I'm very new to QFT so I'm sorry if this is really obvious.
I've read online that one of the defining features that sets QFT apart from previous quantum theories is that it accounts for particle pair creation and annihilation via creation/annihilation operators.
However many of the explanations of these get very bogged down in maths, so I'm wondering if anyone could offer a more physical explanation of how QFT deals with creation and annihilation.
In usual quantum mechanics the Hilbert space $\mathcal{H}$ is the space of states for a single particle. Then if you want to introduce multiple particles into the game you take states which are tensor products of single particle states, appropriately (anti)-symmetrized. So for $n$ particles you'd have $\mathcal{H}_n = \mathcal{H}\otimes\cdots \otimes \mathcal{H}$ $n$ times. If you want to put together all Hilbert spaces of any number of particles, you'd have to take the direct sum of those $\mathcal{H}_0 \oplus \mathcal{H}_1\oplus \mathcal{H}_2\cdots$.
QFT has a completely different paradigm. The Hilbert space $\mathcal{H}$ is the space of states for a field. And you have only one field for each type of particle in the universe. One photon field, one Higgs field and so on. Then having $n$ particles is just a particular state of the field. So you can talk about creation and annihilation because you can study transition amplitudes such as $$ \langle n\;\mathrm{particles}|e^{iHt}|n'\;\mathrm{particles}\rangle\,. $$ Also, you don't have to think of a field as an object having a fixed number of particles. You can also have superpositions in the particle number. The states that are typically prepared in experiments do have a fixed number of particles, but it's not a mathematical necessity.
How is QFT able to pull this off? This is easily seen in the free particle case. In that case the Hilbert space is the tensor product of one harmonic oscillator Hilbert space for each single particle state $$ \mathcal{H} = \mathcal{H}_{p_1} \otimes \mathcal{H}_{p_2} \otimes \cdots\,. $$ These are Hilbert spaces of a harmonic oscillator with frequency $p_i$. This is completely different from $\mathcal{H}_n$ of before: each $\mathcal{H}_{p}$ is not a single particle Hilbert space. It is the Hilbert space of excitations of frequency $p$, but we can have arbitrarily many of those (well, at most one if they are fermions). So already a single factor allows for particle creation/annihilation. The level of the harmonic oscillator is the particle number.
This is called second quantization. And, in very reductive terms, all we did is just rewriting a sum as a product, a bit like $\sum_{n_i} \prod_i a_{n_i,i} = \prod_i\sum_n a_{n,i}$ $$ \begin{aligned} \bigoplus_{n_1,n_2,\ldots=0}^\infty \bigotimes_{i=1}^\infty \mathcal{H}_{\mbox{state $p_i$, $n_i$ particles}} &= \bigotimes_{i=1}^\infty \bigoplus_{n=0}^\infty \mathcal{H}_{\mbox{state $p_i$, $n$ particles}}\,, \end{aligned} $$ and noticed that $\bigoplus_{n=0}^\infty \mathcal{H}_{\mbox{state $p_i$, $n$ particles}} \equiv \mathcal{H}_{p_i}$ for a free field is an harmonic oscillator. If this comment is not clear to you, ignore it.
Then the states are built with the $a_{p_i}$ and $a_{p_i}^\dagger$ operators of each $\mathcal{H}_{p_i}$. So $$ \Psi \equiv \left|\mbox{state with $n_1$ particles at state $p_1$, $n_2$ particles at state $p_2$}\ldots\right\rangle \,, $$ is defined as $$ \Psi = (a_{p_1}^\dagger)^{n_1}(a_{p_2}^\dagger)^{n_2}\cdots |\mbox{state with no particles at all}\rangle\,. $$ And the state with no particles at all we usually call it $|0\rangle$. It is the ground state of the field.
As the comment asked: here's how we can see that $e^{iHt}$ can change the number of particles. It's easy: $H$ may and will contain interaction terms which, when written in terms of $a$ and $a^\dagger$ operators aquire the form that I am going to write below. Let us denote as $a,b,c,\cdots$ the annihilation operators of the particles of species $a,b,c,\cdots$. For instance $a$ may be photons, $b$ electrons and so on. Then $H$ contains terms $$ H \supset (a^\dagger)^{n_a} (b^\dagger)^{n_b} (c^\dagger)^{n_c}\cdots a^{m_a}b^{m_b}c^{m_c}\cdots\,. $$ This term creates $n_a$ particles of type $a$ (and so on) and destroys $m_a$ particles of type $a$ (and so on). So you can have processes that destroy a bunch of particles and create a whole new set of different ones. When you exponentiate $H$ you can get even more interactions. For concreteness, in electroweak there is a term $$ a b^\dagger \bar{c}^\dagger\,, $$ which destroys a $W$ boson ($a$) and creates an electron ($b$) and an antineutrino ($\bar{c}$). This gives rise to the $W$ decay.
