Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Is there a system of interacting quantum spin 1/2 particles (of any topology) whose the states where all spins are up or down are eigenstates of its Hamiltonian and yet does not conserve the total spin polarization (in the z-direction) ?

share|improve this question
add comment

1 Answer

There are plenty, the condition is very weak. For a simple example, with N spin 1/2, all spins up and all spins down both have total angular momentum N/2. You can make a Hamiltonian be zero on the N/2 total angular momentum states, so that all N/2 states are eigenstates, but it can be anything at all to the states of lower total angular momentum, without violating your condition. So, for example letting $\vec\sigma_j$ be the Pauli matrix for the n-th spin, and

$$ P = ({N/2(N/2+1)\over 2} - |\sum_j \vec\sigma_j|^2)$$

be the projection operator which zeros out the total angular momentum N/2 state, then

$$ H = P A P $$

will work, where A is any Hermitian operator at all (except for specially chosen ones). The projection on both sides guarantees that the N/2 total angular momentum states are all eigenvectors with eigenvalue zero.

But I assume you are more interested in a local spin-coupling. The construction above is global--- it requires you to decompose the total angular momentum. In this case, it's still easy to do. Make a matrix element which flips spins only when they are different. It will take adjacent |-+-> to |+-+> and adjacent |+-+> to |-+->. Both of these operations do not conserve total z spin.

The Pauli matrices suffice to expand any operator. In Pauli form $(\sigma^z_i + 1)$, and $(\sigma^z_i-1)$ project out the upper and lower component of a spin, and $\sigma^x_i$ flips a spin. So the Hamiltonian above is:

$$H = \sum_i \sigma^x_{i-1}\sigma^x_{i}\sigma^x_{i+1}( (\sigma^z_{i-1} +1)(\sigma^z_i-1)(\sigma^z_{i+1} +1) + (\sigma^z_{i+1} -1)(\sigma^z_i +1)(\sigma^z_i -1) ) $$

Which is Hermitian (the second term in the sum gives the Hermitian conjugate of the first when multiplied by all the sigma-x's).

The reason you were having trouble is probably because you wanted a two-spin nearest neighbor coupling. To preserve all up and all down, you can't change aligned-spins. This means that the only nonzero matrix elements of H are between antialigned spins, so you only have matrix elements between |+-> and |-+> at every site, and each of these two allowed transitions preserves total spin.

So only for the special case of two-spin interactions, the condition of having all up and all down preserved implies that the total z spin is conserved.

share|improve this answer
    
Thanks Ron Maimon. Are there physically realizable Hamiltonians like the Ising, or Heisenberg systems with long-range interactions? –  physicist Nov 3 '11 at 9:14
    
The long-range interactions might be used to model surface states, although I don't know cases of surface localized spins with a dynamics which are studied (most likely just out of ignorance, somebody please correct me). If you have spin electrons on the surface interacting through a bulk, the effective hamiltonian can be nonlocal. There are also direct nonlocal forces in materials like dielectrics with charged impurities, and you can possibly mock up two state systems in those, like a floppy protein crystal with a charged group in two symmetric conformations. Other people know better than me. –  Ron Maimon Nov 4 '11 at 7:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.