Translationally invariant Hamiltonian and property of the energy eigenstates

Question

If the Hamiltonian of a quantum mechanical system is invariant under spatial translation, then the linear momentum is a constant of motion. Apart from that, can we make some comment about the nature of the energy eigenstates? What if the Hamiltonian is invariant under discrete translation such as in a periodic crystal?

EDIT: For example, if the Hamiltonian is parity invariant, then non-degenerate energy eigenstates are either even or odd. So can we conclude something similar to this? Bloch theorem is about discrete translation. What would happen if the translation symmetry is continuous?

I'm not interested in any specific example of translationally invariant Hamiltonian. I'm interested in the property of the energy eigenstates of a generic translationally invariant Hamiltonian. In particular, I guess translational invariance leads to a energy eigenstate that is delocalised in space. But I'm not being able to show it mathematically.

Answer 1

No, there is no such requirement. It is pretty easy to find counterexamples where you have translation-invariant hamiltonians which have localized energy eigenstates with no such translation invariance. In particular, the statement you make,

translational invariance leads to a energy eigenstate that is delocalised in space,

is false in general, given the reasonable understanding of the above into the more precise statement

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

which does not hold.

To make a simple counterexample to the statement above, consider the hamiltonian for a free particle in two dimensions, $H=\frac12(p_x^2+p_y^2)$, which obviously has translation invariance and translationally invariant eigenfunctions of the form $$ \langle x,y|p_x,p_z\rangle = \frac{1}{2\pi} e^{i(xp_x+yp_y)}. $$ However, there is no requirement that the eigenfunctions be like that, and indeed you can form rotationally invariant wavefunctions that have a clear localization at the origin by taking phased superpositions of plane waves in the form $$ |p,l\rangle = \frac{1}{2\pi} \int_0^{2\pi} e^{il\theta}|p\cos(\theta),p\sin(\theta)\rangle \mathrm d\theta. $$ These are somewhat easier to understand in polar coordinates, where you have \begin{align} \langle r,\theta|p,l\rangle & = \frac{1}{2\pi} \int_0^{2\pi} \langle r,\theta|p\cos(\theta'),p\sin(\theta')\rangle e^{il\theta'}\mathrm d\theta' \\ & = \frac{1}{(2\pi)^2} \int_0^{2\pi} e^{ipr(\cos(\theta)\cos(\theta')+\sin(\theta)\sin(\theta'))} e^{il\theta'}\mathrm d\theta' \\ & = \frac{1}{(2\pi)^2} e^{il\theta}\int_0^{2\pi} e^{ipr\cos(\theta'-\theta)} e^{il(\theta'-\theta)}\mathrm d(\theta'-\theta') \\ & = \frac{i^{l}}{2\pi} e^{il\theta} J_{l}(pr) , \end{align} which are obviously the separable cylindrical-harmonics solutions of the Schrödinger equation in two dimensions. This means that they are legitimate eigenfunctions of $H$, but they have absolutely nothing to do with translation symmetry. Instead, they are eigenfunctions of the rotational symmetry of $H$ - and, in fact, the plane-wave states you started with are excellent examples of how a rotationally-invariant hamiltonian can have eigenfunctions that do not respect that symmetry.

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That said, if you're really looking for an analogue of the initial result you stated,

if the Hamiltonian is parity invariant, then non-degenerate energy eigenstates are either even or odd

then yes, it's possible - but it is absolutely crucial to have non-degenerate eigenvalues. (This is of course also true in the parity case, and if you have even and odd eigenstates at the same eigenvalue then it's trivial to construct mixed-parity eigenstates that do not have any definite symmetry.)

If you do manage to find a translationally invariant hamiltonian $H$ such that $[H,T_a]=0$ and some eigenvalue $p$ is non-degenerate (like e.g. $p=0$ for a free particle as the unique physically relevant case), then yes, the eigenstate $|\psi_p\rangle$ must be translationally invariant, since $T_a|\psi_p\rangle$ must be an eigenstate of the same eigenvalue, and by non-degeneracy it must be proportional to $|\psi_p\rangle$ , i.e. $T_a|\psi_p\rangle = e^{i f(a)}|\psi_p\rangle$, so $|\psi_p\rangle$ is translationally invariant.

However, you're highly unlikely to find any nontrivial, physically meaningful hamiltonians that are translationally invariant but not parity invariant, so you will always have at least a twofold energy degeneracy in all nonzero eigenvalues, making the argument above largely useless.

Answer 2

It seems to me that you would be interested in the following theorem:

If two operators $A$ and $B$ commute, we can find a joint eigenbasis of vectors $|a, b\rangle$ so that $A |a,b \rangle = \lambda_a\, |a,b\rangle$ and $B |a,b \rangle = \mu_b\, |a,b\rangle$.

Let's apply this to the systems you are talking about:

• If the Hamilton is parity-invariant, that means $[H,P] = 0$. Then, by the theorem above, we can choose the eigenbasis of $H$ in such a way that each eigenvector $|n\rangle$ has definite parity, $P|n\rangle = \pm |n\rangle$. From this we conclude $$ \psi_n(x) = \langle x | n \rangle = \pm \langle x | P | n \rangle = \pm \langle -x | n \rangle = \pm \psi_n(-x) \;. $$

• Let $T_x$ be a translation by $x$. If the Hamilton is translation-invariant (continuous symmetry), then $[H, T_x] = 0$ for all $x$. Remember $$ T_x = \exp\left( -\frac {\mathrm i} \hbar x\, p \right) $$ (where $p$ is the momentum operator). If $H$ and $T_x$ commute for all $x$, this must mean that $[H,p] = 0$. By the theorem above, we can find a basis of eigenvectors $|n\rangle$ of the Hamiltonian which are also eigenvectors of $p$ so that $$ \psi_n(x) \sim \mathrm e^{\frac{\mathrm i}\hbar k_n x} $$ for some $k_n$.

• If only $[H, T_x] = 0$ for discrete $x = na$. By applying the theorem again, we get that we can choose a basis where $$ \psi_n(x + a) = C \psi_n(x) \;, $$ where $C$ is a constant with $|C| = 1$. Write $C = \mathrm e^{2\pi\mathrm i\, \theta}$, define $k = \frac{2\pi\theta}a$ and $$ u_n(x) = \mathrm e^{-\mathrm i kx} \psi_n(x) \;. $$ You immediately see that $u_n(x+a) = u_n(x)$, and now we understand how the Bloch theorem follows from the general theorem I quoted above.
  (The Bloch theorem states that $\psi_n(x) = \mathrm e^{\mathrm i kx} u_n(x)$ with periodic $u_n$.)

Important: "we can find an eigenbasis such that..." does not in general mean that all bases are of this form. If the spectrum of $H$ is degenerate, we will in general be able to write down basisvectors of $H$ that do not respect the symmetry of the other operator, be it $P$ or $T_x$. See the answer of Emilio Pisanty.

Answer 3

I work in one spatial dimension for simplicity. If the Hamiltonian is translation invariant, i.e., if

$$\left[H,U\right]=0$$

for the unitary translation operator

$$U = e^{-ip},$$

then we can find simultaneous eigenstates $\left|\varepsilon,\theta\right>$ of both the Hamiltonian and the translation operator, where $H\left|\varepsilon,\theta\right>=\varepsilon\left|\varepsilon,\theta\right>$ and $U\left|\varepsilon,\theta\right>=e^{i\theta}\left|\varepsilon,\theta\right>.$ Consider the spatial wavefunction

$$\psi(x)= \left<x\right|\left.\!\varepsilon,\theta\right>.$$ We see that $$ \begin{align} \psi(x+a) &= \left<x+a\right|\left.\!\varepsilon,\theta\right>\\ &=\left<x\right|U^{-a}\left|\varepsilon,\theta\right>\\ &= e^{-ia\theta}\psi(x) \end{align} $$ for all real $a$. It follows that $$|\psi(x)|^2 = |\psi(x')|^2$$ for all $x$,$x'$, so the probability density described by such wavefunction is spatially constant. Any nonzero function obeying this condition will not have an $L^2$ norm.

Answer 4

For discrete translationally invariant systems Bloch's theorem applies.

For continuous translational invariance of the Hamiltonian the Hamiltonian is a constant ($\forall a:H(r) = H(r+a) \rightarrow H(r)=H=const.$). Hence plane waves are a set of eigenstates (which indeed also agrees with continuum limit of Bloch's theorem).