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With just single qubit gate in your circuit, you can only generate a small subset of quantum states. In fact, the states that you can generate are called separable states. These states have no entanglement in them.

Here is an example to see why you need to be able to generate entangle state to have successful VQE calculation, supposed you have the Hamiltonian $H = \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \end{pmatrix} $

And you running VQE to find the lowest eigenvalue to this Hamiltonian. This means you need to be able to generate the eigenstate that correspond to the lowest eigenvalue. Now, if you look at all the eigenstates of $H$ and their correspondence eigenvalues:

$$|\psi_1 \rangle = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ 1/\sqrt{2} \end{pmatrix} \rightarrow 1, |\psi_2 \rangle = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ -1/\sqrt{2}\end{pmatrix} \rightarrow -1, |\psi_3 \rangle = \begin{pmatrix} 0 \\ 1/\sqrt{2}\\ 1/\sqrt{2} \\ 0\end{pmatrix} \rightarrow 1, |\psi_4 \rangle = \begin{pmatrix} 0 \\ -1/\sqrt{2} \\ 1/\sqrt{2} \\ 0\end{pmatrix} \rightarrow -1$$

Here you should notice that these states are all entangled. So if you want your VQE algorithm to find the lowest eigenvalue (-1) in this case, it must be able to generate an entangled state. That means if you design your Ansatze only made up from $RY$ rotation, for example:

then you can't never generate such entangled state, which means you can never reach the minimum eigenvalue. So the complexity of the Ansatz depends on the complexity of the Hamiltonian that you consider.

But you are right about the fact that given a diagonal Hamiltonian, like those that you are interested in then the eigenstate will be one of the computational basis state! Thus, you do not need entanglement gate or any sort... in fact, you only need $X$ gates to generate such states.

And since you know the solution you are looking for is in one of these states, it is quite easy to classically go through all the calculations of the expectation values (depending on how many terms in your cost function Hamiltonian). Although, you do have to do $2^n$ check, these checks are very efficient or easy... you just look at the parity of the Pauli string and this can be done very quickly.... So this makes me wonder, giving the sophistication of classical computer systems (like this cluster), how many qubits do we need for quantum computer to beat classical computer? Classical computers can go through this check so fast... even if there are $2^n$ terms... these are easy check...

With just single qubit gate in your circuit, you can only generate a small subset of quantum states. In fact, the states that you can generate are called separable states. These states have no entanglement in them.

Here is an example to see why you need to be able to generate entangle state to have successful VQE calculation, supposed you have the Hamiltonian $H = \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \end{pmatrix} $

And you running VQE to find the lowest eigenvalue to this Hamiltonian. This means you need to be able to generate the eigenstate that correspond to the lowest eigenvalue. Now, if you look at all the eigenstates of $H$ and their correspondence eigenvalues:

$$|\psi_1 \rangle = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ 1/\sqrt{2} \end{pmatrix} \rightarrow 1, |\psi_2 \rangle = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ -1/\sqrt{2}\end{pmatrix} \rightarrow -1, |\psi_3 \rangle = \begin{pmatrix} 0 \\ 1/\sqrt{2}\\ 1/\sqrt{2} \\ 0\end{pmatrix} \rightarrow 1, |\psi_4 \rangle = \begin{pmatrix} 0 \\ -1/\sqrt{2} \\ 1/\sqrt{2} \\ 0\end{pmatrix} \rightarrow -1$$

Here you should notice that these states are all entangled. So if you want your VQE algorithm to find the lowest eigenvalue (-1) in this case, it must be able to generate an entangled state. That means if you design your Ansatze only made up from $RY$ rotation, for example:

then you can't never generate such entangled state, which means you can never reach the minimum eigenvalue. So the complexity of the Ansatz depends on the complexity of the Hamiltonian that you consider.

But you are right about the fact that given a diagonal Hamiltonian, like those that you are interested in then the eigenstate will be one of the computational basis state! Thus, you do not need entanglement gate or any sort... in fact, you only need $X$ gates to generate such states. But then this means you have to iterate through $2^n$ states... which is not ideal.

And since you know the solution you are looking for is in one of these states, it is quite easy to classically go through all the calculations of the expectation values (depending on how many terms in your cost function Hamiltonian). You do have to do $2^n$ check, so it's not ideal either, but these checks are very easy... you just look at the parity of the Pauli string and this can be done very quickly.... So this makes me wonder, giving the sophistication of classical computer systems (like this cluster), how many qubits do we need for quantum computer to beat classical computer? Classical computers can go through this check so fast... but since the number of terms to check do scale as $2^n$... we will reach a bottle neck at some point. But how many qubits is that going to take before quantum computer wins out? 100 or few hundreds?

With just single qubit gate in your circuit, you can only generate a small subset of quantum states. In fact, the states that you can generate are called separable states. These states have no entanglement in them.

Here is an example to see why you need to be able to generate entangle state to have successful VQE calculation, supposed you have the Hamiltonian $H = \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \end{pmatrix} $

And you running VQE to find the lowest eigenvalue to this Hamiltonian. This means you need to be able to generate the eigenstate that correspond to the lowest eigenvalue. Now, if you look at all the eigenstates of $H$ and their correspondence eigenvalues:

$$|\psi_1 \rangle = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ 1/\sqrt{2} \end{pmatrix} \rightarrow 1, |\psi_2 \rangle = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ -1/\sqrt{2}\end{pmatrix} \rightarrow -1, |\psi_3 \rangle = \begin{pmatrix} 0 \\ 1/\sqrt{2}\\ 1/\sqrt{2} \\ 0\end{pmatrix} \rightarrow 1, |\psi_4 \rangle = \begin{pmatrix} 0 \\ -1/\sqrt{2} \\ 1/\sqrt{2} \\ 0\end{pmatrix} \rightarrow -1$$

Here you should notice that these states are all entangled. So if you want your VQE algorithm to find the lowest eigenvalue (-1) in this case, it must be able to generate an entangled state. That means if you design your Ansatze only made up from $RY$ rotation, for example:

then you can't never generate such entangled state, which means you can never reach the minimum eigenvalue. So the complexity of the Ansatz depends on the complexity of the Hamiltonian that you consider.

But you are right about the fact that given a diagonal Hamiltonian, like those that you are interested in then the eigenstate will be one of the computational basis state! Thus, you do not need entanglement gate or any sort... in fact, you only need $X$ gates to generate such states.

And since you know the solution you are looking for is in one of these states, it is quite easy to classically go through all the calculations of the expectation values calculation (depending on how many terms in your cost function Hamiltonian). Although, you do have to do $2^n$ check, these checks are very efficient or easy... you just look at the parity of the Pauli string and this can be done very quickly.... So this makes me wonder, giving the sophistication of classical computer systems (like this cluster), how many qubits do we need for quantum computer to beat classical computer? Classical computers can go through this check so fast... even if there are $2^n$ terms... these are easy check...

With just single qubit gate in your circuit, you can only generate a small subset of quantum states. In fact, the states that you can generate are called separable states. These states have no entanglement in them.

Here is an example to see why you need to be able to generate entangle state to have successful VQE calculation, supposed you have the Hamiltonian $H = \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \end{pmatrix} $

And you running VQE to find the lowest eigenvalue to this Hamiltonian. This means you need to be able to generate the eigenstate that correspond to the lowest eigenvalue. Now, if you look at all the eigenstates of $H$ and their correspondence eigenvalues:

$$|\psi_1 \rangle = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ 1/\sqrt{2} \end{pmatrix} \rightarrow 1, |\psi_2 \rangle = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ -1/\sqrt{2}\end{pmatrix} \rightarrow -1, |\psi_3 \rangle = \begin{pmatrix} 0 \\ 1/\sqrt{2}\\ 1/\sqrt{2} \\ 0\end{pmatrix} \rightarrow 1, |\psi_4 \rangle = \begin{pmatrix} 0 \\ -1/\sqrt{2} \\ 1/\sqrt{2} \\ 0\end{pmatrix} \rightarrow -1$$

Here you should notice that these states are all entangled. So if you want your VQE algorithm to find the lowest eigenvalue (-1) in this case, it must be able to generate an entangled state. That means if you design your Ansatze only made up from $RY$ rotation, for example:

then you can't never generate such entangled state, which means you can never reach the minimum eigenvalue. So the complexity of the Ansatz depends on the complexity of the Hamiltonian that you consider.

But you are right about the fact that given a diagonal Hamiltonian, like those that you are interested in then the eigenstate will be one of the computational basis state! Thus, you do not need entanglement gate or any sort... in fact, you only need $X$ gates to generate such states.

And since you know the solution you are looking for is in one of these states, it is quite easy to classically go through all the calculations of the expectation values (depending on how many terms in your cost function Hamiltonian). Although, you do have to do $2^n$ check, these checks are very efficient or easy... you just look at the parity of the Pauli string and this can be done very quickly.... So this makes me wonder, giving the sophistication of classical computer systems (like this cluster), how many qubits do we need for quantum computer to beat classical computer? Classical computers can go through this check so fast... even if there are $2^n$ terms... these are easy check...

With just single qubit gate in your circuit, you can only generate a small subset of quantum states. In fact, the states that you can generate are called separable states. These states have no entanglement in them.

Here is an example to see why you need to be able to generate entangle state to have successful VQE calculation, supposed you have the Hamiltonian $H = \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \end{pmatrix} $

And you running VQE to find the lowest eigenvalue to this Hamiltonian. This means you need to be able to generate the eigenstate that correspond to the lowest eigenvalue. Now, if you look at all the eigenstates of $H$ and their correspondence eigenvalues:

$$|\psi_1 \rangle = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ 1/\sqrt{2} \end{pmatrix} \rightarrow 1, |\psi_2 \rangle = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ -1/\sqrt{2}\end{pmatrix} \rightarrow -1, |\psi_3 \rangle = \begin{pmatrix} 0 \\ 1/\sqrt{2}\\ 1/\sqrt{2} \\ 0\end{pmatrix} \rightarrow 1, |\psi_4 \rangle = \begin{pmatrix} 0 \\ -1/\sqrt{2} \\ 1/\sqrt{2} \\ 0\end{pmatrix} \rightarrow -1$$

Here you should notice that these states are all entangled. So if you want your VQE algorithm to find the lowest eigenvalue (-1) in this case, it must be able to generate an entangled state. That means if you design your Ansatze only made up from $RY$ rotation, for example:

then you can't never generate such entangled state, which means you can never reach the minimum eigenvalue. So the complexity of the Ansatz depends on the complexity of the Hamiltonian that you consider.

But you are right about the fact that given a diagonal Hamiltonian, like those that you are interested in then the eigenstate will be one of the computational basis state! Thus, you do not need entanglement gate or any sort... in fact, you only need $X$ gates to generate such states.

And since you know the solution you are looking for is in one of these states, it is quite easy to classically go through all the calculations of the expectation values calculation (depending on how many terms in your cost function Hamiltonian). Although, you do have to do $2^n$ check, these checks are very efficient or easy... you just look at the parity of the Pauli string and this can be done very quickly.... So this makes me wonder, giving the sophistication of classical computer systems, how many qubits do we need for quantum computer to beat classical computer? Classical computers can go through this check so fast... even if there are $2^n$ terms... these are easy check...

With just single qubit gate in your circuit, you can only generate a small subset of quantum states. In fact, the states that you can generate are called separable states. These states have no entanglement in them.

Here is an example to see why you need to be able to generate entangle state to have successful VQE calculation, supposed you have the Hamiltonian $H = \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \end{pmatrix} $

And you running VQE to find the lowest eigenvalue to this Hamiltonian. This means you need to be able to generate the eigenstate that correspond to the lowest eigenvalue. Now, if you look at all the eigenstates of $H$ and their correspondence eigenvalues:

$$|\psi_1 \rangle = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ 1/\sqrt{2} \end{pmatrix} \rightarrow 1, |\psi_2 \rangle = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ -1/\sqrt{2}\end{pmatrix} \rightarrow -1, |\psi_3 \rangle = \begin{pmatrix} 0 \\ 1/\sqrt{2}\\ 1/\sqrt{2} \\ 0\end{pmatrix} \rightarrow 1, |\psi_4 \rangle = \begin{pmatrix} 0 \\ -1/\sqrt{2} \\ 1/\sqrt{2} \\ 0\end{pmatrix} \rightarrow -1$$

Here you should notice that these states are all entangled. So if you want your VQE algorithm to find the lowest eigenvalue (-1) in this case, it must be able to generate an entangled state. That means if you design your Ansatze only made up from $RY$ rotation, for example:

then you can't never generate such entangled state, which means you can never reach the minimum eigenvalue. So the complexity of the Ansatz depends on the complexity of the Hamiltonian that you consider.

But you are right about the fact that given a diagonal Hamiltonian, like those that you are interested in then the eigenstate will be one of the computational basis state! Thus, you do not need entanglement gate or any sort... in fact, you only need $X$ gates to generate such states.

And since you know the solution you are looking for is in one of these states, it is quite easy to classically go through all the calculations of the expectation values calculation (depending on how many terms in your cost function Hamiltonian). Although, you do have to do $2^n$ check, these checks are very efficient or easy... you just look at the parity of the Pauli string and this can be done very quickly.... So this makes me wonder, giving the sophistication of classical computer systems (like this cluster), how many qubits do we need for quantum computer to beat classical computer? Classical computers can go through this check so fast... even if there are $2^n$ terms... these are easy check...