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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?

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In essence the calculations are correct, though those numbers $𝑎^2_i, 𝑏^2_i$ must be taken with modulus, because they are in fact $a_ia_i^*, b_ib_i^*$. In general $𝑎^2_0$ is a complex number, so you can't write $𝑎^2_0>0$.

Normalization is required when we consider quantum states (also, otherwise, $\operatorname{Tr}(\rho_1^2)$ can be arbitrary close to 0). With normalization you have showed that $\operatorname{Tr}(\rho_1^2) < 1$, hence the state must be entangled.

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  • $\begingroup$ Thank you for the very useful remarks! $\endgroup$ – user929304 Feb 13 at 8:46
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I haven’t checked the content of your calculation, but, yes, conceptually that’s all correct. The idea of testing the purity and comparing to 1 is entirely predicated on notion that your initial state is normalised, so you did exactly the right thing.

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  • $\begingroup$ Thanks very much! $\endgroup$ – user929304 Feb 13 at 8:47

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