SU(3) phase operators: some solutions and properties

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SU(3) phase operators: some solutions and properties. Hubert de Guise Lakehead University. Collaborators:. Luis Sanchez-Soto Andrei Klimov. Summary. Polar decomposition : can be easily generalized but many “free parameters” - PowerPoint PPT Presentation

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SU(3) phase operators:some solutions

and properties

Hubert de Guise

Lakehead University

Collaborators:

Luis Sanchez-Soto Andrei Klimov

Summary

• Polar decomposition:– can be easily generalized but many “free

parameters”– Normally yields non-commuting phase

operators

• Complementarity:

– cannot be easily generalized but no “free parameters”

– Normally yields commuting phase operators

The origin: the classical harmonic oscillator

Classical harmonic oscillator: 2 2 21 1

2 2H p m x

m

Use:2 2

*, 2 2

m i m ia x p a x p

m m

( ) i t i i ta t Ae A e e

Quantize: †*1 1, a a a a

Two approaches

Two approaches

†expa i a a• write operator in polar form:

• think of as the exponential of a hermitian phase operator

exp i

Two approaches

†expa i a a• write operator in polar form:

• think of as the exponential of a hermitian phase operator

exp i

• Use complementarity condition:

, , exp exp

,

N a a N i i

N

What they have in common

• Look at rather than exp i

What they have in common

• Look at rather than

• is assumed unitary: is hermitian

exp i

exp i

What they have in common

• Look at rather than

• is assumed unitary: is hermitian

• Must fix some “boundary” problems by hand

exp i

exp i

SU(2) phase operator

ˆexp , 1 ,m

i j m j m

ˆ ˆ ˆ,exp expzL i i

mod(2j+1)

SU(2) phase operator

0 1

0 1 1ˆˆexp ,

0 1 1

0 0 0*

z

j

j

i L

j

j

ˆexp , 1 ,m

i j m j m

ˆ ˆ ˆ,exp expzL i i

mod(2j+1)

Only one “boundary” condition

An example: j=1

1 0 0 0 1 0

ˆ ˆ0 0 0 , exp 0 0 1 ,

0 0 1 01 0zL i

ˆexp 1,1 1,0 1,0 1, 1 1, 1 1, 2i -2=1mod(3)

An example: j=1

1 0 0 0 1 0

ˆ ˆ0 0 0 , exp 0 0 1 ,

0 0 1 01 0zL i

ˆexp 1,1 1,0 1,0 1, 1 1, 1 1, 2i -2=1mod(3)

0 1 0

ˆ ˆ,exp 0 0 1

0 02zL i

-2=1mod(3)

A short course on su(3)

• There are eight elements in su(3)

† †1 1 1 2 2

† †2 2 2 3 3

ˆ ˆ: 1, 2,3

ˆ ˆ ˆ ˆ:

ˆ ˆ ˆ ˆ:

ij i jC a a i j

h a a a a

h a a a a

A short course on su(3)

• There are now two relative phases

• There are eight elements in su(3)

† †1 1 1 2 2

† †2 2 2 3 3

ˆ ˆ: 1, 2,3

ˆ ˆ ˆ ˆ:

ˆ ˆ ˆ ˆ:

ij i jC a a i j

h a a a a

h a a a a

A short course on su(3)

• There are now two relative phases

• There are eight elements in su(3)

† †1 1 1 2 2

† †2 2 2 3 3

ˆ ˆ: 1, 2,3

ˆ ˆ ˆ ˆ:

ˆ ˆ ˆ ˆ:

ij i jC a a i j

h a a a a

h a a a a

• States are of the form 1 2 3n n n

Commutation relations

Commutation relations

12 , C

Commutation relations

12 23,CC

Commutation relations

12 23 13,C CC

Commutation relations

13 23C 0,C

Geometry of weight space

Geometry of weight space

3,0,0

2,0,1

1,1,1

Geometry of weight space

12C

3,0,0

2,0,1

1,1,1

Geometry of weight space

12C

3,0,0

2,0,1

1,1,1

23C

3-dimensional case

1,0,0

0,1,00,0,1

3-dimensional case

1,0,0

0,1,00,0,1

3-dimensional case

1,0,0

0,1,00,0,1

3-dimensional case

1,0,0

0,1,00,0,1

L

3-dimensional case

1,0,0

0,1,00,0,1

L

3-dimensional case

1,0,0

0,1,00,0,1

NOT

an su(3) system

SU(3) phase operators:polar decomposition

12 23

0 1 0 0 0 0

C 0 0 0 C 0 0 1

0 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0

0 0 1 0

* *

= 0 0 1 0 0 0

0 0 0 0 0 0 0 1

* *

* * * *

Solution 1: commuting solution

Solution 1: commuting solution

12

23

2

2

0 1 0

0

0

0

0

0

0

0

0

0 1

0

i

i

e

e

Solution 1: commuting solution

12

23

2

2

0 1 0

0

0

0

0

0

0

0

0

0 1

0

i

i

e

e

23 13 2312 1212 23 12 23 13, 0i i ii ie e e e e

ComplementaritryThe matrices

12 1

13 2

2 212 1

13 12 23 12 2

0 1 0 1

0 0

0 0

0 0 1 1

0 0

0 0

i ih

i ih

E e H e

E E E e H e

form generalized discrete Weyl pairs, in the sense

kij k k ijE H H E

Solution 2: the SU(2) solution

12

23

1 0

0

0 1 0

0

0

0

0 0 1

1

1 0

0 1 0

i

i

e

e

23 13 2312 1212 23 12 23 13, 0i i ii ie e e e e

Higher-dimensional cases

• No commuting solutions

• No complementarity

Infinite dimensional limit

• The edges are infinitely far

•One can find commuting solutions: the phase operator commute, and have common eigenstates of zero uncertainty

Summary

• Polar decomposition:– can be easily generalized but many “free

parameters”– Normally yields non-commuting phase

operators

• Complementarity:

– cannot be easily generalized but no “free parameters”

– Normally yields commuting phase operators

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