Seeking an explanation of the Hadamard transformation on a 2-qubit system

Question

Edit Here is a link to the notes I referred to. The problem is on page 7.

https://www.eecs.ucf.edu/~dcm/Teaching/COT6600-Fall2010/Literature/CleveQuantumComplexityTheory.pdf

I am a beginner in this field. I am reading Cleves' notes on quantum computing. In these notes, it is written that when you apply a Hadamard transformation on the second qubit of a 2 qubit system, the resulting transformation is

However, prior to this it says that when you apply a Hadamard transformation to a given q-bit $x$ in an $m$-qbit system, only q-bit $x$ is altered. However, as can be seen in the figure above, in the third and fourth basis vector the first q-bit is also altered. In other words, if I apply the one qubit Hadamard to the second q-bit in the two system, I get something different:

What is the reason for my misunderstanding? Thanks for your help.

Answer 1

If you apply Hadamard gate on second qubit in two-qubit system, assuming that no gate is applied on the first qubit, the resulting quantum gate is $$ I \otimes H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 & 0 & 0\\ 1 & -1 & 0 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & 1 & -1 \end{pmatrix}. $$ Now assume that the two-qubit system is in a general quantum state $$ |\psi\rangle = \begin{pmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \end{pmatrix}, $$ where $|a_0|^2+|a_1|^2+|a_2|^2+|a_3|^2=1$.

Application of the gate $I\otimes H$ on state $|\psi\rangle$ then leads to state $$\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 & 0 & 0\\ 1 & -1 & 0 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & 1 & -1 \end{pmatrix} \begin{pmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} a_0 + a_1 \\ a_0 - a_1 \\ a_2 + a_3 \\ a_2 - a_3 \end{pmatrix}. $$

Note that in your case, you apply the gate on two different states. Each state is column of the second matrix. Also note that your states are not normalized. For example the first state $$ \begin{pmatrix} 0\\0\\1\\1 \end{pmatrix} $$ should be $$ \frac{1}{\sqrt{2}}\begin{pmatrix} 0\\0\\1\\1 \end{pmatrix}. $$ This means that $a_0=a_1$ and $a_2=a_3=\frac{1}{\sqrt{2}}$ and the result after application of the gate is state $$ \begin{pmatrix} 0\\0\\1\\0 \end{pmatrix}. $$ You can proceed similarly for the second state. The result will be $$ \frac{1}{2}\begin{pmatrix} 1\\-1\\1\\-1 \end{pmatrix}. $$