# How to write the Fröhlich Hamiltonian in one dimension?

I am currently working on a (functional) analysis problem refining Pekar's Ansatz (or adiabatic approximation, as it is called in his beautiful 1961 manuscript "Research in Electron Theory of Crystals").

Anyways, I have two related questions, which the members of this community may find simple.

The Fröhlich Hamiltonian is given as follows in three dimensions

$$H=\mathbf{p^{2}}+\sum_{k}a_{k}^{\dagger}a_{k}-\biggl(\frac{4\pi\alpha}{V}\biggr)^{\frac{1}{2}}\sum_{k}\biggl[\frac{a_{k}}{|\mathbf{k}|}e^{i\mathbf{k\cdot x}}+\frac{a_{k}^{\dagger}}{|\mathbf{k}|}e^{-i\mathbf{k\cdot x}}\biggr]$$

The physical scenario here is an electron moving in a 3-dimensional crystal. Each $k$ signifies a (vibrational) mode of the crystal.

If we restrict ourselves to just a 1-dimensional crystal, why is it that the Hamiltonian can be written as follows:

$$H=\mathbf{p^{2}}+\sum_{k}a_{k}^{\dagger}a_{k}-\biggl(\frac{4\pi\alpha}{V}\biggr)^{\frac{1}{2}}\sum_{k}\biggl[a_{k}e^{i\mathbf{k\cdot x}}+a_{k}^{\dagger}e^{-i\mathbf{k\cdot x}}\biggr]$$

Namely, why do we drop the $|\mathbf{k}|$ factor in the third term?

Furthermore, I see how the creation and annihilation operators work on the (bosonic) Fock space (referring to the crystal here), especially when we write the creation operator in the form $\sum_{k=0}^{\infty}\frac{(a^{\dagger})^{k}}{\sqrt{k!}}\left|0\right\rangle =\left|k\right\rangle$. Namely, the creation operator is jumping from one tensored state in Fock Space to the next. However, I also see the form $a_{k}=\frac{1}{\sqrt{2}}\bigl(k+\frac{d}{dk}\bigr)$. How are the two forms connected? How do you intuitively think of the latter form? For example, I thought of the former form as the creation operator jumping from one state in fock space to the next, but the latter form I am not quite sure.

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This is a good question, but I think it would be better asked as two separate pieces. As a rule of thumb, you should be able to actually write your question in the title, and (maybe it's just me, but) I don't think your two questions are closely related enough to do that succinctly. I'd suggest removing the second question from this post and posting it separately. – David Zaslavsky♦ Dec 8 '11 at 6:16
Can you give the exact reference that says you can drop the |k|? You might have misread a nontrivial manipulation that might be apparent in context. – Ron Maimon Dec 8 '11 at 9:57

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Also, the equation you have for $\left|k\right\rangle$ makes no sense --- there is no free $k$ index on the left hand side... – genneth Dec 8 '11 at 10:13
@genneth: The "k" in the formula is an occupation number on the right, but is summed over on the left. I was going to fix it to be an "n", but it does indicate a conceptual confusion. r.g.: Each k mode needs to have a seperate integer occupation number n, and the formula you wrote for the action of the creation operator is on one occupation number only. – Ron Maimon Dec 8 '11 at 21:07

## 1 Answer

The answer to your second question is simple. For a Harmonic oscillator, the creation and annihilation operators are related to the x and p operators by (up to a choice of units):

$$a = x+ ip$$ $$a^\dagger = x-ip$$

Writing the x operator as $i{\partial\over\partial p}$ reproduces your formula (up to phases and signs, the above phases and signs are correct in the usual physics convensions), with p playing the role of k. The k in your formula must be interpreted as the k operator.

This is the polaron problem, which was studied heavily in the mid 1950s, after Frohlich deduced from the isotope effect that phonon electron interactions must be responsible for superconductivity. The oscillators are the phonon modes, the 1 over |k| tells you that long wavelength phonons are singular, but I don't know the answer to the first question, because the identity is superficially impossible, because one of the two forms will be dimensionally inconsistent. If you provide a reference to fix the conventions, one can decide which one is correct, and perhaps this is a simple misunderstanding. The phonons in your description have the exact same frequency, for example, which is incorrect--- the dispersion for phonons should makes the second term $\sum_k |k| a^{\dagger} a$\$.

User wsc tells me that the Frohlich Hamiltonian is used to model the interactions with optical phonons. These have a flat dispersion, so the phonon part is ok as you wrote it.

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This is also very common in quantum optics, where the variance in x and y correspond to intensity and phase fluctuations respectively. – Antillar Maximus Dec 8 '11 at 11:53

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Most of this is fine, but Froehlich hamiltonians generally describe an interaction with optical phonons, so the flat dispersion is correct. – wsc Dec 8 '11 at 15:28
@wsc: Sorry, I misread your comment. I will update the answer accordingly. – Ron Maimon Dec 9 '11 at 0:37
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