# Phonons and Second Quantization

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

(B) $$$$p_{n}(t)=P_{0}(t)+\sum_{l \neq 0}\left[-i m \omega_{l} \alpha_{l} e^{-i\left(\omega_{l} t-k_{l} n a\right)}+i m \omega_{l} \alpha_{l}^{\dagger} e^{i\left(\omega_{l} t-k_{l} n a\right)}\right]$$$$ where $$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 $$$$\sum_{n=1}^{N} u_{n} e^{-i k_{l} n a}=\sum_{n} \sum_{l^{\prime}}\left[\alpha_{l} e^{-i\left(k_{l}-k_{l^{\prime}}\right) n a}+\alpha_{l}^{\dagger} e^{-i\left(k_{l}+k_{l^{\prime}}\right) n a}\right]=N\left(\alpha_{l}+\alpha_{-l}^{\dagger}\right)$$$$

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. 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.
• 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. Commented Jun 28, 2020 at 23:42