Скачать книгу

tensor product is defined as:

      

(1.39)

      which creates a 4×4 matrix.

      Suppose we wanted to construct a two-qubit circuit starting in the state |10⟩ with an X gate applied to the left qubit, and a Y gate applied to the other. Mathematically this would be written

      

(1.40)

      

(1.41)

      To see how this would be implemented using the matrix representation, we first construct the X⊗Y matrix:

      upper X circled-times upper Y equals Start 3 By 2 Matrix 1st Row 1st Column 0 Start 2 By 2 Matrix 1st Row 1st Column 0 2nd Column negative i 2nd Row 1st Column i 2nd Column 0 EndMatrix 2nd Column 1 Start 2 By 2 Matrix 1st Row 1st Column 0 2nd Column negative i 2nd Row 1st Column i 2nd Column 0 EndMatrix 2nd Row 1st Column Blank 3rd Row 1st Column 1 Start 2 By 2 Matrix 1st Row 1st Column 0 2nd Column negative i 2nd Row 1st Column i 2nd Column 0 EndMatrix 2nd Column 0 Start 2 By 2 Matrix 1st Row 1st Column 0 2nd Column negative i 2nd Row 1st Column i 2nd Column 0 EndMatrix EndMatrix equals Start 4 By 4 Matrix 1st Row 1st Column 0 2nd Column 0 3rd Column 0 4th Column negative i 2nd Row 1st Column 0 2nd Column 0 3rd Column i 4th Column 0 3rd Row 1st Column 0 2nd Column negative i 3rd Column 0 4th Column 0 4th Row 1st Column i 2nd Column 0 3rd Column 0 4th Column 0 EndMatrix period (1.42)

      Completing the calculation gives the expected result:

      upper X circled-times upper Y Math bar pipe bar symblom 10 mathematical right-angle equals Start 4 By 4 Matrix 1st Row 1st Column 0 2nd Column 0 3rd Column 0 4th Column negative i 2nd Row 1st Column 0 2nd Column 0 3rd Column i 4th Column 0 3rd Row 1st Column 0 2nd Column negative i 3rd Column 0 4th Column 0 4th Row 1st Column i 2nd Column 0 3rd Column 0 4th Column 0 EndMatrix Start 4 By 1 Matrix 1st Row 0 2nd Row 0 3rd Row 1 4th Row 0 EndMatrix equals i Start 4 By 1 Matrix 1st Row 0 2nd Row 1 3rd Row 0 4th Row 0 EndMatrix equals i Math bar pipe bar symblom 01 mathematical right-angle period (1.43)

      A particularly interesting two-qubit circuit is formed by applying a Hadamard gate to each qubit in the ground state: HH|00⟩. Let us first compute HH:

      upper H circled-times upper H equals one-half Start 3 By 2 Matrix 1st Row 1st Column 1 Start 2 By 2 Matrix 1st Row 1st Column 1 2nd Column 1 2nd Row 1st Column 1 2nd Column negative 1 EndMatrix 2nd Column 1 Start 2 By 2 Matrix 1st Row 1st Column 1 2nd Column 1 2nd Row 1st Column 1 2nd Column negative 1 EndMatrix 2nd Row 1st Column Blank 3rd Row 1st Column 1 Start 2 By 2 Matrix 1st Row 1st Column 1 2nd Column 1 2nd Row 1st Column 1 2nd Column negative 1 EndMatrix 2nd Column minus 1 Start 2 By 2 Matrix 1st Row 1st Column 1 2nd Column 1 2nd Row 1st Column 1 2nd Column negative 1 EndMatrix EndMatrix equals one-half Start 4 By 4 Matrix 1st Row 1st Column 1 2nd Column 1 3rd Column 1 4th Column 1 2nd Row 1st Column 1 2nd Column negative 1 3rd Column 1 4th Column negative 1 3rd Row 1st Column 1 2nd Column 1 3rd Column negative 1 4th Column negative 1 4th Row 1st Column 1 2nd Column negative 1 3rd Column negative 1 4th Column 1 EndMatrix period (1.44)

      Completing the calculation gives:

      upper H circled-times upper H Math bar pipe bar symblom 00 mathematical right-angle equals one-half Start 4 By 4 Matrix 1st Row 1st Column 1 2nd Column 1 3rd Column 1 4th Column 1 2nd Row 1st Column 1 2nd Column negative 1 3rd Column 1 4th Column negative 1 3rd Row 1st Column 1 2nd Column 1 3rd Column negative 1 4th Column negative 1 4th Row 1st Column 1 2nd Column negative 1 3rd Column negative 1 4th Column 1 EndMatrix Start 4 By 1 Matrix 1st Row 1 2nd Row 0 3rd Row 0 4th Row 0 EndMatrix equals one-half Start 4 By 1 Matrix 1st Row 1 2nd Row 1 3rd Row 1 4th Row 1 EndMatrix period (1.45)

      Note that the resulting state vector can be decomposed into a sum of all of the two-qubit basis states:

      one-half Start 4 By 1 Matrix 1st Row 1 2nd Row 1 3rd Row 1 4th Row 1 EndMatrix equals one-half Start 4 By 1 Matrix 1st Row 1 2nd Row 0 3rd Row 0 4th Row 0 EndMatrix plus one-half Start 4 By 1 Matrix 1st Row 0 2nd Row 1 3rd Row 0 4th Row 0 EndMatrix plus one-half Start 4 By 1 Matrix 1st Row 0 2nd Row 0 3rd Row 1 4th Row 0 EndMatrix plus one-half Start 4 By 1 Matrix 1st Row 0 2nd Row 0 3rd Row 0 4th Row 1 EndMatrix comma (1.46)

      or alternatively

      upper H circled-times upper H Math bar pipe bar symblom 00 mathematical right-angle equals one-half left-parenthesis Math bar pipe bar symblom 00 mathematical right-angle plus Math bar pipe bar symblom 01 mathematical right-angle plus Math bar pipe bar symblom 10 mathematical right-angle plus Math bar pipe bar symblom 11 mathematical right-angle right-parenthesis period (1.47)

      Although the matrix representation can be helpful in understanding the operations, calculations can often be done more compactly once the effect of the gates are understood. For example, we could write HH |00⟩ = HH |0⟩ |0⟩, apply the Hadamard gates to each qubit, and simplify:

      StartLayout 1st Row 1st Column upper H circled-times upper H Math bar pipe bar symblom 0 mathematical right-angle Math bar pipe bar symblom 0 mathematical right-angle 2nd Column equals upper H Math bar pipe bar symblom 0 mathematical right-angle circled-times upper H Math bar pipe bar symblom 0 mathematical right-angle 2nd Row 1st Column Blank 2nd Column equals StartFraction Math bar pipe bar symblom 0 mathematical right-angle plus Math bar pipe bar symblom 1 mathematical right-angle Over StartRoot 2 EndRoot EndFraction circled-times StartFraction Math bar pipe bar symblom 0 mathematical right-angle plus Math bar pipe bar symblom 1 mathematical right-angle Over StartRoot 2 EndRoot EndFraction 3rd Row 1st Column Blank 2nd Column equals one-half left-parenthesis Math bar pipe bar symblom 00 mathematical right-angle plus Math bar pipe bar symblom 01 mathematical right-angle plus Math bar pipe bar symblom 10 mathematical right-angle plus Math bar pipe bar symblom 11 mathematical right-angle right-parenthesis period EndLayout (1.48)

      We conclude this section

Скачать книгу