Why bosonic field operator in momentum space contains both creation and destruction operator? For fermionic field, the transformation from real space to momemtum space is a simple Fourier transformation $$\psi^\dagger(x)=\sum_{\mathbf{k}}c^\dagger_{\mathbf{k}}e^{ik\cdot x}$$ But in bosonic case, the field operator is 
$$\phi(x)=\sum_{k}\left[b^\dagger_{-k}+b_{k}\right]e^{ik\cdot x}$$
How does this difference come from? What's the physical significance of this difference?
 A: $\def\bk{{\bf k}} \let\dag=\dagger$
I've never seen an expression like your first one.
For a fermion field I would write
$$\psi(x) = \sum_{\bk,s} \left(a_{\bk,s} u_{\bk,s} e^{-ikx} + 
                          b_{\bk,s}^\dag v_{\bk,s} e^{ikx}\right).$$
Leaving apart several details, $a_{\bk,s}$ is a destruction operator for a particle with momentum $\bk$ and helicity $s$, while $b_{\bk,s}^\dag$ is a creation operator for the corresponding antiparticle. Of course
$$\psi^\dag(x) = \sum_{\bk,s} \left(a_{\bk,s}^\dag u^*_{\bk,s} e^{ikx} + 
                   b_{\bk,s} v^*_{\bk,s} e^{ikx}\right)\!.$$
For a charged boson field a quite analogous equation holds, whereas for a neutral one, where particles and antiparticles coincide, you would have $a$'s in place of $b$'s.
A: The Fermi field obeys $\{\psi(x),\psi^\dagger(x')\}=\delta^3(x-x')$ so we don't need both $a_k$ and $a^\dagger_k$ in the field $\psi(x)$ to get this from $\{a_k,a^\dagger_{k'}\}= \delta_{kk'}$ . For the bose field we need $[\phi(x),\partial_t \phi(x')]=i\delta^3(x-x')$ so we need both $b_k$ and $b^\dagger_k$ in the field to get a non-zero commutator.
A: The difference doesn't actually boil down to bosonic vs. fermionic. Instead, the two kinds of fields arise in different contexts. The former kind of field, which only contains a Fourier transform of one kind of ladder operator, typically arises in nonrelativistic situations where there are no antiparticles. The latter kind of field, which contains both a creation and an annihilation operator, tends to come up in relativistic situations, or in non-relativistic situations in which the effective field theory description has emergent Lorentz invariance and/or antiparticles. Either type of field can consist of either bosonic or fermionic ladder operators, but they're useful in different contexts and obey slightly different (anti)commutation relations.
