# Are there translation-invariant hamiltonians that are not parity symmetric?

Are there translation-invariant hamiltonians that are not parity symmetric? I am primarily thinking in terms of the state space of a single massive particle in one or more dimensions, but I would like to deliberately keep the question slightly vague to see just how pathological an example you'd have to bring up to have a system with that kind of symmetry.

Moreover, I'm mostly interested in systems that do not have any symmetry that could be reasonably interpreted as a parity transformation. By this I mean that you can make a 1D hamiltonian that "breaks parity" by having $$H=\frac12p^2 + p_0 p$$ where $p_0$ is a constant, but this is clearly not that interesting since you can come up with a deformed 'parity' operator of the form $P'=e^{-ip_0x}Pe^{ip_0x}$ (i.e. a boost by $p_0$, parity, and a boost back by $p_0$). In these terms, the clearest marker of success would be a translationally invariant hamiltonian with large chunks of nondegenerate spectrum.

Is this possible? How far do you need to bend the normal examples to get there?

Edit: To explain a bit the motivation for this question, this related thread circles around claims of the form

if $H$ is translation-invariant and $|\psi\rangle$ is an eigenfunction of $H$, then $|\psi\rangle$ also needs to be translation invariant

which are normally scuppered by the fact that a translation-invariant $H$ is usually parity symmetric in that direction, which introduces a degeneracy into almost all of the spectrum, and therefore makes the usual non-degeneracy argument useless.

Now, translation invariance and inversion symmetry normally come together in real-world hamiltonians, but they are formally independent and there's no reason the former cannot come without the latter for a 'pathological enough' hamiltonian. The question here is, then, what does 'enough' mean after that pathological? How far off the beaten track do you need to go? And how many of the desirable properties of a hamiltonian (such as e.g. boundedness from below, or the existence of a ground state) can you preserve in the process?

• FWIW, the Hamiltonian of the Standard Model is invariant under translations and breaks the parity symmetry. I guess you have point-mechanics in mind instead of QFT though. – AccidentalFourierTransform Feb 9 '17 at 14:13
• What do you mean by 1D? Is it to exclude field theories in 2D or 4D? – Arnold Neumaier May 5 '17 at 7:16
• @Arnold That was just a simple example in single-particle QM - I do not want to exclude any dimensionality from answers, though I would prefer answers that don't involve QFT if that is possible (and if it's not, the reasons why are an interesting question I in its own right, and this is a good place for it). – Emilio Pisanty May 5 '17 at 7:25
• The highlighted statement in the Edit part is valid if and only if the eigenfunction belongs to a simple eigenvalue. This is simple linear algebra and has nothing to do with parity. – Arnold Neumaier May 5 '17 at 15:50
• @ArnoldNeumaier Yeah, that's sort of the point - the claim is obviously false in general, as discussed in detail in the linked thread. The link to parity is that normally parity is the unavoidable reason why the eigenvalue is not simple and the statement is not applicable. – Emilio Pisanty May 5 '17 at 16:00

OK, so thinking a bit more about this, and suggested in terms of working with the momentum eigenfunctions and where to put their eigenvalues in a way that will avoid both degeneracies and unboundedness-from-below, here is one example. Working in $L_2(\mathbb R)$, consider the hamiltonian $$\hat{H} = h_0 \exp(a\hat{p}/\hbar) = h_0\exp\left(-i a\frac{\mathrm d}{\mathrm dx}\right),$$ where $a$ and $h_0$ are constants with dimensions of length and energy, respectively, and $\hat{p}$ is the usual momentum operator. This operator is translation-invariant but it does not seem to have any easy relationship with its mirror version (and, more importantly for the linked motivation, it does not have any degeneracies).

Moreover, the spectrum is bounded from below, but unfortunately it does not seem to have any clear ground state (since the sequence $\psi_n(x)=e^{-inx/a}$ has eigenenergies $h_0e^{-n}$ which asymptotically approach zero but never reach it), and thinking in terms of hamiltonians of the form $\hat{H} = h_0 f(a\hat{p}/\hbar)$ does not suggest any obvious ways to get a clear ground state without introducing degeneracies into the spectrum.

Thus, I'll consider this a partial answer - hopefully a similar example can come along which does have a ground state.

What about $H=\frac12(p_1^2+p_2^2)+V(p,x_1-x_2)$, where $V(p,r)$ is an arbitrary odd function, i.e., $V(-p,-r)=-V(p,r)$. It should be possible to choose $V$ such that no gauge transformation can simplify the Hamiltonian to a parity invariant one.

There are plenty of such $V$. For example all linear combinations of products of any odd number of variables $p_1,p_2,r$ work. These depend on infinitely many parameters. Whereas the physically natural simplifying transformations that would allow calling the transformed parity to be still sensibly viewed as parity have few parameters only. Thus for most of the $V$ there is no natural parity that would be conserved.

If you want translation invariance in 3D, the same works with vector-valued $p_1,p_2,x_1,x_2$ of diemsnion 3.

• What's the role of $a$ in this expression? The mixture of even kinetic and odd potential terms is interesting, but doesn't the $a$ vanish after a suitable gauge transformation? (Sorry, my brain is a bit sluggish today) – Emilio Pisanty May 5 '17 at 11:48
• I'll explain in my answer. – Arnold Neumaier May 5 '17 at 13:54
• I look forward to that $V$ =) - it doesn't look like a trivial property to me. If you leave it in the form $H = \frac12 (p_1^2+p_2^2)+V(x_1-x_2)$, though, by rotating in the $x_1,x_2$ plane, we can reformulate that hamiltonian as $H = \frac12 (\tilde p_1^2+\tilde p_2^2)+V(\tilde x_1)$, which has no definite parity on $\tilde x_1$ but which has one suitable parity on $\tilde x_2$, introducing degeneracy. – Emilio Pisanty May 5 '17 at 14:02
• (See edited question for a bit more on the motivation.) – Emilio Pisanty May 5 '17 at 14:13
• @EmilioPisanty: well, you completely changed the question. This makes answering a thankless Sisyphus job. -- Note that I had the $a$ and later the more general $V$ just to prevent that such simple reformulations allow to regain a conserved parity. – Arnold Neumaier May 5 '17 at 15:47