tensor product states.pdf

8/18/2019 Tensor Product States.pdf

1/18

Lecture 24:Tensor Product States

Phy851 Fall 2009


2/18

Basis sets for a particle in 3D

• Clearly the Hilbert space of a particle in threedimensions is not the same as the Hilbertspace for a particle in one-dimension

• In one dimension, X and P are incompatible

– If you specify the wave function in coordinate-

space, !x|" #, its momentum-space state iscompletely specified as well:! p|" # = $ dx ! p| x #! x |" #

– You thus specify a state by assigning anamplitude to every possible position OR byassigning and amplitude to every possiblemomentum

• In three dimensions, X, Y, and Z, arecompatible.

– Thus, to specify a state, you must assign anamplitude to each possible position in threedimensions.

– This requires three quantum numbers

– So apparently, one basis set is:

! ! x x =)(

! ! z y x z y x ,,),,( =

or

x, y, z{ }" x, y, z # ( x, y, z)$ R

3

" ( p) = p "


3/18

Definition of Tensor product

• Suppose you have a system with 10 possible

states

• Now you want to enlarge your system byadding ten more states to its Hilbert space.

– The dimensionality of the Hilbert space increasesfrom 10 to 20

– The system can now be found in one of 20possible states

– This is a sum of two Hilbert sub-spaces

– One quantum number is required to specifywhich state

• Instead, suppose you want to combine your

system with a second system, which has tenstates of its own

– The first system can be in 1 of its 10 states

– The second system can be in 1 of its 10 states

• The state of the second system is independent of the state of the first system

– So two independent quantum numbers are

required to specify the combined state

• The dimensionality of the combined Hilbertspace thus goes from 10 to 10x10=100

– This combined Hilbert space is a

(Tensor ) Product of the two Hilbert sub-spaces


4/18

Formalism

• Let H 1 and H 2 be two Hilbert spaces

– We will temporarily ‘tag’ states with a label tospecify which space the state belongs to

• Let the Hilbert space H 12 be the tensor-

product of spaces H 1 and H 2 .

• The Tensor product state |" 12#=|" 1#(1) %|" 2#

(2)

belongs toH

12.

• The KEY POINT TO ‘GET’ IS:

– Bras and kets in the same Hilbert space‘attach’.

– BUT, Bras and kets in different Hilbertspaces do not ‘attach’

21

)1(

2

)1(

1 ! ! ! ! "#$

1

)1(H !" 2

)2(H !"

2112 H H H != H 12 is the ‘tensor product’

of H 1 and H 2

)1(

1

)2(

1

)2(

1

)1(

1 ! " " ! #$%They just slide past

each other


5/18

Schmidt Basis

• The easiest way to find a good basis for a

tensor product space is to use tensor productsof basis states from each sub-space

If: {|n1#(1) } ; n1=1,2,…, N 1 is a basis in H 1

{|n2#(2) } ; n2=1,2,…, N 2 is a basis in H 2

It follows that:

{|n1, n2#(12) }; |n1, n2#

(12)=|n1#(1)%|n2#

(2) is a basis in H 12.

If System 1 is in state:

and System 2 is in state:

Then the combined system is in state:

• Schmidt Decomposition Theorem:All states in a tensor-product space can beexpressed as a linear combination of tensorproduct states

!=

=

1

1

1

1

)1(

1

)1(

1

N

n

n na"

!=

=

2

2

2

1

)2(

2

)2(

2

N

n

n nb"

" 1," 2(12)

= " 1(1)# " 2

(2)= a

n1bn2

n1,n2(12)

n2=1

N 2

$n1=1

N 1

$

!!= =

=

1

1

2

2

21

1 1

)12(

2,1

)12(

12

N

n

N

n

nn nnc"


6/18

Entangled States

• The essence of quantum `weirdness’ lies in

the fact that there exist states in the tensor-product space of physically distinct systemsthat are not tensor product states

• A tensor-product state is of the form

– Tensor-product states are called ‘factorizable’

• The most general state is

– This may or may-not be ‘factorizable’

• Non-factorizable states are called ‘entangled’

– For an `entangled state’, each sub-system has no independent objectivereality

!!= =

=

1

1

2

2

21

1 1

)12(

2,1

)12(

12

N

n

N

n

nn nnc"

)2(

2

)1(

1

)12(

! ! ! "=


7/18

Configuration Space

• The state of a quantum system of Nparticles in 3 dimensions lives in

‘configuration space’

– There are three quantum numbersassociated with each particle

• It takes 3N quantum numbers to specify astate of the full system

– Coordinate Basis:

– Wavefunction:

– To specify a state of N particles in d dimensions requires d·N quantum

numbers

x1, y

1, z

1, x

2, y

2, z

2,..., x

N , y

N , z

N { }

(We don’t know about spin yet)

or we could just write

r

r1,r

r2,...,

r

r N { }

( ) ! ! N N

r r r r r r rrrrrr

,...,,,...,,2121

=

So counting quantum numbers might be a goodway to check if you are using a valid basis

This form is thecoordinate system

independent

representation


8/18

Tensor Products of Operators

• THEOREM:

Let A (1) act in H 1 , and B

(2) act in H 2 ,

Then the tensor product operator C (12) = A(1)% B(2)

acts in H 12 .

• PROOF:

• The action of C (12) on a tensor-product state:

)1()1()1(

mm

m

m aaa A !=

)2()2()2(mm

m

m bbb B !=

A(1)" B

(2)= a

mbn

m,n

# am(1)

am

(1)" b

n

(2)bn

(2)

= ambn

m,n" a

m

(1)# b

n

(2)

( ) a

m

(1)# b

n

(2)

( ))12()12(

,

)12( ,,nmnm

nm

nm bababaC !=

)2(

2

)1(

1

)12(

,

)12(

21

)12( ,, ! ! ! ! nmnm

nm

nm bababaC "=

)2(

2

)1(

1

)12(

21 :, ! ! ! ! "=


9/18

General form of Operators inTensor-product spaces

• The most general form of an operator in H 12is:

– Here |m,n# may or may not be a tensorproduct state. The important thing is thatit takes two quantum numbers to specifya basis state in H 12

• A basis that is not formed from tensor-product states is an ‘entangled-state’ basis

• In the beginning, you should always startwith a tensor-product basis as your ‘physicalbasis’

– Then all operators are well-defined

– Just expand states and operators onto tensor-product states

– Then match up the bras and kets with theirproper partners when taking inner products

)12()12(

,

',';,

)12( ',', nmnmcC nm

nmnm!=

)12()12()12(

',';, ,,: nmC nmc nmnm !!=


10/18

Upgrading Subspace Operators

• Any operator in H1

can be upgraded to an

operator in H12 by taking the tensor product

with the identity operator in H2 :

– If A1 is an observable in H1, then it is also an

observable in H12 (since it remains Hermitianwhen upgraded).

– The spectrum of A1 remains the same afterupgrading

Proof:

Let

Then:

)2()1()12( I A A !=

A(1)" I

(2)am

(1)" #

(2)( ) = A(1) am (1)( ) " I (2) # (2)( )= a

m a

m

(1)" #

(2)

=

am am

(1)

" #

(2)

( )Note that |" 2# is completely

arbitrary, but |a 1 #%|"

2 # is an

eigenstate of A(12)

)1()1()1(

mmm aaa A =


11/18

Product of two Upgraded Operators

• Let A(1) and B(2) be observables in their

respective Hilbert spaces• Let A(12)= A(1)% I (2) and B(12)= I (1)% B(2) .

• The product A(12) B(12) is given by A(1) % B(2)

– Proof:

A1" I

2( ) I 1 " B2( ) = A1 I 1 " I 2 B2= A

1" B

2


12/18

Compatible observables

• Let A(1) and B(2) be observables in their

respective Hilbert spaces• Let A= A(1)% I (2) and B= I (1)% B(2) .

• Theorem: [ A,B]=0

Proof:

• Conclusion: any operator in H 1 , is ‘compatible’

with any operator in H 2 ,.

• I.e. simultaneous eigenstates exist.

– Let A1 |a1#= a1 |a1# and B2 |b2#= b2 |b2#.

– Let a= a1 and b= b2

– Let |ab#= |a1# % |b2#.

– Then AB|ab#=ab|ab#.

[ ]( )( ) ( )( )

( ) ( ) ( ) ( )0

,

2121

22112211

21212121

=!"!=

!"!=

!!"!!=

"=

B A B A

I B A I B I I A

I A B I B I I A

BA AB B A


13/18

‘And’ versus ‘Or’

• The tensor product correlates with a system

having property A and property B– Dimension of combined Hilbert space is product

of dimensions of subspaces associated with Aand B

– Example: start with a system having 4 energylevels. Let it interact with a 2 level system. TheHilbert space of the combined system has 8

possible states.

• Hilbert spaces are added when a system canhave either property A or property B

– Dimension of combined Hilbert space is sum of dimensions of subspaces associated with A and

B– Example: start with a system having 4 energy

levels. Add 2 more energy levels to your model,and the dimension goes from 4 to 6


14/18

Example #1 : Particle in Three Dimensions

• Let H

1 be the Hilbert space of functionsin one dimension– The projector is:

– So a basis is:

• Then H 3 = (H

1)3 would then be the

Hilbert space of square integrablefunctions in three dimensions.

– Proof:

• Note: H 3 is also the Hilbert space of

three particles in one-dimensional space

I 1= dx " x x

x { }

I 3= I

1" I

1" I

1

= dx # x x " dy # y y " dz # z z

= dxdydz # x x " y y " z z

= dxdydz # x " y " z( ) x " y " z( )

= dxdydz # x, y, z x, y, z

),,(,, z y x z y xdxdydz ! ! " =

z y x z y x !!",,

! ! z y x z y x ,,),,( =


15/18

Three-dimensional Operators

• We can define the vector operators:

• Note that: X = X (1)% I (2) % I (3) and P y = I

(1)% P (2)

% I (3)

so that [ X , P y]=0.

• With

• We can use:

z P y P x P P

z Z yY x X R

z y x ˆˆˆ

ˆˆˆ

++=

++=

r

r

( ) z y x z z y y x x z y x R ,,ˆˆˆ,, ++=r

Z R

Y R

X R

=

=

=

3

2

1

z

y

x

P P P P

P P

=

=

=

3

2

1

R j , Rk [ ] = P j ,Pk [ ] = 0

R j

,Pk

[ ]= ih"

jk

r

R

r

r=

r

r

r

ror


16/18

Example #2: Two particles in OneDimension

• For two particles in one-dimensional space,

the Hilbert space is (H 1 )2.

)2(

2

)1(

121, x x x x !=

! ! 2121 ,),( x x x x =

I = dx1dx

2 " x1, x2 x1, x2 =1

X j , X k [ ] = P j ,Pk [ ] = 0

X j

,Pk

[ ]= ih"

jk

...

212

211

etc

P I P

I X X

!=

!=


17/18

Hamiltonians

• One particle in three dimensions:

– Each component of momentum contributesadditively to the Kinetic Energy

• Two particles in one dimension:

( )

)(2

1

),,(2

1 222

RV P P m

Z Y X V P P P m

H z y x

vrr

+!

=

+++=

),(22

21

2

2

2

1

2

1 X X V

m

P

m

P H ++=


18/18

Conclusions

• The ‘take home’ messages are:

– The combined Hilbert space of twosystems, dimensions d 1 and d 2, hasdimension d 1%2=d 1·d 2

– A physical basis set for the combinedHilbert space, H 1%2 can be formed by

taking all possible products of one basisstate from space H 1 with one basis statefrom H 2.

– In a tensor product space, a bra from onesubspace can only attach to a ket fromthe same subspace:

– For N particles (spin 0) in d dimensions,d·N quantum numbers are required tospecify a unit-vector in any basis

n1

(1){ }" H 1, n2 (2){ }" H 2n

1,n

2 := n

1

(1)# n

2

(2)

$ n1,n2{ }" H 12 :=H 1 #H 2

21

)1(

2

)1(

1 ! ! ! ! "#$

)1(

1

)2(

1

)2(

1

)1(

1 ! " " ! #$%

Nz Ny Nx z y x z y x

N

nnnnnnnnn

r r r

,,,,,,,,,

,,

222111

21

K

r

K

rr

tensor product states.pdf

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