Lancaster and Blundell Chapter 17

(17.1) Calculate the retarded field propagator for a free particle in momentum space and the time domain.

$$G^+_0(\mathbf p,t_x,\mathbf q, t_y)=\theta(t_x-t_y)\langle 0|\hat a_{\mathbf p}(t_x)\hat a^\dagger_{\mathbf q}(t_y)|0\rangle$$

We go into the Schrodinger picture:

$$=\theta(t_x-t_y)\langle 0|e^{i\hat Ht_x}\hat a_{\mathbf p}e^{-i\hat Ht_x}e^{i\hat Ht_y}\hat a^\dagger_{\mathbf q}e^{-i\hat Ht_y}|0\rangle$$

Since and , the exponentials are .

$$=\theta(t_x-t_y)\langle \mathbf p|e^{-i\hat Ht_x}e^{i\hat Ht_y}|\mathbf q\rangle\\=\theta(t_x-t_y)e^{-i(E_{\mathbf p}t_x-E_{\mathbf q}t_y)}\langle \mathbf p|\mathbf q\rangle\\=\theta(t_x-t_y)e^{-i(E_{\mathbf p}t_x-E_{\mathbf q}t_y)}\delta^{(3)}(\mathbf p-\mathbf q)$$

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(17.2) Demonstrate that the (1+1-dimensional) free scalar propagator is the Green’s function of the Klein-Gordon equation.

We write

$$\Delta(x,y)=\langle0|T\hat\phi(x)\hat\phi(y)|0\rangle=\int \frac{d^2p}{(2\pi)^2}e^{-ip\cdot(x-y)}\frac{i}{p^2-m^2+i\epsilon}$$

Operating on this with ,

$$\partial_t^2-\partial_x^2+m^2(\Delta(x,y))=\int\frac{d^2p}{(2\pi)^2}\left(-p_0^2+p_x^2+m^2\right)\frac{ie^{-ip\cdot(x-y)}}{p_0^2-p^2-m^2+i\epsilon}\\=-i\int\frac{d^2p}{(2\pi)^2}\frac{p_0^2-p_x^2-m^2}{p_0^2-p^2-m^2+i\epsilon}e^{-ip\cdot(x-y)}\\=-i\delta^{(2)}(x-y)$$

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(17.3) Show that the action for the free scalar field may be written $$S=\frac12\int\frac{d^4p}{(2\pi)^4}\tilde\phi(-p)(p^2-m^2)\tilde\phi(p)$$

We integrate the Lagrangian density by parts to get

$$\mathcal L=\frac12\phi(x)\partial^2\phi(x)-\frac{m^2}2\phi(x)^2=\frac12\phi(x)(\partial^2-m^2)\phi(x)$$

We substitute the Fourier transform for each copy of :

$$\mathcal S=\frac12\int\frac{d^4xd^4pd^4q}{(2\pi)^8}\tilde\phi(p)e^{ip\cdot x}(\partial^2-m^2)\tilde\phi(q)e^{iq\cdot x}=\frac12\int\frac{d^4xd^4pd^4q}{(2\pi)^8}\tilde\phi(p)e^{i(p+q)\cdot x}\phi(q)(-q^2-m^2)$$

The integral over gives , so integrating over gives

$$\frac12\int\frac{d^4p}{(2\pi)^4}\tilde\phi(-p)(p^2-m^2)\tilde\phi(p)$$

We may thus identify as times the inverse of the quadratic term in the momentum-space action ().

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(17.4) Find the Feynman propagator for the quantum harmonic oscillator with spring constant .

We write the Lagrangian as

$$L=\frac{\dot x^2}{2m}+\frac{m\omega_0^2}{2}x$$

Substituting the Fourier transform of , the same steps as before (integrating the first term by parts and identifying the free propagator) gives

$$\tilde G(\omega)=\frac1m\frac{i}{\omega^2-\omega_0^2+i\epsilon}$$

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(17.5) Consider the Lagrangian density $$\mathcal L=\frac12(\partial_x\phi)^2+\frac{m^2}2\phi(x)^2$$ Discretize this theory () and find the momentum space expression for the action.

The derivative terms becomes a finite difference $$\frac{\phi_{j+1}-\phi_j}{a}=\frac{1}{a\sqrt{Na}}\sum_p\tilde\phi_p(e^{ip(j+1)a}-e^{ipja})$$

The integral becomes . Carrying out some tedious sums, we find that the momentum is .

$$S=\frac12\sum_p\tilde\phi_{-p}\left(\frac2{a^2}-\frac{2\cos pa}{a^2}+m^2\right)\tilde\phi_p$$

$$\tilde G(p)=\frac{i}{\frac2{a^2}-\frac{2\cos pa}{a^2}+m^2}$$

The Lagrangian for a 1D elastic string in (1+1)-Minkowski space is 

$$\mathcal L=\frac12\left((\partial_0\phi)^2-(\partial_1\phi)^2\right)$$

Discretize this theory in space only and find the propagator, with .

This is truly unenlightening.

$$\tilde G(\omega,p)=\frac{i}{\omega^2-\omega_0^2(1-\cos pa)}$$

To reason a little, the acts as the effective mass in this theory, leading to the in the denominator, whereas the comes out of the spatial derivative term. There is an extra minus sign on that term since the Lagrangian has the extra minus sign compared to before.