8

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 :)

Sanchayan Dutta
  • 17,945
  • 8
  • 50
  • 112
P_Gate
  • 678
  • 3
  • 17

2 Answers2

9

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.

Niel de Beaudrap
  • 12,522
  • 1
  • 33
  • 73
Josu Etxezarreta Martinez
  • 4,136
  • 15
  • 42
6

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.

DaftWullie
  • 62,671
  • 4
  • 55
  • 140