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

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.

• The Schmidt decomposition asks you to find the states $|{u} \rangle$ and $|v \rangle$ in the subspace such that $|\Psi \rangle$ can be written as a linear combination of $|uu \rangle$ and $|vv \rangle$. The statement "the only requirement is that |i> and |j> are orthonormal" is not correct. Commented Mar 30, 2022 at 11:38
• @MariusLadegårdMeyer The statement "the only requirement is that |i> and |j> are orthonormal" refers to writing $|\Psi>=\sum A_{ij}|ij>$ (which I am fully aware is not the Schmidt decomposition) and then using $A$ to find the Schmidt decomposition Commented Mar 30, 2022 at 12:01
• @MariusLadegårdMeyer I also don't think I made it clear that I meant that $|i>$ is an orthonormal basis, and $|j>$ is too, not that they are orthonormal to each other. Commented Mar 30, 2022 at 12:08
• @NorbertSchuch I haven't worked it out explicitly as it looks like a lot of messy algebra, but I don't think so (did abs(my eigenvalue) - eigenvalue from solution on wolfram alpha and did not get 0). How can you tell they are the same up to a phase just by looking at it> Commented Mar 30, 2022 at 12:16
• If $A$ were Hermitian, an eigendecompostion would be perfectly fine, since you'd get orthogonal eigenvectors and real eigenvalues. But because it's not, the columns of $P$ in $PDP^{1}$ are not orthogonal in general and neither are the eigenvalues real. Furthermore, an eigendecomposition would not generalize to e.g. a qubit+qutrit. TL;DR: You need to do an SVD. Note that the squared singular values are the eigenvalues of $A^*A$. Commented Mar 30, 2022 at 12:39

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.
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 $$\lvert\alpha_k\rangle$$.
• @baker_man Well, as you can see from my answer when you use the SVD of $A$ you get a Schmidt decomposition. If you insert the same formula for the eigenvalue decomposition, you will not get a Schmidt decomposition, unless the basis transformation into the eigenbasis is unitary (i.e. $A$ is a normal matrix). Commented Apr 4, 2022 at 16:42