# Compute the Schmidt Decomposition of a two-qubit state

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): $$\lambda_1[|00\rangle + |01\rangle + |10\rangle + |11\rangle] + \lambda_2[|00\rangle - |01\rangle - |10\rangle + |11\rangle]$$

• WP. Diagonalized right. Normalize. What's the question? – Cosmas Zachos Mar 22 '20 at 14:47

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 $$|v\rangle|w\rangle$$: $$|A\rangle = \frac{1+\sqrt3}{2\sqrt2}|+,+\rangle + \frac{1-\sqrt3}{2\sqrt2}|-,-\rangle.$$