adamcallison
commented
2 years ago I understand that the VQE computes the expectation value of a Hamiltonian operator (or Rayleigh Ritz quotient)
Yes, and these are equivalent, as the state is normalised.
according to the standard variational Ritz method. The trial states are prepared by the reconfigurable circuit by assigning values to phase shifters
The notion of a "phase shifter" is specific to photonic chips like the one in the Peruzzo paper. Other devices will implement different gates, but it is usually the case that only the single-qubit gates are parameterised (here, the phase shifters), while the entangling gates are fixed (here, the CNOT, which I think is implemented through $d_6$ to $d_8$).
according to some parameters fed by a classical optimizer. Each module of the quantum processor actually computes the expectation value of a tensor product of Pauli operators, the Hamiltonian being decomposed as a sum of such products.
Yes. They don't say it explicitly, but it seems likely that, for this experiment, each module is just a separate set of runs on the same chip ( but measuring a different Pauli operators).
Similar to QPE, the computational quantum advantage boils down to the representation of the quantum state, which for a q-bit system may be represented in a superposition of basis states. A classical computer working on the $2^n$ amplitude coefficients need $O(2^n)$ floating operations to evaluate an expectation value.
The source of a quantum advantage is fairly deep philosophical issue, but I don't think it's right to say that quantum speedup is entirely explained by the state representation. For many problems, there is not expected to be any quantum speedup, regardless of how you encode your input. For QPE, the specific, well-crafted algorithm gives you the speedup (although it does of course require the input state to be given as an actual quantum state. If there are advantages in VQE, the discussion of why is still very much open.
On the other hand, the quantum processor directly operates on the q-bit state for which $O(1/p^2)$ repeated measurements are necessary to get an expectation value with precision $p$.
Yes, although perhaps it's worth noting that for a specific set of parameters, performing the repeats can be done quite quickly.
Considering that the minimizer scales linearly with the number of parameters (Nelder-Mead requiring $N+1$ evaluations at each update step), the overall quantum complexity is $O(n^r M |h^2| p^{-2})$ in the quantum case against $O(M2^n(2^n)^r)$ for classical case.
I've never really considered the big-O notation scalings for VQEs. I'm not sure what $M$ and $r$ are here, but I can believe what you've written. However, it's worth nothing that there's usually no formal guarantee of correct outputs for VQEs.
It's also worth noting that, although they've chosen Nelder-Mead here, other classical optimization methods could be used. Since repetitions are being used to estimate expectation values, the so-called "shot noise" that arises from not operating in the infinite-shot limit can sometimes be harmful for gradient-based optimizers. Gradient-free optimizers are sometimes considered, add are stochastic gradient-based optimizers. One that I see a lot in VQE papers is called Adam (no relation).
- I wonder if $n$ is the actual number of q-bit? The prototype works with 2-qbit but 6 parameters are being optimized. I can see from the circuit that the number of physical gates should scale linearly with the number of qbits. This is provided that the number of phase shifters per qbit does not scale with larger systems. Then the number of optimized parameters should indeed scale linearly with the number of qbits.
I don't have much to say here, other than that I agree it's not clear. Ultimately, to have good results from this algorithm, the parameterised circuit should be sufficiently expressible; that is, by varying the parameters, you should be able to generate output states (pre-measurement) that cover much of the Hilbert space (or at least some relevant part of the Hilbert space), so that it's likely to be able to express a state that is reasonably close to the target ground-state. Given that adding more qubits exponentially increases the size of the Hilbert space, it seems intuitive that you would need to increase the number of parameterised gates per qubit (and perhaps also the number of fixed CNOT gates) to maintain expressibility.
This brings me to some questions with respect to the parametrization of the state :
- In the supplementary information, the proposed mapping uses 6 parameters to represent a 2 q-bit state. Do you know (or have references about) how this parametrization evolve for a 3,4,5 q-bit system represented by 8,16,32 amplitudes respectively?
No. By simple counting arguments, one in principle needs $2(2^n - 1)$ real-valued degrees of freedom to be able to express the entire Hilbert space of $n$ qubits. This is because there are $2^n$ complex-valued amplitudes in a (pure) quantum state on $n$ qubits, but one-degree of freedom is fixed by normalisation and another constitutes an irrelevant global phase. This happens to much up with 6 parameters for 2 qubits, but I doubt this is why they chose 6.
- The coupled cluster is mentioned in the paper and references as a special ansatz intractable classically. Is this ansatz somehow used and related to this mapping?
I don't think so. The (unitary) coupled cluster ansatz is an example of a "problem-inspired" ansatz circuit that is particularly suited to quantum chemistry problems. It's thought that this should make the classical optimization easier for some problems, perhaps because states in Hilbert space that are of the right form to be quantum-chemistry ground-states will be closer together in parameter space. By preparing a suitable initial state (typically a classical approximation of the ground-state that ignores quantum correlations), you ansatz is likely to prepare the right sort of states,, and the classical optimization is easier.
Unfortunately, problem-inspired ansatze are often difficult to implement in current devices, so a lot of research looks at "hardware-inspired" anstze instead. This essentially just means building a circuit that the device can easily perform, as has been done here. This approach can lead to the problem of "barren plateaus", where there are regions of parameter space where the gradients become very small, and the optimizer can get lost. Mitigating this effect is an active topic of research.
- 6 parameters are mapped to 8 physical components, the phase shifters, among which 2 are used to account for the tensor product of Pauli operators (in the supplementary information $\phi_5 $ and $\phi_6 $ are set to $\pi $ to actually represent the identity operator). Do you know how the mapping changes for the different tensor products $\sigma_i $ and $\sigma_i \otimes \sigma_j $ ? Would higher order tensor products require more channels and phase shifters?
I don't understand their mapping well enough to talk about the required phases for the measurements. Setting $\phi_5 $ and $\phi_6 $ to $\pi $ is done analytically in their calculation in order to find the mapping between the state parameters and the circuit parameters.
In fact, it seems that actually 4 of the 8 shifters are used to set the measurement basis; $\phi{5} $ and $\phi{6} $ are used exclusively for this, while $\phi_7 $ and $\phi_8 $ are given parameters that combine state preparation settings and measurement settings. The key point is that the physical measurement is only performed in one basis, probably the $\sigma_Z $ basis. If so, then to do a logical measurement in the $\sigma_Z $ basis, the "measurement phases" on the relevant qubit would be set to $\pi $ so that the identity is performed. If a logical measurement is required in the $\sigma_X $ basis, then I'm guessing the measurement phases would be set to $\pi/2 $ which may perform a Hadamard gate. These can be combined to produce measurements of tensor-products of Pauli operators.
To produce higher order products, you would need more qubits for them to be acting on, so yes you would need more channels. However, sometimes Hamiltonians on more than 2 qubits can just be written with one- and two-body terms. In these cases, it's possible to schedule your measurements so that multiple compatible measurement operators can be performed on the same run, in order to reduce the total number of runs.
ronandrevon
I understand that the VQE computes the expectation value of a Hamiltonian operator (or Rayleigh Ritz quotient) according to the standard variational Ritz method. The trial states are prepared by the reconfigurable circuit by assigning values to phase shifters according to some parameters fed by a classical optimizer. Each module of the quantum processor actually computes the expectation value of a tensor product of Pauli operators, the Hamiltonian being decomposed as a sum of such products.
Similar to QPE, the computational quantum advantage boils down to the representation of the quantum state, which for a $n$ q-bit system may be represented in a superposition of $2^n$ basis states. A classical computer working on the $2^n$ amplitude coefficients need $O\bigl((2^n)$ floating operations to evaluate an expectation value. On the other hand, the quantum processor directly operates on the q-bit state for which $O(1/p^2)$ repeated measurements are necessary to get an expectation value with precision $p$. Considering that the minimizer scales linearly with the number of parameters (Nelder-Mead requiring $N+1$ evaluations at each update step), the overall quantum complexity is $O\bigl(n^r M |h^2|p^{-2}\bigr)$ in the quantum case against $O\bigl(M 2^n (2^n)^r \bigr)$ for classical case.
This brings me to some questions with respect to the parametrization of the state :