The SHO in QM with mass $m=1$ has action $$ S[x] = \int dt \frac{1}{2} \dot x^2 + \frac{1}{2}\omega^2 x^2 $$ by integration by parts we see this is the same as 1 dim Klein Gordon QFT action with field $x(t)$ and mass $m=\omega$: $$ S[x] = \int dt \frac{1}{2} x ( -\partial_t^2 + \omega^2 )x $$ Now, as is done in section 2 of this article (http://authors.library.caltech.edu/8383/1/BOOejp07b.pdf) , we can then derive the 1+0D QFT Feynman propagator in the QFT: $$ \langle 0|\mathcal{T}x(t_i) x(t_f) | 0 \rangle = \frac{1}{2\omega} e^{-i\omega|t_i-t_f|} $$ Since the actions are the same, I feel like I should somehow be able to relate this result to the QM amplitude $$ \langle x_f,t_f|x_i,t_i\rangle $$ which is quite a complicated formula with sines and cosines as derived from the path integral in this article (http://web.mit.edu/dvp/www/Work/8.06/dvp-8.06-paper.pdf). Is there a neat way to map this 1D QFT to the QM SHO? Or from the QM SHO to the 1+0D QFT?
1 Answers
Yes, it is possible to match the QM with a QFT in 1+0 dimensions. However, the Fock vacuum $|0\rangle$ (which is annihilated an annihilation operator $a|0\rangle=0$) is naturally related to the coherent states $$\hat{a} |z\rangle~=~ z|z\rangle \tag{1}$$
rather than position eigenstates $$\hat{q} |q\rangle~=~q|q\rangle.\tag{2}$$ [Of course, it is possible to translate between the different operators and eigenstates (1) $\leftrightarrow$ (2).] Therefore it is most easy to consider the coherent state overlap $$\langle z_f^{\ast},t_f | z_i, t_i \rangle \tag{3}$$ rather than the position space overlap $$\langle q_f,t_f | q_i, t_i \rangle \tag{4} .$$ A QFT 2-pt function $$\langle 0^{\ast} | T[\hat{a}(t_1) \hat{a}^{\dagger}(t_2) ]|0 \rangle \tag{5}$$ is closely related to the coherent state overlap (3) with two extra insertions and $z_i=0=z_f$. See e.g. Ref. 1 for details.
References:
- L.S. Brown, QFT; Sections 1.7-1.8.
- 220,844