-1
$\begingroup$

I would like to know if this one qbit quantum gates exists:

2 0
0 0.5

Or any quantum gates with x and y such as :

x>y>0 and

x 0
0 y

The first gate is reversible so i think it makes sense yet every one qbit gate have value of the same norm.

Thank you.

$\endgroup$
0

2 Answers 2

1
$\begingroup$

Quantum gates have to be unitary and yours is not. There are a number of ways to define this, but one is that it preserves the inner product. The norm that you reference is also preserved because the norm is the square root of the inner product. Another way to define a unitary operator is through its matrix. If U is a unitary matrix, then $U^{-1}=U^\dagger$.

$\endgroup$
2
  • $\begingroup$ Thank you for your answer ! $\endgroup$ Jul 28 at 7:15
  • $\begingroup$ NP! @078766554567, please mark it as the accepted answer if you're good with it :) $\endgroup$
    – LeWoody
    Jul 28 at 17:46
0
$\begingroup$

The requirement on a gate is that it ought to be unitary (and linear, although, the linearity condition is satisfied by construction since it's understood that gates are matrices). This requirement comes from the fact that the evolution of a quantum state is unitary in quantum mechanics.

It is true that a unitary gate is reversible, although, not every reversible gate is unitary. In particular, unitarity means $U^\dagger U = I\implies U^{-1}=U^\dagger$, i.e., a unitary matrix is not only invertible but that the inverse of the matrix will be equal to its own adjoint. Thus, unitarity is a stronger condition than reversibility.

Now, your first gate is indeed reversible -- but it is not unitary because its inverse ${\rm diag}(1/2, 2)$ is not equal to its adjoint ${\rm diag}(2,1/2)$. More broadly, if a matrix of the form ${\rm diag}(x,y)$ is unitary then it must be the case that $\vert x\vert=\vert y\vert=1$. You can verify that this is precisely what is needed (and is sufficient) to ensure that $U^\dagger U=I$ for $U={\rm diag}(x,y)$. In other words, the diagonal entries of a diagonalized unitary matrix ought to be of unit norm, i.e., the eigenvalues of a unitary matrix are of unit norm!

PS: To explicitly answer your second question, the only unitary gate of the kind you describe would be the identity matrix. The only non-negative real number of unit norm is $1$, and thus, the only unitary matrix with non-negative real eigenvalues is the identity.

$\endgroup$
1
  • $\begingroup$ Thank you it is really clear! $\endgroup$ Jul 28 at 7:15

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.