Is the Hamiltonian of a string given by a sum of harmonic oscillators?

[spaghetti3451]

Homework Statement

This problem is a continuation of the problem I posted in this thread: https://www.physicsforums.com/threads/equation-of-motion-from-a-lagrangian.867784/

(We have set the mass per unit length in that question to $\sigma$ = 1 to simplify some of the formulae a little.)

The string has classical Hamiltonian given by $H= \sum\limits_{n=1}^{\infty} (\frac{1}{2}p_{n}^{2}+\frac{1}{2}\omega_{n}^{2}q_{n}^{2})$ where $\omega_n$ is the frequency of the $n$th mode.

After quantization, $q_n$ and $p_n$ become operators satisfying $[q_n , q_m]=[p_n , p_m]=0$ and $[q_n , p_m]=i\delta_{nm}$.

Introduce creation and annihilation operators $a_n = \sqrt{\frac{\omega_n}{2}}q_{n}+\frac{i}{\sqrt{2\omega_{n}}}p_{n}$ and $a^{\dagger}_{n}=\sqrt{\frac{\omega_n}{2}}q_{n}-\frac{i}{\sqrt{2\omega_{n}}}p_{n}$.

Show that they satisfy the commutation relations $[a_n , a_m]=[a^{\dagger}_n , a^{\dagger}_m]=0$ and $[a_n , a^{\dagger}_m]=\delta_{nm}$.

Show that the Hamiltonian of the system can be written in the form $H=\sum\limits_{n=1}^{\infty}\frac{1}{2}\omega_{n}(a_{n}a^{\dagger}_{n}+a^{\dagger}_{n}a_{n})$.

Given the existence of a ground state $|0\rangle$ such that $a_{n}|0\rangle = 0$, explain how, after removing the vacuum energy, the Hamiltonian can be expressed as $H=\sum\limits_{n=1}^{\infty}\omega_{n}a^{\dagger}_{n}a_{n}$.

Show further that $[H, a^{\dagger}_{n}] = \omega_{n} a^{\dagger}_{n}$ and hence calculate the energy of the state $|l_{1},l_{2}, \dots , l_{N}\rangle = (a^{\dagger}_{1})^{l_{1}}(a^{\dagger}_{2})^{l_{2}}\dots (a^{\dagger}_{N})^{l_{N}}|0\rangle$.

Homework Equations

The Attempt at a Solution

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Let me solve the problem part by part.


(We have set the mass per unit length in that question to $\sigma$ = 1 to simplify some of the formulae a little.)

The string has classical Hamiltonian given by $H= \sum\limits_{n=1}^{\infty} (\frac{1}{2}p_{n}^{2}+\frac{1}{2}\omega_{n}^{2}q_{n}^{2})$ where $\omega_n$ is the frequency of the $n$th mode.

In the previous problem, $L = \int_{0}^{a} dx \bigg[ \frac{\sigma}{2} \Big( \frac{\partial y}{\partial t}\Big)^{2} - \frac{T}{2} \Big( \frac{\partial y}{\partial x}\Big)^{2} \bigg]= \sum\limits_{n=1}^{\infty} \bigg[ \frac{\sigma}{2} \dot{q}_{n}^{2} - \frac{T}{2} \big( \frac{n \pi}{a}\big)^{2} q_{n}^{2} \bigg]$

under the Fourier expansion $y(x,t) = \sqrt{\frac{2}{a}} \sum\limits_{n=1}^{\infty} q_{n}(t)\text{sin} \big(\frac{n\pi x}{a}\big)$,

so that the displacement profile $y(x,t)$ is decomposed into an infinite number of displacement profiles $y_{n}(x,t)=q_{n}(t)\sqrt{\frac{2}{a}}\text{sin} \big(\frac{n\pi x}{a}\big)$ indexed by $n$.
Now, $p_{m}=\frac{\partial L}{\partial \dot{q}_{m}}= \frac{\partial}{\partial \dot{q}_{m}}\Big(\sum\limits_{n=1}^{\infty} \bigg[ \frac{\sigma}{2} \dot{q}_{n}^{2} - \frac{T}{2} \big( \frac{n \pi}{a}\big)^{2} q_{n}^{2} \bigg]\Big) = \sum\limits_{n=1}^{\infty} \sigma \dot{q}_{n}\ \delta_{nm} = \sigma \dot{q}_{m}$

so that $H = \Big(\sum\limits_{n=1}^{\infty}p_{n}\dot{q}_{n}\Big)-L = \sum\limits_{n=1}^{\infty} \bigg[ \frac{1}{\sigma} p_{n}^{2} -\frac{1}{2\sigma} p_{n}^{2} + \frac{T}{2} \big( \frac{n \pi}{a}\big)^{2} q_{n}^{2} \bigg] = \sum\limits_{n=1}^{\infty} \bigg[ \frac{1}{2\sigma} p_{n}^{2} + \frac{T}{2} \big( \frac{n \pi}{a}\big)^{2} q_{n}^{2} \bigg]$

so that, under the assumption that the mass per unit length $\sigma = 1$,

the string has classical Hamiltonian given by $H= \sum\limits_{n=1}^{\infty} (\frac{1}{2}p_{n}^{2}+\frac{1}{2}\omega_{n}^{2}q_{n}^{2})$ where $\omega_{n} = \sqrt{\frac{T}{\sigma}} \big( \frac{n \pi}{a} \big)$ is the frequency of the $n$th mode.


Would you please comment on my attempt so far?

 

[spaghetti3451]

I think I can solve almost all of the problem by myself, but I just need someone to check my working in case I may made mistakes I can't spot.