Tools / Majorana constellation
A symmetric state as n points
A symmetric state of \(n\) qubits is not a list of \(2^n\) amplitudes: it is \(n\) directions on one sphere. Drag a star, or prepare a known state, and the symmetric state — its Dicke amplitudes, its collective spin and its distance from the nearest product state — follows. The same construction is the standard way to see why symmetric states are the tractable corner of multipartite entanglement.
Majorana constellation explorer
Constellation
Cat, n = 4
- stars
- 4
- \(\lvert\langle\hat J\rangle\rvert/(n/2)\)
- 0.000
- max fidelity
- 0.500
- \(E_G\) (bits)
- 1.000
\(P(z)=0.71z^{4}+0.71\)
The symmetric state behind the stars
The amplitudes are written in the Dicke basis: \(\ket{n,k}\) is the normalized sum of all computational basis states with exactly \(k\) ones. For \(n\) spins there are \(n+1\) of them, which is the whole of the symmetric subspace.
| \(k\) | basis state | amplitude | \(\lvert c_k\rvert\) | phase |
|---|---|---|---|---|
| 0 | \(\ket{4,0}\) | 0.707 | 0.707 | 0° |
| 1 | \(\ket{4,1}\) | 0.000 | 0.000 | — |
| 2 | \(\ket{4,2}\) | 0.000 | 0.000 | — |
| 3 | \(\ket{4,3}\) | 0.000 | 0.000 | — |
| 4 | \(\ket{4,4}\) | 0.707 | 0.707 | 0° |
| 5 | \(\ket{4,5}\) | 0.000 | 0.000 | — |
| 6 | \(\ket{4,6}\) | 0.000 | 0.000 | — |
| 7 | \(\ket{4,7}\) | 0.000 | 0.000 | — |
| 8 | \(\ket{4,8}\) | 0.000 | 0.000 | — |
Why n points are enough
Permuting the qubits of an \(n\)-qubit state leaves it unchanged only for special states. Those states form the symmetric subspace, of dimension \(n+1\): the Dicke states \(\ket{n,k}\) for \(k=0,\dots,n\). Majorana’s observation is that this subspace can also be described by \(n\) points on the sphere. A product of \(n\) identical spin states is symmetric, and every symmetric state is, up to a scalar, the symmetrization of \(n\) single-spin states
\[|\psi\rangle=\mathcal{N}\;\mathcal{S}\bigl(\ket{n_1}\ket{n_2}\cdots\ket{n_n}\bigr),\]
with the \(n\) directions \(n_j\) unique up to permutation. The proof is a factorization. Read the coefficients \(c_k\) in the Dicke basis and assemble the homogeneous polynomial \(P(X,Y)=\sum_k\sqrt{\binom{n}{k}}\,c_k\,X^{n-k}Y^{k}\), which is \(\prod_j(u_jX+v_jY)\) up to a scale and so factors over the complex numbers into \(n\) linear forms; each form is a spin direction. This page labels each star by its root \(z=-u/v\) of the polynomial in one variable, with \(z=\infty\) sent to the north pole, so the north pole is the spin \(\ket0\) and the south pole \(\ket1\).
Three consequences are visible in the picture. Stars that coincide cannot be told apart, so the polynomial has a repeated root and the state is the product \(\ket{\hat n}^{\otimes n}\) — a spin coherent state. For two spins the only maximally entangled symmetric state is the Bell state \((\ket{01}+\ket{10})/\sqrt2\), and its two stars sit at opposite ends of a diameter. And because a global phase multiplies the polynomial without moving its roots, the constellation depends on the state only up to a scalar, exactly as a point on the Bloch sphere does for one qubit.
Reading the numbers
The spin length \(\lvert\langle\hat J\rangle\rvert/(n/2)\) is the length of the mean of the \(n\) star vectors. It is one when all stars coincide and zero when their centroid sits at the centre of the sphere; it does not, by itself, measure entanglement.
The entanglement readout is the geometric measure, \(E_G=-\log_2\max_{\hat n}\lvert\langle\hat n^{\otimes n}|\psi\rangle\rvert^2\), the distance from the state to the nearest product state on a logarithmic scale. The maximum is taken numerically over all product directions, with the same fixed grid in the page and in the build. A value of zero means the state is a product; one bit is the value for the cat state \((\ket{0\dots0}+\ket{1\dots1})/\sqrt2\) at every \(n\), because that state is as far from a product as a Bell pair.
Familiar states have constellations worth memorising. The cat state always lands on \(n\) equally spaced stars on the equator, because its polynomial is \(z^n+1\). The Dicke state \(\ket{n,k}\) puts \(k\) stars at the south pole and \(n-k\) at the north, so its constellation records a single number. The state with one excitation — the symmetric W state — has one star at the south pole and the rest at the north, which is why its spin length is \((n-2)/n\).
Things to try
- Start on the cat: four stars at the equator, spin length 0.000, one bit of entanglement. Raise \(n\): the stars stay on the equator and \(E_G\) stays at 1.000.
- Press Coherent: every star collapses to one point, the spin length becomes 1.000 and \(E_G\) becomes 0.000. A coherent state carries no entanglement at all.
- Choose Dicke with \(k=2\) at \(n=4\): two stars at each pole, spin length 0, \(E_G=1.415\) bits. Set \(k=1\) instead: that is the W state, spin length 0.500.
- From a coherent state, drag one star off the pole: the spin length falls and \(E_G\) rises as soon as the stars separate.
- Set a star to \(\theta=0\) and another to \(\theta=180^\circ\) with the same \(\varphi\), and watch the polynomial lose all but two terms.
- Compare \(n=2\) with the cat: the two stars are antipodal, so the state is the Bell state and \(E_G=1.000\).
This page draws the symmetric subspace only. A general \(n\)-qubit state has \(2^n\) amplitudes and is not a symmetrization of \(n\) spin directions, so \(n\) stars cannot determine it; extending the picture to non-symmetric states needs more data than one point per spin, and is the next step planned for this tool.