I have been reading David Tong's notes on Phonons: http://www.damtp.cam.ac.uk/user/tong/aqm/aqmfour.pdf I am quite interested in Section 4.1.4, where he quantises the vibrations. First, he defines the most general (classical) solution $u_{n}(t)$, for the displacement of the $n^{th}$ atom in the chain:

(A) \begin{equation} u_{n}(t) = X_0(t) + \sum_{l\neq0}\bigg[\alpha_l\ e^{-i(\omega_lt-k_lna)} + \alpha_l^{\dagger}\ e^{i(\omega_lt-k_lna)}\bigg] \end{equation} and the corresponding momentum:

(B) \begin{equation} p_{n}(t)=P_{0}(t)+\sum_{l \neq 0}\left[-i m \omega_{l} \alpha_{l} e^{-i\left(\omega_{l} t-k_{l} n a\right)}+i m \omega_{l} \alpha_{l}^{\dagger} e^{i\left(\omega_{l} t-k_{l} n a\right)}\right] \end{equation} where $l$ indexes the wave-mode ($l = -N/2 , ... , N/2$) and wavenumber: $k_l = 2\pi\ l/Na$, with $N$ the number of unit cells and $a$ the lattice constant. These are treated as operators in the Heisenberg picture, which can be inverted to find the operators $\alpha_{l}$ and $\alpha_{l}^{\dagger}$. My confusion arises in the following step:

We can invert the equations above by setting t = 0 and looking at \begin{equation} \sum_{n=1}^{N} u_{n} e^{-i k_{l} n a}=\sum_{n} \sum_{l^{\prime}}\left[\alpha_{l} e^{-i\left(k_{l}-k_{l^{\prime}}\right) n a}+\alpha_{l}^{\dagger} e^{-i\left(k_{l}+k_{l^{\prime}}\right) n a}\right]=N\left(\alpha_{l}+\alpha_{-l}^{\dagger}\right) \end{equation}

  1. How did the last equality come about and where did $\alpha_{-l}^{\dagger}$ come from?
  2. Is it mathematically wrong to just add $u_{n}(t = 0)$ and $p_n(t = 0)$ and rearrange to find $\alpha_{l}$, rather than looking at $\sum_{n=1}^{N} u_{n} e^{-i k_{l} n a}$? If so, why?

If someone can help me parse this step, I'd be very grateful :)


1 Answer 1

  1. Using the representation of the Kronecker delta $$\delta_{l,l'} = \frac{1}{N} \sum_{n=1}^N e^{2\pi i (l'-l) n/N}$$ we have $$\sum_n \sum_{l'} \alpha_{l'} e^{-i(k_l-k_{l'})na} = \sum_{l'} \alpha_{l'} N \delta_{l,l'} = N \alpha_l$$ and $$\sum_n \sum_{l'} \alpha^\dagger_{l'} e^{-i(k_l+k_{l'})na} = \sum_{l'} \alpha^\dagger_{l'} N \delta_{-l,l'} = N \alpha^\dagger_{-l}$$ as desired.

  2. To find $\alpha_l$ we need to invert the given relations. Since the these relations take the form of a Fourier series, what Tong does is the most natural. Other methods will amount to the same thing.

  • 1
    $\begingroup$ Ah, so there was a typo, $a_l$ should probably have been $a_{l'}$ in the original test, right? Otherwise the wrong index is contracted. $\endgroup$
    – qwyxivi
    Commented Jun 28, 2020 at 23:42
  • $\begingroup$ Yes, there must be a typo either in the text or in your transcription of it. $\endgroup$
    – eric
    Commented Jun 29, 2020 at 1:46

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