How to compute observables from the boson field operator?

I think I understand that if given the two boson wavefunction of two different states \begin{align} \Psi(\boldsymbol{r}_1,\boldsymbol{r}_2) = \dfrac{ \psi_1(\boldsymbol{r}_1)\psi_2(\boldsymbol{r}_2) + \psi_1(\boldsymbol{r}_2)\psi_2(\boldsymbol{r}_1) } {\sqrt{2}} \end{align} the expectation value of an operator $\hat{Q}$ for $\Psi(\boldsymbol{r}_1,\boldsymbol{r}_2)$ is \begin{align} \langle \Psi | \hat{Q} | \Psi \rangle = \langle \psi_1(\boldsymbol{r}_1) | \hat{Q}_1 | \psi_1(\boldsymbol{r}_1) \rangle + \langle \psi_2(\boldsymbol{r}_2) | \hat{Q}_2 | \psi_2(\boldsymbol{r}_2) \rangle \end{align} If I add one more boson to state $\psi_1$, then the total wavefunction changes to \begin{align} \Psi(\boldsymbol{r}_1,\boldsymbol{r}_2,\boldsymbol{r}_3) = \dfrac{ \psi_1(\boldsymbol{r}_1)\psi_1(\boldsymbol{r}_2)\psi_2(\boldsymbol{r}_3) + \psi_1(\boldsymbol{r}_2)\psi_1(\boldsymbol{r}_3)\psi_2(\boldsymbol{r}_1) + \psi_1(\boldsymbol{r}_3)\psi_1(\boldsymbol{r}_1)\psi_2(\boldsymbol{r}_2) } {\sqrt{6}} \end{align} The expectation value of $\hat{Q}$ for $\Psi(\boldsymbol{r}_1,\boldsymbol{r}_2,\boldsymbol{r}_3)$ becomes \begin{align} \langle \Psi | \hat{Q} | \Psi \rangle = 2\langle \psi_1(\boldsymbol{r}_1) | \hat{Q}_1 | \psi_1(\boldsymbol{r}_1) \rangle + \langle \psi_2(\boldsymbol{r}_2) | \hat{Q}_2 | \psi_2(\boldsymbol{r}_2) \rangle \end{align} I am stuck with understanding how to compute the expectation values of $\hat{Q}$ in terms of the boson field operator \begin{align} \Psi(\boldsymbol{r})=\sum_{\nu} \psi_{\nu} \left( \boldsymbol{r} \right) b_{\nu} \end{align} where \begin{align} \left[\Psi(\boldsymbol{r}_1),\Psi^\dagger(\boldsymbol{r}_2)\right]=\delta (\boldsymbol{r}_1-\boldsymbol{r}_2) \\ [b_\alpha^\dagger,b_\beta^\dagger]=[b_\alpha,b_\beta]=0,\quad [b_\alpha,b_\beta^\dagger]=\delta_{\alpha\beta} \end{align} I am assuming that \begin{align} \langle \psi_{\mu}^\dagger \left( \boldsymbol{r} \right) | \psi_{\nu} \left( \boldsymbol{r} \right) \rangle = \int \psi_{\mu}^\dagger \left( \boldsymbol{r} \right) \psi_{\nu} \left( \boldsymbol{r} \right) d^3\boldsymbol{r} = \delta_{\mu\nu} \end{align}

• Could you clarify what's giving you trouble? The computation of expectation values in multi-particle states is no different from computing them from single particle states – ACuriousMind Jul 24 '14 at 21:43
• @ACuriousMind I am going to edit the question to be specific to a particle field. – linuxfreebird Jul 24 '14 at 23:53
• I am still trying to add more detail to the question. I $\boldsymbol{\Psi}$ is an operator I believe. – linuxfreebird Jul 25 '14 at 0:30
• @ACuriousMind I made some edits to the question. I will transfer the post to a new post, because I was told that on hold questions can never be redeemed. – linuxfreebird Jul 25 '14 at 0:54
• They can! Don't post this anew, it will only be closed as duplicate. Since you edited it, it is automatically placed in the reopen queue, and will be reopened if it is clear. But you seem to be conflating QFT and QM objects here, and the question still makes little sense to me. The boson field $\Psi(\vec r)$ is a QFT object, it is an operator distribution, while your wavefunctions are simply functions taking values in $\mathbb{C}$. They are not at all the same, and the boson field does not represent a state, so you cannot calculate any expectation value from it. – ACuriousMind Jul 25 '14 at 1:12

In quantum field theory, you change the space of wavefunctions w.r.t. quantum mechanics. The space is still a Hilbert space, but it is called Fock space, and it takes into account the possibility of having any number of identical "particles" (or excitations of the field). A vector of such Fock space is of this form: $$(\psi_0,\psi_1,\psi_2,\psi_3,\dotsc)$$ where $\psi_1$ is an usual QM wavefunction of one particle, $\psi_2$ is the symmetric (for bosons) completion of products $\psi(x_1)\phi(x_2)$ of single-particle wavefunctions etc... $\psi_0$ is the state with no particles, or vacuum. Mathematically, the Fock space have a particularly nice structure described by means of direct sums and (symmetric) tensor products of Hilbert spaces. Let $\mathscr{H}$ be the one particle Hilbert space, then the symmetric Fock space $\Gamma_s(\mathscr{H})$ is $$\Gamma_s(\mathscr{H})=\bigoplus_{n=0}^\infty \mathscr{H}_n\; ,\; \mathscr{H}_n=\underbrace{\mathscr{H}\otimes_s\dotsc\otimes_s\mathscr{H}}_{n}$$ with the convention $\mathscr{H}_0=\mathbb{C}$.
So the $\Psi$ you have written above is some state belonging to one of the $\mathscr{H}_n$, not an operator by any means. Then you write the observable $Q$ as acting like a one particle operator on each particle. This type of operator is possible in the Fock space and it is called second quantization of $Q$. However it is not the only possible operator. Given $Q$ acting on $\mathscr{H}\equiv\mathscr{H}_1$, the second quantization $d\Gamma(Q)$ acts on $\mathscr{H}_n$ as $$Q\otimes1\otimes\dotsc\otimes1 +1\otimes Q\otimes1\otimes\dotsc\otimes1+\dotsc+1\otimes\dotsc\otimes Q\; ;$$ and on $\mathscr{H}_0$ as zero ($1$ is the identity operator on $\mathscr{H}$). So computing the average of a second quantization $\langle\Psi,d\Gamma(Q)\Psi\rangle$ (where $\Psi\in\Gamma_s(\mathscr{H})$) may be quite easy.
But there are other types of operators on the Fock space as you can imagine, in particulare those who relate $\mathscr{H}_n$ with $\mathscr{H}_{n+1}$ and vice versa. The most famous ones of this type are the creation and annihilation operators $a^*(x)$ and $a(x)$ ($[a(x),a^*(y)]=\delta(x-y)$), and are the fundamental operators of the Fock space, as momentum and position are of the QM space. The boson field, in its simplest form, is just a combination of the two: let $f\in\mathscr{H}$, then the field $\phi(f)$ is $$\phi(f)=\frac{1}{\sqrt{2}}(a^*(f)+a(f))=\frac{1}{\sqrt{2}}\int \bigl(a^*(x)f(x)+a(x)\bar{f}(x)\bigr)dx\; .$$ As you see this operator relates $\mathscr{H}_n$ with $\mathscr{H}_{n+1}$ and $\mathscr{H}_{n-1}$ and it is self-adjoint, so you can think of it as another type of observable (and not a state!) on the Fock space. Also of it you can compute $\langle \Psi, \phi(f)\Psi\rangle$ but it is not an easy calculation as before...You can also see how the field interacts with the second quantization of $Q$, but by means e.g. of calculating their commutator $[\phi(f),d\Gamma(Q)]$. Hope this helps to clarify a little bit ;-)