2
$\begingroup$

I have a very basic question. I have found qubits are represented as complex vectors. I get it totally. I understand bracket notation and vector\matrix algebra. However, I cannot move further from here. It is often referred like the equation below.

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

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

My questions are:

  1. Why is divided by $\frac{1}{\sqrt{2}} $ ?
  2. What does this symbol mean $|+\rangle$, $|-\rangle$? I understand what $|0\rangle$ and $|1\rangle$ means.
$\endgroup$
3
$\begingroup$

#1: the $1/\sqrt{2}$ is a normalization which ensures that the ``length'' of the vector is one.

#2: The notation $|\pm\rangle$ is just a label for the two states defined above. Since the states $|0\rangle, |1\rangle$ are elements of a vector space, you can take linear combinations and therefore construct the states $|\pm\rangle$

| improve this answer | |
$\endgroup$
  • $\begingroup$ regd point #1, but why is needed to normalized. i mean why do we want the length of |0> + |1> to be one ? $\endgroup$ – amarnath chatterjee Jul 16 at 17:45
  • 3
    $\begingroup$ The square of the amplitudes ($\pm 1/\sqrt 2$) correspond to the probability of the qubits being measured in those respective states. This is called the Born rule. The "length" should sum to one, because the probabilities sum to one. $\endgroup$ – Mark S Jul 16 at 17:59
2
$\begingroup$

The $\frac{1}{\sqrt{2}}$ is due to the normalization condition which says that sum of the squares of the amplitudes of the must be equal to one while the square of the amplitude refers to the probability of getting that particular state when the qubits are measured

The vectors $|+⟩$ and $|-⟩$ are known as the eigenvectors for the Hadamard gate. When we apply the $H$ gate on the $|0⟩$ and $|1⟩$, we get $|+⟩$ and $|-⟩$ respectively.

That is, $$H|0⟩=|+⟩$$ and $$H|1⟩=|-⟩$$

| improve this answer | |
$\endgroup$
  • $\begingroup$ The explaination for the vectors |+⟩ and |−⟩ is usefull, thanks $\endgroup$ – amarnath chatterjee Jul 20 at 14:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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