator ${c_{r}\left(\mathbf{p}\right)}$ we get an eigenstate of the number operator ${N_{r}\left(\mathbf{p}\right)}$ with eigenvalue 0:

$\displaystyle N_{r}\left(\mathbf{p}\right)c_{r}\left(\mathbf{p}\right)\left|\psi_{r,\mathbf{p}}\right\rangle =0\times c_{r}\left(\mathbf{p}\right)\left|\psi_{r,\mathbf{p}}\right\rangle \ \ \ \ \ (9)$

That is, the state ${c_{r}\left(\mathbf{p}\right)\left|\psi_{r,\mathbf{p}}\right\rangle }$ is the vacuum state ${\left|0\right\rangle }$. Similarly, the operator ${c_{r}^{\dagger}\left(\mathbf{p}\right)}$ creates a particle with spin ${r}$ and momentum ${\mathbf{p}}$ when it operates on the vacuum state (or on any state that does not already contain a particle with spin ${r}$ and momentum ${\mathbf{p}}$):

$\displaystyle N_{r}\left(\mathbf{p}\right)c_{r}^{\dagger}\left(\mathbf{p}\right)\left|0\right\rangle =1\times c_{r}^{\dagger}\left(\mathbf{p}\right)\left|0\right\rangle \ \ \ \ \ (10)$

so that

$\displaystyle c_{r}^{\dagger}\left(\mathbf{p}\right)\left|0\right\rangle =\left|\psi_{r,\mathbf{p}}\right\rangle \ \ \ \ \ (11)$

However, if we try to create another particle of the same spin and momentum in the same state, we get zero because of 8:

$\displaystyle \left[c_{r}^{\dagger}\left(\mathbf{p}\right)\right]^{2}\left|0\right\rangle =0 \ \ \ \ \ (12)$

Note the subtle distinction between 9 and 12. In the former, we destroy an existing particle thus producing the vacuum state, which is real physical state containing no particles. In the latter, we actually annihilate the vacuum state by attempting to create two identical particles in it, producing zero, that is, nothing at all (not even the vacuum). This is the theory’s way of telling us that it is physically impossible to create two identical fermions, an effect we encountered previously in non-relativistic quantum theory.

Hamiltonian for the Dirac equation

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.24.

Here we continue with the derivation of the total Dirac Hamiltonian. We’ve seen that the Hamiltonian density is given as

$\displaystyle \mathcal{H}_{0}^{1/2}=-i\bar{\psi}\gamma^{j}\partial_{j}\psi+m\bar{\psi}\psi \ \ \ \ \ (1)$

where the general solutions of the Dirac equation are given by

 $\displaystyle \psi$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[c_{r}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)e^{-ipx}+d_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (2)$ $\displaystyle \bar{\psi}$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)\bar{v}_{r}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)e^{ipx}\right] \ \ \ \ \ (3)$

The full Hamiltonian is the integral of the density over all space, that is

$\displaystyle H_{0}^{1/2}=\int\left[-i\bar{\psi}\gamma^{j}\partial_{j}\psi+m\bar{\psi}\psi\right]d^{3}x \ \ \ \ \ (4)$

From section 4.4.1 in Klauber’s book and the previous post, we see that four of the terms in the expansion of this integral (after substituting in 2 and 3) cancel out, leaving us with (implied sum over ${j=1,2,3}$ in the first two lines):

 $\displaystyle H_{0}^{1/2}$ $\displaystyle =$ $\displaystyle -\int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}d_{r}\left(\mathbf{p}\right)\bar{v}_{r}\left(\mathbf{p}\right)\gamma^{j}p^{j}v_{s}\left(\mathbf{p}\right)d_{s}^{\dagger}\left(\mathbf{p}\right)d^{3}x+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)\gamma^{j}p^{j}u_{s}\left(\mathbf{p}\right)c_{s}\left(\mathbf{p}\right)d^{3}x+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}d_{r}\left(\mathbf{p}\right)\bar{v}_{r}\left(\mathbf{p}\right)mv_{s}\left(\mathbf{p}\right)d_{s}^{\dagger}\left(\mathbf{p}\right)d^{3}x+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)mu_{s}\left(\mathbf{p}\right)c_{s}\left(\mathbf{p}\right)d^{3}x \ \ \ \ \ (5)$

Klauber shows that the sum of the first and third lines gives us

$\displaystyle \sum_{r,\mathbf{p}}E_{\mathbf{p}}d^{\dagger}\left(\mathbf{p}\right)d_{r}\left(\mathbf{p}\right)-1\equiv\sum_{r,\mathbf{p}}E_{\mathbf{p}}\bar{N}_{r}\left(\mathbf{p}\right)-1 \ \ \ \ \ (6)$

By following similar steps to Klauber’s equations 4-70 to 4.73, we can work out the sum of the second and fourth lines, as follows. We can use the inner products of the spinors:

$\displaystyle u_{r}^{\dagger}\left(\mathbf{p}\right)u_{s}\left(\mathbf{p}\right)=v_{r}^{\dagger}\left(\mathbf{p}\right)v_{s}\left(\mathbf{p}\right)=\frac{E_{\mathbf{p}}}{m}\delta_{rs}=\frac{p_{0}}{m}\delta_{rs} \ \ \ \ \ (7)$

The adjoint spinors are defined as

 $\displaystyle \bar{u}_{r}$ $\displaystyle =$ $\displaystyle u_{r}^{\dagger}\gamma^{0} \ \ \ \ \ (8)$

so combined with the identity ${\gamma^{0}\gamma^{0}=1}$ we have

 $\displaystyle u_{r}^{\dagger}\left(\mathbf{p}\right)u_{s}\left(\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle \bar{u}_{r}\left(\mathbf{p}\right)\gamma^{0}u_{s}\left(\mathbf{p}\right)\ \ \ \ \ (9)$ $\displaystyle \bar{u}_{r}\left(\mathbf{p}\right)\gamma^{0}p_{0}u_{s}\left(\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle \frac{\left(p_{0}\right)^{2}}{m}\delta_{rs}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{E_{\mathbf{p}}^{2}}{m}\delta_{rs} \ \ \ \ \ (11)$

The sum of the second and fourth lines of 5 can then be written as

 $\displaystyle \int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)\left[-\gamma^{j}p_{j}+m-\gamma^{0}p_{0}+\gamma^{0}p_{0}\right]u_{s}\left(\mathbf{p}\right)c_{s}\left(\mathbf{p}\right)d^{3}x$ $\displaystyle =$ $\displaystyle \int\frac{d^{3}x}{V}\sum_{r,\mathbf{p}}\frac{m}{E_{\mathbf{p}}}c_{r}^{\dagger}\left(\mathbf{p}\right)\left[\bar{u}_{r}\left(\mathbf{p}\right)\left(-\gamma^{\mu}p_{\mu}+m\right)u_{r}\left(\mathbf{p}\right)+\frac{E_{\mathbf{p}}^{2}}{m}\right]c_{r}\left(\mathbf{p}\right) \ \ \ \ \ (12)$

Using Klauber’s equation B4-3.3:

$\displaystyle \left(\gamma^{\mu}p_{\mu}-m\right)u_{r}\left(\mathbf{p}\right)=0 \ \ \ \ \ (13)$

and the fact that

$\displaystyle \int d^{3}x=V \ \ \ \ \ (14)$

that is, the integral over the entire volume is just the volume ${V}$ itself, we see that 12 reduces to

 $\displaystyle \sum_{r,\mathbf{p}}\frac{m}{E_{\mathbf{p}}}c_{r}^{\dagger}\left(\mathbf{p}\right)\frac{E_{\mathbf{p}}^{2}}{m}c_{r}\left(\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle \sum_{r,\mathbf{p}}E_{\mathbf{p}}c_{r}^{\dagger}\left(\mathbf{p}\right)c_{r}\left(\mathbf{p}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \sum_{r,\mathbf{p}}E_{\mathbf{p}}N_{r}\left(\mathbf{p}\right) \ \ \ \ \ (16)$

Combining this result with 6 we get the final Hamiltonian as

$\displaystyle H_{0}^{1/2}=\sum_{r,\mathbf{p}}E_{\mathbf{p}}\left(N_{r}\left(\mathbf{p}\right)-\frac{1}{2}+\bar{N}_{r}\left(\mathbf{p}\right)-\frac{1}{2}\right) \ \ \ \ \ (17)$

The operators are interpreted as number operators so that ${N_{r}\left(\mathbf{p}\right)}$ is the operator whose eigenvalue is the number of particles with spin ${r}$ and momentum ${\mathbf{p}}$, and ${\bar{N}_{r}\left(\mathbf{p}\right)}$ is the operator whose eigenvalue is the number of antiparticles with spin ${r}$ and momentum ${\mathbf{p}}$. The two ${\frac{1}{2}}$ terms give rise to an infinite negative vacuum energy.

Hamiltonian density for the Dirac equation

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.23.

The Lagrangian density which gives the Dirac equation is

$\displaystyle \mathcal{L}_{0}^{1/2}=\bar{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi \ \ \ \ \ (1)$

We can follow the same procedure that we used for scalar fields to derive the conjugate momenta for the Dirac spin-${\frac{1}{2}}$ field. For the conjugate momenta we have from 1

 $\displaystyle \pi^{1/2}$ $\displaystyle \equiv$ $\displaystyle \frac{\partial\mathcal{L}_{0}^{1/2}}{\partial\psi_{,0}}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\bar{\psi}\gamma^{0}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\psi^{\dagger}\gamma^{0}\gamma^{0}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\psi^{\dagger} \ \ \ \ \ (5)$

where we’ve used the definition of the adjoint field ${\bar{\psi}\equiv\psi^{\dagger}\gamma^{0}}$.

Because the Lagrangian is not symmetric with respect to ${\psi}$ and ${\bar{\psi}}$ (it contains derivatives of ${\psi}$ but not of ${\bar{\psi}}$), the adjoint momentum turns out to be zero:

$\displaystyle \bar{\pi}^{1/2}\equiv\frac{\partial\mathcal{L}_{0}^{1/2}}{\partial\bar{\psi}_{,0}}=0 \ \ \ \ \ (6)$

From the conjugate momenta and the Lagrangian, we can derive the Hamiltonian density:

 $\displaystyle \mathcal{H}_{0}^{1/2}$ $\displaystyle =$ $\displaystyle \pi_{r}^{1/2}\phi_{,0}^{r}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \pi^{1/2}\dot{\psi}+\bar{\pi}^{1/2}\dot{\bar{\psi}}-\mathcal{L}_{0}^{1/2}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\psi^{\dagger}\dot{\psi}+0-\bar{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\bar{\psi}\gamma^{0}\dot{\psi}-i\bar{\psi}\gamma^{0}\dot{\psi}-i\bar{\psi}\gamma^{j}\partial_{j}\psi+m\bar{\psi}\psi\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\bar{\psi}\gamma^{j}\partial_{j}\psi+m\bar{\psi}\psi \ \ \ \ \ (11)$

where in the last 2 lines, the ${j}$ index is summed over spatial coordinates only.

The general solution of the Dirac equation for discrete momenta is

 $\displaystyle \psi$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[c_{r}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)e^{-ipx}+d_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (12)$ $\displaystyle \bar{\psi}$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)\bar{v}_{r}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)e^{ipx}\right] \ \ \ \ \ (13)$

We can find the total Hamiltonian ${H_{0}^{1/2}}$ by substituting these into 11 and integrating over space:

 $\displaystyle H_{0}^{1/2}$ $\displaystyle =$ $\displaystyle \int\mathcal{H}_{0}^{1/2}d^{3}x\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\left[-i\bar{\psi}\gamma^{j}\partial_{j}\psi+m\bar{\psi}\psi\right]d^{3}x \ \ \ \ \ (15)$

As you might guess, this is a messy operation, and Klauber goes through (most) of the gory details in his section 4.4.1. Basically, we substitute 12 and 13 into 15, using different indices for the sums in each case (${r,\mathbf{p}}$ for ${\psi}$ and ${s,\mathbf{p}^{\prime}}$ for ${\bar{\psi}}$). We then use the fact that the integral of ${e^{\pm i\mathbf{p}\cdot\mathbf{x}}}$ over all space gives zero unless ${\mathbf{p}=0}$ (because we’re assuming that the volume ${V}$ is such that an integral number of wavelengths fits exactly within its boundaries). This allows us to collapse the double sum over ${\mathbf{p}}$ and ${\mathbf{p}^{\prime}}$ to a single sum over ${\mathbf{p}}$, although we must still retain the double sum over the spins ${r}$ and ${s}$. We get a set of four integrals resulting from the derivative term in 15 and another four integrals resulting from the mass term in 15. We need to add each integral in the first set of four to the corresponding integral in the second set. We’ll consider the sum of the fourth integral from each set here (the derivation is similar to that done by Klauber in his Box 4-3, where he adds the first integral from each set). The fourth integral from the derivative term is (implied sum over ${j=1,2,3}$)

$\displaystyle I_{d4}\equiv-\int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}c_{r}^{\dagger}\left(-\mathbf{p}\right)\bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{j}p^{j}v_{s}\left(\mathbf{p}\right)d_{s}^{\dagger}\left(\mathbf{p}\right)e^{2iE_{\mathbf{p}}t}d^{3}x \ \ \ \ \ (16)$

The fourth integral from the mass term is

$\displaystyle I_{m4}\equiv\int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}c_{r}^{\dagger}\left(-\mathbf{p}\right)\bar{u}_{r}\left(-\mathbf{p}\right)mv_{s}\left(\mathbf{p}\right)d_{s}^{\dagger}\left(\mathbf{p}\right)e^{2iE_{\mathbf{p}}t}d^{3}x \ \ \ \ \ (17)$

The only differences between the two integrals are the minus sign in ${I_{d4}}$ and the replacement of ${\gamma^{j}p^{j}}$ in ${I_{d4}}$ by ${m}$ in ${I_{m4}}$.

[In Klauber’s Box 4-3, he uses a symbol with a slash through it to indicate a sum of that symbol multiplied by ${\gamma^{\mu}}$. Unfortunately, the wordpress.com Latex server doesn’t support symbols with a slash through them, so I’ll need to resort to the longhand notation.]

Suppose we have a single eigensolution which is a single term from the last term in 12, that is (no sum over ${r}$; we can take ${r}$ to be either spin):

$\displaystyle \psi=\sqrt{\frac{m}{VE_{\mathbf{p}}}}d_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)e^{ipx} \ \ \ \ \ (18)$

Since this is a solution of the Dirac equation we must have

 $\displaystyle \left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (19)$ $\displaystyle \sqrt{\frac{m}{VE_{\mathbf{p}}}}d_{r}^{\dagger}\left(\mathbf{p}\right)e^{ipx}\left(-\gamma^{\mu}p_{\mu}-m\right)v_{r}\left(\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (20)$

Remember that this is a matrix equation, and note that the terms ${\sqrt{\frac{m}{VE_{\mathbf{p}}}}d_{r}^{\dagger}\left(\mathbf{p}\right)e^{ipx}}$ are not matrices, nor are they zero, so we must have

$\displaystyle \left(\gamma^{\mu}p_{\mu}+m\right)v_{r}\left(\mathbf{p}\right)=0 \ \ \ \ \ (21)$

Returning to 16, we can convert the integrand using the inner product of the spinor terms

$\displaystyle u_{r}^{\dagger}\left(-\mathbf{p}\right)v_{s}\left(\mathbf{p}\right)=0 \ \ \ \ \ (22)$

From the definition of the adjoint spinor ${\bar{u}_{r}}$ and the identity ${\gamma^{0}\gamma^{0}=1}$, we have

 $\displaystyle u_{r}^{\dagger}\left(-\mathbf{p}\right)v_{s}\left(\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle u_{r}^{\dagger}\left(-\mathbf{p}\right)\gamma^{0}\gamma^{0}v_{s}\left(\mathbf{p}\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{0}v_{s}\left(\mathbf{p}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (25)$

Since this term is zero, we can multiply it by anything without changing it, so we multiply it by ${p_{0}}$ to get

$\displaystyle u_{r}^{\dagger}\left(-\mathbf{p}\right)v_{s}\left(\mathbf{p}\right)=\bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{0}p_{0}v_{s}\left(\mathbf{p}\right)=0 \ \ \ \ \ (26)$

The middle part of 16 can be written as

 $\displaystyle \bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{j}p^{j}v_{s}\left(\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle \bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{j}p_{j}v_{s}\left(\mathbf{p}\right)\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{0}p_{0}v_{s}\left(\mathbf{p}\right)+\bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{j}p_{j}v_{s}\left(\mathbf{p}\right)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\bar{u}_{r}\left(-\mathbf{p}\right)\gamma^{\mu}p_{\mu}v_{s}\left(\mathbf{p}\right)\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \bar{u}_{r}\left(-\mathbf{p}\right)mv_{s}\left(\mathbf{p}\right) \ \ \ \ \ (30)$

where we subtracted 26 (which is zero, so it doesn’t change anything) in the second line and used 21 to get the last line. Putting this final result into 16 we get

$\displaystyle I_{d4}=-\int\sum_{r,s,\mathbf{p}}\frac{m}{VE_{\mathbf{p}}}c_{r}^{\dagger}\left(-\mathbf{p}\right)\bar{u}_{r}\left(-\mathbf{p}\right)mv_{s}\left(\mathbf{p}\right)d_{s}^{\dagger}\left(\mathbf{p}\right)e^{2iE_{\mathbf{p}}t}d^{3}=-I_{m4} \ \ \ \ \ (31)$

so ${I_{d4}+I_{m4}=0}$ and the two integrals cancel each other.

Dirac equation in quantum field theory: Lagrangian density

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.22.

Our treatment of the Dirac equation so far has been restricted to relativistic quantum theory, and has not touched on field theory. To make the transition to field theory, we start with the same equation, namely

$\displaystyle \left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi=0 \ \ \ \ \ (1)$

except that now ${\psi}$ is interpreted as a field, rather than a quantum state. Since this equation is mathematically identical to the Dirac equation that we solved for quantum states, its solutions are the same as well

 $\displaystyle \psi^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (2)$ $\displaystyle \psi^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (3)$ $\displaystyle \psi^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (4)$ $\displaystyle \psi^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (5)$

The adjoint Dirac equation also remains the same form

$\displaystyle i\partial_{\mu}\bar{\psi}\gamma^{\mu}+m\bar{\psi}=0 \ \ \ \ \ (6)$

and has solutions

 $\displaystyle \bar{\psi}^{\left(1\right)}$ $\displaystyle =$ $\displaystyle u_{1}^{\dagger}\gamma^{0}e^{ipx}\equiv\bar{u}_{1}e^{ipx}\ \ \ \ \ (7)$ $\displaystyle \bar{\psi}^{\left(2\right)}$ $\displaystyle =$ $\displaystyle u_{2}^{\dagger}\gamma^{0}e^{ipx}\equiv\bar{u}_{2}e^{ipx}\ \ \ \ \ (8)$ $\displaystyle \bar{\psi}^{\left(3\right)}$ $\displaystyle =$ $\displaystyle v_{2}^{\dagger}\gamma^{0}e^{ipx}\equiv\bar{v}_{2}e^{-ipx}\ \ \ \ \ (9)$ $\displaystyle \bar{\psi}^{\left(4\right)}$ $\displaystyle =$ $\displaystyle v_{1}^{\dagger}\gamma^{0}e^{ipx}\equiv\bar{v}_{1}e^{-ipx} \ \ \ \ \ (10)$

The general solutions for ${\psi}$ and ${\bar{\psi}}$ are linear combinations of these solutions, which are written as

 $\displaystyle \psi$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[c_{r}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)e^{-ipx}+d_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \psi^{+}+\psi^{-}\ \ \ \ \ (12)$ $\displaystyle \bar{\psi}$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)\bar{v}_{r}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)\bar{u}_{r}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \bar{\psi}^{+}+\bar{\psi}^{-} \ \ \ \ \ (14)$

where ${V}$ is the volume containing the particles.

Again, these solutions are mathematically equivalent to the solutions of the quantum state Dirac equation, except that the coefficients ${c_{r}}$ and ${d_{r}}$ now turn out to be operators rather than just numbers.

To proceed further with field theory, we need the Lagrangian and Hamiltonian densities for the Dirac equation. As with the Klein-Gordon equation, we’ll pull the Lagrangian density out of thin air (Klauber says it is obtained by ‘trial and error’):

$\displaystyle \mathcal{L}_{0}^{1/2}=\bar{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi \ \ \ \ \ (15)$

The ${\frac{1}{2}}$ superscript on ${\mathcal{L}}$ signifies that we’re dealing with spin-${\frac{1}{2}}$ particles (and is not a square root!), while the subscript ${0}$ indicates that we’re dealing with free particles. [It’s worth remembering at this point that this is a matrix equation; ${\bar{\psi}}$ is a 4-d row vector, ${\psi}$ is a 4-d column vector, ${\gamma^{\mu}}$ is a ${4\times4}$ matrix and ${m}$ is multiplied by the ${4\times4}$ identity matrix. After multiplying all these matrices together, though, the Lagrangian density is a scalar.] We can verify that this Lagrangian density does give the Dirac equation and its adjoint by plugging it into the Euler-Lagrange equation. We have

$\displaystyle \frac{d}{dx^{\mu}}\left(\frac{\partial\mathcal{L}}{\partial\phi_{,\mu}^{n}}\right)-\frac{\partial\mathcal{L}}{\partial\phi^{n}}=0 \ \ \ \ \ (16)$

where the index ${n}$ indicates which field we’re considering. Here, we have two fields: ${\phi^{1}=\bar{\psi}}$ and ${\phi^{2}=\psi}$. For ${\phi^{1}}$ we have

 $\displaystyle \frac{d}{dx^{\mu}}\left(\bar{\psi}i\gamma^{\mu}\right)-\left(-\bar{\psi}m\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (17)$ $\displaystyle i\partial_{\mu}\bar{\psi}\gamma^{\mu}+m\bar{\psi}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (18)$

This is adjoint Dirac equation 6 above.

For ${\phi^{2}}$ we get, since there are no derivatives of ${\bar{\psi}}$ in the Lagrangian 15,

 $\displaystyle \frac{d}{dx^{\mu}}\left(\frac{\partial\mathcal{L}}{\partial\bar{\psi}_{,\mu}}\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (19)$ $\displaystyle \frac{\partial\mathcal{L}}{\partial\bar{\psi}}$ $\displaystyle =$ $\displaystyle \left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi=0 \ \ \ \ \ (20)$

which is just the original Dirac equation 1.

Dirac equation in relativistic quantum mechanics: summary

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.21.

At this point, it’s useful to summarize the Dirac equation and its solutions as found in relativistic quantum theory. First, the theory that we’ve studied so far is not field theory; the solutions ${\psi}$ or ${\left|\psi\right\rangle }$ are generalizations of the solutions of the Schrödinger equation, in which the position ${\mathbf{x}}$ of a particle is an operator and is itself a function of time. In field theory, ${\mathbf{x}}$ and ${t}$ are placed on an equal footing and both serve as nothing more than labels of points in spacetime.

The Dirac equation was postulated to have the same form as the Schrödinger equation, that is

$\displaystyle i\frac{\partial}{\partial t}\left|\psi\right\rangle =H\left|\psi\right\rangle \ \ \ \ \ (1)$

To make the equation consistent with relativity, it was postulated that for a free particle, the Hamiltonian had the form

$\displaystyle H=\boldsymbol{\alpha}\cdot\mathbf{p}+\beta m \ \ \ \ \ (2)$

Requiring that the square of this Hamiltonian satisfies the relativistic energy-momentum relation ${E^{2}=p^{2}+m^{2}}$ leads to conditions on the four matrices ${\boldsymbol{\alpha}}$ and ${\beta}$:

 $\displaystyle \left\{ \alpha_{i},\alpha_{j}\right\}$ $\displaystyle =$ $\displaystyle \left\{ \alpha_{i},\beta\right\} =0\mbox{ if }i\ne j\ \ \ \ \ (3)$ $\displaystyle \alpha_{i}^{2}=\beta^{2}$ $\displaystyle =$ $\displaystyle I \ \ \ \ \ (4)$

Working with the Dirac equation is easier if we introduce the gamma matrices which allows the Dirac equation to be written in the compact form

$\displaystyle \left(i\gamma^{\mu}\partial_{\mu}-m\right)\left|\psi\right\rangle =0 \ \ \ \ \ (5)$

Remember that the ${\gamma^{\mu}}$ constitute four ${4\times4}$ matrices, and that ${\left|\psi\right\rangle }$ is a 4-d column vector, so this simple form of the Dirac equation actually represents four coupled PDEs in the four spacetime variables.

At this point, we are given one set of solutions to 5, written in the form

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (6)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (7)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (8)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (9)$

Here the ${u_{r}}$ and ${v_{r}}$ are the 4-d spinor components, and the exponentials ${e^{\pm ipx}}$ contain the spacetime dependence. [Again, remember that these solutions are only for a free particle; there is no potential term and hence no interactions.] The inner products of the spinors satisfy several normalization and orthogonality conditions, such as

 $\displaystyle u_{r}^{\dagger}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle v_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)=\frac{E}{m};\;\;r=1,2\ \ \ \ \ (10)$ $\displaystyle u_{1}^{\dagger}\left(\mathbf{p}\right)u_{2}\left(-\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (11)$ $\displaystyle u_{1}^{\dagger}\left(\mathbf{p}\right)v_{2}\left(-\mathbf{p}\right)$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (12)$

The overall solutions satisfy the orthogonality conditions

$\displaystyle \left\langle \psi^{\left(r\right)}\left(\mathbf{p}\right)\left|\psi^{\left(s\right)}\left(\mathbf{p}\right)\right.\right\rangle =\frac{EV}{m}\delta_{rs} \ \ \ \ \ (13)$

where ${V}$ is the volume containing the particle.

The adjoint solutions are defined by

$\displaystyle \left\langle \bar{\psi}^{\left(n\right)}\right|\equiv\left\langle \psi^{\left(n\right)}\right|\gamma^{0} \ \ \ \ \ (14)$

These solutions satisfy the adjoint Dirac equation

$\displaystyle i\partial_{\mu}\left\langle \bar{\psi}\right|\gamma^{\mu}+m\left\langle \bar{\psi}\right|=0 \ \ \ \ \ (15)$

If we defined a probability current by

$\displaystyle j^{\mu}\equiv\left\langle \bar{\psi}\left|\gamma^{\mu}\right|\psi\right\rangle \ \ \ \ \ (16)$

we find that it satisfies the conservation law

$\displaystyle \partial_{\mu}j^{\mu}=0 \ \ \ \ \ (17)$

Using the ${\mu=0}$ component of ${j^{\mu}}$ as a probability density ${\rho}$ leads to positive probabilities for both particles and antiparticles. However, antiparticles still end up with negative energy.

The spin and helicity properties of the Dirac equation are summarized in Klauber’s Wholeness Chart 4-1.

Helicity operator in the Dirac equation

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.20.

In the Dirac equation, the spin of a particle is described using the spin operator

$\displaystyle \Sigma_{i}=\frac{1}{2}\left[\begin{array}{cc} \sigma_{i} & 0\\ 0 & \sigma_{i} \end{array}\right];\;i=x,y,z \ \ \ \ \ (1)$

(using natural units where ${\hbar=1}$) and where the Pauli matrices are

 $\displaystyle \sigma_{x}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\ \ \ \ \ (2)$ $\displaystyle \sigma_{y}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle \sigma_{z}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right] \ \ \ \ \ (4)$

and the 0 components in 1 are ${2\times2}$ zero matrices. Spin can be oriented in any direction for particles travelling at a velocity ${v<1}$, although at the speed of light (${v=1}$), the spin is always aligned with the velocity due to length contraction effects. The relation between the directions of spin and the particle’s velocity is given by the helicity. If the 3-momentum ${\mathbf{p}}$ and spin both point in the same direction, the helicity has its maximum value (a positive quantity), while if they point in opposite directions, the helicity has its maximum negative value. If ${\mathbf{p}}$ and spin are at right angles, the helicity is zero.

These relations suggest that a helicity operator can be defined in terms of the scalar product of ${\boldsymbol{\Sigma}}$ and ${\mathbf{p}}$. The helicity operator is

 $\displaystyle \Sigma_{\mathbf{p}}$ $\displaystyle \equiv$ $\displaystyle \boldsymbol{\Sigma}\cdot\frac{\mathbf{p}}{\left|\mathbf{p}\right|}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Sigma_{1}\frac{p^{1}}{\left|\mathbf{p}\right|}+\Sigma_{2}\frac{p^{2}}{\left|\mathbf{p}\right|}+\Sigma_{3}\frac{p^{3}}{\left|\mathbf{p}\right|} \ \ \ \ \ (6)$

Since the ${\Sigma_{i}}$ are each a ${4\times4}$ matrix, the helicity ${\Sigma_{\mathbf{p}}}$ is also a ${4\times4}$ matrix, but with reference to 3-d space, it is a scalar matrix.

The solutions to the Dirac equation consist of a 4-element column spinor and a spacetime component:

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (7)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (8)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (9)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (10)$

For a particle moving in the ${z}$ direction, ${p^{3}>0}$, ${p^{1}=p^{2}=0}$ and we’ve seen that the states ${\left|\psi^{\left(1\right)}\right\rangle }$ and ${\left|\psi^{\left(3\right)}\right\rangle }$ are eigenstates of ${\Sigma_{3}}$ with eigenvalue ${\frac{1}{2}}$. In this case, ${p^{3}/\left|\mathbf{p}\right|=1}$ so from 6

$\displaystyle \Sigma_{\mathbf{p}}\left|\psi^{\left(1\right)}\right\rangle =\frac{1}{2}\left|\psi^{\left(1\right)}\right\rangle \ \ \ \ \ (11)$

Thus for a particle whose spin and velocity both point in the same direction, the maximum helicity value of ${\frac{1}{2}}$ is obtained. Similarly, if ${p^{3}<0}$, ${p^{1}=p^{2}=0}$ and ${p^{3}/\left|\mathbf{p}\right|=-1}$ so

$\displaystyle \Sigma_{\mathbf{p}}\left|\psi^{\left(1\right)}\right\rangle =-\frac{1}{2}\left|\psi^{\left(1\right)}\right\rangle \ \ \ \ \ (12)$

With velocity and spin pointing in opposite directions, the maximum negative value of ${-\frac{1}{2}}$ is obtained for the helicity.

Now suppose we have a particle in state ${\left|\psi^{\left(2\right)}\right\rangle }$ and that ${p^{1}\ne0}$, ${p^{2}=p^{3}=0}$. This is not a helicity eigenstate, since ${\left|\psi^{\left(2\right)}\right\rangle }$ is an eigenstate of ${\Sigma_{3}}$ with eigenvalue ${-\frac{1}{2}}$, so the spin is in the ${-z}$ direction while the velocity is in the ${x}$ direction, so the spin is not parallel to the velocity.

Mathematically, in order for ${\left|\psi^{\left(2\right)}\right\rangle }$ to be a helicity eigenstate, the spinor in ${\left|\psi^{\left(2\right)}\right\rangle }$ would have to be an eigenstate of ${\Sigma_{1}}$, which is not true, since

 $\displaystyle \Sigma_{1}\left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right]\sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}}{E+m}\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ 0\\ \frac{p^{1}}{E+m} \end{array}\right]e^{-ipx} \ \ \ \ \ (14)$

If ${p^{3}\ne0}$, ${p^{1}=p^{2}=0}$, however, we would expect ${\left|\psi^{\left(2\right)}\right\rangle }$ to be a helicity eigenstate since the spin and velocity are parallel in this case. Since ${\left|\psi^{\left(2\right)}\right\rangle }$ is an eigenstate of ${\Sigma_{3}}$ with eigenvalue ${-\frac{1}{2}}$, we have

$\displaystyle \Sigma_{\mathbf{p}}\left|\psi^{\left(1\right)}\right\rangle =\pm\frac{1}{2}\left|\psi^{\left(1\right)}\right\rangle \ \ \ \ \ (15)$

with the + corresponding to ${p^{3}>0}$ and the ${-}$ to ${p^{3}<0}$.

Dirac equation: non-relativistic limit

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.19.

The solutions to the Dirac equation consist of a 4-element column spinor and a spacetime component:

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (1)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (2)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (3)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (4)$

In the non-relativistic limit, the relative velocity of the particle satisfies ${v\ll1}$, which means that the momentum components all satisfy ${p^{j}\ll E\approx m}$. Thus ${\sqrt{\frac{E+m}{2m}}\approx1}$ and the solutions reduce to

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (5)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (6)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 0\\ 1\\ 0 \end{array}\right]e^{ipx}\ \ \ \ \ (7)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right]e^{ipx} \ \ \ \ \ (8)$

Comparing this with the free particle solutions to the non-relativistic Schrödinger equation, we see that the ${e^{\pm ipx}=e^{\pm Et}e^{\mp\mathbf{p}\cdot\mathbf{x}}}$ factor is just what we’d get in that case. The first two components of the spinors in ${\left|\psi^{\left(1\right)}\right\rangle }$ and ${\left|\psi^{\left(2\right)}\right\rangle }$ also correspond to the eigenstates in the non-relativistic spin ${\frac{1}{2}}$ theory.

In the 4-d case, we can operate on these solutions with the 4-d spin operator ${\Sigma_{z}}$ to get

 $\displaystyle \Sigma_{z}\left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right]\left[\begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left|\psi^{\left(1\right)}\right\rangle \ \ \ \ \ (11)$ $\displaystyle \Sigma_{z}\left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right]\left[\begin{array}{c} 0\\ 1\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left[\begin{array}{c} 0\\ 1\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left|\psi^{\left(2\right)}\right\rangle \ \ \ \ \ (14)$

Similarly, we get

 $\displaystyle \Sigma_{z}\left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left|\psi^{\left(3\right)}\right\rangle \ \ \ \ \ (15)$ $\displaystyle \Sigma_{z}\left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left|\psi^{\left(4\right)}\right\rangle \ \ \ \ \ (16)$

These are the same results that we get by applying the Pauli spin matrices to the 2-d spin space spinors in the non-relativistic theory.

Dirac equation: spinors near the speed of light

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.18.

The solutions to the Dirac equation consist of a 4-element column spinor and a spacetime component:

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (1)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (2)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (3)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (4)$

For a particle at rest (${\mathbf{p}=0,\;E=m}$), or for a particle moving in the ${z}$ direction, all four solutions are eigenstates of the spin operator ${\Sigma_{z}}$, with eigenvalues (spins) of ${\pm\frac{1}{2}}$. If the particle is moving in the ${x}$ or ${y}$ direction, the individual spinors above aren’t eigenstates of any of the spin operators. As the speed of the particle approaches ${c}$, however, we can get some eigenstates of ${\Sigma_{x}}$ and ${\Sigma_{y}}$.

First, suppose the particle is moving in the ${x}$ direction at a speed approaching ${c}$, with any motion in the ${y}$ and ${z}$ directions much smaller by comparison. In this case, ${E\rightarrow p^{1}\rightarrow\infty}$ and the spinor components (which we’ll call ${s^{\left(n\right)}}$ for ${n=1,\ldots,4}$) of the solutions above become

 $\displaystyle s^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ 1 \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle s^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (6)$ $\displaystyle s^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (7)$ $\displaystyle s^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ 1 \end{array}\right] \ \ \ \ \ (8)$

The ${x}$ spin operator is

$\displaystyle \Sigma_{x}=\frac{1}{2}\left[\begin{array}{cccc} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right] \ \ \ \ \ (9)$

Multiplying ${\Sigma_{x}}$ into ${s^{\left(1\right)}+s^{\left(2\right)}}$, we get

 $\displaystyle \Sigma_{x}\left(s^{\left(1\right)}+s^{\left(2\right)}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ 1\\ 1\\ 1 \end{array}\right]\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(s^{\left(1\right)}+s^{\left(2\right)}\right) \ \ \ \ \ (11)$

Similarly

 $\displaystyle \Sigma_{x}\left(s^{\left(3\right)}+s^{\left(4\right)}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ 1\\ 1\\ 1 \end{array}\right]\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(s^{\left(3\right)}+s^{\left(4\right)}\right) \ \ \ \ \ (13)$

Thus the sums ${u_{1}+u_{2}}$ and ${v_{1}+v_{2}}$ are both eigenstates of ${\Sigma_{x}}$ with eigenvalue ${\frac{1}{2}}$.

Now suppose the particle is moving in the ${y}$ direction with ${v^{y}\rightarrow1}$ so that ${E\rightarrow p^{2}\rightarrow\infty}$. The four spinors now become

 $\displaystyle s^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ i \end{array}\right]\ \ \ \ \ (14)$ $\displaystyle s^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ -i\\ 0 \end{array}\right]\ \ \ \ \ (15)$ $\displaystyle s^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ i\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (16)$ $\displaystyle s^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} -i\\ 0\\ 0\\ 1 \end{array}\right] \ \ \ \ \ (17)$

The ${y}$ spin operator is

$\displaystyle \Sigma_{y}=\frac{1}{2}\left[\begin{array}{cccc} 0 & -i & 0 & 0\\ i & 0 & 0 & 0\\ 0 & 0 & 0 & -i\\ 0 & 0 & i & 0 \end{array}\right] \ \ \ \ \ (18)$

In this case

 $\displaystyle \Sigma_{y}\left(s^{\left(1\right)}+s^{\left(3\right)}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 0 & -i & 0 & 0\\ i & 0 & 0 & 0\\ 0 & 0 & 0 & -i\\ 0 & 0 & i & 0 \end{array}\right]\left[\begin{array}{c} 1\\ i\\ 1\\ i \end{array}\right]\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ i\\ 1\\ i \end{array}\right]\ \ \ \ \ (20)$ $\displaystyle \Sigma_{y}\left(s^{\left(2\right)}+s^{\left(4\right)}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 0 & -i & 0 & 0\\ i & 0 & 0 & 0\\ 0 & 0 & 0 & -i\\ 0 & 0 & i & 0 \end{array}\right]\left[\begin{array}{c} -i\\ 1\\ -i\\ 1 \end{array}\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} -i\\ 1\\ -i\\ 1 \end{array}\right] \ \ \ \ \ (22)$

Thus the sums ${u_{1}+v_{2}}$ and ${v_{1}+u_{2}}$ are both eigenstates of ${\Sigma_{y}}$ with eigenvalue ${\frac{1}{2}}$. [As the ${u_{j}}$ spinors are supposed to represent particles and the ${v_{j}}$ antiparticles, I’m not sure what a mixture of the two is supposed to represent.]