Quantum computing is hard because we want long coherence but fast control by 
Ciro Santilli 37 Updated 2025-07-16
Ciro Santilli 37 Updated 2025-07-16Mentioned e.g. at:
These are two conflicting constraints:
- long coherence times: require isolation from external world, otherwise observation destroys quantum state
- fast control and readout: require coupling with external world
Since a matrix can be seen as a linear map , the product of two matrices can be seen as the composition of two linear maps:One cool thing about linear functions is that we can easily pre-calculate this product only once to obtain a new matrix, and so we don't have to do both multiplications separately each time.
Most commonly, boundary conditions such as the Dirichlet boundary condition are taken to be fixed values in time.
But it also makes sense to think about cases where those values vary in time.
For humans specifically: en.wikipedia.org/wiki/List_of_systems_of_the_human_body
At Section "Quantum computing is just matrix multiplication" we saw that making a quantum circuit actually comes down to designing one big unitary matrix.
Instead, they use quantum logic gates.
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 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:
1 0 0 0
0 1 0 0
0 0 0 1
0 0 1 0The 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:
0 ----+----- 0
|
1 ---CNOT--- 1
2 ---------- 2TODO 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.
The Hadamard gate takes or (quantum states with probability 1.0 of measuring either 0 or 1), and produces states that have equal probability of 0 or 1.
Equation 1.
Hadamard gate matrix
. 
The quantum NOT gate swaps the state of and , i.e. it maps:As a result, this gate also inverts the probability of measuring 0 or 1, e.g.
- if the old probability of 0 was 0, then it becomes 1
- if the old probability of 0 was 0.2, then it becomes 0.8
Equation 2.
Quantum NOT gate matrix
. 
One is reminded of Nick Leeson.
These minorities actually had different legal statuses, e.g. they were exempt from the One Child Policy.
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