Mentioned e.g. at:

These are two conflicting constraints:

Ciro Santilli distinctly remembers being taught that at basic electrical engineering school during Ciro Santilli's undergrad studies at the University of São Paulo.

It really allows you to do alternating current calculations much as you'd do DC calculations with resistors, quite poweful. It must have been all the rage in the 1950s.

Synonym to gate-based quantum computer/digital quantum computer?

TODO confirm: apparently in the paradigm you can choose to measure only certain output qubits.

This makes things irreversible (TODO what does reversibility mean in this random context?), as opposed to Circuit-based quantum computer where you measure all output qubits at once.

TODO what is the advantage?

As of 2022, this tends to be the more "default" when you talk about a quantum computer.

But there are some serious analog quantum computer contestants in the field as well.

A quantum circuit is a graph of quantum logic gates.

Quantum circuits are the prevailing model of quantum computing as of the 2010's - 2020's

We don't need to understand a super generalized version of tensor products to know what they mean in basic quantum computing!

Intuitively, taking a tensor product of two qubits simply means putting them together on the same quantum system/computer.

When we write the bra-ket notation: $|00⟩$ that is the same as $|0⟩\otimes |0⟩$.

The tensor product is called a "product" because it distributes over addition.

E.g. consider:

Intuitively, in this operation we just put a Hadamard gate qubit together with a second pure $|0⟩$ qubit.

And the outcome still has the second qubit as always 0, because we haven't made them interact.

The quantum state $\frac{|00⟩+|10⟩}{\sqrt{2}}$ is called a separable state, because it can be written as a single product of two different qubits. We have simply brought two qubits together, without making them interact.

If we then add a CNOT gate to make a Bell state:we can now see that the Bell state is non-separable: we've made the two qubits interact, and there is no way to write this state with a single tensor product. The qubits are fundamentally entangled.

At Section "Quantum computing is just matrix multiplication" we saw that making a quantum circuit actually comes down to designing one big unitary matrix.

We have to say though that that was a bit of a lie.

Quantum programmers normally don't just produce those big matrices manually from scratch.

Instead, they use quantum logic gates.

The following are the main reasons for that:

One key insight, is that the matrix of a non-trivial quantum circuit is going to be huge, and won't fit into any amount classical memory that can be present in this universe.

This is because the matrix is exponential in the number qubits, and $2^{100}$ is more than the number of atoms in the universe!

Therefore, off the bat we know that we cannot possibly describe those matrices in an explicit form, but rather must use some kind of shorthand.

But it gets worse.

Even if we had enough memory, the act of explicitly computing the matrix is not generally possible.

This is because knowing the matrix, basically means knowing the probability result for all possible $2^N$ outputs for each of the $2^N$ possible inputs.

But if we had those probabilities, our algorithmic problem would already be solved in the first place! We would "just" go over each of those output probabilities (OK, there are $2^N$ of those, which is also an insurmountable problem in itself), and the largest probability would be the answer.

So if we could calculate those probabilities on a classical machine, we would also be able to simulate the quantum computer on the classical machine, and quantum computing would not be able to give exponential speedups, which we know it does.

To see this, consider that for a given input, say 000 on a 3 qubit machine, the corresponding 8-sized quantum state looks like:and therefore when you multiply it by the unitary matrix of the quantum circuit, what you get is the first column of the unitary matrix of the quantum circuit. And 001, gives the second column and so on.

As a result, to prove that a quantum algorithm is correct, we need to be a bit smarter than "just calculate the full matrix".

Which is why you should now go and read: Section "Quantum algorithm".

This type of thinking links back to how physical experiments relate to quantum computing: a quantum computer realizes a physical experiment to which we cannot calculate the probabilities of outcomes without exponential time.

So for example in the case of a photonic quantum computer, you are not able to calculate from theory the probability that photons will show up on certain wires or not.

One direct practical reason is that we need to map the matrix to real quantum hardware somehow, and all quantum hardware designs so far and likely in the future are gate-based: you manipulate a small number of qubits at a time (2) and add more and more of such operations.

While there are "quantum compilers" to increase the portability of quantum programs, it is to be expected that programs manually crafted for a specific hardware will be more efficient just like in classic computers.

TODO: is there any clear reason why computers can't beat humans in approximating any unitary matrix with a gate set?

This is analogous to what classic circuit programmers will do, by using smaller logic gates to create complex circuits, rather than directly creating one huge truth table.

The most commonly considered quantum gates take 1, 2, or 3 qubits as input.

The gates themselves are just unitary matrices that operate on the input qubits and produce the same number of output qubits.

For example, the matrix for the CNOT gate, which takes 2 qubits as input is:

The final question is then: if I have a 2 qubit gate but an input with more qubits, say 3 qubits, then what does the 2 qubit gate (4x4 matrix) do for the final big 3 qubit matrix (8x8)? In order words, how do we scale quantum gates up to match the total number of qubits?

The intuitive answer is simple: we "just" extend the small matrix with a larger identity matrix so that the sum of the probabilities third bit is unaffected.

More precisely, we likely have to extend the matrix in a way such that the partial measurement of the original small gate qubits leaves all other qubits unaffected.

For example, if the circuit were made up of a CNOT gate operating on the first and second qubits as in:

then we would just extend the 2x2 CNOT gate to:

TODO lazy to properly learn right now. Apparently you have to use the Kronecker product by the identity matrix. Also, zX-calculus appears to provide a powerful alternative method in some/all cases.

Bibliography:

Just like as for classic gates, we would like to be able to select quantum computer physical implementations that can represent one or a few gates that can be used to create any quantum circuit.

Unfortunately, in the case of quantum circuits this is obviously impossible, since the space of N x N unitary matrices is infinite and continuous.

Therefore, when we say that certain gates form a "set of universal quantum gates", we actually mean that "any unitary matrix can be approximated to arbitrary precision with enough of these gates".

Or if you like fancy Mathy words, you can say that the subgroup of the unitary group generated by our basic gate set is a dense subset of the unitary group.

The first two that you should study are: