There appears to be an apparent dichotomy between the interpretation of second quantized operators in condensed matter and quantum field theory proper. For example, if we look at Peskin and Schroeder, the Hamiltonian for the quantized Dirac field Hamiltonian is given by eqn. 3.104 on p. 58:
$$H =\int \frac{d^3 p}{(2\pi)^3}\sum_s E_p (a_p^{s\dagger}a_p^s+b_p^{s\dagger}b_p^s). \tag{3.104}$$
Here, $a_p^{s\dagger}$ and $a_p^s$ are the creation and annihilation operators for fermions, while $b_{p}^{s\dagger}$ and $b_p^s$ are the creation and annihilation operators for anti-fermions. Note that particles and antiparticles are described by the same field, which we will call $\psi$. Also note that these operators satisfy the relations
$$a_p^{s\dagger}=b_{p^*}^s,\qquad a_p^s=b_{p^*}^{s\dagger}$$
In condensed matter notation, we might similarly have creation and annihilation operators for some many-body state, which we will call $c_i^\dagger$ and $c_i$ (we consider spinless fermions for simplicity). In numerous texts and in my classes, I have learned that $c_i$ might be interpreted as the creation operator of a hole, which might be thought of as the condensed matter equivalent of an antiparticle.
Why do we need separate creation and annihilation operators for fermions and anti-fermions in quantum field theory, but in condensed matter we can simply treat the annihilation operator as the creation operator of the "antiparticle"? It is my understanding that there is a fundamental difference between the Dirac field second quantized operators and the condensed matter second quantization operators, but I can't find any references that discuss this. Any explanation would be appreciated, as well as any references at the level of Peskin and Schroeder.
Best Answer
It's just a convention to write the Dirac Hamiltonian in terms of electron and positron operators. The operator $b_p^{s\dagger}$ creates a "hole", i.e. a positron, which is the same as annihilating an electron. So we could just as well define $c_p^s = b_p^{s\dagger}$ and write the Hamiltonian in terms of $a$ and $c$ instead of $a$ and $b$. Then everything is expressed in terms of electron operators.
So we are reduced to asking why you need two sets of operators $a$ and $c$. The reason is simply that the dispersion relation of the Dirac equation is $E = \pm \sqrt{p^2 + m^2}$, so at any given momentum there are two bands. Hence, you need one operator that creates/annihilates an electron in the top band, and one which creates/annihilates an electron in the bottom band. This would equally well be true in a condensed matter system treated in momentum space, if there are multiple bands.
(As for why this convention is chosen: in relativistic quantum theory, if you write everything in terms of electrons, you are forced to conclude that the vacuum $|0\rangle$ has all the negative-energy states occupied, since $c_p^{s\dagger}|0\rangle = 0$. This was, in fact, Dirac's original picture, but conceptually it's easier to imagine that the vacuum contains no particles. On the other hand, in condensed matter physics, we are used to thinking about our solid-state materials as containing a lot of electrons, so it's actually less confusing to take the opposite point of view.)