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?


3 Answers 3


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.

  • $\begingroup$ 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

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.cx(0, 1)

qc2 = QuantumCircuit(2)
qc2.cx(0, 1)

enter image description here

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

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.


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