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Is the following state entangled?

$\left| \psi \right> = \alpha_0 \beta_0 \left| 00 \right> + \alpha_0 \beta_1 \left| 01 \right> + 0 \left| 10 \right>+ \alpha_1 \beta_1 \left| 11 \right>$

I 'know' it is an NOT an entangled state, but I here's where I am a bit confused. It says $ \alpha_1 \beta_0 = 0$ which implies either $ \alpha_1 = 0$ or $\beta_0 =0$ , but that would be a contradiction. (since we assume $\left| 00 \right>$ and $\left| 11 \right>$ have non-zero probability amplitudes)

Also, if this is NOT an entangled state, how can I factor it to $\left| \phi_1 \right> \otimes \left| \phi_2 \right> $?

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    $\begingroup$ "It says $\alpha_1 \beta_0=0$". No, it doesn't. What it actually says is that the coefficient for $|10\rangle$ is zero, as you have written. You're probably confused because you've written the coefficients as if the state were a product state. $\endgroup$
    – DanielSank
    Oct 13, 2014 at 19:00

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It is an entangled state, as you concluded yourself it cannot be written as a direct tensor product. To recap again: $$ \left( \begin{array}{c} \alpha_0\\ \alpha_1\\ \end{array} \right) \otimes \left( \begin{array}{c} \beta_0\\ \beta_1\\ \end{array} \right) = \left( \begin{array}{c} \alpha_0 \beta_0\\ \alpha_0 \beta_1\\ \alpha_1 \beta_0\\ \alpha_1 \beta_1\\ \end{array} \right) $$ If $\alpha_1 \beta_0$ is set to 0, and the rest kept non-zero, writing $\left| \psi \right>$ as a direct tensor product fails, as there no $\alpha$ or $\beta$ satisfying these conditions at the same times.


More generally, when it comes to questions of type "is this state entangled or separable", it is much simpler to tackle it at the level of density matrices. For instance, given any pure 2-qubit state, such as yours, a simple criterion indicating a separable state would be to calculate the density matrix $\rho$, trace out (integrating out undesired degrees of freedom) $\rho$ w.r.t. the basis of one of the qubits to obtain the reduced density matrix of the other qubit, then calculate $\operatorname{Tr}(\rho_{\rm reduced}^2).$ The state is separable iff the latter is equal to $1,$ otherwise it is entangled. In other words, if the reduced density matrix has rank $1,$ then it is a pure state (as opposed to mixed), and thus the composite 2-qubit state must be a separable one.

Let's take a separable state and verify the above scheme. Suppose we have $|\psi\rangle=|\phi_1\rangle|\phi_2\rangle,$ then $\rho=|\psi\rangle \langle \psi|.$ The reduced density matrix of the first qubit is then obtained by tracing out in the basis of the second: $\rho_1=\operatorname{Tr_2}(\rho)=|\phi_1\rangle \langle \phi_1|.$ Squaring and tracing $\operatorname{Tr}(\rho_1^2)=1,$ thus, confirming that $\psi$ is indeed a separable state.

Specifically for your example, the normalization condition gives you a relation between the coefficients, namely, $|\alpha_0\beta_0|^2+|\alpha_0\beta_1|^2+|\alpha_1\beta_1|^2=1$ $:(1).$ Then by performing the above calculation (well-encouraged exercise), you should get to an equation of type $Tr(\rho_1^2)=|\alpha_0|^2|\beta_0|^2+|\alpha_1|^2|\beta_2|^2 <1,$ clearly smaller than $1,$ which directly follows from the normalization criterion $(1).$ Hence, $|\psi\rangle$ describes indeed an entangled state of the 2-qubit system.

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I'm not a fan of explicit calculations, so let us answer a bit more general question:

Given two two-dimensional Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ with basis $\lvert \uparrow \rangle_i$, $\lvert \downarrow \rangle_i$, when is a state in $\mathcal{H}_1 \otimes \mathcal{H}_2$ entangled?

Observe that $\lvert \psi \rangle_i = u_i\lvert \uparrow \rangle_i + d_i\lvert \downarrow \rangle_i$ is the most general state in a single space. Now take

$$\lvert \psi \rangle_1 \otimes \lvert \psi \rangle_2 = u_1u_2\lvert \uparrow \rangle_1\lvert \uparrow \rangle_2 + u_1d_2\lvert \uparrow \rangle_1\lvert \downarrow \rangle_2 + d_1u_2\lvert \downarrow \rangle_1\lvert \uparrow \rangle_2 + d_1d_2\lvert \downarrow \rangle_1\lvert \downarrow \rangle_2 $$

and take the most general state in the tensor product:

$$\lvert \phi \rangle = a_{uu}\lvert \uparrow \rangle_1\lvert \uparrow \rangle_2 + a_{ud}\lvert \uparrow \rangle_1\lvert \downarrow \rangle_2 + a_{du}\lvert \downarrow \rangle_1\lvert \uparrow \rangle_2 + a_{dd}\lvert \downarrow \rangle_1\lvert \downarrow \rangle_2 $$

Now, just set $\lvert \phi \rangle \overset{!}{=} \lvert \psi \rangle_1 \otimes \lvert \psi \rangle_2$ and obtain a system of equations by comparing coefficients. We thus obtain restrictions on the ratios of the coefficients of $\lvert \phi \rangle$ to be an entangled state (as long as the $a_{ii}$ in the denominator aren't zero), for example:

$$ \frac{a_{uu}}{a_{ud}} = \frac{a_{du}}{a_{dd}}$$

where is it obvious that your given state does not fulfill that for $\alpha_0\beta_0 \neq 0$. The general constraint that these non-entangled states (necessarily and sufficiently) fulfill is (convince yourself that there isn't an additional one)

$$ a_{uu}a_{dd} - a_{ud}a_{du} = 0$$

Note that this is not an equation defining a subspace - the non-entangled states do not form a subspace of the tensor product.


It is worth noting that we are here, as always as physicists, doing the "wrong" thing - we are thinking of Hilbert spaces when we should be thinking of projective Hilbert spaces, since states are only determined up to rays. The projective space associated with two-dimensional spaces is also called the Bloch sphere (it's a normal 2-sphere).

For projective spaces, the non-entangled spaces are precisely the image of the Segre embedding, in this case

$$ \iota: \mathbb{C}P^1 \times \mathbb{C}P^1 \to \mathcal{P}(\mathcal{H}_1 \otimes \mathcal{H}_2) \cong \mathbb{C}P^3, ([x_1 : y_1],[x_2 : y_2]) \mapsto [x_1x_2:x_1y_2:y_1x_2:y_1y_2] $$

whose image is seen to be the solution set (a variety in general, and a quadric in this case) to the equation

$$Z_0Z_3 - Z_1Z_2 = 0$$

where $[Z_0:Z_1:Z_2:Z_3] \in \mathbb{C}P^3$, reproducing our result from above.

Even in higher dimensional cases, the image of the Segre embedding remains a variety, with its defining equation yielding a sufficient and necessary criterion for non-entanglement. That projective spaces are the more natural setting for this discussion is evidenced by the fact that the Segre embedding makes the non-entanglement variety a proper subvariety of the projective tensor product.

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  • $\begingroup$ I think this could be clearer. What does $\stackrel{!}{=}$ mean? If it means "not equal" why not use $\neq$? Is OP's state entangled or not? $\endgroup$
    – DanielSank
    Oct 14, 2014 at 2:25
  • $\begingroup$ @DanielSank: $\overset{!}{=}$ is just a sign for an equality that is demanded rather than derived (I was under the impression it was commonplace, but it's not really necessary as its own sign, an ordinary $=$ will suffice.). I remark that OP's state is entangled after I state one of the ratios of coefficents one may write down. $\endgroup$
    – ACuriousMind
    Oct 14, 2014 at 10:34
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    $\begingroup$ The "where it is obvious" sentence is not so easy to parse. Furthermore, what I think you are saying in that sentence is not "obvious" because (since you used different notation than OP) it almost looks like you have a divide by zero. I understand it but OP and others who are learning things may not. $\endgroup$
    – DanielSank
    Oct 14, 2014 at 11:38
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This problem can be solved using Peres–Horodecki criterion (or PPT criterion). But the simple way to do this is to compute directly.

Suppose $|\phi⟩ = (A|0⟩ + B|1⟩)_A (C|0⟩ + D|1⟩)_B$. Compare the coeffient, we must have: $AC = α_0β_0, BC=0, AD= \alpha_0 \beta_1, BD= \alpha_1 \beta_1$.

If $\alpha_1, \beta_0, \beta_1$ all non-zero, the equations above cannot be satisfied because $ACADBD = \alpha_0^2 \alpha_1 \beta_0 \beta_1^2 \neq 0$ but $ACADBD = A^2 D^2 BC =0$, which is a contradiction. Thus it is a entangled state.

If $\alpha_0 =0$, the state above is obviously an product state.

If $\alpha_0 \neq 0, \beta_1=0$, the state above is obviously an product state.

If $\alpha_0 \neq 0, \beta_0=0, \beta_1 \neq 0$, the state above is obviously an product state.

The above discussion have exhausted all the situations. Thus, the state is an entangled state if and only if $\alpha_1, \beta_0, \beta_1$ all non-zero.

By the way, the general theorem of Peres can be used to solve this problem. You can consult Peres–Horodecki criterion (or PPT criterion). Although in general this is just a necessary condition, it can be proved in $2 \times 2$ or $2 \times 3$ cases this is also a sufficient condition.

The projection on bell state cannot be used to justify whether a state is entangled or not. For example, the state $|00⟩ + |01⟩ \sim |\Phi_+⟩ - |\Phi_-⟩ + |\Psi_+⟩ -|\Psi_-⟩$. But it is obviously a product state. This is because bell states are the complete basis thus ANY state can be expressed as the linear conbination of bell state.

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The spin is entangled if $\alpha_1\beta_1\neq0$ because a measurement on the first spin gives you information on the expected measurement values on the second. In particular if you get a 1 in the first spin you know you will get a 1 in the second spin. And knowing that you got a 0 in the first spin will changes the probablility of the measurement of spin 2. This is formalized by the reduced density matrix. If you trace out the degree of freedom associated with spin 1, then you get a non-diagonal density matrix for the spin 1. $$\rho_1=|\alpha_0|^2(|\beta_0|^2+|\beta_1|^2)|0\rangle\langle 0|+\alpha_0\alpha_1^*|\beta_1|^2|0\rangle\langle 1|+\alpha_0^*\alpha_1|\beta_1|^2|1\rangle\langle 0|+|\alpha_1|^2|\beta_1|^2|1\rangle\langle 1|$$ This tells you that the state cannot be factored into a pure state in bit 1 times a pure state in bit 2.

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