Undergrad Replace a Toffoli gate with an equivalent circuit

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The discussion revolves around the challenge of demonstrating the equivalence of a Toffoli gate with an equivalent quantum circuit. The user has attempted to show this by calculating the output for the input state |111⟩, but has not reached the expected conclusion of |110⟩. There is a suggestion that a mistake may have occurred in the calculations, particularly regarding the application of the V* gate when the second qubit is |0⟩. The user also notes that using bra-ket notation simplifies the process compared to matrix representations. Additionally, there is a mention of a potential error in the problem statement regarding the equivalent Toffoli gate as described in a referenced paper.
Peter_Newman
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Hello,

I am currently working on this task here. Given is a Toffoli gate. Now it is stated that there is an equivalent quantum circuit. I tried to show the equivalence, for example by looking at case ##|111\rangle##.

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However, with what I have calculated so far, I do not come to the conclusion that the circuits are equivalent.
I have rewritten the term V first:

$$V=(1-i)\frac{I+iX}{2}=\begin{pmatrix}0.5-0.5i&0.5+0.5i\\0.5+0.5i&0.5-0.5i\end{pmatrix}$$
Now I just start with the case ##|111\rangle##. This should, because it comes to the equivalence dof the Toffoli gate at the end of ##|110\rangle## come out.

$$|111\rangle\xrightarrow{V} |11\rangle\begin{pmatrix}0.5-0.5i&0.5+0.5i\\0.5+0.5i&0.5-0.5i\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=|11\rangle\begin{pmatrix}0.5+0.5i\\0.5-0.5i\end{pmatrix}\xrightarrow{CNOT}|10\rangle\begin{pmatrix}0.5+0.5i\\0.5-0.5i\end{pmatrix}$$

$$\left(V^T\right)^{*}=\begin{pmatrix}0.5+0.5i&0.5-0.5i\\0.5-0.5i&0.5+0.5i\end{pmatrix}$$

$$|10\rangle\begin{pmatrix}0.5+0.5i\\0.5-0.5i\end{pmatrix}\xrightarrow{\left(V^T\right)^{*}} |10\rangle\begin{pmatrix}0.5+0.5i&0.5-0.5i\\0.5-0.5i&0.5+0.5i\end{pmatrix}\begin{pmatrix}0.5+0.5i\\0.5-0.5i\end{pmatrix}=|10\rangle|1\rangle$$

$$|10\rangle|1\rangle\xrightarrow{CNOT} |11\rangle|1\rangle\xrightarrow{V}|11\rangle\begin{pmatrix}0.5-0.5i&0.5+0.5i\\0.5+0.5i&0.5-0.5i\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=|11\rangle\begin{pmatrix}0.5+0.5i\\0.5-0.5i\end{pmatrix}\neq |110\rangle$$

It could be that I made a mistake in my calculation. What do you all mean?
 
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Peter_Newman said:
$$|10\rangle\begin{pmatrix}0.5+0.5i\\0.5-0.5i\end{pmatrix}\xrightarrow{\left(V^T\right)^{*}} |10\rangle\begin{pmatrix}0.5+0.5i&0.5-0.5i\\0.5-0.5i&0.5+0.5i\end{pmatrix}\begin{pmatrix}0.5+0.5i\\0.5-0.5i\end{pmatrix}=|10\rangle|1\rangle$$
In this step since the second qubit has a value of ##\left |0\right>##, the ##V^*## gate should not be applied according to your circuit diagram.
 
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tnich said:
In this step since the second qubit has a value of ##\left |0\right>##, the ##V^*## gate should not be applied according to your circuit diagram .
By the way, it seems much easier to keep the bra-ket notation in this case rather than revert to matrices. For example, the first step becomes
$$|111\rangle\xrightarrow{V} |11\rangle\frac 1 2 (1-i)(I+iX)\left|1\right> = \frac 1 2 (1-i)|11\rangle(\left|1\right>+i\left|0\right>)$$
Also, I think there is an error in the problem statement. In paper that described this method, Elementary gates for quantum computation (PhysRevA.52.3457), the equivalent Toffoli gate is given as

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where in this case ##U=V^2=X##. Note that the last controlled(V) gate is controlled by a different bit than the first one.
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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