Graduate Overlap of Ground States in Quantum Field Theory

[hilbert2]

I was reading Peskin&Schroeder's QFT book, and there was some discussion about how $\left|0\right>$, the ground state of a free field and $\left|\Omega \right>$, the ground state of an interacting field differ from each other, and they outlined how the latter can be obtained by propagating the former in imaginary time with the full interacting Hamiltonian... There was a mention that in this calculation it was assumed that there is some overlap between these two ground states: $\left< 0 | \Omega \right> \neq 0$.

Is it actually even possible for the ground states of two systems that have the same degrees of freedom to be orthogonal to each other? If you consider a simple 1D single particle system in nonrelativistic QM, it seems that the ground state wavefunction $\psi_0 (x)$ can always be chosen to be a nonnegative real-valued function. The only examples of nonoverlapping ground states of two such systems would be something like two particle-in-a-box systems where the infinite potential wells don't have any overlap between them (e.g. one is the interval $[0,L]$ and the other is the interval $[2L,3L]$). Then the wavefunctions wouldn't differ from zero at any common point, but also the difference between the two box potentials couldn't be seen as a "perturbation" because it wouldn't even be a proper finite-valued function.

However, if we have two hydrogenic atom problems with a particle moving in a $1/r$ potential, and the centers of the potentials in the two systems are separated by a large (macroscopic) distance, there would be practically no overlap between the ground states because the hydrogenic ground state wavefunction decreases exponentially with increasing distance from the attraction center.

Is it possible for two systems of several fermions to have nonoverlapping ground states because of the antisymmetry requirement?

 

[stevendaryl]

First of all, I'm not sure that looking at an example from quantum mechanics will give much insight into the corresponding problem in quantum field theory. Haag's theorem, for example, which suggests (I don't say "proves", just because there is no consensus about what the theorem implies for QFT) that an interacting QFT has a different Hilbert space than noninterating QFT, has no analog that I know of for single or multi-particle quantum mechanics.

Second, here's what I think might be an example of two different Hamiltonians with non-overlapping ground states: The top diagram shows a potential (dark black curve) that has a bump in the middle, and goes to infinity as x \Rightarrow \pm \infty. I think that the ground state for such a potential (if the bump in the middle is high enough) will be qualitatively like the thin curve: It will be an odd function.

The bottom diagram shows the same potential without the bump. The ground state will be an even function.

The overlaps of these two ground states will be zero, by the fact that their symmetries are opposite.

An exactly solvable case would be to use an infinite square-well with a "delta-function" potential in the middle as the bump.

 

[vanhees71]

Why should in the first case the ground-state wave function have an odd parity and a node? Shouldn't the ground state have no node and thus should have even parity (given the potential is symmetric under spatial reflection)?

 

[Orodruin]

Shouldn't the ground state have no node and thus should have even parity (given the potential is symmetric under spatial reflection)?

This. The ground state should be symmetric. However, it should be almost degenerate with the odd parity state shown in #2.

 

[stevendaryl]

Why should in the first case the ground-state wave function have an odd parity and a node? Shouldn't the ground state have no node and thus should have even parity (given the potential is symmetric under spatial reflection)?

Okay, I worked out the limiting case of a particle in a box (confined to the region -L &lt; x &lt; L) with a delta-function potential in the middle, and you guys are right, the even eigenstate is lower energy than the odd eigenstate (with the difference disappearing in the limit as the strength of the delta-function potential goes to infinity).

My reasoning (which was wrong) was that an even eigenstate would be nonzero in the region in the center where the potential is high, and that this would contribute to a higher expectation value of the potential energy. There is such an effect, but it isn't enough to overcome the difference in expectation value of the kinetic energy between the even and odd states.

 

[hilbert2]

Thanks for the answers, people.