You have a tensor product space $H_1 \otimes H_2$. Any vector $w$ in this space has a Schmidt decomposition:

$$ \mathbf{w} = \sum_{i} \alpha_i \mathbf{u_i}\otimes\mathbf{v_i} $$

Vector $w$ is not entangled if the Schmidt rank is 1, i.e., only one possible $\alpha_i$.

Now, in Physics you consider the eigenvectors of a Hermitian matrix (the Hamiltonian). According to the Spectral Theorem, this matrix is always diagonalizable.

So given the eigenvectors, you can go to a basis where these eigenvectors are just the trivial basis for $\mathbb{C}^n$ (by trivial basis I mean the columns of the identity matrix $\mathbb{1}_n$).

We see that in this basis the eigenvectors are clearly not entangled...

But entanglement is basis independent! And I know as a fact that the eigenvectors of some Hamiltonians are entangled. So whats going on here? What have I forgotten?

  • $\begingroup$ You seem to think that only the eigenvectors play any role in physics. Why? That assertion is simply false. $\endgroup$
    – ACuriousMind
    Commented Jul 18, 2017 at 16:59
  • $\begingroup$ It is simply false that I think this. My question is what it is. I'm pretty sure some Hamiltonians have entangled eigenvectors. I cant explain this mathematically. An example of an entangled Hamiltonian would be the Heisenberg model with disorder( a random magnetic field). $\endgroup$
    – DLV
    Commented Jul 18, 2017 at 17:00
  • $\begingroup$ The subtlety occurs when the Hamiltonian has degenerate eigenvalues. Then some combinations of eigenvectors with the same eigenvalues might be physically meaningful. The simplest example are the $M=0$ states of system containing 2 spin-1/2 particles. They are separable when written as $|+>|->$ and $ |->|+>$ but one usually works with the singlet or triplet states (for angular momentum purposes) where the states are, up to a normalization, $|+>|-> \pm |->|+>$ $\endgroup$ Commented Jul 18, 2017 at 17:04
  • $\begingroup$ Thanks Zero for your answer. But I still think my argument still holds. We can choose an orthonormal basis for the degenerate subspace. And therefore, my new basis will still be the columns of the identity matrix, which is not entangled. $\endgroup$
    – DLV
    Commented Jul 18, 2017 at 17:07
  • $\begingroup$ The dynamics or some other constraint might truncate the full tensor-product Hilbert space to a subspace of definite angular momentum. In the case of EPR for instance, this restriction comes about from conservation of angular momentum so the outgoing states must have $L=0$. $\endgroup$ Commented Jul 18, 2017 at 17:09

1 Answer 1


One simple way to see why this doesn't make sense is to consider an arbitrary vector $|v \rangle$ in $H_1 \otimes H_2$. You can always construct a basis which contains this vector as the first element, so that its coordinate representation is $$(1, 0, \ldots, 0).$$ So if we trusted your argument, we would conclude that entanglement doesn't exist, period.

The problem is that you can't tell if a state is entangled by just looking at its components in some basis -- it matters whether the basis vectors themselves are entangled or not too. Most standard bases we use contain separable basis vectors by construction; yours does not.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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