A Feynman propagator for a simple harmonic oscillator

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Glenn Rowe

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I'm reading through Lancaster & Blundell's Quantum Field Theory for the Gifted Amateur and have got to Chapter 17 on calculating propagataors. In their equation 17.23 they derive the expression for the free Feynman propagator for a scalar field to be $$\Delta\left(x,y\right)=\int\frac{d^{4}p}{\left(2\pi\right)^{4}}e^{-ip\cdot\left(x-y\right)}\frac{i}{\left(p^{0}\right)^{2}-E_{\boldsymbol{p}}^{2}+i\epsilon}$$ where ##p^0=E## represents an energy that is not on the mass shell, so that in general ##p^{0}\ne\sqrt{E_{\boldsymbol{p}}^{2}+m^{2}}##. I'm able to follow their derivation (I think), but then in Exercise 17.4, they ask us to show that the Feynman propagator for the quantum simple harmonic oscillator with spring constant ##m\omega_{0}^{2}## is given by $$\tilde{G}\left(\omega\right)=\frac{i}{m\left(\omega^{2}-\omega_{0}^{2}+i\epsilon\right)}$$
It seems to me that the energy of the harmonic oscillator in its "one-particle" state is ##\omega_0##, and the general energy (off the mass shell) is given by ##\omega## so that the position-space propagator would be given by $$G\left(x,y\right)=\int\frac{d^{4}p}{\left(2\pi\right)^{4}}e^{-ip\cdot\left(x-y\right)}\frac{i}{\omega^{2}-\omega_{0}^{2}+i\epsilon}$$
From there, we can read off the momentum-space Fourier component as $$\tilde{G}\left(\omega\right)=\frac{i}{\left(\omega^{2}-\omega_{0}^{2}+i\epsilon\right)}$$
I can't figure out where the extra factor of ##m## in the denominator comes from. Introducing the extra ##m## seems to mess up the units as well, since their general expression for the momentum-space propagator is $$\tilde{\Delta}\left(p\right)=\frac{i}{\left(p^{0}\right)^{2}-E_{\boldsymbol{p}}^{2}+i\epsilon}$$. I'm guessing I'm missing something simple (since pretty well all the exercises in the book aren't too complex once you understand the principles), but I just can't see it.
 

Glenn Rowe

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I've seen a derivation similar to that somewhere else (in English as well, I think) but I don't think that's what Lancaster & Blundell have in mind, since it doesn't give the answer they ask us to prove.

I have to admit I don't understand how you would introduce the harmonic oscillator into the methods they describe in their chapter 17, since their derivation is be done for free Klein-Gordon fields, which I don't think are the same as harmonic oscillators.

Maybe a better question is: is the expansion of the field $$\phi\left(x\right)=\frac{1}{\left(2\pi\right)^{3/2}}\int\frac{d^{3}p}{\left(2\omega\right)^{1/2}}\left(a_{\boldsymbol{p}}^{\dagger}e^{ip\cdot x}+a_{\boldsymbol{p}}e^{-ip\cdot x}\right)$$ valid for a harmonic oscillator, where ##a_{\boldsymbol{p}}^{\dagger}## and ##a_{\boldsymbol{p}}## are the creation and annihilation operators for the harmonic oscillator? If so, does ##a_{\boldsymbol{p}}^{\dagger}\left|0\right\rangle =\left|\boldsymbol{p}\right\rangle## where ##\left|\boldsymbol{p}\right\rangle## represents a single oscillator with momentum p?
If all that is true, then it would seem that the rest of their derivation would follow in the same way for a harmonic oscillator which would give the result $$\tilde{G}\left(\omega\right)=\frac{i}{\left(\omega^{2}-\omega_{0}^{2}+i\epsilon\right)}$$ that I got originally (without the extra ##m## in the denominator).

What makes me wonder if their answer is correct is that I've seen the derivation of the German result several times, but I've never seen anyone give Lancaster & Blundell's answer anywhere.
 

Glenn Rowe

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Never mind - I figured it out and have posted my solution here. However, It requires using the technique developed in the previous exercise 17.3, although I have to say I'm not sure why that method works, but anyway.
 

vanhees71

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For the harmonic oscillator you can of course define creation and annihilation operators. These always refer to a basis. In this case it's the energy eigenbasis. For the 1D harmonic oscillator that's all you need to define a complete orthonormal basis. You can define the normal-ordered Hamiltonian setting the arbitrary value for the ground-state energy to 0. Then you have
$$\hat{H}=\omega \hat{a}^{\dagger} \hat{a}=\omega \hat{N}.$$
The eigenvalues of ##\hat{N}## are ##n \in \{0,1,2,\ldots \}##. The ground state is defined by
$$\hat{a} |\Omega \rangle=0,$$
and the eigenstates with eigenvalue ##n## are
$$|n \rangle=\frac{1}{\sqrt{n!}} (\hat{a}^{\dagger})^n |\Omega \rangle.$$
All this can be algebraically derived from the commutator
$$[\hat{a},\hat{a}^{\dagger}]=\hat{1} .$$
That given you can reconstruct the position and momentum operators as appropriate linear Hermitean linear combinations:
$$\hat{x}=A \hat{a} + A^* \hat{a}^{\dagger}, \quad \hat{p}=\mathrm{i} (B \hat{a}-B^* \hat{a}^{\dagger}).$$
To get the Heisenberg algebra you need
$$[\hat{x},\hat{p}]=\mathrm{i} \hat{1},$$
but now
$$[\hat{x},\hat{p}] = \mathrm{i} [A \hat{a} + A^* \hat{a}^{\dagger},B \hat{a}-B^* \hat{a}^{\dagger}] = \mathrm{i} (A B^* + A^* B) \hat{1}\; \Rightarrow \; A B^* + A^*B=2 \mathrm{Re}(A B^*) \stackrel{!}{=}1.$$
The usual choice is ##A,B \in \mathrm{R}##

$$\hat{x}=\sqrt{\frac{1}{2m\omega}}(\hat{a}+\hat{a}^{\dagger}), \quad \hat{p}=\mathrm{i} \sqrt{\frac{m \omega}{2}} (\hat{a}^{\dagger}-\hat{a}).$$
In analogy to the field-theory case you simply have in the Heisenberg picture (!)
$$\hat{x}_H(t)=\sqrt{\frac{1}{2 m \omega}} [\hat{a} \exp(-\mathrm{i} \omega t)+\hat{a}^{\dagger} \exp(\mathrm{i} \omega t)].$$
It's easy to check that ##\hat{p}_H(t)=m \dot{\hat{x}}##.

I have assumed that Heisenberg and Schrödinger picture coincide at time ##t=0##.

I can only guess what the exercise may be asking. It's perhaps to calculate the time-ordered vacuum time-ordered Green's function
$$G(t)=-\mathrm{i} \langle \Omega |\mathcal{T} \hat{x}_H(t) \hat{x}(0)|\Omega \rangle$$
or its Fouier transform [corrected in view of #6]
$$\tilde{G}(k_0)=\int_{\mathbb{R}} \mathrm{d} t \exp(\mathrm{i} k_0 t) G(t).$$
This is of course a propagator for the classical-mechanics equation of motion,
$$-m \ddot{G}(t)- m (\omega^2- \mathrm{i} 0^+) G(t) = \delta(t).$$
The Fourier transform thus obeys
$$m (k_0^2-\omega^2+\mathrm{i} 0^+) \tilde{G}(\omega)=1.$$
Obviously your book has a slightly different definition of the Green's function. Obviously
$$G_{\text{book}}=\mathrm{i} G.$$
 
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Glenn Rowe

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Thanks for the explanation. I'm assuming you left out a factor of ##G(t)## in the integrand of ##\tilde{G}(k_0)##?
 

vanhees71

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Argh. Of course, I'll edit it in the original posting.
 

vanhees71

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I'd rather say it's a nice mathematical exercise in introductory QFT boiling it down to the simple harmonic oscillators. In QFT the various "auto-correlation functions" are very important for calculations. They are called Green's functions.
 

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