2
$\begingroup$

I am currently reading Shankar's "Bosonization: How to make it work for you in condensed matter" (http://inspirehep.net/record/408901/). In page 9, I am stuck with computing the correlation function in (39).

The part of the article is as follows:

enter image description here enter image description here enter image description here

My question is about deriving (40). Following the guide given in few lines below the article, I found out $$:e^A: = :e^{A^++A^-}:=e^{A^+}e^{A^-}$$ but how can I proceed to (40). Moreover, if I just assume (40), I cannot understand (42). Letting $A=i\beta\phi(x), \quad B=i\beta\phi(0)$ and using (40) gives $$e^{i\beta\phi(x)}+e^{-i\beta\phi(0)}=:ie^{i\beta(\phi(x)-\phi(0))}: \exp\left(\beta^2\langle \phi(x)\phi(0)-\frac{\phi(x)^2+\phi(0)^2}{2}\rangle\right).$$ How can I match this with (42)?

$\endgroup$
1
$\begingroup$

First of all there is a typo on the left-hand side of eq. (40) in Ref. 1. It should read $$T(e^Ae^B)~=~\ldots.\tag{40'}$$ Here the time-order $T$ is often not explicitly written in the notation. E.g. there is also an implicitly written time-order $$G_{\beta}~\equiv~\langle T\left(\ldots \right)\rangle\tag{39'}$$ inside the correlator function (39). Under the time-order symbol the operator ordering doesn't matter so we can write eq. (40') as $$ T(e^C)~=~:e^C:~ e^{\frac{1}{2}\langle C^2 \rangle},\tag{40"}$$ where $$C~=~A+B.$$ Eq. (40") is a standard form of Wick's theorem, see e.g. my related Phys.SE answer here.

References:

  1. R. Shankar, Bosonization: How to make it work for you in condensed matter, Acta Phys.Polon. B26 (1995) 1835-1867.
$\endgroup$
1
$\begingroup$

Equation 40 should read (a correct version can be found can be found in Shankars book) $$ e^Ae^B = :e^{A+B}: e^{\left\langle AB + \frac{1}{2}\left(A^2+B^2\right)\right\rangle} $$ The procedure is essentiall laid out in the paragraphs under eq.(40); given $A$ and $B$ commute with $[A,B]$, we have eq.(41) (this can be derived as a special case of the Baker-Campbell-Haausdorff formula) \begin{align} e^{A+B} &= e^A e^B e^{-\frac{1}{2}[A,B]}\\ &= e^B e^A e^{\frac{1}{2}[A,B]}\;. \end{align} We can now write $A = A^+ + A^-$ interms of creation and destruction parts (and similarly $B$) and use the above expression to reorder terms until we arrive at the normal ordered expression $:e^{A+B}:$ multiplied by a bunch of commutators.

Finally we note that \begin{align} \langle C D \rangle &= \langle (C^++C^-)(D^++D^-) \rangle\\ &=\langle C^-D^+\rangle \\ &= [C^-,D^+] \end{align} and use this to write the commutators in terms of the vacuum expectation values to obtain the final result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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