July 22, 2024
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Quantum – Theory – Basis & Born’s Rule

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> = \begin {bmatrix} 1^h \\ 0 \end{bmatrix} and |jv> = \begin {bmatrix} 0 \\ 1^h \end{bmatrix}

which are orthogonal because their inner product = 0:

               <j^|jv> = [1^ 0].\begin {bmatrix} 0  \\ 1^h \end{bmatrix}   = 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, \begin {bmatrix} 1^h \\ 0 \end{bmatrix} represents a 100% chance of the qubit being a |j^>  and \begin {bmatrix} 0 \\ 1^h \end{bmatrix}  a 100% chance of it measuring as |jv>.  

1/√2\begin {bmatrix} 1^h \\ 1^h \end{bmatrix} represents a 50% chance of either the qubit measuring as |j^> or |jv>.

\begin {bmatrix} 1^v \\ 0 \end{bmatrix} also represents a 100% chance of the qubit being a |j^> and  \begin {bmatrix} 0 \\ 1^v \end{bmatrix} a 100% chance of it measuring as |jv>.

1/√2\begin {bmatrix} 1^h \\ 1^v \end{bmatrix} , 1/√2\begin {bmatrix} 1^v \\ 1^v \end{bmatrix}  and   1/√2\begin {bmatrix} 1^v \\ 1^h \end{bmatrix}represent a 50% chance of either the qubit measuring as |j^>   or |jv>  .

Examples

For the state |ψ> = |j^> = \begin {bmatrix} 1^h \\ 0 \end{bmatrix} Born’s rule implies:

  • P(|j^>) = |(1^ 0].\begin {bmatrix} 1^h \\ 0 \end{bmatrix} |2
  • = |1^*1^ + 0*0)|2 = 1

Secondly, for the state |ψ> = 1/√3(|j^> + √2|jv>) = 1/√3\begin {bmatrix} 1^h \\ \sqrt{2}^v \end{bmatrix} Born’s rule implies:

  • P(|j^>) = |[1^ 0].\begin {bmatrix} 1^h/\sqrt{3} \\ \sqrt{2^h}/\sqrt{3} \end{bmatrix} |2
  • = |1^/√3  + 0 )|2  = 1/3
  • => P(jv) = 2/3

Thirdly, for the state example: |ψ> = 1/√2(|j^>  + |jv>) = 1/√2\begin {bmatrix} 1^h \\ ^-1^h \end{bmatrix} Born’s rule implies:

  • P(|j^>) = |[1^ 0].\begin {bmatrix} 1^h/\sqrt{2} \\ 1^v/\sqrt{2} \end{bmatrix} |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  =  \begin {bmatrix} 1^h /\sqrt{2} \\ 1^h /\sqrt{2} \end{bmatrix} 

  • |v> = (|jˆ> + |jv>)/√2  =  \begin {bmatrix} 1^h /\sqrt{2} \\ 1^v /\sqrt{2} \end{bmatrix} which are orthogonal because the inner product of their bra-ket is 0:
    • <^|v> = [1^/√2^ 1^/√2].\begin {bmatrix} 1^h /\sqrt{2}  \\ 1^v /\sqrt{2} \end{bmatrix}  
      • = 1^/√2*1^/√2 + 1^/√2*1v/√2
      • = 1^/2 +  1v/2 = 0

Example

For the state |ψ> = (|j^>  + |jv>)/√2 = 1/√2 \begin {bmatrix} 1^h \\ 1^v \end{bmatrix}   (The state |v>)

  • P(|^>) = |[1^/√2 1^/√2].\begin {bmatrix} 1^h /\sqrt{2} \\ 1^v /\sqrt{2} \end{bmatrix}  |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 =   \begin {bmatrix} 1^h /\sqrt{2} \\ i^h /\sqrt{2} \end{bmatrix} 

  • |iv> = (|jˆ> + iv|jv>)/√2 =    \begin {bmatrix} 1^h /\sqrt{2} \\ i^v /\sqrt{2} \end{bmatrix} which are orthogonal because the inner product of their bra-ket is 0:
    • <i^|iv> = [1^/√2 iv/√2].\begin {bmatrix} 1^h /\sqrt{2}  \\ i^v /\sqrt{2} \end{bmatrix} 
      • = 1^/√2*1^/√2 + iv/√2*iv/√2
      • = 1^/2  + 1v/2        = 0

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 \begin {bmatrix} 1^h \\ i^h \end{bmatrix}   (The state |i^>)

  • P(|i^>) = |[1^/√2 iv/√2].\begin {bmatrix} 1^h /\sqrt{2} \\ i^h /\sqrt{2} \end{bmatrix}  |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

 

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