-1
$\begingroup$

I can't understand how a state measure on the Hadmard basis collapses to $$ |+\rangle =\dfrac{|0\rangle+|1\rangle}{\sqrt{2}} $$ or $$ |-\rangle =\dfrac{|0\rangle-|1\rangle}{\sqrt{2}} $$

as they dont see to be deterministic outputs as $ |0\rangle $ or $ |1\rangle $.

$\endgroup$
1
$\begingroup$

When a measurement is performed in a certain basis, an orthogonal projection is performed on the state in question. So given a state $|\psi\rangle$, a measurement in the hadamard basis would result in the following:

$$\frac{|+\rangle\langle+|\psi\rangle}{|\langle+|\psi\rangle|}$$ with probability $|\langle+|\psi\rangle|^{2}$ or $$\frac{|-\rangle\langle-|\psi\rangle}{|\langle-|\psi\rangle|}$$ with probability $|\langle-|\psi\rangle|^{2}$.

So give, for example, a state $|0\rangle$, it would result in one of the above two states with equal probabilty $\frac{1}{2}$

$\endgroup$
0
$\begingroup$

That's the beauty of the Hadamard gate or the $\text{H}$ gate which is represented by the following matrix: H gate matrix

It puts the qubit in what is known as superposition. So for example if we apply the $\text{H}$ gate on $|0⟩$ or $|1⟩$, we get the output qubit $|+⟩$ and $|-⟩$ respectively with $$ |+\rangle=\frac{1}{\sqrt{2}} ( |0\rangle+ |1\rangle) $$

$$ |-\rangle=\frac{1}{\sqrt{2}} ( |0\rangle- |1\rangle) $$

That is in the $|+⟩$ state or the $|-⟩$,when we "measure" our qubit there is $\frac{1}{2}$ probability of the state collapsing into $|0⟩$ and $\frac{1}{2}$ probabilty of the state collapsing into $|1⟩$

$\endgroup$
1
  • $\begingroup$ Thanks, I get the Hadamard gate, what I don't understand is when the "measuring " of a state is done on a Hadamard basis rather than on computation basis, in this basis the QBIT should the collapse to |+⟩ or |−⟩ and not |0⟩ or |1⟩ $\endgroup$ Jun 3 at 7:42
0
$\begingroup$

The main idea is that when we apply measurement to qubit by default in qiskit it measures in Z bases. So applying a H gate before measurement changes the measurement bases to X bases where we have |+> and |-> bases instead of |0> and |1> bases as like in normal Z measurement.

So based on what state you measure we can have the below results.

For |0> state when measured in X basis (which is H followed by measurement):

enter image description here

enter image description here

The Hadamard action on |0> states brings it to |+> state and when we measure along Z (meaning that you are projecting on Z basis) it may collapse 50% to 0 and 50% to 1.

For |1> state when measured in X basis

enter image description here

enter image description here

As like you would expect we get 50% probability for 0 and 1. (with 1 phase shifted by Pi).

Now when you measure |+> the states in H bases which is X measurement.

enter image description here

enter image description here

Since we measure |+> on |+> and |-> bases we get 100% probability of |+> state and since we don't have direct X measurement to see |+> state we take a H gate and perform Z measurement. The output |0> denotes that its 100% in |+> state. Since we don't have any visualization for X measurement, we use Z bases to visualize the outcome by applying H gate. The reason for operation is that H distinguishes between + and - state by 0 and 1. If the outcome of this kind of measurement is 0 it denotes that we are in + state.

Now when you measure the states |-> in H bases which is X measurement.

enter image description here

enter image description here

The reason being that here we have initialized the qubit to |-> state and since we measure the qubit in |+> and |-> bases we ought to get |-> bases as the result of measurement. Since we are in the X view we are not able to see the 50% probability of 0 and 1, and so it is collapsed to 1 based on the H operation.

$\endgroup$

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