# How to find the output state after evolution through a unitary?

I was reading about quantum postulates, and I have a few questions about the second postulate that describes the evolution of a quantum system.

For a system S, we describe the evolution after applying the unitary $$U$$ by the following:

$$|S'\rangle = U|S\rangle$$

I understand the idea a bit, I saw a couple of simple examples like H, but I would like to see it in action on a slightly more complex unitary (with a tensor product).

For example, can someone please give the detailed steps to calculate the following evolution for $$|S\rangle = |00\rangle$$.

• What would be the state for $$|S'\rangle$$ with $$U = CNOT\circ (H\otimes I)$$, I'm bit confused by the tensor product...
• What happens if we apply $$|S''\rangle = U'|S'\rangle$$ with $$U'$$ this time $$U'=(H\otimes I) \circ CNOT$$ (different order).

Thanks in advance and happy new year :)

First note that $$H = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}$$ and $$|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$ and $$|1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$

Hence $$H|0\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$$

Thus, we have that: $$\big( H\otimes I \big) |00 \rangle = H|0\rangle \otimes I|0\rangle = \bigg(\dfrac{|0\rangle + |1\rangle}{\sqrt{2} } \bigg)\otimes |0\rangle = \dfrac{|00\rangle + |10\rangle}{\sqrt{2} }$$

Now note that $$CNOT = |0\rangle\langle 0| \otimes I +|1 \rangle \langle 1| \otimes X =\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{pmatrix}$$. Thus, the CNOT gate here just apply the bit flip gate, $$X$$ gate, to the target qubit when the controlled-qubit is $$|1\rangle$$ and do nothing otherwise.

Therefore, if you apply $$U = CNOT \cdot H\otimes I$$ to the state $$|S\rangle = |00\rangle$$ then you get $$\big( CNOT \cdot H\otimes I\big) |00\rangle = CNOT \cdot \bigg( \big(H \otimes I \big) |00\rangle \bigg) = CNOT \cdot \bigg(\dfrac{|00\rangle + |10\rangle}{\sqrt{2} } \bigg) = \dfrac{|00\rangle + |11\rangle}{\sqrt{2} }$$

That is, $$U |00\rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2} }$$. On a quantum circuit, this can be realized as: Note that for the figure above, $$|\psi_0 \rangle = |S\rangle = |00\rangle$$, $$|\psi_1 \rangle = \dfrac{|00\rangle + |10\rangle}{\sqrt{2} }$$, and $$|\psi_2 \rangle = |S'\rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2} }$$.

Now, if you apply $$\big( H \otimes I \big) \cdot CNOT$$ to the state $$|S'\rangle$$ then this is equivalent to applying $$U_1 = CNOT \cdot \big(H\otimes I\big) \cdot \big( H \otimes I \big) \cdot CNOT$$ to $$|S\rangle = |00\rangle$$ since \begin{align} \bigg( \big( H \otimes I \big) \cdot CNOT \bigg) |S'\rangle &= \bigg( \big( H \otimes I \big) \cdot CNOT \bigg) \bigg( CNOT \cdot \big( H\otimes I\big) |00\rangle \bigg) \\ &= \big( H \otimes I \big) \cdot CNOT \cdot CNOT \cdot \big(H\otimes I\big) |00\rangle \\ &= \big( H \otimes I \big) \cdot \big(H\otimes I\big) |00\rangle \\ &= |00\rangle \end{align}

Note: $$CNOT \cdot CNOT = I$$ since $$CNOT$$ is its own inverse. Similarly, $$H \otimes I$$ is its own inverse hence $$\big( H \otimes I \big) \cdot \big( H \otimes I \big) = I$$.

You can see this on the quantum circuit as follows: • Thank you very much for taking the time to write such detailed explanation! I really appreciate it. I'm learning these on my own and generous people like you make our lives easier! thank you Jan 1 at 15:13