How to compute observables from the boson field operator?

Question

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}

Answer 1

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 ;-)