1
$\begingroup$

I have these 2 quantum circuits:

Are they equivalent?I think they are but I cannot understand how this could be possible.Lets assume that the initial condition of the first circuit is:]

Lets assume that gate A transforms any qubit $ \begin{pmatrix} f\\ e \end{pmatrix} $ into: $ \begin{pmatrix} f'\\ e' \end{pmatrix} $ and gate B transforms any 2 qubit system from: $ \begin{pmatrix} gi\\ gj\\ hi\\ hj \end{pmatrix} $ into: $ \begin{pmatrix} gi\\ gj\\ hi''\\ hj'' \end{pmatrix} $

Lets take the first system.From a initial condition of:

$ \begin{pmatrix} ac\\ ad\\ bc\\ bd \end{pmatrix} $ after gate A it becomes: $ \begin{pmatrix} a'c\\ a'd\\ b'c\\ b'd \end{pmatrix} $ and after controlled gate B it becomes: $ \begin{pmatrix} a'c\\ a'd\\ b'c''\\ b'd'' \end{pmatrix} $ Now lets take the second system.From a initial condition of: $ \begin{pmatrix} ac\\ ad\\ bc\\ bd \end{pmatrix} $ after gate B it becomes:

$ \begin{pmatrix} ac\\ ad\\ bc''\\ bd'' \end{pmatrix} $ and after gate A it becomes: $ \begin{pmatrix} a'c\\ a'd\\ b'c''\\ b'd'' \end{pmatrix} $ so the 2 are equivalent?But this cannot be correct , intuitively it is not correct.Where am I wrong here?

$\endgroup$

3 Answers 3

3
$\begingroup$

Your problem is that your notation is leading you astray. In your first way of writing the circuit, you have the gate $B$ achieving the change in amplitudes $b'c\rightarrow b'c''$ while in the second you have $bc\rightarrow bc''$ assuming that those two $c''$ should be the same. There is no reason that they should be.

To illustrate this more concretely, consider $A$ and $B$ both being the not gate, and both qubits start in $|0\rangle$. In the first case, you get $$ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}\rightarrow\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}\rightarrow \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} $$ while in the second case, you get $$ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}\rightarrow\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}\rightarrow \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}. $$ The initial conditions for the controlled-B gate are different between the two cases, so give different outputs.

$\endgroup$
4
  • $\begingroup$ But the gate is the same so they should be the same.Well obviously the initial condition is the same and the gates are the same. $\endgroup$ Mar 28 at 16:05
  • $\begingroup$ but the initial conditions are not the same. $b'c$ and $bc$ are different initial conditions. $\endgroup$
    – DaftWullie
    Mar 28 at 16:26
  • $\begingroup$ u mean before the gate B is acted on the system?True but I dont see why it would make such a big difference. $\endgroup$ Mar 28 at 16:29
  • $\begingroup$ It makes all the difference. $\endgroup$
    – DaftWullie
    Mar 28 at 16:32
1
$\begingroup$

The circuits are not equivalent in general. As an example, take $A = B = X$. The unitary matrices representing the circuits are different:

from qiskit import QuantumCircuit
from qiskit.quantum_info import Operator

qc1 = QuantumCircuit(2)
qc1.x(0)
qc1.cx(0, 1)
display(qc1.draw())
display(Operator(qc1).draw('latex'))


qc2 = QuantumCircuit(2)
qc2.cx(0, 1)
qc2.x(0)
display(qc2.draw())
display(Operator(qc2).draw('latex'))

enter image description here

$\endgroup$
1
  • $\begingroup$ Hi.Can you spot my mistake in the calculations? $\endgroup$ Mar 28 at 16:19
1
$\begingroup$

No, these two circuits won't be equivalent in general.

Reasoning with states might not be the easiest way to understand it, I found expressing $A$ and $B$ in terms of matrices more clear.

Consider $A = \begin{pmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{pmatrix}$ and $B = \begin{pmatrix}b_{11}&b_{12}\\ b_{21}&b_{22}\end{pmatrix}$.

The $A$ gate applied to the first qubit correspond to the operator (written in block matrix notation):

$$ \begin{pmatrix} a_{11}I_2 & a_{12}I_2 \\ a_{21}I_2& a_{22}I_2 \end{pmatrix} $$

where $I_2$ is the $2\times2$ identity matrix.

The controlled-$B$ gate has matrix representation:

$$ \begin{pmatrix} I_2 & 0 \\ 0& B \end{pmatrix} $$

Hence, the first circuit seen overall as a matrix is:

$$ \begin{pmatrix} a_{11}I_2 & a_{12}I_2 \\ a_{21}B& a_{22}B \end{pmatrix} $$

Whereas the second circuit is:

$$ \begin{pmatrix} a_{11}I_2 & a_{12}B \\ a_{21}I_2& a_{22}B \end{pmatrix} $$

As a result, the circuits can only be equivalent if $a_{21}I_2 = a_{21}B$ and $a_{12}I_2 = a_{12}B$.

This condition is never met unless $a_{21} = a_{12} = 0$ i.e. the matrix $A$ is diagonal (or in the trivial case $B=I_2$).

You can check that taking $A=B=X$ doesn't give you an equivalence between the circuit, but taking $A=Z$ and $B=X$ does as $Z$ is a diagonal matrix.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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