# A What does "solving a quantum mechanics problem" mean?

#### jordi

In analogy to classical mechanics, I thought a good definition to "What does "solving a quantum mechanics problem" mean?" was to give the propagator (aka the Green function, or the 2-point correlation function):

In classical mechanics, solving a problem means to give the path of the particle over time, given some initial conditions.

It seemed to make sense to me to define "solving a QM problem" as giving the evolution in time of the wave function, given some initial wave function. And this is what the propagator does.

However, the propagator is the second derivative of the generating functional. In principle, if we think in probabilistic terms (QM is a probabilistic theory), one needs to give all the moments/cumulants of the probability distribution in order to properly define the probability distribution. And in probabilistic terms, DEFINING a probability theory means to give the probability distribution.

As a consequence, giving only the second derivative of the generating functional is not enough to "solve a QM problem", if we understand this problem in probabilistic terms. We should give all the derivatives of the generating functional. In physical terms, it also seems to make sense that higher-order derivatives of the generating functional need to be provided, since there are many-body interactions that cannot be accounted for the 2-point correlation function alone.

In QFT, all this is obvious, and all books work on computing n-point correlation functions. Through the LSZ theorem, we use these n-point correlation functions to calculate S-matrix elements. So, solving a QFT needs all n-point correlation functions, not only the propagator (even though, of course, the propagator is very important).

Also, I note it is not usual to speak about the wave function of QFT. I recall having heard about it, and that in some formalism of QFT there is something called a wave functional. But it is not often books speak about it. If there were a wave functional in QFT, then we could have the same "problem" we have in QM (why don't we define "solving a QFT problem" as giving the propagator only?). But since it seems that this is not the usual language in QFT, everything fits in QFT: what matters are all n-point correlation functions, of which the propagator is only one of them. All n-point correlation functions are needed to define the QFT problem.

But in QM, it seems that with the propagator is enough to solve the QM problem.

Question: why QM is not treated in analogy with QFT, by defining the solution as giving all n-point correlation functions? And if so, why is not enough to give the propagator to solve the QM problem, since the "wave function is all there is", and with the propagator we have enough to completely specify the wave function at all times, given an initial condition?

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#### A. Neumaier

Science Advisor
What does "solving a quantum mechanics problem" mean?
It depends on the problem.

Often (e.g., in most of quantum chemistry) one just wants to find the ground state. Sometimes one wants to know the spectrum. Sometimes one wants to know the dynamics of the state. Sometimes one wants to know some statistics. Sometimes one wants to know time correlations, etc. In each case, solving the problem means computing the required information.

QFT is no different from QM in this respect. In each case, there are many things you may want to compute, and getting them computed solves the problem. The n-point functions (for small n) are usually just the starting point for getting more interesting information.

• jordi and George Jones

#### jordi

It depends on the problem.

Often (e.g., in most of quantum chemistry) one just wants to find the ground state. Sometimes one wants to know the spectrum. Sometimes one wants to know the dynamics of the state. Sometimes one wants to know some statistics. Sometimes one wants to know time correlations, etc. In each case, solving the problem means computing the required information.

QFT is no different from QM in this respect. In each case, there are many things you may want to compute, and getting them computed solves the problem. The n-point functions (for small n) are usually just the starting point for getting more interesting information.
Thank you. Is there any interesting information from a system described by a QFT, which genuinely cannot be understood, in some way or another, as a (function of) n-point correlation functions?

In particular, the ground state, the spectrum, the dynamics of the state and the time correlations, AFAIK, they can be computed from n-point correlation functions.

#### A. Neumaier

Science Advisor
Thank you. Is there any interesting information from a system described by a QFT, which genuinely cannot be understood, in some way or another, as a (function of) n-point correlation functions?
Having the vacuum n-point correlation functions implies knowing the QFT completely, in the sense that everything else can in principle be computed from them. But ....

...it may still be highly nontrivial. For example,
• to get an exact scattering cross section you need to compute from them time-ordered correlation functions (which is nontrivial because of the singularities), and then compute an infinite sum (which is likely to diverge...), followed by integration. Computing directly the time-orderded correlations may be simpler.
• it helps almost nothing to computing the finite temperature correlation functions.

• mattt, dextercioby and jordi

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