0
$\begingroup$

The question on the book is:

The Hadamard operator on one qubit may be written as $$\frac{1}{\sqrt{2}}[(|0\rangle + |1\rangle)\langle0| + (|0\rangle - |1\rangle)\langle1|]$$ Show explicitly that the Hadamard transform on $n$ qubits, $H^{\otimes n}$, may be written as $$H^{\otimes n}=\frac{1}{\sqrt{2^n}}\sum_{x,y}(-1)^{x\cdot y}|x\rangle\langle y|.\quad\quad \textrm{(2.55)}$$ Write out an explicit matrix representation for $H^{\otimes 2}$.

$H^{\otimes 2}$ is easy. It is $\frac{1}{2}\begin{bmatrix} 1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1\\ 1 & 1 & -1 & -1\\ 1 & -1 & -1 & 1 \end{bmatrix} $. It can be obtained as the Kronecker product of two $\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1\\ 1 & -1\\ \end{bmatrix}$. However, I am not sure how this is linked to (2.55). Say $|11\rangle\langle11|$. This term should has a coefficient of 1 according to the matrix of $H^{\otimes 2}$. (I think it is the bottom right corner.) However, according to (2.55), the coefficient seems to be -1 since $x=y=3$ and $x\cdot y=9$. In my understanding, the ranges for both $x$ and $y$ are $\{0, 1, 2, 3\}$.

Am I wrong with something? Do I understand the meanings of $x$, $y$, and $\cdot$ correctly? Or, is (2.55) wrong?

$\endgroup$
1
  • $\begingroup$ Thanks for the help of @thomas and @zeeshan! After getting the keyword modulo, I find out that the dot operator was introduced in (1.50), a formula very similar to (2.55). Glad to join the community! $\endgroup$
    – Jintao Yu
    Jun 5, 2022 at 4:07

2 Answers 2

0
$\begingroup$

Welcome to quantum computing's stackexchange!

Your tensor product is correct, the issue is with your understanding of $x\cdot y$. The value is calculated through an element-wise dot product rather than converting it into a decimal and then multiplying. Which means that $$ 11 \cdot 11 = [(1 \cdot 1) + (1\cdot 1)] \text{ mod } 2 = 0 $$ Sidenote: You can further understand this notation by thinking of what the power of $-1$ signifies, if you notice that the single Hadamard gate can also be written in the same way as you have written, try expanding the tensor product of two qubit Hadamard using the $x$ and $y$ notation to see why it's element-wise dot product

$\endgroup$
0
$\begingroup$

Be careful : in expression (2.55), the term $x \cdot y$ is in fact the sum (modulo 2) of the bitwise products : $$x \cdot y=x_{0} y_{0} \oplus x_{1} y_{1} \oplus \cdots \oplus x_{n-1} y_{n-1}$$

so, for $x = y = 3$, you have $x \cdot y = 1*1 \oplus 1*1 = 0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.