Does a Quantum Field Creation Operator Create Particles at a Given Location?

[acegikmoqsuwy]

Hi,
It appears that the definition of a quantum field creation operator is given by $$\Psi^{\dagger}(\mathbf r) = \sum\limits_{\mathbf k} e^{-i\mathbf k\cdot \mathbf r} a^{\dagger}_{\mathbf k}.$$

But then if we examine how this operator acts on the vacuum state, we get $$\Psi^{\dagger}(\mathbf r) |vac\rangle = \sum\limits_{\mathbf k} e^{-i\mathbf k\cdot \mathbf r} |\mathbf k\rangle.$$

I thought this operator was supposed to create a particle at the given location, but we also have $$|\mathbf r\rangle = \sum\limits_{\mathbf k} e^{i\mathbf k\cdot \mathbf r} |\mathbf k\rangle.$$

These are both different, so in particular, it won't be the case that $$\langle \mathbf r'| \Psi^{\dagger}(\mathbf r)| vac\rangle = \delta^{(3)}(\mathbf r -\mathbf r ').$$

What went wrong?

 

[acegikmoqsuwy]

Never mind, I figured it out. The expansion of an eigenstate of position should instead be $$|\mathbf r\rangle = \sum\limits_{\mathbf k} e^{-i\mathbf k\cdot \mathbf r} |\mathbf k\rangle.$$