Given the quantum state $$|\chi\rangle=\dfrac{1}{||A||}\sum_{i=0}^{m-1}||A_i|||A_i\rangle|i\rangle,$$ how can we obtain the partial trace operation on the first register, i.e., $$\begin{align}\text{tr}_1(|\chi\rangle\langle\chi|)=&\dfrac{1}{||A||^2}\sum_{i,j=0}^{m-1}||A_i||\cdot||A_j|| \langle A_j|A_i\rangle|i\rangle\langle j|,\\\underbrace{=}_{??}&\dfrac{A^TA}{\text{tr}(A^TA)}\end{align}$$

As I understand, we have $$\sum_{i,j=0}^{m-1}||A_i||\cdot||A_j|| \langle A_j|A_i\rangle|i\rangle\langle j|=\sum_{i,j=0}^{m-1}c_{ij}|i\rangle\langle j|=C,$$ where $C=[c_{ij}]$ is the matrix whose elements are $c_{ij}$. So, I have to show that the elements of the matrix $(A^TA)_{ij}$ is equal to $||A_i||\cdot||A_j||\langle A_j|A_i\rangle$? Is this correct? How to do this?


1 Answer 1


You don't state this explicitly, but I'm guessing this is the crucial part: How do $|A_i\rangle$ relate to $A$?

I assume that $|A_i\rangle$ correspond to normalized rows, $i$ of matrix $A$, while $\|A_i\|$ is the weight of the row $i$ such that $\|A_i\||A_i\rangle$ would have all the elements corresponding to the $i^{th}$ row of matrix $A$.

In other words, $$ |A_i\rangle=\frac{1}{\|A_i\|}\sum_jA_{ij}|j\rangle. $$ Now, let's consider your specific calculation $$ \sum_{i,j}\|A_i\|\|A_j\|\langle A_j|A_i\rangle|i\rangle\langle j|=\sum_{i,j}\sum_kA_{ik}A_{jk}^*|i\rangle\langle j|=\sum_{i,j}\langle i|A A^\dagger|j\rangle|i\rangle\langle j|=AA^\dagger $$

If you want the outcome to be $A^\dagger A$, you might want to carefully check my assumption about the definition of $|A_i\rangle$. If your matrix $A$ is real, the hermitian conjugate becomes a transpose.


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