Deriving the Single Body Hamiltonian in QFT

Question

I'm having some trouble with a few steps of the reasoning.

My first issue is why kinetic energy is diagonal in momentum representation, and why that means the Hamiltonian as a whole will be diagonal in this representation (as far as I can understand, the potential can cause scattering which manifests in off diagonal terms). The book (Condensed Matter Field Theory (2nd Ed) by Alexander Altland and Ben Simons) says to start with the momentum representation, and to generalise to the position basis. I assume this is because the Hamiltonian is diagonal in the momentum basis, which is what the book says to use as a starting point. Is it simply because in momentum space the kinetic energy operator behaves like a c-number (i.e. $\frac{p^2}{2m}$ where $p$ is a c-number).

The second part that I'm having some issue is with the actual derivation. My steps are as follows. Starting with what I think is the second quantised representation of the Hamiltonian $\hat{H}_1$ in momentum representation $$\hat{H}_1=\sum^{\infty}_{p=0}\langle p|\frac{p^2}{2m}+U(p)|p\rangle a_p^\dagger a_p$$ where $a^\dagger$ is the creation operator, $a$ is the annihilator operator, and $p$ is the momentum in momentum space representation.

Using the projection operator as a unitary transformation, we change the basis to position representation $$\hat{H}_1=\sum_{p=0}^\infty\int\langle p|x\rangle\langle x|\frac{p^2}{2m}+U(p)| x\rangle\langle x|p\rangle a_p^\dagger a_pd^dr$$ This I'm not quite sure about, since I think $\langle p|\frac{p^2}{2m}+U(p)|p\rangle$ should be treated as a constant, so I'm not sure whether I'm allowed to insert the identity $I=\int |x\rangle\langle x|d^dr$ inside the expected value. I'm also not completely sure on why we can take the sum over the same range, since in general for a sincle body operator $\hat{O}_1$ $$\hat{O}_1=\sum_{\mu\nu}\langle\mu|\hat{o}|\nu\rangle a_\mu^\dagger a_\nu$$ Is the domain of the integral the same because the projection operator is unique, and we are projecting onto the same range? The solution given is $$\hat{H}=\int a^\dagger(\textbf{r})\left[ \frac{\hat{\textbf{p}}^2}{2m}+U(\textbf{r})\right]a(\textbf{r})d^dr$$ which I can get by using the identity $$a(x)=\sum_k\langle x|k\rangle a_k$$

Any help is much appreciated, thanks!

Answer 1

The expression for $O_1$ does not depend on in which single-particle basis $\{|\nu\rangle\}_{\nu}$ you express it.

For example, you can start from your expression of $O_1$ and then use of the relation $$a_{\nu} \equiv \int \mathrm d x \,\varphi^*_{\nu}(x)\, a(x) \quad ,$$ where $\varphi_{\nu}(x)$ is the single-particle wave function corresponding to the single-particle state $|\nu\rangle$. By making use of the orthonormality of these functions and evaluating the matrix element $\langle x^\prime|p^2/ 2m + U(x)|x\rangle$, which should look familiar from non-relativistic (single-particle) quantum mechanics, you will arrive at the desired expression.

Alternatively, and I think this is what you're trying: Try to insert the completeness relation $$ \mathbb I = \int \mathrm d x\, |x\rangle \langle x|$$ twice (but with different indices) in the matrix element and then make use of $$a(x) = \sum\limits_{\nu} \varphi{_\nu}(x)\, a_{\nu} \quad .$$

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To see why the kinetic energy is diagonal in the momentum representation, just use $o=p^2/2m$ and express $O_1$ in that basis. However, if $o=p^2/2m + U(x)$, then in general $O_1$ will not be diagonal in the momentum basis.

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As an example, let us demonstrate how to change the representation of $O_1$ from an orthonormal single-particle basis $\{|\nu\rangle\}_{\nu}$ to $\{|n\rangle\}_{n}$. To start, we see that $$O_1 = \sum\limits_{\mu\nu} \langle \mu|o|\nu\rangle \, a^\dagger_{\mu}\, a_{\nu} = \sum\limits_{mn} \sum\limits_{\mu \nu} \langle \mu|m\rangle\langle m|o|n\rangle \langle n|\nu\rangle \,a^\dagger_{\mu}\, a_{\nu} \quad , $$

where we have inserted an identity operator (of the single-particle Hilbert space) $ \displaystyle \mathbb I = \sum\limits_n |n\rangle \langle n|$ twice. Now we have to note that \begin{align} a^\dagger_m &= \sum\limits_{\mu} \langle \mu |m\rangle \,a^\dagger_{\mu}\\ a_n &= \sum\limits_{\nu} \langle n |\nu\rangle \,a_{\nu} \quad , \end{align} which eventually yields $$O_1 = \sum\limits_{mn} \langle m|o|n\rangle \, a^\dagger_m a_n \quad .$$