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Hadamard and Toffoli Gates form a universal set of gates, thus all quantum circuits can be created out of them. My question is assume any circuit with some $N$ inputs and $K$ gates.

For any such circuit we add gates one by one to extend this circuit. Let us add a $N+1^{th}$ gate which by definition can either be a Hadamard or Toffoli gate. These gates have the form:

  • Hadamard: Input:1 Output:1

  • Toffolli: Input:3 Output:3

Since each gates inputs 1 or 3 amplitudes, and modify 1 amplitude in both cases:

Query 1: Does that mean if we were to measure any single/subgroup of qubits in the rest of the system (other than the qubit whose amplitude is being modified) before and after the application of $N+1^{th}$ gate the probability of collapse to each of the states stays unchanged?

If not can someone explain what happens with an explicit non trivial example (let $N>3$) consisting of the application of a single $N+1^{th}$ gate?

I am trying to get a conceptual idea if/how application of a gate on a single qubit does to the rest of the system (with an example). I am new to quantum computing.

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First, let me try and nudge a bit of the terminology in the right direction...

Since each gates inputs 1 or 3 amplitudes, and modify 1 amplitude in both cases

Each gate acts on a certain number of input qubits. They can affect every amplitude of the whole system (depending on the basis in which you express it). For example, if I start with a state of 3 qubits $$ \sum_{x\in\{0,1\}^3}\alpha_x|x\rangle $$ and act Hadamard on just the first qubit, I get an output state with amplitudes $\beta_y$ satisfying $$ \beta_y=(\alpha_{0y_2y_3}+(-1)^{y_1}\alpha_{1y_2y_3})/\sqrt{2}. $$ Generically, every single one of the 8 amplitudes changes.

Now, as for your actual question:

Query 1: Does that mean if we were to measure any single/subgroup of qubits in the rest of the system (other than the qubit whose amplitude is being modified) before and after the application of $N+1$th gate the probability of collapse to each of the states stays unchanged?

Yes, that is correct. Formally, the best way to describe "the rest of the system" is to calculate the density matrix of the remaining system by taking the partial trace over the bits you're not measuring. This partial trace effectively removes the action of any unitary that just acts on the bit you're tracing out. (How is that consistent with my first statement? The partial trace adds together multiple amplitudes that cancel out any individual localised changes.)

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