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Consider a free Bosonic system. The Hamiltonian is given by $$ H=\sum_k \frac{k^2}{2m}a_k^\dagger a_k. $$ Since the spectrum is gapless, the ground state can be of any particle number (or even superposition of different particle numbers) as long as all particles are in state $\vert k=0 \rangle$.

Now consider a particular bosonic condensed state with $N$ particles $\vert\Phi_0\rangle=\frac{(a_{k=0}^\dagger\ \ )^N}{\sqrt{N!}}\vert vac\rangle$. What is the time-ordered density correlation function defined as the following? $$ \langle\Phi_0\vert \hat{T}[\rho(x,t)\rho(0,0)]\vert\Phi_0\rangle $$ (This is Problem 3.1.2 in Wen's book)

I have done the problem by brute force (by which I mean I spanned all the states and operators and then evaluate them one by one.) My result is: $$ \begin{eqnarray} \langle\Phi_0\vert \hat{T}[\rho(x,t)\rho(0,0)]\vert\Phi_0\rangle&=&\theta(t)\langle \Phi_0\vert e^{iHt}a^\dagger(x)a(x)e^{-iHt}a^\dagger(0)a(0)\vert\Phi_0\rangle\\\ &+&\theta(-t)\langle \Phi_0\vert a^\dagger(0)a(0)e^{iHt}a^\dagger(x)a(x)e^{-iHt}\vert\Phi_0\rangle\\\ &=&\frac{N(N-1)}{V^2}+\frac{N}{V}\int\frac{\mathrm{d}k}{2\pi}[\theta(t)\exp(ikx-it\frac{k^2}{2m})+\theta(-t)\exp(-ikx+it\frac{k^2}{2m})], \end{eqnarray} $$ which indeed shows the off-diagonal long-range order of the system.

The calculation is a little bit painful. But as implied in the book, it should be obtained with Wick's theorem. However, I have some troubles:

  1. The expectation value is evaluated in state $\Phi_0$ rather than vacuum. How do I use Wick's theorem in this case?
  2. The book says "$a(x,t)$ is a linear combination of $a_k$ and $a_k^\dagger$" but I failed to prove it. Is it even right? I mean, using a Fourier transformation, $a(x)$ should only be involved with $a_k$ but no $a_k^\dagger$, and with the time-evolution operator $a(x,t)$ is more than just linear terms of $a_k$ and $a_k^\dagger$.
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  • $\begingroup$ If you would like to use Wick theorem, it seems that it is convenient to use field-operators (see field operators of 2nd quantized system) and then find how density-density correlation function can be expressed in terms of fields operators. May be this step is not strictily required but for me it is simpler to work in terms of fields rather then ladder operators. Then, you will have time-ordering average and can use Wick theorem to obtain all the possible contractions of fields operators, this contractions will produce something like $(N/V)^2-G_{0,X}G_{X,0}$ where $G$ is Green function. $\endgroup$
    – Artem Alexandrov
    Commented Aug 29, 2019 at 22:04
  • $\begingroup$ @ArtemAlexandrov Thank you for comment. But I think for a Schrodinger field, $a(x), a^\dagger(x)$ is the same as $\phi(x), \phi^\dagger(x)$. And $\rho(x,t)$ is just $a^\dagger(x) a(x)$. We are not considering a Klein-Gordon field here. $\endgroup$
    – Prongs
    Commented Aug 29, 2019 at 22:38
  • $\begingroup$ Actually, what is bothering me is that $\vert \Phi_0\rangle$ is not the vacuum state so that $\langle \Phi_0 \vert : (\cdots) : \vert \Phi_0 \rangle$ cannot be simply taken as $0$. But pulling things out from that state gives $\langle vac \vert a\cdots a a(x,t)^\dagger a(x,t) a(0,0)^\dagger a(0,0) a^\dagger \cdots a^\dagger\vert vac\rangle$. And that is too many field operators to contract! $\endgroup$
    – Prongs
    Commented Aug 29, 2019 at 22:53

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I have figured this out myself... The crucial point is that the non-equal time commutation relation $[a(t), a(t')]=0$ for free Bosonic fields (the same for $a^\dagger$).

(In the context of high energy physics, people usually use field operators $\phi(x)$ which involves with both $a_k$ and $a_k^\dagger$. This is because the dispersion relation is given by $E^2=p^2+m^2$. However, in our case, we will consider the non-relativistic limit $E=\frac{p^2}{2m}$ thus the field operator is just $\psi(x)=a(x)$ which is only related to $a_k$ but not $a_k^\dagger$.)

Now comes to the Wick's theorem. Things are perfectly nice when we are dealing with the vacuum state. However, here we are considering a ground state with $N$ particles. Of course, one may pull them out as $\vert\Phi_0\rangle=\frac{(a_0^\dagger)^N}{\sqrt{N!}}\vert vac\rangle$, but that is too many operator to contract! Let's use Wick Theorem here: $$ \begin{eqnarray} \hat{T}[a^\dagger(x,t)a(x,t)a^\dagger(0,0)a(0,0)]&=&:a^\dagger(x,t)a(x,t)a^\dagger(0,0)a(0,0):\\ &+&:a^\dagger(x,t)a(0,0):\langle vac\vert \hat{T}a(x,t)a^\dagger(0,0)\vert vac\rangle\\ &+&:a(x,t)a^\dagger(0,0):\langle vac\vert \hat{T}a(0,0)a^\dagger(x,t)\vert vac\rangle\\ &+&\langle vac \vert \hat{T}a^\dagger(x,t)a(0,0)\vert vac\rangle\langle vac\vert\hat{T}a(x,t)a^\dagger(0,0)\vert vac\rangle. \end{eqnarray} $$ (Terms containing $\langle vac\vert\hat{T}a^\dagger(x,t)a(x,t)\vert vac\rangle$ and $\langle vac\vert\hat{T}a^\dagger(0,0)a(0,0)\vert vac\rangle$ vanishes).

The last term containing $\theta(t)\theta(-t)$ so that it vanishes as well. So we are left with three terms.

The second and the third are easy to compute. I won't do it here (it is the first question under this problem in the book). What was confusing me is the first term:

How do I normal ordering it?

Should I first evolve $a(0,0)$ to $a(0,t)$ to make them at the same time then do the normal ordering? This will be too much compicated. Or should I just put the $a^\dagger$'s to the left of $a$'s ? If so, shall it be $a^\dagger(x,t)a^\dagger(0,0)a(x,t)a(0,0)$ or $a^\dagger(0,0)a^\dagger(x,t)a(x,t)a(0,0)$ or $a^\dagger(x,t)a^\dagger(0,0)a(0,0)a(x,t)$, or ...? The point is, I didn't realise it that $[a(x,t),a(0,0)]=0$ , since the Bosons are free, so that the order doesn't actually matter.

It is easy to see in $k$-space, $$ [H,a_k]=-\frac{k^2}{2m}a_k\Longrightarrow e^{iHt}a_k e^{-iHt}=\exp(-i\frac{k^2}{2m}t)a_k, $$ that the evolution of free Bosonic operator only gives rise to a phase factor so that $$ [a_k(t),a(0)]=0 $$ for all $t$. (That is, non-equal-time commutation relation is trivial as the equal-time one. This is only true if the field is free.) And then performing a Fourier transformation shows that $[a(x,t),a(0,0)]=0$ as well. I myself think this is not that trivial but few books talk about it. They just ignore all the time label when doing normal-ordering. A good reference about this is David Tong's lecture notes on QFT, Page 37. Moreover, non-equal-time commutation relation for $a$ and $a^\dagger$ is just a $c$-number rather than an operator for a free field. (We don't need this point here)

I think this means that the Wich's theorem is only a pertubative tool, which is not exact if interaction is turned on.

Now evaluate the operators in $\vert \Phi_0\rangle$, yields $$ \begin{eqnarray} \langle\Phi_0\vert\hat{T}\rho(x,t)\rho(0,0)\vert \Phi_0\rangle&=&\langle\Phi_0\vert (a(x,t)a(0,0))^\dagger (a(x,t)a(0,0))\vert \Phi_0\rangle\\ &+&\langle \Phi_0\vert a^\dagger(x,t)a(0,0)\vert \Phi_0\rangle\langle vac\vert\hat{T}a(x,t)a^\dagger(0,0)\vert vac\rangle\\ &+&\langle \Phi_0\vert a(x,t)a^\dagger(0,0)\vert \Phi_0\rangle\langle vac\vert\hat{T}a^\dagger(x,t)a(0,0)\vert vac\rangle. \end{eqnarray} $$ The first term can be evaluated by some trick: $(a(x)a(0))^\dagger$ can be somehow regarded as a pair creation operator so that $(aa)^\dagger(aa)$ is a pair-number operator which count how many pairs are in the state. Now that we have got $N$ Bosons in $\vert \Phi_0\rangle$, for each Boson there are $N-1$ to pair thus the number of pairs is $N(N-1)$ and the pair density is given by $\frac{N(N-1)}{V^2}$.

And then we end up with the same result: $$ \begin{eqnarray} \langle\Phi_0\vert\hat{T}\rho(x,t)\rho(0,0)\vert \Phi_0\rangle&=& \frac{N(N-1)}{V^2}\\ &+&\frac{N}{V}\int\frac{\mathrm{d}k}{2\pi}[\theta(t)\exp(ikx-it\frac{k^2}{2m})\\ &+&\theta(-t)\exp(-ikx+it\frac{k^2}{2m})]. \end{eqnarray} $$

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