Can a partial transpose map have negative eigenvalues?
The fact that can have negative eigenvalues is another way of saying that the transpose map is positive but not completely positive, and thus plays a big role in entanglement theory. In this post we consider the question of how exactly the partial transpose map can transform the eigenvalues of :
Which is the eigenvalue of the equation sin?
Recall that we are assuming that λ > 0 λ > 0 here and so this will only be zero if c 2 = 0 c 2 = 0. Now, the second boundary condition gives us, sin ( 2 π √ λ) = 0 ⇒ 2 π √ λ = n π n = 1, 2, 3, … sin ( 2 π √ λ) = 0 ⇒ 2 π √ λ = n π n = 1, 2, 3, …
How to find the eigenvalues of a BVP?
Now, the second boundary condition gives us, sin ( 2 π √ λ) = 0 ⇒ 2 π √ λ = n π n = 1, 2, 3, … sin ( 2 π √ λ) = 0 ⇒ 2 π √ λ = n π n = 1, 2, 3, … Solving for λ λ and we see that we get exactly the same positive eigenvalues for this BVP that we got in the previous example. λ n = ( n 2) 2 = n 2 4 n = 1, 2, 3]
When to drop the constant in an eigenfunction?
For eigenfunctions we are only interested in the function itself and not the constant in front of it and so we generally drop that. Let’s now move into the second case. Here, unlike the first case, we don’t have a choice on how to make this zero. This will only be zero if c 2 = 0 c 2 = 0.
When do we only apply the transpose to one half of a quantum state?
However, strange things can happen if we instead only apply the transpose to one half of a quantum state. That is, if then its eigenvalues in general will be very different from the eigenvalues of , where is the identity map on and is the transpose map on (the map is called the partial transpose ).
Is the transpose of the density matrix the same as the quantum state?
It is a simple fact that, given any density matrix (i.e., quantum state) , the eigenvalues of are the same as the eigenvalues of (the transpose of ). However, strange things can happen if we instead only apply the transpose to one half of a quantum state.