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

## 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?

## Answers and Replies

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.