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I'm trying to compute the schmidt composition of $$ |A\rangle = \frac{1}{2 \sqrt{2}}(|00\rangle + \sqrt{3}|01\rangle + \sqrt3 |10\rangle + |11\rangle) $$ I've calculated the eigenvalues to be $$ 1+\sqrt{3} ; 1-\sqrt{3} $$ And eigenvectors to be $(1, 1)$ and $(1, -1)$ respectively. I'm struggling turning theory to an actual solution to this problem and am unsure where to go from here being that (due to school shutdown and viral things) I have no real examples of solving these problems. I am unsure where to go from here, if I could have some inspiration, that would be very helpful (ie. you don't have to solve the problem necessarily, though it would be helpful to see a solution).

Edit: I think I got it, feel free to correct if you see something off. I'm mostly doing this because I didn't find many written out walk throughs. Using $\lambda_1$ and $\lambda_2$ to represent the two eigenvalues (given above), I found the formula: $$ |A _{schmidt}\rangle = \lambda_1|u_1\rangle|v_1\rangle + \lambda_2|u_2\rangle|v_2\rangle $$ Where $UDV = A$; $D$ is the diagonalized form of $A, U$ and $V$ are conjugate transposes of one another $$ U= \begin{matrix} U_{11} & U_{12} \\ U_{21} & U_{22} \\ \end{matrix} $$ $V$'s construction is analogous, except with "$V$'s" $$ |u_i\rangle = U_{1i}|0\rangle + U_{2i}|1\rangle \\ |v_i\rangle = V_{i1}|0\rangle + V_{i2}|1\rangle \\ i \in {1,2} $$ Going by this methodology, I got the schmidt decomposition to be (with lambdas being the eigenvalues): $$ \lambda_1[|00\rangle + |01\rangle + |10\rangle + |11\rangle] + \lambda_2[|00\rangle - |01\rangle - |10\rangle + |11\rangle] $$

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  • $\begingroup$ WP. Diagonalized right. Normalize. What's the question? $\endgroup$ – Cosmas Zachos Mar 22 at 14:47
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Writing the state in the form $|A\rangle=\sum_{ij}c_{ij}|ij\rangle$, you can associate to each set of coefficients a matrix $C\equiv (c_{ij})_{ij}$. In your case, this is $$C=\frac{1}{2\sqrt2}\begin{pmatrix}1 & \sqrt3 \\ \sqrt3 & 1\end{pmatrix}.$$ You want to find the SVD of this matrix. In this case this amounts to diagonalising it, because the matrix is Hermitian. You have $$C=\frac{1+\sqrt3}{2\sqrt2}|+\rangle\!\langle+| + \frac{1-\sqrt3}{2\sqrt2}|-\rangle\!\langle-|.$$ The Schmidt decomposition can now be obtained directly by converting each operator $|v\rangle\!\langle w|$ into $|v\rangle|w\rangle$: $$ |A\rangle = \frac{1+\sqrt3}{2\sqrt2}|+,+\rangle + \frac{1-\sqrt3}{2\sqrt2}|-,-\rangle. $$

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