I have the Hamiltonian evolution operator for a two qubit system :

\begin{equation} \hat{U}= e^{-i\gamma H}\;,\quad H = \frac{1}{2}(I - Z\otimes Z) \end{equation} where the $Z$ gate is applied on both qubits, ($I=I\otimes I$).

I would like to express this evolution operator in terms of CNOT gate and rotations.

I was told that $e^{i \frac{\gamma}{2}Z\otimes Z}$ could be decomposed as follows : \begin{equation} e^{i \frac{\gamma}{2}Z\otimes Z}=\text{CX}\;(I\otimes R_z(\gamma))\;\text{CX}\;,\quad R_z(\gamma) = e^{-i\frac{\gamma}{2}Z} \end{equation} where the CNOT gate operates on the second qubit each time (first qubit being the control one) and the rotation applies to the second one.

I found a related question here but I wonder if this can be simply demonstrated with the CNOT gate decomposition : \begin{equation} \text{CX}= e^{i\frac{\pi}{4}(I -Z)\otimes (I-X)} \end{equation}

The development is the following : \begin{equation} \begin{split} \text{CX}\;(I\otimes e^{-i\frac{\gamma}{2}Z})\;\text{CX} &= e^{i\frac{\pi}{4}(I -Z)\otimes (I-X)}\;(I\otimes e^{-i\frac{\gamma}{2}Z})\; e^{i\frac{\pi}{4}(I -Z)\otimes (I-X)}\\ &=e^{i\frac{\pi}{4}(I -Z)\otimes (I-X)}\;e^{-i\frac{\gamma}{2}I\otimes Z}\; e^{i\frac{\pi}{4}(I -Z)\otimes (I-X)} \end{split} \end{equation}

Then, I am not sure how to proceed, it seems to me not trivial to end up with $e^{i \frac{\gamma}{2}Z\otimes Z}$. This is the question. If the matrices within exponential do not commute, I don't think I can go further.

  • $\begingroup$ You seem to have already answered your own question, so what are you actually asking? $\endgroup$
    – DaftWullie
    Feb 23 at 12:09
  • $\begingroup$ I tried to demonstrate the second formula by expanding CX gates with its decomposition in Pauli basis, but i was neither sure of the developments nor of if it is really achievable in this way. The related question I mentioned takes another way by using another expression of the CNOT gate. $\endgroup$
    – deb2014
    Feb 23 at 12:58

1 Answer 1


To verify the action of this circuit, I suggest computing how it acts on each of the 4 basis states $|00\rangle,|01\rangle,|10\rangle$ and $|11\rangle$.

For example $$ |00\rangle\xrightarrow{CX}|00\rangle\xrightarrow{R_Z}e^{-i\gamma/2}|00\rangle\xrightarrow{CX}e^{-i\gamma/2}|00\rangle. $$ Work this out for all 4 basis states. Is this consistent with what your gate is supposed to do?

You've asked for an algebraic method. While I don't entirely understand what is sought, here is another method...

Let $P_0=|0\rangle\langle 0|$ and $P_1=|1\rangle\langle 1|$. Then $$ CX=P_0\otimes I+P_1\otimes X $$ and $$ I\otimes R_Z=(P_0+P_1)\otimes R_Z. $$ It is important to recognise that $P_0P_1=0$. Hence, when I multiply together the 3 circuit elements, I get $$ P_0\otimes R_Z+P_1\otimes(XR_ZX)=P_0\otimes R_Z+P_1\otimes(R_Z^\dagger)=\frac12(I+Z)\otimes R_Z+\frac12(I-Z)\otimes R_Z^\dagger $$ Regroup the terms, $$ I\otimes(R_Z+R_Z^\dagger)+Z\otimes (R_Z-R_Z^\dagger)=I\otimes I\cos(\gamma/2)-i\sin(\gamma/2)Z\otimes Z=e^{-i\gamma Z\otimes Z/2} $$ but this all seems rather painful!

  • $\begingroup$ Yes, this is the way I was told the equality. My question was : is there an algebraic way to demonstrate this equality. $\endgroup$
    – deb2014
    Feb 23 at 14:27
  • $\begingroup$ But what are the bounds on what you mean by that? Why not simply write out the matrices and multiply them? $\endgroup$
    – DaftWullie
    Feb 23 at 14:38

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