Apparent spacetime dependence of creation and annihilation operators

Question

I'm currently going through An Introduction to Quantum Field Theory by Hartmut Wittig I've stumbled upon. Having trouble with equation (2.29), I'm asking the question:

Do creation and annihilation operators ($\hat a^\dagger(k)$ and $\hat a(k)$) depend on spacetime?

At first, it seems they shouldn't, because time dependence is being put in $e^{ik\cdot x}$ and $e^{-ik\cdot x}$, however, after using the commutation relations \begin{split} [\hat p^\mu,\hat a^\dagger(k)]=k^\mu \hat a^\dagger(k)\\ [\hat p^\mu,\hat a(k)]=-k^\mu \hat a(k) \end{split} and generalized Heisenberg equation \begin{equation} \frac{\partial}{\partial x^\mu}\hat A = i[\hat p^\mu,\hat A] \end{equation} it seems that they do depend on spacetime.

In equation (2.29), in the first line derivation was applied, while in second line the generalized Heisenberg equation was applied. It seems that in the first line the constancy of $\hat a^\dagger(k)$ and $\hat a(k)$ was used, while in the second line it seems that $x_\mu$ (without hat) and $\hat p_\nu$ commute (I also fail to understand why is this so, because $\hat p_\mu$ is essentially $\frac{\partial}{\partial x^\mu}$ up to some factor).

So, are $\hat a^\dagger(k)$ and $\hat a(k)$ constant? If they are, how to reconcile this with apparent contradiction with commutation relations with $\hat p_\mu$? If they aren't, what is the correct form of equation (2.29)?

Answer 1

In Hamiltonian formalism, operators such as $a(k)$ and $a^{\dagger}(k)$ can depend on time (in Heisenberg picture), or they can not (in Schroedinger picture).

They don't depend on space in sense that they are given for any 3-momentum $k$. But there is a Fourier transform $\tilde{a}(x)$ and $\tilde{a}^{\dagger}(x)$, which is given for any spacial point $x$. So you are confusing actual creation-annihilation operators and their Fourier images.

To avoid being confused again, think of the generalized Heisenberg relation as applicable only to space-dependent operators.

Answer 2

To say that a theory has Poincare-symmetry, means that for every Poincare-transformation $(\Lambda,a)$, there is a unitary $U(\Lambda,a)$ acting on Hilbert space. Under some assumptions there are generators $P^\mu$ (focussing on the translational part) such that

$$ U(1,a) = \exp(-i a_\mu P^\mu ) \ . $$

Now the creation operator $a_k^\dagger$ creates a particle in an unitary representation and thus satisfies the commutation relation (for bosons)

$$ [ P^\mu, a_k^\dagger ] = k^\mu a_k^\dagger \ . $$

Note that so far, there is no mentioning of space-time coordinates, so i.p. we cannot take any derivatives. This changes when introducing the notion of a field: this is a linear combination of creation and annihilation operators, all of which transform in an unitary, infinite dimensional representation $U$, such that the field transforms like a tensor in a finite-dimensional representation $S$:

$$ U(\Lambda,a) \psi(x) U(\Lambda,a)^{-1} = S(\Lambda,a) \psi(\Lambda^{-1} x - a) \ .$$

From there you may derive your Heisenberg equation

$$ [P^\mu ,\psi(x) ] = \frac{\partial \psi}{\partial x^\mu}(x) \ .$$

Crucial in deriving this relation was the assumption that $\psi(x)$ was a field, i.e. transforming in a finite-dimensional representation of the Poincare group. This is not true for the creation and annihilation operators $a_k$ and thus the equation does not hold.