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Let's consider a system composed of the tensor product of two qubits in the following states:

$$|\Psi\rangle = \alpha |00\rangle - \beta |11\rangle$$

$$\rho_{AB} = \rho_1|\Psi^{-}\rangle \langle \Psi^{-}| + \rho_2|\Psi^{+}\rangle \langle \Psi^{+}| + \rho_{3}|\psi^{-}\rangle \langle \psi^{-}| + \rho_{4}|\psi^{+}\rangle \langle \psi^{+}|$$

$$\rho_{AB} = \rho|\psi^{-}\rangle\langle \psi^{-}| + (1-\rho)|00\rangle\langle 00|$$

Where

$$|\Psi^{\pm}\rangle = \frac{1}{\sqrt{2}}\biggr(|00\rangle\pm\langle 11|\biggr)$$

$$|\psi^{\pm}\rangle = \frac{1}{\sqrt{2}}\biggr(|01\rangle\pm\langle 10|\biggr)$$

are different entangled states. If the two qubits are separate, as in Bell inequality experiment, how could we find the state of Alice's subsystem in the three states?

If I am not mistaken, $|\Psi^{\pm}\rangle = \frac{1}{\sqrt{2}}\biggr(|00\rangle\pm\langle 11|\biggr)$, $|\psi^{\pm}\rangle = \frac{1}{\sqrt{2}}\biggr(|01\rangle\pm\langle 10|\biggr)$ form a Bell basis for the quantum system $\mathcal{H}_{A}\otimes\mathcal{H}_{B}$. Then, could we approach this problem by representing $\rho_{AB}$ as a density matrix (by acting on Bell basis), and then by taking the partial trace $Tr_{A}(\rho_{AB})$?

Yau
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1 Answers1

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Your exposition of the problem isn’t terribly clear. I gather that the first index in the states and the subscript $A$ refer to Alice’s subsystem and the second index and the subscript $B$ presumably to Bob’s. The state of Alice’s subsystem is obtained by tracing over Bob’s state, not over Alice’s. I’m not sure why you want to do this in a Bell basis – that doesn’t seem particularly practical, as you already have the states expressed in an unentangled basis for the individual subsystems.

As an example, for the third state we have

\begin{eqnarray} \operatorname{Tr}_B(\rho_{AB}) &=& \operatorname{Tr}_B(\rho|\psi^-\rangle\langle\psi^-|+(1-\rho)|00\rangle\langle00|) \\ &=& \operatorname{Tr}_B\left(\frac\rho2(|01\rangle-|10\rangle)(\langle01|-\langle10|)+(1-\rho)|00\rangle\langle00|)\right) \\ &=& \operatorname{Tr}_B\left(\frac\rho2(|01\rangle\langle01|-|10\rangle\langle01|-|01\rangle\langle10|+|10\rangle\langle10|)+(1-\rho)(|00\rangle\langle00|\right) \\ &=& \frac12\rho(|0\rangle_A\langle0|_A+|1\rangle_A\langle1|_A)+(1-\rho)(|0\rangle_A\langle0|_A \\ &=& \left(1-\frac\rho2\right)|0\rangle_A\langle0|_A+\frac\rho2|1\rangle_A\langle1|_A\;, \end{eqnarray}

where the subscript $A$ indicates the state of Alice’s subsystem. Tracing over Bob’s state eliminates the off-diagonal terms $|10\rangle\langle01|$ and $|01\rangle\langle10|$.

joriki
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  • Yes, but does my way not also work? – Yau Jan 27 '23 at 18:50
  • @Yau: I'm not sure I understand what your way is. How would you trace over Bob's state in the Bell basis? It would help if you showed in the question how far you got in trying to apply it, and where you got stuck. – joriki Jan 27 '23 at 18:51
  • Yes, sure, I'll be doing so now. – Yau Jan 27 '23 at 18:52
  • I updated my question – Yau Jan 27 '23 at 19:04
  • I am simply trying to come up with the matrix representation of $\rho_{AB}$, and then take the partial trace $tr_{B}(\rho_{AB})$. – Yau Jan 27 '23 at 19:05
  • @Yau: What is "the" matrix representation of $\rho_{AB}$? You have $\rho_{AB}$ represented with respect to the $AB$ basis, which is exactly what you need in order to form the partial trace (as I did in the answer above). You could transform to a (not "the") different matrix representation, but that will only make it more difficult to take the trace – why not perform the trace in the matrix representation that you've already got? – joriki Jan 27 '23 at 19:08
  • Sure, that makes perfect sense! – Yau Jan 27 '23 at 22:19
  • @Yau: It may become clearer if you explicitly write the density in matrix form. If you write it with the rows and columns in the order $|00\rangle$, $|10\rangle$, $|01\rangle$, $|11\rangle$, then you have $2\times2$ blocks, and you can think of each block as a density matrix for Alice's subsystem. The trace over Bob's subsystem is performed by adding the two diagonal blocks and discarding the two off-diagonal blocks. – joriki Jan 27 '23 at 22:20
  • @Yau: Here it is explicitly:

    $$ \left(\begin{array}{cc|cc}1-\rho\&\frac\rho2&-\frac\rho2\\hline&-\frac\rho2&\frac\rho2\&&&0\end{array}\right);, $$

    so performing the trace by adding the diagonal blocks yields

    $$ \pmatrix{1-\rho\&\frac\rho2}+\pmatrix{\frac\rho2\&0}=\pmatrix{1-\frac\rho2\&\frac\rho2};. $$

     – joriki Jan 27 '23 at 22:29