[QFT] David Tong Ch1 Exercises Solutions (Part 1)

Ch1. Classical Field Theory

1. String as a set of harmonic oscillators

 1) Derive the Lagrangian.

  The partial derivatives of $y(x,t)$ are,

 $$ \frac{\partial y}{\partial t}= \sqrt{\frac{2}{a}} \sum_{n=1}^{\infty} \sin{\left( \frac{n\pi x}{a}\right)} \dot{q_n} \quad \text{,} \quad \frac{\partial y}{\partial x} = \sqrt{\frac{2}{a}} \sum_{n=1}^{\infty} \left(\frac{n \pi}{a}\right) \cos{\left( \frac{n\pi x}{a}\right)}q_n $$
 and inserting the derivatives into the given Lagrangian,

$$ \begin{split} L &= \int^{a}_{0} dx \left[ \frac{\sigma}{2} \left( \frac{\partial y}{\partial t} \right)^2 - \frac{T}{2} \left(\frac{\partial y}{\partial x} \right)^2 \right] \\ &= \int^{a}_{0} dx \left[ \frac{\sigma}{a} \left( \sum_{n=1}^{\infty} \sin{\left( \frac{n\pi x}{a}\right)} \dot{q_n} \right)^2 - \frac{T}{a} \left(\frac{n\pi}{a}\right)^2 \left( \sum_{n=1}^{\infty} \cos{\left( \frac{n\pi x}{a}\right)} q_n \right)^2\right] \\ &= \sum_{n=1}^{\infty} \left[ \frac{\sigma}{2} \dot{q_n}^2 - \frac{T}{2} \left( \frac{n \pi}{a} \right)^2 q_{n}^{2}\right] \end{split} $$
 Thus the Lagrangian shows up to be

$$ L=\sum_{n=1}^{\infty} \left[ \frac{\sigma}{2} \dot{q_n}^2 - \frac{T}{2} \left( \frac{n \pi}{a} \right)^2 q_{n}^{2}\right] $$

 2) Derive the equations of motion. 

 The partial derivatives of the Lagrangian are,

$$ \frac{\partial L}{\partial q_n}= -T  \left( \frac{n \pi}{a} \right)^2 q_n \quad \text{,} \quad \frac{\partial L}{\partial \dot{q_n}}=\sigma \dot{q_n}$$
 and inserting the results into the Euler-Lagrange equation,

$$ \ddot{q_n} = \frac{T}{\sigma} \left( \frac{n \pi}{a} \right)^2 q_n  $$
 and solving this differential equation, it yields an infinite set of decoupled harmonic oscillators with frequencies of

$$\omega_{n} = \sqrt{\frac{T}{\sigma}} \left( \frac{n \pi}{a}\right)$$

2. Lorentz invariance and KG eq.

Under a Lorentz transformation $x^{\mu} \rightarrow \Lambda^{\mu}_{\nu} x^{\nu}$, $\phi(x)$ transforms as $\phi(x) \rightarrow \phi(\Lambda^{-1}x)$. Thus the term $\eta_{\mu\nu} \partial^{\mu}  \partial^{\nu} \phi(x)$ transforms as, 

$$\begin{split} \eta_{\mu\nu} \partial^{\mu} \partial^{\nu} \phi(x) \rightarrow \eta_{\mu\nu} \partial^{\mu} \partial^{\nu} \phi(\lambda^{-1}x) &= \eta_{\mu\nu} \left(\Lambda^{-1} \right)^{\mu}_{\alpha} \left(\Lambda^{-1} \right)^{\nu}_{\beta} \partial^{\alpha} \partial^{\beta} \phi(y) \\ &= \eta_{\alpha\beta}\partial^{\alpha} \partial^{\beta} \phi(y) \end{split} $$

Which shows that the term $\partial_{\mu}\partial^{\mu}\phi=\eta_{\mu\nu} \partial^{\mu} \partial^{\nu} \phi$ is invariant under $x^{\mu} \rightarrow \Lambda^{\mu}_{\nu} x^{\nu}$. Thus the Klein-Gordon equation $\partial_{\mu}\partial^{\mu}\phi+m^2\phi=0$ is invariant under any Lorentz transformation $\Lambda$. 

3. Complex Scalar Fields

 1)  Write down the Euler-Lagrangian field equations. 

  The E-L field equation with respect to $\psi^*$ is,

$$\frac{\partial \mathcal{L}}{\partial \psi^*}  = -m^2 \psi - \lambda \psi^2 \psi^* \quad \text{,} \quad \partial_\mu \left(\frac{\partial \mathcal{L}}{\partial \partial_{\mu}\psi^*)} \right) = \partial_{\mu} \partial^{\mu} \psi$$

$$\partial _{\mu} \partial ^{\mu} \psi + m^2 \psi + \lambda |\psi|^2 \psi = 0$$

and with respect to $\psi$, 

$$\frac{\partial \mathcal{L}}{\partial \psi} = -m^2 \psi^* - \lambda \psi (\psi^*)^2 \quad \text{,} \quad \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial \partial_{\mu}\psi)}\right) = \partial_{\mu} \partial^{\mu} \psi^*$$

$$\partial _{\mu} \partial ^{\mu} \psi ^ * + m^2 \psi ^* + \lambda |\psi|^2 \psi^*= 0$$

which are complex conjugates to each other.

 2) Verify that the Lagrangian is invariant under such infinitesimal transformation.

  Under the infinitesimal transformation,

$$\delta \psi = i \alpha \psi \quad \text{,} \quad \delta\psi^{*} = -i\alpha \psi^{*}$$

 the Lagrangian transforms as, 

 $$\delta \mathcal{L} = \partial_\mu\delta \psi^* \partial^\mu \psi + \partial_\mu \psi^* \partial^\mu \delta \psi - m^2 (\delta \psi^* \psi + \psi^* \delta \psi ) - \lambda ( \psi^* |\psi| \delta \psi + \psi |\psi| \delta \psi^*)$$

As $\delta \psi = i\alpha \psi$ and $\delta \psi^* = (\delta \psi )^*$,  \delta \mathcal{L} shows up to be 0. Thus $\delta \mathcal{L} = \partial_\mu(0) = 0$, which makes the Lagrangian invariant under such infinitesimal transformation.