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There have been a few other questions about this section of Nielsen and Chuang, but when working through the output of the circuit, there are some inconsistencies that are probably due to some mistep/false assumption of mine, and I can't seem to figure it out.

Suppose we have the Hamiltonian $$ H = Z_1 ⊗ Z_2 ⊗ \cdots ⊗ Z_n,\tag{4.113}$$ which acts on an $n$ qubit system. Despite this being an interaction involving all of the system, indeed, it can be simulated efficiently. What we desire is a simple quantum circuit which implements $e^{-iH\Delta t}$, for arbitrary values of $\Delta t$. A circuit doing precisely this, for $n = 3$, is shown in Figure 4.19. The main insight is that although the Hamiltonian involves all the qubits in the system, it does so in a

classicalmanner: the phase shift applied to the system is $e^{-i\Delta t}$ if theparityof the $n$ qubits in the computational basis is even; otherwise, the phase shift should be $e^{i\Delta t}$. Thus, simple simulation of $H$ is possible by first classically computing the parity (storing the result in an ancilla qubit), then applying the appropriate phase shift conditioned on the parity, then uncomputing the parity (to erase the ancilla).

Considering the case of n=3, let's say we had $|\psi\rangle = a|000\rangle + b|110\rangle$. The above circuit is supposed to apply $e^{-iZ\otimes Z\otimes Z \Delta t}$ to the three qubits. Using what I've read on tensor products,

$$e^{-iZ\otimes Z\otimes Z t}|\psi\rangle = e^{Z\otimes Z\otimes (-iZ \Delta t)}(a|000\rangle + b|110\rangle)$$

$$= a(e^Z|0\rangle e^Z|0\rangle e^{-iZ \Delta t}|0\rangle)+b(e^Z|1\rangle e^Z|1\rangle e^{-iZ \Delta t}|0\rangle)$$

Using Taylor Series, I found that $$e^Z|0\rangle = e|0\rangle$$ $$e^Z|1\rangle = \frac{1}{e}|1\rangle$$ $$e^{-iZ \Delta t}|0\rangle = e^{-i\Delta t}$$

$$e^{-iZ\otimes Z\otimes Z t}|\psi\rangle = a(e|0\rangle e|0\rangle e^{-i \Delta t}|0\rangle)+b(\frac{1}{e}|1\rangle \frac{1}{e}|1\rangle e^{-i \Delta t}|0\rangle)$$ $$= ae^2e^{-i \Delta t}(|0\rangle |0\rangle |0\rangle)+b\frac{1}{e^2}e^{-i \Delta t}(|1\rangle |1\rangle |0\rangle) \tag{eq. 1}$$

Once I did this, I attempted to evaluate what the circuit above, presented in Nielsen and Chuang, does to the vector $|\psi\rangle$.

$$|\psi \rangle = (a|000\rangle + b|110\rangle)|0\rangle$$

Since both $|000\rangle$ and $|110\rangle$ are even, we apply $e^{-iZ\Delta t}$ to the aux 4th qubit, so we get:

$$|\psi \rangle = (a|000\rangle + b|110\rangle)e^{-iZ\Delta t}|0\rangle$$

And since we know that $e^{-iZ\Delta t}|0\rangle =e^{-i\Delta t}|0\rangle$

$$|\psi \rangle = (a|000\rangle + b|110\rangle)e^{-i\Delta t}|0\rangle$$ $$= a(e^{-i\Delta t})(|000\rangle) + b(e^{-i\Delta t})(|110\rangle) \tag{eq. 2}$$

But this equation 2 isn't equivalent to equation 1! What exactly am I missing here? The output of the circuit isn't matching up with what happens when I actually put the state $|\psi\rangle$ through that transformation.

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Another way of applying this operator (without ancillary qubit) can be found here https://quantumcomputing.stackexchange.com/a/11373/9459.

– Davit Khachatryan – 2020-04-26T10:41:23.367Related question: https://quantumcomputing.stackexchange.com/q/1875/9459

– Davit Khachatryan – 2020-04-26T11:05:11.410