I came across a quantum circuit very similar to the phase estimation circuit, which is shown below:

enter image description here

In the phase estimation algorithm we assume, that we can efficiently implement an operator $U$, which performs the following operation:

$$ U|u\rangle \equiv e^{2 \pi i \phi} |u\rangle, $$

where $|u\rangle$ is the eignevector of $U$ and $e^{2 \pi i \phi}$ is the corresponding eigenvalue. Such an operator can be denoted in the matrix form as

$$ U \equiv \begin{bmatrix} e^{2 \pi i \phi} & 0 \\ 0 & e^{2 \pi i \phi}\end{bmatrix} $$

and its controlled version can be written as

$$ CU \equiv \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & e^{2 \pi i \phi} & 0 \\ 0 & 0 & 0 & e^{2 \pi i \phi}\end{bmatrix} $$

(the $CU^j$ gate will only have $e^{2 \pi i \phi j}$ in places of $e^{2 \pi i \phi}$). For such a case I was still able to express the resulting vector (after going through the $CU^j$ gate) as the tensor product of two vectors. Thanks to this, I was able to see what is the effect of applying the inverse quantum Fourier transform on one of these vectors.

Now let's say, that $U$ is not diagonal and its each entry is different from 0. In such a case, I think there appears strong entanglement between qubits in the first and the second register. Because of this (from the very definition of entanglement) I wasn't able to express the resulting vector as tensor product of some compound vectors and I don't know, what will be the result of applying the inverse quantum Fourier transform on the first register.

All of the above is just one example showing my real problem - how to analyze quantum circuits, where qubits (or registers) are highly entangled?


1 Answer 1


The $U$ used in phase estimation is not only a diagonal matrix with the same diagonal elements. Instead, it is an arbitrary unitary matrix.

The way that you analyse it, instead, is that the input $|u\rangle$ is specifically chosen to be an eigenvector of $U$. That means $U|u\rangle=e^{i\phi}|u\rangle$. But there are different eigenvectors with different eigenvalues. For example, perhaps $|v\rangle$ with $U|v\rangle=e^{i\theta}|v\rangle$.

Now, the way that you've analysed it tells you that $|0\rangle|u\rangle\mapsto|\phi\rangle|u\rangle$ (assuming $\phi$ is of the form $2\pi k/2^t$ for integer $k$). But it tells you that you get the same effect for every eigenvector, so $|0\rangle|v\rangle\mapsto|\theta\rangle|v\rangle$.

So, what happens for an arbitrary input that is not an eigenvector of $U$? The eigenvectors form a basis, so we can write any input state as a superposition of the different eigenvectors. And by linearity we know how each of those components evolve. For example, $$ |0\rangle(\alpha|u\rangle+\beta|v\rangle)\rightarrow \alpha|\phi\rangle|u\rangle+\beta|\theta\rangle|v\rangle. $$ This will, typically, be highly entangled, but the entanglement itself does not necessarily cause a problem in the analysis.

More generally, what's the answer? Well, ultimately, you might not be able to analyse these circuits. Part of the point of quantum circuits is that you cannot easily simulate them on a classical computer. But there are plenty of strategies such as the one above (picking a basis of states which are easy to analyse), or the Gottesman-Knill theorem, that let you deal with specific situations when there's lots of entanglement present.


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