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Rules of a Quantum World
The Stern-Gerlach Experiment

                              N            ½
Electron gun
                                           ½
                              S
               Ignore
             horizontal           Beam splits into
          deflection as per         two! Not a
           Fleming’s Left           continuous
             Hand Rule                spread
Abstract Representation

                                        ½
                                   UP
  Electron gun
                                   DN
                                        ½




                 Arrow points to
                 the North pole
Cascading Devices Z Z          Z




                           ½    ½
Electron gun

                       ½                      -Z



                                    Only UP
Z
          Cascading Devices Z -Z


                     ½
Electron gun                             -Z
                                ½
                      ½


                               All DN!
Z
               Cascading Devices Z X                          X



                              Arrow goes into
                                 the screen


                                 ¼                  -X
                      ½
Electron gun                    ¼                        -Z
                      ½


                               Half UP, Half DN!!
Z
               Cascading Devices Z Xθ                    X

                                                    θ




                                  At angle θ
                                               -X

                                                    -Z
                               Cos2 (θ/2)/2
                       ½
Electron gun                   Sin2 (θ/2)/2

                       ½
Z
                                                                  X
               Cascading Devices Z X Z

                                                  -X

                                                             -Z
                                                       1/8
                               ¼
                          ½                            1/8
Electron gun                  ¼
                      ½


                                   Down along Z
                                    reappears!
How do we model this behaviour?
Starting Point
• Electrons must have an intrinsic state




• This state differs with orientation in 3d space



     • states along different orientations are dependent
Describing State

                    Prob of being in
                      the UP state
          p                                p

         1-p                               1-p
                        Prob of being in
                         the DN state




p changes with
the orientation
Transformations

                      p         q
                                     • Tzx must be a Stochastic
         Tzx               =
                                       Transformation
                     1-p       1-q

                                        – Non-negative entries

Transforms state                        – Each column sums to 1
  along Z axis to
state along X axis
Stochastic Transformations

                            • Can two stochastic
                              matrices multiply to
Txz Tzx      =    I           yield an identity
                              matrix?

                              – All matrix entries are non-
                                negative
 Transforms state along
                              – So NO, unless each matrix is I!
  Z axis to state along X
 axis and then transform
            back                         Stochastic
                                      Transformations
                                         ruled out
Revisiting the State Description

        Can we allow for
        negative values
 a           here?
                           a2 +b2 =1

 b



                               Points on a unit
          How do we
                                    circle
        translate these
       to probabilities?
Transformations

                     a
                                  • Tzx must be preserve
                             a’
         Tzx         b
                         =
                                    Euclidean length
                             b’


                                     – (Tzx)T Tzx = I


Transforms state                         Cosθ   -Sinθ
  along Z axis to
state along X axis                       Sinθ   Cosθ


                                                        For any θ
Z
            Explanations I                                                     X

                                                                     θ




                                                           -X
           1        0       1/√2                                      -Z
           0        1       1/√2    Cos(θ/2) -Sin(θ/2)     1
TZZ
                                     Sin(θ/2) Cos(θ/2)     0

      Initial state along
                Z

                                                         Initial state along
                                                                   Z
                                   TZXθ
Z
                                                                                                          X
                               Explanations II

                                                                                     -X

           1        0                                                                            -Z
                            1/√2

           0        1       1/√2          1/√2 -1/√2           1
TZZ                                                                         1/√2     1/√2             1
                                         1/√2      1/√2        0            -1/√2 1/√2                0

      Initial state along
                Z                                                  TXZ = Inverse
                                   TZX                                of TZX

                                         Initial state along
                                                   Z                       Initial state along
                                                                                     X
Z
          X

              Bringing in the Y Dimension
-Y            Y                                                      Initial state along
                                       1              1
                          TZXTYZ           = TYX
                                                                     Y transformed to
                                                                       state along X
                                       0              0
-X

     -Z

                    1/√2 -1/√2         a     c        1            +/- 1/√2
                                                              =
                   1/√2     1/√2       b     d        0            +/- 1/√2



                                                                             All UPs along Y
                    All UPs along Y
                                                                           translate to equal
                  translate to equal       +/- 1/√2                           UPs and DNs
                     UPs and DNs
                                                                                 along X
                        along X            +/- 1/√2
                                                          NOT POSSIBLE!!
Revisiting the State Description
            Yet Again
       Can we allow for
       complex values
a           here?
                                           Complex
b                                         conjugate


                          |a|2 +|b|2 =a a + b b = 1

         How do we
       translate these
      to probabilities?
Revisiting Transformations
                                                               Conjugate
                                                               Transpose


                     a
                                  • Tzx must be preserve |a|2
                             a’
        TYX          b
                         =
                             b’     +|b|2

                                     – (Tzx)† Tzx = I


Transforms state                         eiεCosθ -ei(ψ – φ+ ε) Sinθ
  along Z axis to
state along X axis                       ei(φ+ ε) Sinθ ei(ψ+ ε) Cosθ



                                                           For any θ,ψ,ε
Z
          X

              Bringing in the Y Dimension
-Y            Y                                                             Initial state along
                                            1                1
                          TZXTYZ                  = TYX
                                                                            Y transformed to
                                                                              state along X
                                            0                0
-X

     -Z

                  1/√2 -1/√2            1/√2 -e-iφ 1/√2      1             eiφ’’ 1/√2
                                                                     =
                  1/√2    1/√2         eiφ 1/√2   1/√2       0             eiφ’ 1/√2



                                                                                      All UPs along Y
                    All UPs along Y
                                                                                    translate to equal
                  translate to equal               1/√2                                UPs and DNs
                     UPs and DNs
                                                                                          along X
                        along X                   eiφ 1/√2       Φ=π/2, Φ’’=-π/4,
                                                                    Φ’=π/4!!
The Final Transformations
                       1/√2 -1/√2                   1/√2 i/√2
             TZX       1/√2 1/√2
                                      TYZ           i/√2 1/√2



                                     1/√2 -1/√2         1/√2 i/√2
           TYX=TZXTYZ=               1/√2 1/√2          i/√2 1/√2



                                        e-iπ/4/√2      -e-iπ/4/√2

                                        eiπ/4/√2       eiπ/4/√2


Can you write the transformation from Z to a general direction in 3D space?
Summary
• State vector v has complex entries and satisfies

   – |v|2 = v†v = Σ |vi|2 = 1

       • vi’s are called Amplitudes


• Transformations T satisfy T†T = I

   – T’s are called Unitary Transformations


• When we measure a system in state v

   – We get i with Probability |vi|2
Contrast with Classical States
• Take 2 bits, so state vector [p1 p2 p3 p4]
  corresponding to 00, 01, 10, 11 resp.

• Suppose you replace the first bit by an AND
  of the 2 bits with prob p and by an OR with
  prob 1-p?

  – Show this can be written as a stochastic
    transformation.
Our Two Worlds
                 Σ vi = 1 , 0<=vi<=1

       T is stochastic (non-neg, col sums 1)

Classical
                   Measurement
Quantum

               |v|2 = v†v = Σ |vi|2 = 1

                T is Unitary T†T = I
What does this mean for
    computation?

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Rules of a Quantum World

  • 1. Rules of a Quantum World
  • 2. The Stern-Gerlach Experiment N ½ Electron gun ½ S Ignore horizontal Beam splits into deflection as per two! Not a Fleming’s Left continuous Hand Rule spread
  • 3. Abstract Representation ½ UP Electron gun DN ½ Arrow points to the North pole
  • 4. Cascading Devices Z Z Z ½ ½ Electron gun ½ -Z Only UP
  • 5. Z Cascading Devices Z -Z ½ Electron gun -Z ½ ½ All DN!
  • 6. Z Cascading Devices Z X X Arrow goes into the screen ¼ -X ½ Electron gun ¼ -Z ½ Half UP, Half DN!!
  • 7. Z Cascading Devices Z Xθ X θ At angle θ -X -Z Cos2 (θ/2)/2 ½ Electron gun Sin2 (θ/2)/2 ½
  • 8. Z X Cascading Devices Z X Z -X -Z 1/8 ¼ ½ 1/8 Electron gun ¼ ½ Down along Z reappears!
  • 9. How do we model this behaviour?
  • 10. Starting Point • Electrons must have an intrinsic state • This state differs with orientation in 3d space • states along different orientations are dependent
  • 11. Describing State Prob of being in the UP state p p 1-p 1-p Prob of being in the DN state p changes with the orientation
  • 12. Transformations p q • Tzx must be a Stochastic Tzx = Transformation 1-p 1-q – Non-negative entries Transforms state – Each column sums to 1 along Z axis to state along X axis
  • 13. Stochastic Transformations • Can two stochastic matrices multiply to Txz Tzx = I yield an identity matrix? – All matrix entries are non- negative Transforms state along – So NO, unless each matrix is I! Z axis to state along X axis and then transform back Stochastic Transformations ruled out
  • 14. Revisiting the State Description Can we allow for negative values a here? a2 +b2 =1 b Points on a unit How do we circle translate these to probabilities?
  • 15. Transformations a • Tzx must be preserve a’ Tzx b = Euclidean length b’ – (Tzx)T Tzx = I Transforms state Cosθ -Sinθ along Z axis to state along X axis Sinθ Cosθ For any θ
  • 16. Z Explanations I X θ -X 1 0 1/√2 -Z 0 1 1/√2 Cos(θ/2) -Sin(θ/2) 1 TZZ Sin(θ/2) Cos(θ/2) 0 Initial state along Z Initial state along Z TZXθ
  • 17. Z X Explanations II -X 1 0 -Z 1/√2 0 1 1/√2 1/√2 -1/√2 1 TZZ 1/√2 1/√2 1 1/√2 1/√2 0 -1/√2 1/√2 0 Initial state along Z TXZ = Inverse TZX of TZX Initial state along Z Initial state along X
  • 18. Z X Bringing in the Y Dimension -Y Y Initial state along 1 1 TZXTYZ = TYX Y transformed to state along X 0 0 -X -Z 1/√2 -1/√2 a c 1 +/- 1/√2 = 1/√2 1/√2 b d 0 +/- 1/√2 All UPs along Y All UPs along Y translate to equal translate to equal +/- 1/√2 UPs and DNs UPs and DNs along X along X +/- 1/√2 NOT POSSIBLE!!
  • 19. Revisiting the State Description Yet Again Can we allow for complex values a here? Complex b conjugate |a|2 +|b|2 =a a + b b = 1 How do we translate these to probabilities?
  • 20. Revisiting Transformations Conjugate Transpose a • Tzx must be preserve |a|2 a’ TYX b = b’ +|b|2 – (Tzx)† Tzx = I Transforms state eiεCosθ -ei(ψ – φ+ ε) Sinθ along Z axis to state along X axis ei(φ+ ε) Sinθ ei(ψ+ ε) Cosθ For any θ,ψ,ε
  • 21. Z X Bringing in the Y Dimension -Y Y Initial state along 1 1 TZXTYZ = TYX Y transformed to state along X 0 0 -X -Z 1/√2 -1/√2 1/√2 -e-iφ 1/√2 1 eiφ’’ 1/√2 = 1/√2 1/√2 eiφ 1/√2 1/√2 0 eiφ’ 1/√2 All UPs along Y All UPs along Y translate to equal translate to equal 1/√2 UPs and DNs UPs and DNs along X along X eiφ 1/√2 Φ=π/2, Φ’’=-π/4, Φ’=π/4!!
  • 22. The Final Transformations 1/√2 -1/√2 1/√2 i/√2 TZX 1/√2 1/√2 TYZ i/√2 1/√2 1/√2 -1/√2 1/√2 i/√2 TYX=TZXTYZ= 1/√2 1/√2 i/√2 1/√2 e-iπ/4/√2 -e-iπ/4/√2 eiπ/4/√2 eiπ/4/√2 Can you write the transformation from Z to a general direction in 3D space?
  • 23. Summary • State vector v has complex entries and satisfies – |v|2 = v†v = Σ |vi|2 = 1 • vi’s are called Amplitudes • Transformations T satisfy T†T = I – T’s are called Unitary Transformations • When we measure a system in state v – We get i with Probability |vi|2
  • 24. Contrast with Classical States • Take 2 bits, so state vector [p1 p2 p3 p4] corresponding to 00, 01, 10, 11 resp. • Suppose you replace the first bit by an AND of the 2 bits with prob p and by an OR with prob 1-p? – Show this can be written as a stochastic transformation.
  • 25. Our Two Worlds Σ vi = 1 , 0<=vi<=1 T is stochastic (non-neg, col sums 1) Classical Measurement Quantum |v|2 = v†v = Σ |vi|2 = 1 T is Unitary T†T = I
  • 26. What does this mean for computation?