1
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.