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From the Qiskit textbook I read about Simon's algorithm. There are two n-wide quantum registers, so the general state is given by

$$|x\rangle_n|y\rangle_n$$

where x and y are the $2^n-1$ binary representations. A function from the n-subspace into n-subspace id defined by

$$f: |x\rangle\mapsto|f(x)\rangle$$

Now, the "query function" is given by the operation

$$|x\rangle|a\rangle \rightarrow |x\rangle |a \oplus f(x)\rangle$$

All building blocks in quantum computation shall be unitary transformations, so this mapping is unitary too - but how can I prove that this is really the case?

Is it enough to show, that norm is conserved like this:

Denoting

$$U|x\rangle|a\rangle = |x\rangle |a \oplus f(x)\rangle$$

I would have

$$\langle x| \langle a|U^\dagger = \langle x| \langle a \oplus f(x)|$$

so $$\langle x| \langle a|U^\dagger U |x\rangle |a\rangle= \langle x|x\rangle \langle a \oplus f(x)|a \oplus f(x)\rangle = 1 \cdot 1 = 1$$

But this appears a bit too trivial to be a prove...

glS
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MichaelW
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1 Answers1

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The easiest way to prove this is to apply the operator twice $$ |x\rangle|a\rangle\mapsto |x\rangle|a\oplus f(x)\rangle\mapsto |x\rangle|a\oplus f(x)\oplus f(x)\rangle=|x\rangle|a\rangle. $$ So, two applications of the function is the identity operation. Thus, the function is its own inverse, $U=U^{-1}$. Moreover, the eigenvalues must all be $\pm 1$, which shows that, for example, $U=U^\dagger$, and hence $UU^\dagger=I$, the standard relation for a unitary.

Incidentally, for what you tried to prove, is it clear that because, in a certain basis, all the lengths of those basis states are preserved, that the lengths of all possible superpositions of those states are also preserved?

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DaftWullie
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    Those two conditions combined clearly gives $U U^\dagger=I$. The first, $U=U^{-1}$ is clear. But why eigenvalues of this operation are $\pm 1$ from which one follows $U=U^\dagger$? – MichaelW Oct 25 '21 at 11:01
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    Oh yes, of course: $U^2|\psi\rangle = \lambda^2 |\Psi\rangle$. From this I get trivially $\lambda = \pm 1$. – MichaelW Oct 25 '21 at 11:15
  • Why is my original approach wrong? When the norm is conserved in one orthogonal base, I thought, by insertion of the unit-operator, this is true in any base. The property "unitary" cannot depend on the base I use - where am I wrong? – MichaelW Oct 25 '21 at 19:45
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    Consider the operator $P=\sqrt{2}|+\rangle\langle +|$. This has $P|0\rangle=|+\rangle$ and $P|1\rangle=|+\rangle$. For both computational basis states of a qubit, the length is preserved. But for other states, such as $|+\rangle$, it is not. It should be sufficient to additionally prove that orthogonality of your outputs is preserved. – DaftWullie Oct 26 '21 at 06:39
  • OK - now I got it... My mistake was by assuming, that length conservation of base vectors means automatically length conservation of a general vector. By your example I can see that this is not the case. However, in my case, this assumption is incidentally true, because there is a functional one-to one mapping from computational base to computational base, so $\langle x_i, a_j+f(x_i) | x_{i'}, a_{j'}+f(x_{i'})\rangle = \delta_{i,i'}\delta_{j j'} $ – MichaelW Oct 27 '21 at 09:11
  • I still do not fully understand your other approach: From Eigenvalues of U equal to ±1 one cannot automatically deduce, that U is unitary. E.g for the matrix $\left(\begin{array}{cr}1 & 1 \ 0 & -1\end{array}\right)$ this is not the case, although Eigenvalues are ±1. – MichaelW Oct 27 '21 at 09:19
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    It's true I skipped over that (it wasn't obvious to me from the question what the appropriate level of detail was). It might be easiest to take a look at https://quantumcomputing.stackexchange.com/questions/5779/how-to-prove-that-the-query-oracle-is-unitary – DaftWullie Oct 27 '21 at 09:48
  • Yes, this link describes exactly my question. Thanks to you, my question is already demystified, but in terms of "learning" its always better to find out something by yourself in the hard way instead of reading in a convenient way, what other people thought about ;-) – MichaelW Oct 27 '21 at 10:34