I'm going through an introduction to many-body theory and I am getting tripped up on the formalism. I understand quantities such as $\hat {N} = \sum_{i}\hat{n}_{i}=\sum_{i}\hat{a}_{i}^{\dagger}\hat{a}_{i}=\int d^{3}x\psi^{\dagger}(x)\psi(x)$ but struggling with things interpreting things like the kinetic energy of the system
Specifically, how is it that one goes from the creation/annihilation formalism to the field operators? If you have a general many-body Hamiltonian, I can't really see the transformation (nor do I have a good intuition for it) between the creation and annihilation operators and the field operators formalism for the Hamiltonian. Why do we have the two-particle interaction "sandwiched" between the field operators, but the annihilation/creation operators do not follow the same pattern?
I am aware of basic quantum mechanics, commutation rules, as well as the Fourier transform. I need help developing an intuition for writing down a field operator Hamiltonian. When I read the field operator Hamiltonian, the story that I get is: There are some field operators that create and annihilate, and integrating over them with a energy density yields a total energy term.
But I get lost in the details. For instance, although it has been simplified by IBP above, the kinetic energy term acts on the annihilation operator before the creation operator acts on it. What is the meaning of the motif $H=\int d^{3}x\psi^{\dagger}(\mathbf{x})\hat{h}\psi(\mathbf{x})$?
Best Answer
Concerning the bit about:
They do follow the same pattern, only when using field operators you are dealing with a continuous variable (i.e. $\textbf{x}$) which makes easier to write the relations in another way. To see this consider for example the kinetic energy operator $T$.
Discrete basis case:
Let $c_k^\dagger, c_k$ are the creation/annihilation operators of momentum eigenstates, and let $a_i^\dagger,a_i$ creation/annihilation operators for a generic (complete) set of states $\{ | i \rangle \}_i$. We have $$ \tag{D} T = \sum_{\textbf{k}} \frac{| \textbf{k} |^2}{2m} c_\textbf{k}^\dagger c_\textbf{k} = \sum_{i,j} t_{ij} a_i^\dagger a_j $$ In both cases you are kind of expanding $T$ in terms of its matrix elements between Fock states. The "kind of" here is important because these are not really expansions in terms of an orthonormal basis for a number of reasons, for example because $a_i^\dagger a_j$ are not projection operators. However, it is still true (as you can readily verify using the (anti)commutation relations) that $$ \langle \textbf{k} | T | \textbf{k}' \rangle \equiv \langle 0 | c_\textbf{k} T c_{\textbf{k}'}^\dagger | 0 \rangle = \delta_{\textbf{k}\textbf{k}'} \frac{ | \textbf{k} |^2}{2m}$$ $$ \langle i | T | j \rangle \equiv \langle 0 | a_i T a_j^\dagger | 0 \rangle = t_{ij}$$ Note that the "expanding $T$" interpretation starts to break down when you consider many-particle states, and that in these cases you start appreciating the difference between fermions and bosons. For example for two particle states you have for bosons: $$ \langle \textbf{k}\textbf{q} | T | \textbf{k}\textbf{q} \rangle \equiv \langle 0 | c_\textbf{q}c_\textbf{k} T c_\textbf{k}^\dagger c_\textbf{q}^\dagger | 0 \rangle = \left( \frac{| \textbf{k} |^2}{2m} + \frac{| \textbf{q} |^2}{2m} \right) (1+\delta_{\textbf{k}\textbf{q}})$$ while for fermions: $$ \langle \textbf{k}\textbf{q} | T | \textbf{k}\textbf{q} \rangle \equiv \langle 0 | c_\textbf{q}c_\textbf{k} T c_\textbf{k}^\dagger c_\textbf{q}^\dagger | 0 \rangle = \left( \frac{| \textbf{k} |^2}{2m} + \frac{| \textbf{q} |^2}{2m} \right) (1-\delta_{\textbf{k}\textbf{q}})$$
Continuous basis case
Using the field operators $\psi(\textbf{x})$ and $\psi^\dagger(\textbf{x})$ we write $T$ as: $$ \tag{C} T = \int d^3 x \psi^\dagger(\textbf{x}) \left (\frac{-1}{2m} \nabla^2 \right) \psi(\textbf{x}) $$ so how is this similar to something like (D)? To better see this lets assume we are in a 1D space (so $\textbf{x} \approx x_n$, $\psi(\textbf{x}) \approx \psi_n $) and switch to discrete varibles. We then have $$ \nabla^2 \psi(\textbf{x}) \approx ( \psi_{n+2} - 2 \psi_{n+1} + \psi_n ) $$ and using this $T$ becomes $$ \tag{R} T = \frac{-1}{2m} \sum_n \psi_n^\dagger ( \psi_{n+2} - 2 \psi_{n+1} + \psi_n ) = \sum_{n,m} \psi^\dagger_n T_{nm} \psi_m = \sum_{n,m} T_{nm} \psi^\dagger_n \psi_m $$ where we have defined $$ T_{nm} \equiv \frac{-1}{2m} ( \delta_{n,m+2} - 2 \delta_{n,m+1} + \delta_{n,m} ) $$ as you can see (R) is again a form like (D), and the kinetic operator is no longer "sandwiched" between field operators (but it has acquired a more cumbersome form).