# 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

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.