I am implementing a simulation of Simon's Algorithm, and am at the part of applying the second n-qbit Hadamard transformation, $H^{\otimes n}$, to the first of the two n-qbit registers, leaving the second unchanged. I am wondering if I can implement this step as applying the operation $H^{\otimes n} \otimes I$ (Kronecker product) to the combined state of the two registers. I want to do this because after the oracle, the two registers are entangled and cannot be represented as a tensor product of the states of each register (is my understanding correct here?).

So I am wondering if the following is true

$$(H^{\otimes n} \otimes I)|x⟩|y⟩ = H^{\otimes n}|x⟩ \otimes I|y⟩ = H^{\otimes n}|x⟩ \otimes |y⟩$$

or more generally is the following true of the kronecker/tensor product?

$$(A \otimes B) |x⟩|y⟩ = A|x⟩ \otimes B|y⟩$$


1 Answer 1



The tensor product of two linear maps $S: V \to X$ and $T: W \to Y$ is the linear map

$$S \otimes T: V \otimes W \to X \otimes Y \ni (v \otimes w) \mapsto S(v) \otimes T(w).$$


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