# A Does the effective action make sense in Quantum Mechanics?

#### jordi

I think the effective action should make sense also in Quantum Mechanics, not only in QFT. But I have never seen described in a QM book as such. Could there be a QM book that uses effective actions? Or maybe in QM effective actions are called another name?

I think effective actions in QM could be interesting, since there are several QM models which are solved exactly, so I assume the effective action could be calculated explicitly.

But now that I think about it, in QFT the effective action gives the correlation functions calculating only "tree level". But in QM, there are no loops (which are relativistic) ...

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#### DarMM

Gold Member
But now that I think about it, in QFT the effective action gives the correlation functions calculating only "tree level". But in QM, there are no loops (which are relativistic) ...
You can have loops in QM. Just expand the path integral in terms of the coupling and evaluate against the free theory measure as you do in QFT. It's really no different in that sense.

The only real difference is that the Feynman diagrams converge without renormalization. This is due to the fact that the QM path integral measures have $C^{0}$ functions as their support rather than distributions.

#### jordi

You can have loops in QM. Just expand the path integral in terms of the coupling and evaluate against the free theory measure as you do in QFT. It's really no different in that sense.

The only real difference is that the Feynman diagrams converge without renormalization. This is due to the fact that the QM path integral measures have $C^{0}$ functions as their support rather than distributions.
Sure, you are right.

As a consequence, an effective action for QM could make sense, right?

Why are QM books not using this language, which is closer to QFT? (at least, the QM books I have read).

#### Demystifier

2018 Award
As a consequence, an effective action for QM could make sense, right?

Why are QM books not using this language, which is closer to QFT?
If QFT was used as a theory of fields in the sense in which QM is used as a theory of particles, there would be no much use of effective action in QFT just as there is no much use of effective action in QM. But in practice, QFT is usually not used as a theory of fields. Instead, QFT is usually used as a theory of particles, as a tool to compute the particle scattering amplitude. But as theories of particles, QM and QFT are quite different theories. In particular, particles are fundamental in QM but emergent in QFT. That's, I think, is a deeper reason why effective action is useful in QFT but not so much useful in QM.

Let us briefly explain it at a more technical level. In general, the effective action is a tool to compute the n-point function. In QFT, the n-point function is something like
$$\langle \phi({\bf x}_1,t_1)\cdots \phi({\bf x}_n,t_n)\rangle$$
From this one can compute the particle scattering matrix via the LSZ formalism. Essentially the n-point function above is related to particles because $\phi({\bf x},t)$ can be thought of as an operator that (by acting on the vacuum) creates a particle at position ${\bf x}$ at time $t$.

In QM, on the other hand, the n-point function is something like
$$\langle x(t_1)\cdots x(t_n)\rangle$$
where $x(t)$ is a particle position operator in the Heisenberg picture. The QM n-point function above is not related to the scattering matrix of particles. Unlike $\phi({\bf x},t)$, the operator $x(t)$ is not something that creates a particle at position $x$ at time $t$. When $x$ acts on the 1-particle harmonic-oscillator ground state $\psi_0(x)$ , it creates the first excited state $\psi_1(x)$ of the 1-particle harmonic oscillator - the number of particles remains unchanged.

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#### DarMM

Gold Member
In QM, on the other hand, the n-point function is something like
$$\langle x(t_1)\cdots x(t_n)\rangle$$
where $x(t)$ is a particle position operator in the Heisenberg picture. The QM n-point function above is not related to the scattering matrix of particles. Unlike $\phi({\bf x},t)$, the operator $x(t)$ is not something that creates a particle at position $x$ at time $t$. When $x$ acts on the 1-particle harmonic-oscillator ground state $\psi_0(x)$ , it creates the first excited state $\psi_1(x)$ of the 1-particle harmonic oscillator - the number of particles remains unchanged.
Under the LSZ formula in QM it would be a way of calculating transitions between Energy levels. As the analogue of particles from QFT in QM are the energy levels of the system.
However it's not as good as other methods we have of doing this in QM like the Fermi approximation and higher order corrections to it.

Also even if you were to try it, because in QM all the terms are relevant the effective action would just be more complex than the original one without terms dying off.

So you'd essentially have a more complicated Lagrangian for computing scattering between energy levels via a method less effective than the Born series.

#### stevendaryl

Staff Emeritus
In QM, on the other hand, the n-point function is something like
$$\langle x(t_1)\cdots x(t_n)\rangle$$
where $x(t)$ is a particle position operator in the Heisenberg picture. The QM n-point function above is not related to the scattering matrix of particles. Unlike $\phi({\bf x},t)$, the operator $x(t)$ is not something that creates a particle at position $x$ at time $t$. When $x$ acts on the 1-particle harmonic-oscillator ground state $\psi_0(x)$ , it creates the first excited state $\psi_1(x)$ of the 1-particle harmonic oscillator - the number of particles remains unchanged.
Well, in the case of real scalar fields, it's not the case that $\phi$ is a creation operator.

If you consider a free real scalar field in $D+1$ dimensions ($D$ spatial dimensions and 1 time dimension), then the Lagrangian is:

$\frac{1}{2} (\frac{\partial \phi}{\partial t}^2 - \frac{\partial \phi}{\partial x^j}^2 - m^2 \phi^2)$

To the extent that it makes sense to consider $D=0$, this becomes

$\frac{1}{2} (\frac{\partial \phi}{\partial t}^2 - m^2 \phi^2)$

which is the same as the quantum-mechanical Lagrangian for a harmonic oscillator with mass $1$ and spring constant $m^2$, if you re-interpret $\phi$ as position. So the quantum-mechanical harmonic oscillator can be interpreted as a degenerate (0 spatial dimensions) quantum field theory. In which case, the $n^{th}$ energy level of the harmonic oscillator is interpreted as $n$ particles in the corresponding field theory.

Quantum mechanics with other sorts of potentials can similarly be interpreted as a degenerate quantum field theory with self-interaction.

"Does the effective action make sense in Quantum Mechanics?"

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