$|\psi\rangle = 1/ \sqrt{3}( |0\rangle| 0\rangle + |0\rangle |1\rangle + |1\rangle |1\rangle) $

I absolutely cannot figure out the Schmidt decomposition of this state. I have looked at a ton of examples and have done the calculation multiple times, but my final result does not seem to be correct, as the coefficients do not add up to 1.

My attempt looks like this:

I calculated the density matrix to be: $\rho_{AB}=\frac{1}{3}(|00\rangle\langle00|+|00\rangle\langle01|+|00\rangle\langle11|+\\ |01\rangle\langle00|+|01\rangle\langle01|+|01\rangle\langle11|+|11\rangle\langle00|+|11\rangle\langle01|+|11\rangle\langle11|)$

Which gives me the reduced density matrices: $\rho_{A}=\rho_{B}=\frac{1}{3}(|0\rangle\langle0|+|1\rangle\langle0|+|1\rangle\langle0|+|1\rangle\langle1|)=\begin{pmatrix} \frac{1}{3} & 0\\ \frac{2}{3} & \frac{1}{3}\\ \end{pmatrix}$

This gives me the eigenvalue and corresponding eigenvector: $\lambda=\frac{1}{3}$ and $v=|1\rangle$

Can someone tell me where I am going wrong here? Because the Schmidt decomposition $|\psi\rangle=\frac{1}{\sqrt{3}}|1\rangle|1\rangle$ I would get from this is obviously not correct.


  • 6
    $\begingroup$ you should check your calculation for $\rho_A$, since the matrix you have is not hermitian and cannot be a density operator. $\endgroup$
    – forky40
    Commented May 2 at 15:17

1 Answer 1


One way of computing the decomposition is through density matrices, but then you will have to diagonalize those density matrices. This requires the eigenvalue decomposition of each density matrix. Overall, the general approach to Schmidt decomposition is its mathematical equivalent, Singular Value Decomposition (SVD).

Schmidt Decomposition is simply SVD applied to a quantum state. Given a state $$|\psi\rangle = \sum_{ij} a_{ij} |i\rangle|j\rangle,$$ we want to find the SVD of the matrix $a_{ij}$. In your case, we have $a_{00}=\frac{1}{\sqrt{3}}, a_{01}=\frac{1}{\sqrt{3}}, a_{10}=0$ and $a_{11}=\frac{1}{\sqrt{3}}$. So, we have: $$ A=\begin{pmatrix} \frac{1}{\sqrt{3}} && \frac{1}{\sqrt{3}}\\ 0 && \frac{1}{\sqrt{3}}\\ \end{pmatrix}. $$ Performing SVD on $A$ yields a decomposition $$A = U \Sigma V^{\dagger}.$$ where $U$ and $V$ are unitary matrices and $\Sigma$ is a diagonal matrix of singular values of $A$. Having found the SVD decomposition, you can now find the Schmid decomposition.

Let $\{|0_\mathcal{A}\rangle\, |1_\mathcal{A}\rangle\}$ and $\{|0_\mathcal{B}\rangle\, |1_\mathcal{B}\rangle\}$ be orthonormal eigenbases for qubit $\mathcal{A}$ and qubit $\mathcal{B}$. Then, using the SVD above we can compute the bases: $$|i_\mathcal{A}\rangle = u_{0i}|0\rangle+u_{1i}|1\rangle, \text{ for } i =0,1$$ and $$|i_\mathcal{B}\rangle = v_{i0}|0\rangle+v_{i1}|1\rangle, \text{ for } i =0,1$$ The Schmid-decomposed state is then $$|\psi\rangle = \sum_{i=0}^1 \sigma_{i}|i_\mathcal{A}\rangle|i_\mathcal{B}\rangle,$$ where $\sigma_i = \Sigma_{ii}$ are singular values of the matrix $A$. As the Schmidt decomposition requires, we will have $\sum_{i=0}^1|\sigma_i|^2 = 1$.

Computing the SVD for the given $A$ is similar to computing the eigenvalue decomposition, i.e. you need to find eigenvalues of $A^{T}A$ and then find corresponding eigenvectors.

If you really want to know the answer, you can go to the Wolfram Alpha website and type in Singular value decomposition {{1/sqrt{3}, 1/sqrt{3}},{0,1/sqrt{3}}}.

If you follow the above steps, you will learn that the purity of $Tr((\rho_\mathcal{A})^2)$ and $Tr((\rho_\mathcal{B})^2)$ is $\frac{7}{9}$ (see page 109 of Nielsen and Chuang), which is obtained by performing $Tr(\Sigma^4)$.

For reference, see Nielsen and Chuang, page 109.


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