# Writing down many particle Hamiltonian

We are given that

\begin{align}\mathrm{tr} e^{-\frac{i}{\hbar}\hat{H}t}&= \int D[a_1,\dots,a_n]\times\\&\qquad\exp\left[\int_0^t dt' \left(\frac{1}{2}\sum_j (a_j(t')-a_j^*(t'))(\dot{a}_j(t')-\dot{a}^*_j(t'))- \\ \frac{i}{\hbar}H(a_1(t'),a_2(t'),\dots,a_1^*(t'),a_2^*(t')\,\dots)\right) \right].\end{align}

and asked to write down the path integral for $\mathrm{tr}e^{-\frac{i}{\hbar}\hat{H}t}$ for the many particle Hamiltonian $$\hat{H}=\sum_{i,j}h(i,j)a_i^{\dagger}a_j+\frac{1}{2}\sum_{i,j}U_{\mathrm{int}}(i,j)a_i^{\dagger}a_j^{\dagger}a_ja_i.$$

I cannot see the connection between these two expressions, particularly with respect to the $a_i,a_i^{\dagger}$'s and the $a_j(t'),a_j^*(t')$'s

I cannot see how to make the connection between these two expressions taken as a total. I cannot see the justification for just swapping them directly and it seems like there is more to this.

• What do you mean "pass between" the two? – Kyle Kanos Mar 11 '15 at 16:09
• How to go from $a_j$'s and $a_j^*$'s to $a_i^{\dagger}$'s and $a_i$'s. – Trajan Mar 11 '15 at 16:10
• So your question is real what is the relationship between $a,a^*$ and $a,a^\dagger$? – Kyle Kanos Mar 11 '15 at 16:11
• @KyleKanos I believe my edit makes this a better question – Trajan Mar 11 '15 at 18:07
• That is a better question, I think. I added an align case to make the equation look slightly better, but I think such a large equation cannot help but be this way! – Kyle Kanos Mar 11 '15 at 18:17

I can give an outline of the arguments that show how to arrive at the path integral representation of the expression $\text{tr}\, e^{-i \hat{H}t}$. For simplicity I set $\hbar=1$ and only consider the 1-particle case. The multi-particle case is technically more demanding but conceptionally the same. So the Hamiltonian I consider is simply $$\hat{H}=h a^\dagger a+ U a^\dagger a^\dagger a a.$$
The first step is to see that there is an overcomplete set of eigenstates of the annihilation operators $a$ called the Coherent States $|\alpha\rangle$. They satisfy $$a|\alpha\rangle=\alpha|\alpha\rangle \qquad\text{and}\qquad \mathbb{I}=\frac{1}{\pi}\int d\alpha|\alpha\rangle\langle\alpha|$$ where $\alpha$ is a complex number and the integral is over all complex values for $\alpha$. Another crucial property is the scalar product $$\langle\beta|\alpha\rangle=e^{-\frac{1}{2}(|\beta|^2+|\alpha|^2-2\beta^*\alpha)}.$$
Given these eigenstates of the annihilation operators, we can perform the usual steps in finding a path integral representation of a transition amplitude: \begin{align} \text{tr}\, e^{-i \hat{H}t} &= \frac{1}{\pi}\int d\alpha_1\,\langle\alpha_1|e^{-i \hat{H}t}|\alpha_1\rangle\\ &=\frac{1}{\pi^N}\int d\alpha_1\ldots d\alpha_N\,\langle\alpha_1|e^{-\frac{i \hat{H}t}{N}}|\alpha_2\rangle \langle\alpha_2|e^{-\frac{i \hat{H}t}{N}}|\alpha_3\rangle\langle\alpha_3|\ldots|\alpha_N\rangle \langle\alpha_N|e^{-\frac{i \hat{H}t}{N}}|\alpha_1\rangle \end{align} where we have introduced $N$ sets of full sets of eigenstates at intermediate times dividing the time $t$ into $N$ segments. In the end, we will want to take the limit of $N\to\infty$.
Now come the step where the operators $a$ and $a\dagger$ vanish. They are just replaced by numbers as they are evaluated on the inserted eigenstates. Note that $a|\alpha\rangle=\alpha|\alpha\rangle$ implies $\langle\alpha|a^\dagger=\langle\alpha|\alpha^*$. Therefore, \begin{align} \langle\alpha_1|e^{-\frac{i \hat{H}t}{N}}|\alpha_2\rangle &= \langle\alpha_1|e^{-\frac{it}{N}(h a^\dagger a+ U a^\dagger a^\dagger a a)}|\alpha_2\rangle\\ &= \langle\alpha_1|\alpha_2\rangle e^{-\frac{it}{N}(h \alpha_1^* \alpha_2+ U \alpha_1^* \alpha_1^* \alpha_2 \alpha_2)} \end{align}
$\langle\alpha_1|\alpha_2\rangle$ can be written according to the scalar product given above. In the limit $N\to\infty$ one writes $$\lim_{N\to\infty}\frac{1}{\pi^N}\int d\alpha_1\ldots d\alpha_N=\int Da$$ The discrete $\alpha_1$, $\alpha_2$, $\ldots$, $\alpha_N$ become a function, which we call $a(t)$. The scalar products $\langle\alpha_{i}|\alpha_{i+1}\rangle$, that appear from each of the inserted sets of eigenstates gives rise to the first line in your expression for the path integral. The remaining exponential expressions just give the Hamiltonian, but now simply as a function of $a(t)$ rather than an operator expression.