The standard usage cases for the Hadamard gate seem to be passing a pure $|0\rangle$ or $|1\rangle$ state in to get an even amplitude (probability) mixed state which has either matching or mismatching phase.

The other use of course, is to take such an even amplitude mixed state, pass it through the Hadamard gate, and get back a pure $|0\rangle$ or $|1\rangle$ state.

This of course scales up to multiple qubits.

Are there any other usage cases for the Hadamard Gate? I'm wondering if say you had an uneven amplitude qubit, would passing it through a Hadamard Gate do anything useful? Or the same question with multiple qubits, that had varying amplitudes of states.



1 Answer 1


I don't know of any explicit use of the Hadamard gate in circuits other than to introduce or take away superposition, but here are some interesting applications:

  1. The Hadamard gate is the single-qubit version of the Fourier Transform*, and the Fourier Transform is very useful.
  2. Instead of using the Hadamard at the end of a computation to get back the original basis, you can measure the system in the Hadamard basis instead of the standard basis.
  3. The set of $\{\text{CNOT}, \text{Hadamard}, \text{rotation by 45 degrees}, \frac{\pi}{8}\text{-gate} \}$ is a universal quantum gate set. That is to say, any quantum gate can be approximated arbitrarily well by repeated application of gates from this set.
  4. If you prepend and append two Hadamard gates to a CNOT gate, you reverse the role of control bit and target bit.

*Demonstration: The quantum Fourier transform is defined as follows. Take a system of $n$ particles in the standard basis, and put it in state $|k\rangle, 0 \leq k \leq 2^n-1$. Then applying the Fourier Transform $F$ yields a big superposition:

$$ F|k\rangle = \sum_{u=0}^{2^n-1}e^{i2\pi\frac{u\cdot k}{2^n-1}}|u\rangle $$

Take $n=1$ and $k\in\{0,1\}$, and you see that $F|0\rangle = |+\rangle$ and $F|1\rangle=|-\rangle$. As you can see, it is performs exactly the same transformation as a Hadamard gate. Moreover, for any number of qubits $n$, the Fourier still behaves as the Hadamard only for the state $0$, on which it acts as follows:

$$ F|0\rangle_n = \sum_{u=0}^{2^n-1}e^{i2\pi\frac{u\cdot 0}{2^n-1}}|u\rangle_n = \sum_{u=0}^{2^n-1}e^{0}|u\rangle_n = \sum_{u=0}^{2^n-1}|u\rangle_n $$

I have not normalized my states in these examples, but you should definitely do so on an exam.

  • $\begingroup$ Amazing answer! I can ask a new question if you think it's more appropriate but was use does a single qubit Fourier transform have? And does that by chance scale up for more wires? $\endgroup$
    – Alan Wolfe
    Oct 16, 2015 at 15:01
  • $\begingroup$ Well, you answer your own question: you've already found uses for the Hadamard gate, as you've indicated. This single-qubit FT has many uses! For example, you can use it to get in and out of superpositions, you can... And yes, the Fourier scales up for more wires just like the Hadamard. I've added a demonstration of how exactly to my answer. $\endgroup$ Oct 16, 2015 at 15:14
  • $\begingroup$ Thank you! Final question... Does the conjugate transpose of the hadamard gate (unitary matrix) represent the inverse quantum Fourier transform? $\endgroup$
    – Alan Wolfe
    Oct 16, 2015 at 15:19
  • $\begingroup$ Amost. The conjugate transpose of the Hadamard gate is just the Hadamard gate (verify this on paper: take the Hadamard gate [1 1; 1 -1], transpose it, see what happens). Here's a useful lemma: the inverse of any unitary matrix is its conjugate transpose, i.e. $U^{-1} = U^{\dagger}$. Corrollary: $U\cdot U^{\dagger} = I = U^{\dagger}\cdot U$, so you can apply gate and undo it again very easily. While $H=H^{\dagger}$, this is not true of the general Fourier transform, so you can undo a Hadamard by doing another Hadamard, but to undo a Fourier, you have to use $F^{\dagger}$. $\endgroup$ Oct 16, 2015 at 15:47

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