This question has been fussy me for two hours so I'd appreciate some help.

The Hadamard operator on one qubit may be written as

$H = \frac{1}{\sqrt{2}}[(|0\rangle + |1\rangle)\langle0| + (|0\rangle - |1\rangle)\langle 0|]$

For this transform to act on a two qubits, my working leads me to $\frac{1}{2}[|0\rangle\langle 0| + |1\rangle \langle 0| + |0\rangle \langle 1| + |1\rangle \langle 1| + |0\rangle \langle 0| - |1\rangle \langle 0| - |0\rangle \langle 1| + |1\rangle \langle 1|]$

to give

$|0\rangle\langle0| + |1\rangle\langle 1|$.

But this does not conform to the general expression

$H^{\otimes n} = \frac{1}{\sqrt{2^{n}}}\sum_{x,y}(-1)^{xy}|x\rangle \langle y|$

Where have I gone wrong?


1 Answer 1


You've computed $H^2$ (which acts on one qubit), not $H \otimes H$ (acting on two). As an aside, your computation of $H^2$ is correct since $|0\rangle\langle0| + |1\rangle\langle 1|$ is the identity.

To show the desired result, it's convenient to expand your expression for $H$ (although in your first line, note your final $\langle 0|$ should be a $\langle 1|$), $$H = \frac{1}{\sqrt{2}} \left[ |0 \rangle \langle 0| + |1 \rangle \langle 0 | + |0\rangle \langle 1| - |1\rangle \langle 1| \right]$$ Then we can compute $H \otimes H$, somewhat tediously expanding all 16 terms (and taking the half to the other side), \begin{aligned} 2H \otimes H &= \left[ |0 \rangle \langle 0| + |1 \rangle \langle 0 | + |0\rangle \langle 1| - |1\rangle \langle 1| \right] \otimes \left[ |0 \rangle \langle 0| + |1 \rangle \langle 0 | + |0\rangle \langle 1| - |1\rangle \langle 1| \right] \\ &= |00\rangle \langle00| + |01\rangle \langle00| + |00\rangle \langle01| - |01\rangle \langle01| \\ &+|10\rangle \langle00| + |11\rangle \langle00| + |10\rangle\langle01|-|11\rangle\langle01| \\ &+|00\rangle \langle10| + |01\rangle \langle10| + |00\rangle \langle11| - |01\rangle \langle11| \\ &-|10\rangle \langle10| - |11\rangle \langle10| - |10\rangle \langle11| + |11\rangle \langle11| \end{aligned} where the each of the four subsequent lines comes from $|0\rangle\langle0| \otimes H, \dots, |1\rangle\langle1| \otimes H$ respectively, using the fact that $$|i\rangle \langle j| \otimes |k\rangle \langle l| = |i, k\rangle\langle j, l|$$ where I'm using the convention that the left-most index in $\langle j, l|$ corresponds to the qubit in the first slot - that is to say, $\langle j, l|_{12} = \langle j |_1 \otimes \langle l |_2$ (where the subscript $1$ and $2$ denote the qubit on which to act).

One can verify that this is as required: each $|x\rangle\langle y|$ for $x, y \in B_2$ appears once, with the correct factor of $(-1)^{x \cdot y}$ - you don't explicitly write this dot product in your expression, but I find it best to explicitly identify $x$ and $y$ as elements of $B_2 \cong\mathbb{Z}_2^2$, with $$x \cdot y = x_1 y_1 + x_2 y_2$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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