# Does applying Hadamard gate is identical to applying a measurement gate that is perpendicular to the current qubit state?

Cause applying H (hadamard gate) on a state |0> will evolve to 50% |0> and 50% |1>, assuming we are meausring along the vertical axis (Z-axis).

While applying a measurement along a horizontal axis (e.g. X-axis), will force the state to "evolve" (destroied) and let it be perpendicular to the vertical axis. In another word the state now will have a 50% chance to be measured along the vertical axis to be |0> and 50% to be |1>.

I know that measuring a state will destroy the state, while H gate is unitary and conserves the state from destruction. But I really need to know how the physics behind the Quantum Computer gates.

• You answered the question yourself — measurement collapses the state, whereas the Hadamard gate doesn't. – Yuval Filmus Nov 15 '16 at 16:11

## 1 Answer

As Yuval points out, you have already answered your own question. If I have a state $|q\rangle = a|0\rangle + b|1\rangle$ and I apply a Hadamard gate to that state, I end up with a new state: $$H\cdot|q\rangle = \frac{1}{\sqrt{2}}(a+b)|0\rangle + \frac{1}{\sqrt{2}}(a-b)|1\rangle$$

However, if I measure the state $|q\rangle$ in the standard basis, I get either $|0\rangle$ with probability $|a|^2$ or $|1\rangle$ with probability $|b|^2$. Notice that both $|0\rangle$ and $|1\rangle$ are different states than what you would have gotten had you applied the Hadamard transform instead of a measurement. I admit that a Hadamard-transformed state of 50% $0$ and 50% $1$ looks very much like having done a measurement, but this is not true: you haven't done the measurement yet, and the state is still in superposition. This matters if you do other transformations afterward, or if you work in a multi-qubit system. Then when measuring later, it matters whether you did a Hadamard transform or a measurement.

A good thing to do to acquaint yourself with transformations and measurements is to just compute some by hand on one qubit and on two qubits, and to write out in full the intermediate states in Deutsch's algorithm. You'll get the hang of it.