Second quantised momentum operator in position basis

Question

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?

Answer 1

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.