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I am relatively new and interested in quantum computing. Specifically, I am interested in transforming an equation that I found on Wikipedia. But I did not quite understand the transformation.

$ \frac{1}{\sqrt{2}}(\left|0\right>_x\left|0\right>_y-\left|1\right>_x\left|1\right>_y) = \frac{1}{\sqrt{2}}(|+\rangle_x|-\rangle_y+|-\rangle_x|+\rangle_y) $

My idea is so far to use Hadamard transform for the two qubits:

$ \frac{1}{\sqrt{2}}(H(\left|0\right>_x\left|0\right>_y)-H(\left|1\right>_x\left|1\right>_y)) $

I have used the Hadamard transformation and now come to this:

$ = \frac{1}{\sqrt{2}}(\frac{1}{2}[(\left|00\right>+\left|10\right>+\left|01\right>+\left|11\right>) -(\left|00\right>-\left|10\right>-\left|01\right>+\left|11\right>)]) $

If I simplify that a bit now then I have that as a result:

$ = \frac{1}{\sqrt{2}}(\left|1_x0_y\right>+\left|0_x1_y\right>) $

But the result looks different now than the equation I wrote down at the beginning:

$ = \frac{1}{\sqrt{2}}(\left|1_x0_y\right>+\left|0_x1_y\right>) = \frac{1}{\sqrt{2}}(|+\rangle_x|-\rangle_y+|-\rangle_x|+\rangle_y) $

I do not know if the forming is allowed that way. If somebody knows how the transformation of the equation works, so that I get what I wrote in the beginning, I would be very happy if somebody could explain it!

I hope that my question is understandable :)

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You are not proving the equality in a correct way. By multiplying by the Hadamard matrix, you are changing the state you are trying to calculate, not demonstrating the equality you want to prove.$\def\ket#1{\lvert#1\rangle}$

In order to prove what you state at the beginning of the question, I would use the facts that $$ \ket+ =\frac{1}{\sqrt{2}}\bigl(\ket0 + \ket1\bigr), \qquad\ket-=\frac{1}{\sqrt{2}}\bigl(\ket0 - \ket1\bigr),$$ and then develop the equality in the inverse order. Consequently: $$ \begin{alignat}{2} \frac{1}{\sqrt{2}}&\bigl(\ket+_x \ket-_y - \ket-_x \ket+_y\bigr)\mspace{-128mu} \\[1ex]=\;& \frac{1}{\sqrt{2}}\Bigl(\frac{1}{2}\bigl(&&\ket0_x+\ket1_x\bigr)\bigl(\ket0_y-\ket1_y\bigr)+\bigl(\ket0_x-\ket1_x\bigr)\bigl(\ket0_y+\ket1_y\bigr)\Bigr) \\[1ex]=\;& \frac{1}{\sqrt{2}}\Bigl(\frac{1}{2}\bigl(&&\ket0_x\ket0_y+\ket1_x\ket-_y-\ket0_x\ket1_y-\ket1_x\ket1_y \\[-1ex]&&&+\ket0_x\ket0_y+\ket0_x\ket1_y-\ket1_x\ket0_y-\ket1_x\ket1_y\bigr)\Bigr) \\=\;& \frac{1}{\sqrt{2}}\bigl(\ket0_x\ket0_y - \ket1_x\ket1_y\bigr).\mspace{-128mu} \end{alignat} $$

And so you prove the equality you were trying to solve at the beginning of the question.

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  • $\begingroup$ It might be interesting to expand on how this property is called rotational invariance and has some extremely important implications for the EPR paradox and so forth. $\endgroup$ – ahelwer Oct 31 '18 at 17:14
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You've started wanting to talk about $$ (|00\rangle-|11\rangle)/\sqrt{2}, $$ but you've then gone ahead and calculated $$ (H\otimes H)\cdot(|00\rangle-|11\rangle)/\sqrt{2}. $$ You should not expect these to be equal.

On the other hand, the calculation that you've done is sensible if you understand what you're doing. What you actually want to see, in order to verify your original statement is that $$ (H\otimes H)\cdot(|00\rangle-|11\rangle)/\sqrt{2}=(H\otimes H)\cdot(|+-\rangle+|-+\rangle)/\sqrt{2}. $$ You've already calculated the left-hand side. The calculation on the right-hand side is, I suspect, the whole reason why you've chosen to apply the Hadamard transform - you know that the Hadamard converts $|+\rangle$ to $|0\rangle$, and $|-\rangle$ to $|1\rangle$. So, you can immediately read that $$ (H\otimes H)\cdot(|+-\rangle+|-+\rangle)/\sqrt{2}=(|01\rangle+|10\rangle)/\sqrt{2}, $$ so you can see that the left-hand side is equal to the right-hand side, as desired.

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  • $\begingroup$ Ok, so if I understand that correctly, I would just have to apply the Hadamard transformation on the right as well, and I would have achieved what I wanted to show? $\endgroup$ – P_Gate Oct 31 '18 at 9:41
  • $\begingroup$ @P_Gate Yes, that's correct. It's like starting with a statement 1=1. If you multiply just one side by 5, the statement becomes untrue; you have to do it to both sides. $\endgroup$ – DaftWullie Oct 31 '18 at 9:49
  • $\begingroup$ Thanks for the clarification. I was just too stupid. It is an equation, i. I also have to do what I do on the left side on the right side. Stupid to forget something like that. :) Sorry that I can not upvote your answer :( (not enough rep) $\endgroup$ – P_Gate Oct 31 '18 at 9:51

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