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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,

$$ H=\sum_{i,j}(t_{ij}+U_{ij})\hat{a}_{i}^{\dagger}\hat{a}_{i}+\frac{1}{2}\sum_{i,j,k,m}\left\langle i,j \right | f^{(2)} \left| k,m \right\rangle\hat{a}_{i}^{\dagger}\hat{a}_{j}^{\dagger}\hat{a}_{m}\hat{a}_{k} $$

Where we have for the total kinetic energy and the two-particle interaction term

$$ T=\sum_{i,j}t_{ij}\hat{a}_{i}^{\dagger}\hat{a}_{j} \\ t_{ij}=\left\langle i \right | t \left| j \right\rangle \\ F= \frac{1}{2}\sum_{\alpha \neq \beta}f^{(2)}(\mathbf{x}_{\alpha},\mathbf{x}_{\beta}) $$

In terms of field operators, the Hamiltonian is given as

$$ H=\int d^{3}x\left(\frac{\hbar^{2}}{2m}\nabla\psi^{\dagger}(\mathbf{x})\nabla\psi(\mathbf{x})+U(\mathbf{x})\psi^{\dagger}(\mathbf{x})\psi(\mathbf{x})\right)+\frac{1}{2}\int d^{3}xd^{3}x'\psi^{\dagger}(\mathbf{x})\psi^{\dagger}(\mathbf{x}')V(\mathbf{x},\mathbf{x}')\psi(\mathbf{x}')\psi(\mathbf{x}) $$

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 $H=\int d^{3}x\psi^{\dagger}(\mathbf{x})\hat{h}\psi(\mathbf{x})$?

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Sorry but you should study basics of quantum mechanics. For example, the order of operators is the most important new things that matters in quantum mechanics. $xp$ isn't the same thing as $px$ - this fact is known as the uncertainty principle and it is the cornerstone of quantum mechanics and any theory obeying its postulates. The same holds for $aa^\dagger$ and $a^\dagger a$ or any product of operators. The operators on the right simply act on the ket vectors before those on the left and the order matters because they don't commute. –  Luboš Motl Jan 25 at 8:00
    
The transformation between $\psi(x)$ and $a_i$ should be written in any textbook you are learning these things from. It is something like $\psi(x) = \sum_k a_k \exp(ikx)$ and/or $a_k = \int \exp(-ikx) \psi(x)$ with a lots of extra factors, similar formulae for daggers, and so on. Mostly, the transformation is a Fourier transformation if you use the momentum-based indices $i$ for $a_i$. For other identification of the states annihilated by $a_i$, the transformations are different and reflect the transformation of the one-particle Hilbert space's bases. –  Luboš Motl Jan 25 at 8:03
    
Thanks--perhaps I could revise my question and note my understanding better. Edited to add background –  2πf Jan 25 at 19:04
    
I don't really understand what you don't understand. Have you read about how to go from 1st quantization to 2nd quantization ? Concerning the sandwiching, of the two-body term $V(x-x')$, this not necessary, because $V$ is a c-number. You can put all creation/annihilation fields on the right or on the left of $V$ (if you keep the order of the operators). –  Adam Jan 25 at 19:28
    
I've read about the procedure to go from 1st quantization to 2nd. I was really hoping that $V(x-x')$ was a c-number. Thanks for clearing that up. But what is the significance of the order when it is not (such as in the kinetic energy term)? I notice it is similar in form to an expectation value in 1st quant.--is there any analogy I can draw from this? –  2πf Jan 25 at 20:24

1 Answer 1

Concerning the bit about:

Why do we have the two-particle interaction "sandwiched" between the field operators, but the annihilation/creation operators do not follow the same pattern?

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).

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