# Dimensions in the Second Quantization of an Operator

Consider the one-particle operator $\hat A_{1p}$. As given in e.g. (Altland and Simons, 2nd ed, 2010; pg47) the second quantized version of this is given by: $$\hat A=\sum_{\mu,\nu} \left< \mu \right| \hat A_{1p} \left| \nu \right> a_\mu^\dagger a_\nu$$ for some basis $\{\left| \mu \right>\}$. On the next page the authors give the second quantized version of the "one-body Hamiltonian" as: $$\hat H=\int d^d r\; a^\dagger (\vec r) \left[ \frac{\hat p^2}{2m} +V(\vec r)\right] a(\vec r)\tag{1}$$ According to this answer on a related PSE question the ladder operators are dimensionless. How then do the dimensions of (1) work out? as the LHS appears to have dimensions of $\text{energy}$ whilst the RHS of $\text{volume} \times \text{energy}$.

(note: it is assumed we are working in the limit of infinite volume)

• If you wrote down the commutation relations of the $a(\vec{r})$s, you'd infer their dimension (which, for this infinity of them, cannot be 0). Commented Apr 12, 2018 at 21:41

To see this consider a continuous quantity $\vec q$ which could be position, momentum etc. let the commutation relation between $a^\dagger(\vec q)$ and $a(\vec q')$ be:
$$[a(\vec q'), a^\dagger (\vec q)]=\mathcal{C} \; \delta^d(\vec q-\vec q')$$
for some constant $\mathcal{C}$ (usually, but not always, $\mathcal{C} =1$). As stated in (Altland and Simons, 2nd ed, 2010; pg47) the second quantization of an operator $\hat A_{1p}$ in its eigenbasis: $$\hat A=\sum_{\lambda=0}^\infty \left<\lambda \right| \hat A_{1p} \left| \lambda \right> \hat n_\lambda\tag{A1}$$ Let us look at how this generalizes to the continuous case. Here: $$(\text{number of states \in d^d\vec q})=\hat a^\dagger (\vec q)\; a(\vec q) \;\frac{d^d \vec q}{\mathcal{C} }$$ this can be motiviated by acting on e.g. $a^\dagger(\vec q) \left| 0\right>$ and integrating over $d^d\vec q$ to see you get $1$. Thus we get: $$\hat A=\int \frac{d^d \vec q}{\mathcal{C} } \left<\vec q \right| \hat A_{1p} \left| \vec q \right>\hat a^\dagger (\vec q)\; a(\vec q)$$
So this explains the presence of the form when the eigenbasis is continuous. We will now look at the case where it is discrete but we want to move to a continuous one. Our starting point is thus (A1). Here we have that: $$\hat n_\lambda= \hat a_\lambda^\dagger \hat a_\lambda$$ The resolution of the identity for the states $\left| \vec q\right>$ is (due to the $\mathcal{C}$): $$I=\int \frac{d^d\vec q}{\mathcal{C}} \left|\vec q\right>\left<\vec q \right|$$ Thus performing a change of basis on $\hat n_\lambda$ we get: $$\hat n_\lambda=\int \frac{d\vec q d\vec q'}{\mathcal{C}^2}\left<\vec q|\lambda \right>\left<\lambda|\vec q' \right>\hat a_{\vec q}^\dagger \hat a_{\vec q'}$$ Subbing this into (A1) and using that: $$\sum_{\lambda=0}^\infty \left<\lambda \right| \hat A_{1p} \left|\lambda\right> \; \left| \lambda\right>\left<\lambda\right|=\hat A_{1p}$$ we get: $$\hat A=\int \frac{d\vec q d\vec q'}{\mathcal{C}^2} \left<\vec q \right| \hat A_{1p} \left| \vec q' \right>\hat a^\dagger (\vec q)\; a(\vec q')$$ which is the generalized form of the equation in the question.