This post describes Born’s rule that defines the probability of the collapse of qubits to a quantum basis.
z-axis Basis
Define the z-axis basis states, {|j^>, |jv>}, as:
- |jˆ> =
and |jv> =
which are orthogonal because their inner product = 0:
<j^|jv> = [1^ 0]. = 1^*0 + 0*1^ = 0
Born’s Rule
In Wave Numbers, express probability as a Counter as it is a magnitude. Born’s rule states that the probability that a state |ψ> collapses when measured onto the basis {|x>, |x┴>} (x and its orthogonal complement) is:
- P(x) = |<x|ψ>|2 and ∑P(xi) = 1
In other words, the probability of the state <x| is the square of the dot product of <x| and |ψ>. The sum of all probabilities = 1.
The probability of the qubit measuring as |j^> or |jv> depends on the state equation. Given a state equation of |ψ> = a + bi, then the probability of |j^> is a2 and the probability of |jv> is b2 and a2 + b2 = 1.
The top row of the matrix that represents a state gives the probability of the qubit measuring as |j^> and the bottom the probability of the qubit measuring as |jv>.
So, represents a 100% chance of the qubit being a |j^> and
a 100% chance of it measuring as |jv>.
1/√2 represents a 50% chance of either the qubit measuring as |j^> or |jv>.
also represents a 100% chance of the qubit being a |j^> and
a 100% chance of it measuring as |jv>.
1/√2 , 1/√2
and 1/√2
represent a 50% chance of either the qubit measuring as |j^> or |jv> .
Examples
For the state |ψ> = |j^> = Born’s rule implies:
- P(|j^>) = |(1^ 0].
|2
- = |1^*1^ + 0*0)|2 = 1
Secondly, for the state |ψ> = 1/√3(|j^> + √2|jv>) = 1/√3 Born’s rule implies:
- P(|j^>) = |[1^ 0].
|2
- = |1^/√3 + 0 )|2 = 1/3
- => P(jv) = 2/3
Thirdly, for the state example: |ψ> = 1/√2(|j^> + –|jv>) = 1/√2 Born’s rule implies:
- P(|j^>) = |[1^ 0].
|2
- = |1^/√2 + 0*√2v )|2 = 1/2
- => P(jv) = 1^/2
x-axis Basis and Born’s Rule
In Wave Numbers, define the x-axis basis states, {|^>, |v> }, as:
- |^> = (|jˆ> + |jv>)/√2 =
- |v> = (|jˆ> + –|jv>)/√2 =
which are orthogonal because the inner product of their bra-ket is 0:
- <^|v> = [1^/√2^ 1^/√2].
- = 1^/√2*1^/√2 + 1^/√2*1v/√2
- = 1^/2 + 1v/2 = 0
- <^|v> = [1^/√2^ 1^/√2].
Example
For the state |ψ> = (|j^> + –|jv>)/√2 = 1/√2 (The state |v>)
- P(|^>) = |[1^/√2 1^/√2].
|2
- = |(1^/√2)*(1^/√2) + (1^/√2)*(1v/√2)|2
- = | 1^/2 + 1v/2| = 0
This makes sense as |ψ> is |v> and the probability of the qubit being |^> is 0.
y-axis Basis and Born’s Rule
Defined the y-axis basis states, {|i^>, |iv> }, as:
- |i^> = (|jˆ> + iˆ|jv>)/√2 =
- |iv> = (|jˆ> + iv|jv>)/√2 =
which are orthogonal because the inner product of their bra-ket is 0:
- <i^|iv> = [1^/√2 iv/√2].
- = 1^/√2*1^/√2 + iv/√2*iv/√2
- = 1^/2 + 1v/2 = 0
- <i^|iv> = [1^/√2 iv/√2].
Note that changing |i^> to <i^|results in the Opposite Sign of i being flipped as shown in earlier post on bra-kets.
Example
For |ψ> = (|j^> + iˆ|jv>)/√2 = 1/√2 (The state |i^>)
- P(|i^>) = |[1^/√2 iv/√2].
|2
- = |(1^/√2)*(1^/√2) + (iv/√2)*(i^/√2)|2
- = | 1^/2 + 1^/2| = 1
Note that changing |i^> to <i^|results in the Opposite Sign of i being flipped as shown in earlier post on bra-kets.
This makes sense as |ψ> is |i^> and the probability of the qubit |i^> is 1.
Next: Single and Multipart Qubits
Previous: Bra-Kets