The number operator in second quantization

I have a question about the number operator as applied to a quantum gas containing a mixture of different spins. Let us say the total number operator $hat{N}$ counts the total number of particles in a state, which we define in second quantization by the usual expression, $$hat{N} = sum_{rs}sum_{alpha beta} langle r,alpha |n| s, betarangle hat{a}_{r,alpha}^{dagger} hat{a}_{s, beta}.$$ Here $n$ is the number operator in first quantization. The states $|rrangle,|srangle$ are states with definite momentum, and $|alpharangle, |betarangle$ are states with definite spin. The operator $hat{a}$ is the annihilation operator. This expression for $hat{N}$ can be reworked as follows, $$ hat{N} = int int d^3 mathbf{x} d^3 mathbf{x’} sum_{alpha beta} hat{psi}^{dagger}_alpha(mathbf{x}) langle mathbf{x},alpha |n|mathbf{x’} ,betarangle hat{psi}_beta(mathbf{x’}).$$ Here $hat{psi}(mathbf{x})$ is the field operator corresponding with $hat{a}$ and the wavefunction of the momentum states. Now this is where I encounter some confusion. I believe that the final second quantized form of the number operator looks like this,$$hat{N} = int d^3 mathbf{x} sum_{alpha} hat{psi}^{dagger}_alpha(mathbf{x}) hat{psi}_alpha(mathbf{x}).$$ This makes logical sense to me since it simply counts all the particles in the system. However getting to this form from the form derived above requires that the matrix element $langle mathbf{x},alpha |n|mathbf{x’} ,betarangle = delta(mathbf{x}’-mathbf{x})delta_alpha^{beta}$. I cannot find a way to fully understand why this is true. Is my derivation okay, or have I made a mistake somewhere? Any help would be greatly appreciated!