How to compute the Schmidt decomposition of a two-qubit state?

Question

Trying a to do the Schmidt decomposition of $|\Psi\rangle = \frac{1}{2}(|00\rangle+|01\rangle+|10\rangle+e^{i\phi}|11\rangle)$. The solutions I'm looking at do it by first finding the partial density matrix and diagonalising that, but from reading the textbook I thought it'd be easier to just diagonalise the matrix $A$ where $|\Psi\rangle = \sum A_{ij}|ij\rangle$. As far as I can tell, the only requirement is that $|i\rangle$ and $|j\rangle$ are orthonormal basis, which is true for the $|\Psi\rangle$ given. I think $A$ would be

$$A= \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & e^{i \phi} \end{pmatrix}.$$ However, when I find the eigenvalues for this matrix, they aren't real and they don't agree with the solutions in the textbook (which got the answer from the other method). I tried putting it into wolfram alpha, and the eigenvalues there agree with what I get.

Why doesn't this method work for finding the Schmidt decomposition? I think it should, and I've even seen other Stack Exchange posts giving this method as a way of doing the decomposition. Most likely, I'm making a mistake, or an assumption in my working that is wrong and I'd like some help figuring out what it is.

Answer 1

Given $\lvert\psi\rangle = \sum_{ij} A_{ij} \lvert i,j\rangle$, you want to write $A$ in a singular value decomposition, $$ A_{ij} = \sum_k U_{ik}\sigma_k V^*_{jk}\ , $$ with isometries $U$ and $V$. Then, \begin{align} \lvert\psi\rangle &= \sum_{ij} A_{ij} \lvert i,j\rangle \\ & = \sum_{k} \sigma_k \left(\sum_ i U_{ik} \lvert i\rangle\right)\left(\sum_j V^*_{jk}\lvert j\rangle\right) \\ & = \sum_k \sigma_k \lvert\alpha_k\rangle\lvert \beta_k\rangle\ , \end{align} where $$\lvert \alpha_k\rangle := \sum_ i U_{ik} \lvert i\rangle$$ and $$\lvert \beta_k\rangle := \sum_ i V^*_{jk} \lvert j\rangle$$ are orthonormal bases, since $U$ and $V$ are isometries.

$\lvert\psi\rangle = \sum_k \sigma_k \lvert\alpha_k\rangle\lvert \beta_k\rangle$ is thus the Schmidt decomposition.

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Of course, if $A$ is unitarily diagonalizable (i.e., it is a normal matrix), then the singular value decomposition coincides with the eigenvalue decomposition, and you can use the latter. But in general, the transformation in the eigenvalue decomposition is not unitary, and will thus not give rise to an orthonormal basis $\lvert\alpha_k\rangle$.