# Quantisation of the dynamics of a string via quantisation of the Fourier-moded harmonic oscillators

1. Apr 29, 2016

### spaghetti3451

1. The problem statement, all variables and given/known data

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 $|l_{1},l_{2}, \dots , l_{N}\rangle = (a^{\dagger}_{1})^{l_{1}}(a^{\dagger}_{2})^{l_{2}}\dots (a^{\dagger}_{N})^{l_{N}}|0\rangle$.

2. Relevant equations

3. The attempt at a solution

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?

2. May 4, 2016

### Staff: Admin

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. May 6, 2016

### 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.

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