-2
$\begingroup$

I am trying to understand Schmidt decomposition. I am stuck in one sentence here. See the example picture.

See the picture here

Here, I can understand everything except the line "For both HA and HB the Schmidt basis is diagonal (Hadamard) basis." Can anyone help to understand what the author tried to tell here by using this line?

Improve this question
$\endgroup$
6
  • 3
    $\begingroup$ Please do not post images of text. $\endgroup$
    – Tobias Fünke
    Commented May 12, 2023 at 15:09
  • 1
    $\begingroup$ Where is this excerpt from? $\endgroup$
    – Norbert Schuch
    Commented May 12, 2023 at 15:22
  • $\begingroup$ @TobiasFünke why? Is it against the community rule? I was not aware of it. I have no other option to present my doubts here otherwise sharing the snip of the part... $\endgroup$
    – INDRANIL MAITI
    Commented May 12, 2023 at 20:29
  • $\begingroup$ @NorbertSchuch Elements of Quantum Computation and Quantum Communication by Anirban Pathak $\endgroup$
    – INDRANIL MAITI
    Commented May 12, 2023 at 20:30
  • 3
    $\begingroup$ @INDRANILMAITI transcribing such a short text is not a problem, especially when as you said it's only a very specific part that you don't understand $\endgroup$
    – Amit
    Commented May 12, 2023 at 20:31

1 Answer 1

Reset to default
2
$\begingroup$

"diagonal basis" is non-standard terminology.

On the other hand, "Hadamard basis" usually refers to the basis given by $$ \lvert +\rangle =\frac{\lvert 0\rangle + \lvert 1\rangle }{\sqrt{2}} \ ,\quad \lvert -\rangle =\frac{\lvert 0\rangle - \lvert 1\rangle }{\sqrt{2}}\ , $$

which is indeed the Schmidt basis of the state in the example for both Alice (with Hilbert space $H_A$) and Bob (with Hilbert space $H_B$).

Improve this answer
$\endgroup$

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