I am currently trying to understand section 3.3 from this article, about how to use second quantization techniques in statistical mechanics. Here I have the creation and annihilation operators defined as:

$$ \hat a \left |n \right > = n \left |n-1 \right >$$ $$ \hat a ^\dagger \left |n \right > = \left |n+1 \right >$$ $$ \langle n | m\rangle = m! \delta _{nm} $$

The definition is slightly different from the usual in quantum mechanics because comes from the quantization of a classical system in a lattice (see the article for more details). The coherent states are then defined by:

$$ \left | \alpha \right > = \exp(-|\alpha|^2/2+ \alpha\hat a ^\dagger ) \left |0 \right > = \exp(-|\alpha|^2/2) \sum_{n=0} ^{+\infty} \dfrac{\alpha ^n}{n!} \left |n \right >$$

Question: I want to demonstrate the formula (26) of the article, but I don't know how. I translate it here in a more easy notation:

$$ \langle \alpha_1 |\exp(-\lambda H)| \alpha_2 \rangle = \langle \alpha_1| \alpha_2 \rangle \exp(-\lambda\langle \alpha_1 |H| \alpha_2 \rangle) $$

which should be valid as an expansion for $\lambda \rightarrow 0$.

My attempt at the solution:

First I expand the exponential for low $\lambda$,

$$\langle \alpha_1 |\exp(-\lambda H)| \alpha_2 \rangle = \langle \alpha_1| \alpha_2 \rangle -\lambda \langle \alpha_1 |H| \alpha_2 \rangle + \lambda^2 \dfrac{\langle \alpha_1 |H^2| \alpha_2 \rangle}{2!} - \ldots$$

Then we have to work term by term. If, as the text same, there is constant number of particles, then I may assume that the Hamiltonian is a polynomial in number operators, $H=\sum_k H_k \hat n^k$. Now we work the first term, using the definition of the coherent states:

$$ \langle \alpha_1 |H| \alpha_2 \rangle = \exp(-|\alpha_1|^2/2-|\alpha_2|^2/2) \sum_{nm} \dfrac{\left ( \alpha_1 ^{*} \right )^n \alpha_2^m}{n!m!} \langle n|H| m\rangle$$

However, when I include the expression of the Hamiltonian to compute the matrix element, the result is

$$\langle n|H| m\rangle = \sum_k H_k n^k \delta_{nm} m!$$

leading to:

$$\langle \alpha_1 |H| \alpha_2 \rangle = \exp(-|\alpha_1|^2/2-|\alpha_2|^2/2) \sum_{nmk} \dfrac{\left ( \alpha_1 ^{*} \alpha_2\right )^n}{n!} H_k n^k$$

And I cannot write this as an scalar product between the two states $\alpha_1$ and $\alpha_2$. I should extract this as a common factor to recover the exponential from the series... I think I am missing something important, but I don't get it.

Thank you!


1 Answer 1


The authors made an error in the definition of the function (which is called the operator lower symbol) $H(\alpha_1 , \alpha_2 )$ (in your notation), in the line following equation (26). This error did not formally influence their further work, because they did not return to their original definition and they used the symbol as a classical Hamiltonian in the path integral.

The correct equations in your notation should read: $$ \langle \alpha_1 |\exp(-\lambda \hat{H})| \alpha_2 \rangle = \langle \alpha_1| \alpha_2 \rangle \exp(-\lambda H(\alpha_1 , \alpha_2 )) $$ With $$H(\alpha_1 , \alpha_2 ) = \frac{ \langle \alpha_1 |H| \alpha_2 \rangle}{\langle \alpha_1| \alpha_2 \rangle }$$ You can see now that in the first order of $\lambda = \Delta t$, both sides of the equation are equal.

Having said that, and without contesting the authors results which I am aware are important according to the number of citations they received; the method they chose to construct the coherent state path integral, which is very popular in the physics literature, is not entirely correct.

The reason is as follows: As we checked, equation (26) is correct to the first order of $\lambda$, but there is already a difference between the right and left sides in the second order of $\lambda$. It is true that $\lambda$ is taken to zero, but at the same time the number of elements in the product is made infinite. The fact is that the combined effect results in a non-vanishing contribution which amounts changing the lower symbol given in the second equation to what is called an upper symbol implicitly defined by:

$$\hat{H} = \int \mathcal{H}(\alpha, \alpha) |\alpha\rangle\langle \alpha| e^{-|\alpha|^2} d\alpha d\bar{\alpha}$$

Where $ \mathcal{H}$ is the upper symbol and it is much harder to compute than the lower symbol.

If you are interested in the derivation of this result, please ask and I can add it to the answer.

  • $\begingroup$ If you take the effort to put the derivation of the result, it would be very helpful, thanks! $\endgroup$ Jul 16, 2017 at 19:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.