In second quantization we use Hamiltonian in form: $$H=\int d^3x [ \psi^{\dagger}(x) h \psi(x)],$$ where $h$ is Hamiltonian density. The field operators have following form: $$\psi = \sum\limits _{i} \phi(x) a(t),$$ $$\psi^{\dagger}=\sum\limits_{i}\phi^{*}(x)a^{\dagger}(t),$$ where $a$ and $a^{\dagger}$ are creation and annihilation operators. I want to try this formalism for Hamiltonian: $$H=\sum\limits_{i}{H}_{i}+\sum\limits_{i,j}V(\vec{r}_{i}-\vec{r}{_j}).$$ In the Mahan book, there is following expression: $$H=\sum\limits_{lm}a^{\dagger}_{l}a_{m}H_{lm}+\sum\limits_{lmij}a^{\dagger}_{j}a^{\dagger}_{m}a_{l}a_{i} \cdot \int d^3 r_{1}d^{3}r_{2}\phi^{*}_{m}(r_{1})\phi_{l}(r_{2})V(r_{1}-r_{2})\phi^{*}_{j}(r_{2})\phi_{i}(r_2).$$ I don't understand how he obtains this formula, especially why there are two integrals in the second term. Is there a simple way to get this form of Hamiltonian?
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1$\begingroup$ There's two integrals because the interaction potential depends on the position of the two particles. $\endgroup$– LagerbaerApr 10, 2013 at 19:07
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$\begingroup$ I heard this argument, but it's not convincing for me. If in the second term there are two integrals, they must be also in first term - and they are in matrix element $H_{lm}$, but why in first term we have only two operators rather than four? I will be very gratefull if somebody show me calculations which lead to appropriate result. $\endgroup$– SolitonApr 10, 2013 at 19:31
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1$\begingroup$ In the first line you assume that the Hamiltonian only contains 1-body operators. That is too restrictive. Interesting Hamiltonians, like the one you consider later, contain 2-body, 3-body etc operators. $\endgroup$– ThomasApr 10, 2013 at 23:31
1 Answer
$ \newcommand{\ket}[1]{\left| #1 \right\rangle} \newcommand{\Psih}{\hat{\Psi}} \newcommand{\Psihd}{\hat{\Psi}^\dagger} \newcommand{\bx}{\mathbf{x}} \newcommand{\Hh}{\hat{H}} \newcommand{\Hsp}{\Hh_{\mathrm{sp}}} $
The short answer is that the "Hamiltonian density" of your question for a system with two-body interactions is $\hat{h}=-\frac{\hbar^2\nabla^2}{2m} + U(\bx) + \frac12 \int d \bx' \Psihd(\bx') V(\bx-\bx')\Psih(\bx')$. That is where the other integral comes from. The other integral is there because the two-body Hamiltonian depends on the position of two particles (as @Lagerbaer says in a comment).
The Hamiltonian is then $\Hh =\int d\bx \Psihd(\bx)\hat{h}\Psih(\bx) = \Hsp + \hat{V}$ where
$$\begin{align}
\Hsp &= \int d\bx \Psihd(\bx) \left[-\frac{\hbar^2\nabla^2}{2m} + U(\bx)\right]\Psih(\bx)\\
\hat{V} &= \frac12 \int d\bx d \bx' \Psihd(\bx) \Psihd(\bx') V(\bx-\bx')\Psih(\bx')\Psih(\bx)
\end{align}$$
We can think of this as an ansatz of quantum field theory, to be tested by experiment, but we can check that it agrees with ordinary quantum mechanics (below for the two-body interaction term). The best approach is to satisfy yourself that this makes sense once and for all, then use it as your starting point. From $\Hh$, it is easy to get your final Hamiltonian by substituting in your basis expansion of the field operators.
Now, I'll show that applying $\hat{V}$ to an $n$-body ket $$\ket{\chi_n} = \int d\bx_1 \dots d\bx_n \chi(\bx_1,\dots\bx_n) \prod_k \Psihd(\bx_k)\ket{0}$$ is equivalent to the interaction $\frac12 \sum_{i,j} V(\bx_i-\bx_j)\chi(\bx_1,\dots\bx_n)$ you give in your question (note that in your question, you are missing a factor of a half, see Mahan Eq. 1.134). The following is given in a slightly different form in Greiner, Quantum Mechanics Special Chapters, Ex. 6.1.
We use the commutation relations $[\Psih(\bx),\Psihd(\bx')]_\pm = \delta(\bx-\bx')$ to move the two annihilation operators in $\hat{V}\ket{\chi_n}$ to $\ket{0}$, then we move the creation operators $\Psihd(\bx)$ and $\Psihd(\bx')$ to fill the gaps in $\ket{\chi_n}$. Each time we commute, we pick up a term $\delta(\bx-\bx_i)$ and for fermions we flip the sign of the rest, until finally we can annihilate the vacuum. We find, progressively \begin{align} \Psih(\bx)\prod_k \Psihd(\bx_k)\ket{0} &= \sum_i (\pm 1)^{i-1}\delta(\bx-\bx_i) \prod_{k:k \ne i} \Psihd(\bx_k)\ket{0} \\ \Psihd(\bx')\Psih(\bx')\Psih(\bx)\prod_k \Psihd(\bx_k)\ket{0} &= \sum_{i,j:i\ne j} (\pm 1)^{i-1}\delta(\bx-\bx_i) \delta(\bx'-\bx_j)\prod_{k:k\ne i} \Psihd(\bx_k)\ket{0}\\ \Psihd(\bx) \Psihd(\bx')\Psih(\bx')\Psih(\bx)\prod_k \Psihd(\bx_k)\ket{0} &= \sum_{i,j:i\ne j}\delta(\bx-\bx_i) \delta(\bx'-\bx_j)\prod_{k} \Psihd(\bx_k)\ket{0} \end{align} The middle equation above follows because however many steps it takes $\Psih(\bx')$ to get to $\bx_j$, $\Psihd(\bx')$ takes the same number of steps, so the factors of $\pm 1$ cancel.
The last equation above is the long answer to your question: there are two particles (two pairs of field operators) leading to two Dirac-delta functions, so we're going to need two integrals to reproduce ordinary quantum mechanics. The similar derivation for the single particle Hamiltonian, $\Hsp\ket{\chi_n}$, involves only one pair of field operators, so only needs one integral.
So,
\begin{align*} \hat{V}\ket{\chi_n} &= \frac12 \int d\bx d \bx'd\bx_1 \dots d\bx_n V(\bx-\bx')\chi(\bx_1,\dots\bx_n)\Psihd(\bx) \Psihd(\bx') \Psih(\bx')\Psih(\bx)\prod_k \Psihd(\bx_k)\ket{0} \\ &= \frac12 \int d\bx d \bx'd\bx_1 \dots d\bx_n V(\bx-\bx') \sum_{i,j:i\ne j}\delta(\bx-\bx_i) \delta(\bx'-\bx_j) \chi(\bx_1,\dots\bx_n)\prod_k \Psihd(\bx_k)\ket{0}\\ &= \frac12 \int d\bx_1 \dots d\bx_n \sum_{i,j:i\ne j} V(\bx_i-\bx_j) \chi(\bx_1,\dots\bx_n)\prod_k \Psihd(\bx_k)\ket{0}\\ \end{align*} which is the interaction we expect.
