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In quantum field theory, one can redefine the particle creation and annihilation operators using Bogoliobov transformations, which can give rise to a different vacuum state, using a new set of creation and annihilation operators. To arrive at a consistent and unique vacuum state, one imposes the criteria that the mode functions in the scalar field operator must be positive frequency mode functions, i.e. $e^{-i\omega_{p}t}$.

My question is, how is this consistent with a Dirac antiparticle having negative energy? And can we impose such criteria for a Dirac field?

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    $\begingroup$ One clarification about the background: The vacuum state is defined to be the state of minimum energy, not just any state that happens to be annihilated by some operator $a$ that happens to satisfy relations like $[a,a^\dagger]=1$, whether or not it comes from a Bogoliubov transform. The vacuum state is observer-dependent because the energy operator (which generates time-translations) is observer-dependent. I assume this is what you meant by "To arrive at a consistent and unique vacuum state, one imposes the [positive frequency] criteria...". Just wanted to clarify for other visitors. $\endgroup$ Nov 19, 2021 at 15:30
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    $\begingroup$ Does this answer your question? What's the role of the Dirac vacuum sea in quantum field theory? $\endgroup$ Nov 19, 2021 at 15:33
  • $\begingroup$ Also see Why Negative Energy States are Bad. $\endgroup$ Nov 19, 2021 at 15:44
  • $\begingroup$ And Does Dirac's idea of filled negative energy states make sense? $\endgroup$ Nov 19, 2021 at 15:49

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The whole notion of negative-energy particles in the Dirac equation comes from the rather pathological limit that one needs to take to go from the correct quantum field theory of Dirac fermions to the "one-particle" approximation which Dirac was (unknowingly) working in. This answer follows the treatment in Merzbacher's quantum mechanics textbook.

In any truly relativistic treatment of quantum mechanics, one must allow for particles to be created and destroyed. Trying to write down a normalized wave function for a single particle which evolves unitarily doesn't allow this, which is a first hint that the wave function formalism of the Dirac equation isn't a fully correct one. But using the Dirac equation does give accurate answers, so one should be able to figure out what approximation is being done.

The correct Hamiltonian for free Dirac fermions is the quantum field theory $$ H = \int d^d x \, \Psi^{\dagger}_a(t,x) \left[ - i \hbar c \, \vec{\alpha} \cdot \vec{\nabla} + \beta m c^2 \right]_{ab} \Psi_b(t,x), $$ where the $\beta_{ab}$ and $\alpha^i_{ab}$ (i = 1,2,3) matrices are the same appearing in Dirac's original paper, and $$ \{ \Psi_a(t,x),\Psi^{\dagger}_b(t,x') \} = \delta_{a b} \delta^d(x - x'). $$ This is diagonalized by an expansion in creation and annihilation operators of the form $$ \Psi_a(t,x) = \int \frac{d^d k}{(2 \pi)^d \sqrt{2 \omega_k}} \sum_s \left[ u_a(k,s) b_s(k) e^{- i \omega_k t + i k \cdot x} + v_a(k,s) c^{\dagger}_s(k) e^{i \omega_k t - i k \cdot x} \right], $$ with anticommutation relations $\{ b_s(k),b_{s'}^{\dagger}(k') \} = \{ c_s(k),c^{\dagger}_{s'}(k') \} = \delta_{s s'} \delta(k - k')$, and by choosing the $u_a(k,s)$ and $v_a(k,s)$ judiciously we find $$ H = E_0 + \int \frac{d^d k}{(2 \pi)^d} \sum_s \omega_k \left[ b^{\dagger}_s(k)b_s(k) + c_s^{\dagger}(k)c_s(k) \right]. $$ This Hamiltonian has its energy bounded below. One can define a conserved U(1) charge for this theory, and the particles created by the $c$ operators have opposite charge to the $b$ particles: they are antiparticles of each other. But they have the same positive energy! (At least when energy is measured relative to the vacuum energy $E_0$, which I'll ignore hereafter.)

How do we relate this to Dirac's original formalism? What we want is a one-particle approximation, where we track a single particle without allowing it to be created or destroyed - this is, after all, what Dirac's wave function is describing. In developing this, one is tempted to associate the state $$ \Psi^{\dagger}(t,x)| 0 \rangle = |x \rangle $$ as analogous to the position-space wave function of a single electron, but this does not work, since it is not normalizable. The problem is the $\Psi$ does not annihilate the vacuum - it also creates an antiparticle.

The solution leading to Dirac's theory is to define an electron vacuum by the relation $$ \Psi(t,x) | 0 \mathbf{e} \rangle = 0. $$ Without thinking, let's first assume we can construct the state $|0 \mathbf{e}\rangle$ satisfying this. Then constructing $$ | \Psi_e \rangle = \int d^dx \, \psi_e(t,x) \Psi^{\dagger}(t,x) | 0 \mathbf{e} \rangle, $$ we have the normalization $\langle \Psi_e| \Psi_e \rangle = 1$. More importantly, when we consider the properties of the function $$ \psi_e(t,x) = \langle 0 \mathbf{e} | \Psi(t,x) | \Psi_e \rangle $$ under time evolution, we find $$ i \hbar \partial_t \psi_e(t,x) = \left[ - i \hbar c \, \vec{\alpha} \cdot \vec{\nabla} + \beta m c^2 \right] \psi_e(t,x). $$ This is precisely the Dirac's one-particle wave function!

But what is this electron vacuum $| 0 \mathbf{e}\rangle$? By looking at its definition, it needs to be annihilated by not just the $b_s(k)$, but also by $c^{\dagger}_s(k)$ for all $k$ and $s$: $$ \int \frac{d^d k}{(2 \pi)^d} \sum_s c^{\dagger}_s (k) | 0 \mathbf{e}\rangle= 0. $$ It is pretty intuitive how to construct such a state since $(c^{\dagger}_s)^2 = 0$: we just fill up the entire (real) vacuum up with an infinite sea of antiparticles! That is, $$ | 0 \mathbf{e}\rangle "=" \prod_{k,s} c^{\dagger}_s(k) |0 \rangle. $$

Once one knows that this somewhat pathological limit is the "vacuum" you are working with in the one-particle theory, all of the "paradoxes" you saw when you studied the Dirac equation disappear. By looking at the original QFT Hamiltonian, this state has an infinitely positive energy compared with the actual QFT vacuum, as well as infinite charge. The one-particle states you obtain by taking $\Psi^{\dagger}| 0 \mathbf{e}\rangle$ either add a single electron to this "Dirac sea" of antiparticles, increasing the energy, or they remove a single positron from the sea, which only gives a negative energy relative to the infinitely positive energy you've already added to the system. (When compared to the ground state of the original QFT Hamiltonian, the energy is still (infinitely) positive).

Note that the negative energy states you obtain in this theory should not be interpreted as antiparticles, since they carry the same charge as the $b$ particles. These states involve removing one of the antiparticles in the filled Dirac sea. If one wants to deal with an antiparticle wave function, you should define a "positron vacuum" by $\Psi^{\dagger}(t,x)| 0 \mathbf{p}\rangle = 0$ and work with that instead. (As an exercise, couple the U(1) Noether current of the original Dirac QFT to a gauge field, and keep track of the relative sign that $\psi_e$ and $\psi_p$ couple to it.)

In working in this approximation, one doesn't allow any transitions to states with a different number of particles. Also, operators will generically couple the one-electron and one-hole states as time evolves, even at low energies (you can think of annihilating the electron with one of the positrons in the Dirac sea). So the theory is necessarily approximate, and you should go back to the QFT for the full picture.

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Negative-energy particles are a pathology of a naive approach to relativistic quantum mechanics. If negative-energy antiparticles were to exist, there would be no ground state. This is because we could lower the energy of any state (including any alleged ground state) by adding more antiparticles. In Quantum Field Theory, models are designed so that all particles, including the antiparticles, have positive energy.

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    $\begingroup$ This is not an answer especially because the Dirac theory does predict negative energy for positrons. It is from experiment that it is clear that positrons in reality have positive energy. $\endgroup$
    – my2cts
    Nov 19, 2021 at 15:24
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    $\begingroup$ @my2cts Isn't that the answer's point? The question seems to be mixing two different theories. One is QFT, where the Dirac equation is the time-evolution equation for a field operator. The other is what the answer calls relativistic QM, where the the Dirac equation is the time-evolution equation for a state-vector. The answer correctly points out that those are two different theories, and only one of them predicts negative-energy particles. The question asks why they're not consistent with each other, and the answer is that they're different theories. $\endgroup$ Nov 19, 2021 at 16:41
  • $\begingroup$ @chiralanomaly You can disagree with for example Sakurai, but you can't call his textbook "a naive approach". This pathology is mature, respectable and wide spread and cannot be discarded so easily. For example, relativistic quantum chemistry is based on it and gives excellent results. In this sense, it dies not answer the question: discarding is not the answer. Note that I fully agree that negative energy is a mistake. $\endgroup$
    – my2cts
    Nov 19, 2021 at 17:49
  • $\begingroup$ @my2cts The answer doesn't say that relativistic QM is altogether naive, it just refers to "a naive approach to relativistic QM." Relativistic quantum chemists (and presmably Sakurai) use the Dirac equation correctly, by discarding the negative-energy part and then interpreting the remaining positive-energy part as the electron. They do that because they're not naive, and the answer doesn't say that they are. $\endgroup$ Nov 19, 2021 at 18:03
  • $\begingroup$ @chiralanomaly Also, discarding negative frequencies is probably not possible. I have to consult my expert friends to be sure. We should distinguish energy and frequency. I believe there are nistakes in Dirac theory, not in the equation, that have to be fixed. Negative energy is not naive, it is wrong. $\endgroup$
    – my2cts
    Nov 19, 2021 at 18:05

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