A Feynman propagator for a simple harmonic oscillator

Glenn Rowe

Gold Member
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.

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vanhees71

Gold Member
The Feynman propagator for free Klein-Gordon fields is of course correct, but where does the one for the harmonic oscillator come from? See Sect. 3.5.2 of

It's, however, in German...

Glenn Rowe

Gold Member
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

Gold Member
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

Gold Member
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

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

vanhees71

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

HomogenousCow

G(t)=−i⟨Ω|T^xH(t)^x(0)|Ω⟩
Just curious, is this a useful quantity for anything?

vanhees71

Gold Member
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.

"Feynman propagator for a simple harmonic oscillator"

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