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?
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2$\begingroup$ The actions are not the same! In the QFT case you have an integral over position too. In general QM is a "0D QFT" and doesn't have nearly enough information to describe actual QFT. $\endgroup$– knzhouCommented Apr 20, 2016 at 17:00
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$\begingroup$ Well not if it is a 1D QFT: if one defines a 1 dim QFT, it should only depend on one parameter ( in my case i call this $ t $ ) , and then I define one bosonic field which I call $ x $, dependent on $ t $, i.e. $ x(t) $ is my bosonic field, living in 1 dimension (t). The action then integrates over the base space of the quantum fields, and that is just $ t $. So I'm pretty sure this is the correct 1D QFT action for a Bosonic field. $\endgroup$– 123hoedjevanCommented Apr 20, 2016 at 20:18
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$\begingroup$ Sorry, I should have been more clear. The QHO is a "0+1 dimensional" QFT, with one time dimension. You want to match it up to a "1+1 dimensional" QFT, which doesn't work. $\endgroup$– knzhouCommented Apr 20, 2016 at 20:29
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$\begingroup$ It's easier to see the difference in the classical limit. A classical harmonic oscillator is defined by a function $x(t)$, there is only one parameter at each time. A classical 1D field is defined by a function $\phi(x, t)$, which is totally different. QFT is much more complex. $\endgroup$– knzhouCommented Apr 20, 2016 at 20:30
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$\begingroup$ Okay I'm a bit confused. A classical 1D field should just have a base space of just 1 coordinate right? Not 2. In the same way that 4D fields depend on 4 coordinates. The $ x (t) $ is my bosonic field, so I redefining $ x \to \phi $, I just have 1D field $ \phi(x) $ obeying the action above. EDIT: when I say dimension, I mean $ either $ spatial or temporal $\endgroup$– 123hoedjevanCommented Apr 20, 2016 at 20:53
1 Answer
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.
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$\begingroup$ Related: physics.stackexchange.com/q/735453/2451 $\endgroup$– Qmechanic ♦Commented Nov 20, 2022 at 14:33