Eigenvector structure of a Hamiltonian invariant under binary tree symmetries.

Question

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1. What the question is asking (in plain language)

We are given

a finite binary tree (every internal node has exactly two children),
a Hilbert space whose orthonormal basis consists of the edge states of that tree, e.g.

[ |e_1\rangle =|N\rangle ,\;|e_2\rangle =|P\rangle ,\;|e_3\rangle =|J\rangle ,\dots ,|e_{15}\rangle =|H\rangle , ]

a unitary (hence normal) Hamiltonian (H) that commutes with every symmetry of the tree.
The symmetry group is the group of graph‑automorphisms of the tree – i.e. the set of all permutations of the edges that can be obtained by swapping the left and right sub‑trees at any internal node.

The student wants to know

What can be said about the eigenvectors of (H) only from the knowledge that (H) is invariant under the tree symmetries?
In particular, how are the eigenvectors organised into independent families (a basis)?
If we force the two “parent” components (the edges attached to the root) to be zero, do the remaining eigenvectors automatically become “localised’’ on one of the sub‑trees?
And is there a systematic way to build a basis that respects the tree symmetry?

2. Symmetry → representation theory

2.1 The symmetry group of a binary tree

For a binary tree with (L) internal nodes (including the root) the automorphism group is the wreath product

[ \mathcal G \;=\; C_{2}\wr C_{2}\wr\cdots\wr C_{2}\;(L\ \text{times}), ]

i.e. a direct product of a copy of the two‑element group (C_{2}) (swap left/right) for each internal node, together with the obvious way those swaps compose.
Concretely:

At the root we may exchange the whole left subtree with the whole right subtree.
Inside each of those sub‑trees we may again exchange its own left and right children, etc.

Hence every element of (\mathcal G) is a product of independent swaps at the various nodes.

2.2 What invariance means

The Hamiltonian satisfies

[ [H,\,U_g]=0 \qquad\forall g\in\mathcal G, ]

where (U_g) is the unitary operator that permutes the edge basis exactly as the graph automorphism (g) does.
Since (H) commutes with all (U_g), it belongs to the commutant (centraliser) of the representation of (\mathcal G) on the edge Hilbert space.

The Hilbert space therefore decomposes into isotypic components (direct sums of irreducible representations, irreps) of (\mathcal G):

[ \mathcal H = \bigoplus_{\lambda} \bigl(\mathbb C^{m_\lambda}\otimes V_\lambda\bigr), ]

where

(V_\lambda) is an irrep of (\mathcal G),
(m_\lambda) is its multiplicity (how many copies appear).

Because (H) commutes with every group element, Schur’s Lemma tells us that on each isotypic component (H) acts as

[ H\big|{\mathbb C^{m\lambda}\otimes V_\lambda}= \mathbf{1}{V\lambda}\otimes h_\lambda, ]

i.e. it is block‑diagonal: one block for every irrep, and within a block it is the same operator (h_\lambda) on every copy.

Consequences:

We can choose a basis that first picks an irrep label (\lambda), then a copy index (a=1,\dots,m_\lambda), then a vector inside that irrep.
Each eigenvector of (H) is either entirely symmetric or antisymmetric (or more generally transforms according to a definite irrep) under the swaps at every node.

3. Concrete construction for a binary tree

Below we give a recursive way to build an orthonormal basis that respects the symmetry. The construction works for any depth; we illustrate it on the picture in the question (a tree of depth 3).

3.1 Basis at a single internal node

Consider a node with two child edges (e_L) and (e_R). Define

[ \begin{aligned} |e_{\text{sym}}\rangle &= \frac{1}{\sqrt{2}}\bigl(|e_L\rangle+|e_R\rangle\bigr) ,\[2pt] |e_{\text{asym}}\rangle&= \frac{1}{\sqrt{2}}\bigl(|e_L\rangle-|e_R\rangle\bigr) . \end{aligned} ]

Under the swap that interchanges the left and right child, the first vector is even (trivial irrep), the second is odd (sign irrep).

These two vectors are orthonormal and span the same two‑dimensional subspace as ({

e_L\rangle,

e_R\rangle}).

3.2 Recursion down the tree

Starting at the leaves and moving upward:

Leaves have no children, so each leaf edge stays as a one‑dimensional trivial irrep.
At each internal node we replace the pair of child basis vectors by the two combinations (sym, asym) defined above.
The sym combination becomes the effective edge that feeds into the next higher node; the asym combination is already a complete irrep that never mixes with the rest of the tree, because any higher‑level swap leaves it unchanged (it lives entirely within that node’s two children).

Repeating this procedure yields a global orthonormal set

[ \bigl{\,|\,\text{root}\,\rangle,\;|\,\text{sym}_1\rangle,\;|\,\text{asym}_1\rangle,\; |\,\text{sym}_2\rangle,\;|\,\text{asym}_2\rangle,\dots\bigr}, ]

where the subscript tells at which node the antisymmetric combination was taken.

3.3 Block‑diagonal form of (H)

Because each antisymmetric vector is already an eigenvector of all swaps that involve its own node (it picks up a minus sign) and is invariant under all other swaps, the Hamiltonian cannot couple an antisymmetric vector of node (v) to any vector that is symmetric at that node. Therefore the matrix of (H) in the above basis is block diagonal:

Block	Irrep	Dimension	Physical meaning
Root block	trivial (all swaps +)	1	the fully symmetric mode that lives on the whole tree
For every internal node (v)	sign irrep of the swap at (v)	1	a mode that lives only on the two edges attached to (v) (difference between left and right child)
For every level (k) (except the leaves)	trivial irrep of all swaps below that level	(#)	modes that are symmetric inside each subtree but may differ between subtrees of the same parent. These are the “global” modes that propagate up the tree.

Hence the independent eigenvectors can be taken as the vectors belonging to each block; inside a block we diagonalise the (usually small) matrix (h_\lambda).

4. Answer to the specific questions

4.1 What can we know about the structure of the eigenvectors?

Each eigenvector transforms according to a single irrep of the tree‑automorphism group.
Practically this means that for every internal node the eigenvector is either even (same amplitude on the two children) or odd (opposite amplitude).
The Hilbert space splits into orthogonal invariant subspaces labelled by the pattern of “even/odd’’ choices on the nodes.
The number of subspaces equals (2^{L}) (all possible patterns), but many of them are equivalent because sub‑trees of the same shape give rise to identical copies (multiplicity).
Within each subspace the Hamiltonian is the same; therefore the eigenvalues are degenerate with a multiplicity equal to the number of copies of that irrep.

In short, the eigenbasis can be chosen so that each basis vector is a product of

a symmetry label (which pattern of swaps is even/odd), and
a local wavefunction inside the corresponding invariant subspace (found by diagonalising the small block (h_\lambda)).

4.2 If we set the root edges (N) and (P) to zero, do the remaining eigenvectors localise on sub‑trees?

Setting the first two components to zero forces the global trivial irrep (the completely symmetric mode that has non‑zero weight on the root edges) to be absent.
The remaining subspace decomposes into a direct sum of the sign irreps at the root swap and the irreps that are symmetric under the root but have at least one antisymmetric choice deeper in the tree.

The sign irrep at the root is precisely the vector

[ |\,\psi_{\text{root-asym}}\rangle =\frac{1}{\sqrt{2}}\bigl(|N\rangle-|P\rangle\bigr), ]

which lives only on the two root edges – it does not extend to deeper edges, so it is already fully localised.
All other irreps have the property that they are symmetric under the root swap (so they have zero amplitude on the root edges) but may be antisymmetric at lower nodes.
Consequently, each such eigenvector has support only on the edges belonging to the subtree where the first antisymmetric node occurs.
In other words, after we delete the root amplitudes, every eigenvector either
- is confined to a single pair of sibling edges (an “odd’’ mode of some internal node), or
- lives on an entire isomorphic collection of sub‑trees, being identical on each copy.

Hence localisation is guaranteed only for the modes that are odd at the first node where the pattern becomes odd. Purely symmetric modes (odd nowhere) are impossible once the root edges are forced to zero – they would be the global symmetric mode that we have removed.

4.3 How to construct a symmetry‑adapted basis

Start from the leaves and assign each leaf edge a basis vector ( \ell\rangle).

Proceed upward: at every internal node with children (c_L,c_R) form

[ \begin{aligned} |\text{sym}_v\rangle &=\frac{1}{\sqrt{2}}\bigl(|c_L\rangle+|c_R\rangle\bigr),
|\text{asym}_v\rangle&=\frac{1}{\sqrt{2}}\bigl(|c_L\rangle-|c_R\rangle\bigr). \end{aligned} ]

Keep (	\text{asym}_v\rangle) as a stand‑alone basis vector (it will never mix with anything else).
Replace the pair ((	c_L\rangle,	c_R\rangle)) by the single vector (	\text{sym}_v\rangle) for the purpose of constructing the parent node.

Iterate until you reach the root. At the root you obtain one final symmetric vector ( \

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