[Solved]: Does a Hadamard Gate have uses outside of pure and evenly mixed states?

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 $|krangle, 0 leq k leq 2^n-1$. Then applying the Fourier Transform $F$ yields a big superposition: $$ F|krangle = sum_{u=0}^{2^n-1}e^{i2pifrac{ucdot k}{2^n-1}}|urangle $$ Take $n=1$ and $kin{0,1}$, and you see that $F|0rangle = |+rangle$ and $F|1rangle=|-rangle$. 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: $$ F|0rangle_n = sum_{u=0}^{2^n-1}e^{i2pifrac{ucdot 0}{2^n-1}}|urangle_n = sum_{u=0}^{2^n-1}e^{0}|urangle_n = sum_{u=0}^{2^n-1}|urangle_n $$ I have not normalized my states in these examples, but you should definitely do so on an exam.