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We’ve started studying path integrals and perturbative expansion.

We wrote the action as $S[x]= S_0[x] +S_{int}[x]$ where the first term is the action for the model which we can solve exactly, while the represent the perturbative part, due to interaction (example: anharmonic oscillator, $S_0$: harmonic oscillator, $S_{int}$: terms of order $>2$.

The corresponding path integral(generating functional of correlation functions) $$Z[J]=\int Dx \exp{\{iS_{int}[x]\}} \exp{\{iS_0[x]+i\int dt J(t)x(t)\}}. $$ Or in an alternative way $$Z[J]=\exp{\{iS_{int}[-i\frac{\delta}{\delta J}]\}}Z_0[J]$$ Where $\frac{\delta}{\delta J}$ is the functional derivative and the exponential term is interpreted as a differential operator acting on the generating functional of the free theory.

But what is that operator? How did it act? My wind guess is something like the exponential of a matrix. Can an operator be the argument of a functional?

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    $\begingroup$ It makes sense perturbatively. $\endgroup$ – Qmechanic Jan 23 at 0:11
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It might help to consider a scalar example to start. If you are familiar with statistical mechanics, consider the partition function of a system that consists of three spin 1/2 particles (so there are 8 states total), and let us suppose that the Hamiltonian can be written \begin{align*} H &= -h\sum_{i=0}^2\sigma_i-J\sum_{i=0}^2\sigma_i\sigma_{\overline{i+1}},\quad \overline{i}\equiv i\mod 3\\ &\equiv H_0(\sigma_0,\sigma_1,\sigma_2) +H_{int}(\sigma_0,\sigma_1,\sigma_2) \end{align*} The partition function, $Z(h,J) = \sum_{\sigma_0,\sigma_1,\sigma_2}e^{- H(\sigma_0,\sigma_1,\sigma_2)}$ can be computed directly: \begin{align*} Z(h,J) = e^{3(h + J)} + 3e^{(h-J)} + 3e^{-(h-J)}+e^{-3(h+J)} \end{align*}

Recall that in quantum mechanics, the derivative $-i\partial_x$ (often used to represent the momentum operator) can be viewed as a generator of translations: $\exp(-m\partial_x)\Psi(x) = \Psi(x-m)$. (This is essentially because $f(x+h)\approx f(x)+h\nabla f(x)$.) The exact same principles may be applied in the example above: to evaluate $Z(h,J-m)$, simply apply the operator $\exp(-m\partial_J)$ to $Z(h,J)$.

Now let's explore various ways of interpreting $\exp(-m\partial_J)$, or $\partial_J$ even more simply. Just to give a small taste the full set of available approaches (especially in lattice models), we'll start with a somewhat tangential example.

First, notice in \begin{align*} \partial_JZ(h,J) &= -\sum_\sigma M_{int}(\sigma)e^{-hM_0(\sigma) - JM_{int}(\sigma)}\\ &=\sum_\sigma (\sigma_0\sigma_1 + \sigma_1\sigma_2 + \sigma_2\sigma_0)e^{-h\sum_{i=0}^2\sigma_i-J\sum_{i=0}^2\sigma_i\sigma_{\overline{i+1}}} \end{align*} that the expression $\sigma_0\sigma_1+\sigma_1\sigma_2+\sigma_2\sigma_0$ would match $\frac{1}{2}(\sigma_0 + \sigma_1 + \sigma_2)^2$ if it weren't for the extra $\frac{1}{2}(\sigma_0^2+\sigma_1^2+\sigma_2^2)$ term. In fact, because the $\sigma_i$ are limited to $\{-1, 1\}$, the last expression ($\frac{1}{2}\sum_i\sigma_i^2$) is just a constant; $\frac{3}{2}$ in this case. More generally, however, we could introduce a new term to the Hamiltonian, $-P\sum_{i}\sigma_i^2\equiv P M_{sq}(\sigma)$, so that $M_{int}(\sigma) = \frac{1}{2}(M_0(\sigma)^2 - M_{sq}(\sigma))$. Hence, \begin{align*} \partial_J Z(h,J) = -\sum_\sigma \frac{1}{2}(M_0(\sigma)^2 - M_{sq}(\sigma))e^{-hM_0(\sigma)-JM_{int}(\sigma)-PM_{sq}(\sigma)} \end{align*} (where the right hand side is understood as being evaluated at $P = 0$.) Notice that $-M_{sq}(\sigma)e^{-PM_{sq}(\sigma)}$ can be replaced with $\partial_P e^{-PM_{sq}(\sigma)}$, and $M_0(\sigma)^2 e^{-hM_0(\sigma)}$ can be replaced with $\partial_h^2 e^{-hM_0(\sigma)}$. We have just derived the relation \begin{align*} \partial_J Z(h,J) = -\frac{1}{2}\partial_h^2 Z(h,J,P) - \partial_P Z(h,J,P)\Big|_{P=0} \end{align*}

More possibilities present themselves if we allow the parameters $h$, $J$, $P$ etc. to assume different values for each spin (for example, the spins might be impurities positioned variously in a solid, and form an irregular triangle so that pairwise neighbor couplings are distinct.) In this case, the Hamiltonian would read

\begin{align*} H(\sigma) = -\sum_{i=0}^2(h_i \sigma_i - J_i \sigma_i\sigma_{\overline{i+1}}) \end{align*}

with the partition function modified appropriately.

Instead of having to introduce an auxiliary $M_{sq}(\sigma)$ term, we can now simply write \begin{align*} \sigma_0\sigma_1 e^{-\sum_i h_i\sigma_i} = \partial_{h_0}\partial_{h_1} e^{-\sum_i h_i\sigma_i} \end{align*} and \begin{align*} \partial_J Z(h,J) = \sum_{\sigma} \sum_{i=0}^2 \partial_{h_i} \partial_{h_{\overline{i+1}}} e^{-H_0(\sigma, h) - H_{int}(\sigma, J)}. \end{align*} (where $h$ is now understood as a three component vector or tuple.)

You may recognize $\sum_{i=0}^2\partial_{h_i}\partial_{h_{\overline{i+1}}}$ as $M_{int}(\partial_h)$, and moreover that $M_{int}(\partial_h)$ is independent of the spin degrees of freedom and can be moved outside the sum: written explicitly, \begin{align*} \partial_J Z(h_0, h_1, h_2, J) = -M_{int}(\partial_{h_0}, \partial_{h_1}, \partial_{h_2}) Z(h_0, h_1, h_2, J). \end{align*}

Finally, having established that the linear differential operators $\partial_J$ and $M_{int}(\partial_h)$ are equivalent, we can substitute one for the other in the exponential: \begin{align*} \exp(-m\partial_J) = \exp(-m M_{int}(\partial_h)), \end{align*} and hence, \begin{align*} Z(h,J-m) = \exp(-m M_{int}(\partial_h)) Z(h,J). \end{align*}

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