In order to figure out if a given pure 2-qubit state is entangled or separable, I am trying to compute: the density matrix, then the reduced density matrix by tracing out with respect to one of the qubits, squaring the resulting reduced matrix, and finally taking its trace. Then, if the trace is $=1,$ I know the state is separable and entangled otherwise.
So I am trying this method for the following state (which we know to be entangled):
$$ |\psi\rangle = a_0b_0 |00\rangle+a_0b_1|01\rangle + a_1b_1|11\rangle \tag{1} $$ nothing much else is really said about the coefficients other than neither of the above rhs terms have a coefficient $0.$
My attempt:
- I computed $\rho=|\psi\rangle \langle\psi|$, then took the partial trace in the basis of the 2nd qubit $\rho_1=\operatorname{Tr}_2(\rho).$
The obtained $\rho_1$ is:
$$ \rho_1 = \begin{pmatrix} a_0^2b_0^2 & a_0 a_1^* b_1^2 \\ a_0^* a_1 b_1^2 & a_1^2b_1^2\end{pmatrix} \tag{2} $$
- Then computing $\rho_1^2$ then taking its trace, I obtain:
$$\operatorname{Tr}(\rho_1^2)=a_0^4b_0^4+a_1^4b_1^4+2a_0^2a_1^2b_1^4=(a_0^2b_0^2+a_1^2b_1^2)^2 \tag{3}$$
So for $|\psi\rangle$ to be an entangled state, we need to show $(3)$ is $\neq 1,$ i.e., $(a_0^2b_0^2+a_1^2b_1^2)\neq 1$
The only way I can make progress with the latter, is to assume that our given state $|\psi\rangle$ in $(1)$ is normalized, then I can assert that $(a_0^2b_0^2+a_1^2b_1^2)<1$ since by normalization $a_0^2b_0^2+a_1^2b_1^2+a_0^2 b_1^2=1$ must be unity, thus $(a_0^2b_0^2+a_1^2b_1^2)<1$ must be true and our state is entangled.
Question:
Do my calculations make any sense? Admittedly, I am not very confident about it, and if I am correctly applying the described approach at the start.
Is my assumption that $(1)$ is a normalized ket a necessary one in order to solve the problem of whether $(1)$ is entangled or separable?
In case the assumption is needed, is my end result correctly interpreted then?