# Second quantised momentum operator in position basis

The Problem
I was trying to show that the momentum operator is given by $$\sum_r c^{\dagger}_r (-i\hbar)\nabla c_r$$.This is how I tried showing it

$$\hat{P}=\sum_{r,r'}c_r^{\dagger} \langle r|\hat{P}|r'\rangle c_{r'}=\sum_{r,r'}c_r^{\dagger}(i\hbar\frac{d\delta(r-r')}{dr})c_{r'}=\sum_{r,r'}-i\hbar \frac{dc_r^{\dagger}}{dr}c_{r'}\delta(r-r') = \sum_r \nabla(c_r^{\dagger})(-i\hbar)c_r$$

As you can see, something has gone wrong. The formula derived doesn't even look hermitian! ($$(\nabla(c_r^{\dagger})c_r)^{\dagger}=c_r^{\dagger}c_r\nabla)$$. What went wrong here?

My Thoughts
I believe the issue here is with the behaviour of the matrix elements $$\langle r|\hat{P}|r'\rangle =i\hbar\frac{d\delta(r-r')}{dr}$$. Is there some shenanigans happening in there due to the derivative of the delta function? If the matrix element had instead $$\frac{d}{dr'}\delta(r-r')$$, then I believe things would have been fine. However, did I make a mistake in evaluating this?

Your mistake is more trivial than you think: you used $$\hat{P} = i \hbar \nabla$$, but the correct definition that ensures canonical commutation relations is $$\hat{P} = - i \hbar \nabla$$. If you had used the correct definition, your last formula would obviously read $$\hat{P}= \sum_r \nabla(c_r^{\dagger})(i\hbar)c_r \tag{A}$$ And this is just one step from the formula that you want to prove: just use once again that "intregration-by-parts trick" to switch the derivation, that is $$\nabla$$, from $$c_r^\dagger$$ to $$c_r$$ and involving a change of sign (I mean the same trick that you used in switching the derivative $$\frac{d}{dr}$$ from the delta to the creation operator).
Even if it is correct, equation $$(\mathrm{A})$$ may look wrong to you because, as you wrote, $$(\nabla(c_r^\dagger)c_r)^\dagger = c_r^\dagger c_r \nabla$$, and hence it looks like our $$\hat{P}$$ is not hermitian. But this last equality is wrong. The thing is, you treat $$\nabla$$ as an operator on the Fock space just like $$c_r$$ and $$c_r^\dagger$$, but it isn't. If you think about your procedure, you'll realize that $$\nabla$$ acts on $$c_r^\dagger$$, turning it in a different operator by deriving it with respect to $$r$$ (and not by plain composition of operators). It is the operator $$c_r^\dagger$$ itself that depends on $$r$$ and is derived with respect to it, not the vector resulting from applying $$c_r^\dagger$$ to some vector.
So, the expression $$\nabla(c_r^\dagger)c_r$$ is not the composition of three operators, but rather of only two of them, namely $$\nabla(c_r^\dagger)$$ and $$c_r$$. Therefore, its hermitian conjugate is $$(\nabla(c_r^\dagger)c_r)^\dagger = c_r^\dagger \nabla(c_r)$$. Use this last formula along with another "integration-by-parts trick", and you will see that our $$\hat{P}$$ is indeed hermitian.