Lecture 24:Tensor Product States

Phy851 Fall 2009

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|ψ〉, its momentum-space state iscompletely specified as well:〈p|ψ〉 =∫dx〈p|x〉〈x|ψ〉

– 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:

ψψ xx =)(

ψψ zyxzyx ,,),,( =

or

€

x,y,z{ }

€

∃ x,y,z ∀ (x,y,z)∈ R3

€

ψ(p) = p ψ

Definition of Tensor product

• Suppose you have a system with 10 possiblestates

• Now you want to enlarge your system byadding ten more states to its Hilbert space.– The dimensionality of the Hilbert space increases

from 10 to 20– The system can now be found in one of 20

possible states– This is a sum of two Hilbert sub-spaces– One quantum number is required to specify

which state

• Instead, suppose you want to combine yoursystem 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 ofthe state of the first system

– So two independent quantum numbers arerequired 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

Formalism

• Let H1 and H2 be two Hilbert spaces

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

• Let the Hilbert space H12 be the tensor-

product of spaces H1 and H2 .

• The Tensor product state |ψ12〉=|ψ1〉(1) ⊗|ψ2〉(2)

belongs to H12.

• 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 ψψψψ ⇒←→

1

)1( H∈ψ 2

)2( H∈φ

2112 HHH ⊗= H12 is the ‘tensor product’of H1 and H2

)1(

1

)2(

1

)2(

1

)1(

1 ψϕϕψ ⇒←→ They just slide pasteach other

Schmidt Basis

• The easiest way to find a good basis for atensor product space is to use tensor productsof basis states from each sub-space

If: {|n1〉(1) } ; n1=1,2,…,N1 is a basis in H1

{|n2〉(2) } ; n2=1,2,…,N2 is a basis in H2

It follows that:

{|n1, n2〉(12) }; |n1, n2〉(12)=|n1〉(1)⊗|n2〉(2) is a basis in H12.

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

11

)1(

1

)1(

1

N

nn naψ

∑=

=2

2

21

)2(

2

)2(

2

N

nn nbψ

€

ψ1,ψ2(12)

= ψ1(1)⊗ ψ2

(2)= an1bn2 n1,n2

(12)

n2=1

N2

∑n1=1

N1

∑

∑∑= =

=1

1

2

2

211 1

)12(

2,1

)12(

12

N

n

N

nnn nncψ

Entangled States

• The essence of quantum `weirdness’ lies inthe 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

211 1

)12(

2,1

)12(

12

N

n

N

nnn nncψ

)2(

2

)1(

1

)12(ψψψ ⊗=

Configuration Space

• The state of a quantum system of Nparticles in 3 dimensions lives in‘configuration space’– There are three quantum numbers

associated with each particle• It takes 3N quantum numbers to specify a

state of the full system

– Coordinate Basis:

– Wavefunction:

– To specify a state of N particles in ddimensions requires d·N quantumnumbers

€

x1,y1,z1,x2,y2,z2,...,xN ,yN ,zN{ }

(We don’t know about spin yet)

or we could just write

€

r r 1,

r r 2,...,

r r N{ }

( ) ψψ NN rrrrrrrrrrrr

,...,,,...,, 2121 =

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

This form is thecoordinate system

independentrepresentation

Tensor Products of Operators

• THEOREM:Let A

(1) act in H1 , and B(2) act in H2 ,

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

acts in H12 .

• PROOF:

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

)1()1()1(mm

mm aaaA ∑=

)2()2()2(mm

mm bbbB ∑=

€

A(1) ⊗ B(2) = ambnm,n∑ am

(1) am(1)⊗ bn

(2) bn(2)

€

= ambnm,n∑ am

(1)⊗ bn

(2)( ) am(1)⊗ bn

(2)( ))12()12(

,

)12( ,, nmnmnm

nm bababaC ∑=

)2(

2

)1(

1

)12(

,

)12(

21)12( ,, ψψψψ nmnm

nmnm bababaC ∑=

)2(

2

)1(

1

)12(

21 :, ψψψψ ⊗=

General form of Operators inTensor-product spaces

• The most general form of an operator in H12

is:

– 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 H12

• 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( ',', nmnmcCnm

nmnm∑=

)12()12()12(

',';, ,,: nmCnmc nmnm ′′=

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 anobservable in H12 (since it remains Hermitianwhen upgraded).

– The spectrum of A1 remains the same afterupgrading

Proof:Let

Then:

)2()1()12( IAA ⊗=

€

A(1) ⊗ I(2) am(1)⊗ ψ

(2)( ) = A(1) am(1)( ) ⊗ I(2)ψ (2)( )

= am am(1)⊗ ψ

(2)

= am am(1)⊗ ψ

(2)( )Note that |ψ2〉 is completelyarbitrary, but |a1〉⊗|ψ2〉 is aneigenstate of A(12)

)1()1()1(mmm aaaA =

Product of two Upgraded Operators

• Let A(1) and B(2) be observables in theirrespective 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 ⊗ I2( ) I1 ⊗ B2( ) = A1I1 ⊗ I2B2= A1 ⊗ B2

Compatible observables

• Let A(1) and B(2) be observables in theirrespective Hilbert spaces

• Let A= A(1)⊗ I(2) and B= I(1)⊗ B(2) .

• Theorem: [A,B]=0Proof:

• Conclusion: any operator in H1 , is ‘compatible’

with any operator in H2 ,.

• 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

=⊗−⊗=

⊗−⊗=

⊗⊗−⊗⊗=

−=

BABA

IBAIBIIA

IABIBIIA

BAABBA

‘And’ versus ‘Or’

• The tensor product correlates with a systemhaving 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 8possible 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 andB

– Example: start with a system having 4 energylevels. Add 2 more energy levels to your model,and the dimension goes from 4 to 6

Example #1 : Particle in Three Dimensions

• Let H1 be the Hilbert space of functionsin one dimension– The projector is:

– So a basis is:

• Then H3= (H1)3 would then be the

Hilbert space of square integrablefunctions in three dimensions.– Proof:

• Note: H3 is also the Hilbert space ofthree particles in one-dimensional space

€

I1 = dx∫ x x

€

x{ }

€

I3 = I1 ⊗ I1 ⊗ I1= 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

),,(,, zyxzyxdxdydz ψψ ∫=

zyxzyx ⊗⊗≡,, ψψ zyxzyx ,,),,( =

Three-dimensional Operators

• We can define the vector operators:

• Note that: X = X(1)⊗I(2) ⊗I(3) and Py = I(1)⊗P

(2) ⊗I(3)

so that [X, Py]=0.

• With

• We can use:

zPyPxPP

zZyYxXR

zyx ˆˆˆ

ˆˆˆ

++=

++=r

r

( ) zyxzzyyxxzyxR ,,ˆˆˆ,, ++=r

ZR

YR

XR

=

=

=

3

2

1

z

y

x

PP

PP

PP

=

=

=

3

2

1

€

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

R j ,Pk[ ] = ihδ jk

€

r R

r r =

r r

r r or

Example #2: Two particles in OneDimension

• For two particles in one-dimensional space,the Hilbert space is (H1 )2.

)2(

2

)1(

121, xxxx ⊗=

ψψ 2121 ,),( xxxx =

€

I = dx1dx2∫ x1,x2 x1,x2 =1

€

X j ,Xk[ ] = Pj ,Pk[ ] = 0

X j ,Pk[ ] = ihδ jk

...212

211

etc

PIP

IXX

⊗=

⊗=

Hamiltonians

• One particle in three dimensions:– Each component of momentum contributes

additively to the Kinetic Energy

• Two particles in one dimension:

( )

)(2

1

),,(2

1 222

RVPPm

ZYXVPPPm

H zyx

vrr+⋅=

+++=

),(22 21

2

22

1

21 XXV

m

P

m

PH ++=

Conclusions

• The ‘take home’ messages are:– The combined Hilbert space of two

systems, dimensions d1 and d2, hasdimension d1⊗2=d1·d2

– A physical basis set for the combinedHilbert space, H1⊗2 can be formed bytaking all possible products of one basisstate from space H1 with one basis statefrom H2.

– 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 2

n1,n2 := n1(1)⊗ n2

(2)

∴ n1,n2{ }⊇ H 12 := H 1 ⊗H 2

21

)1(

2

)1(

1 ψψψψ ⇒←→)1(

1

)2(

1

)2(

1

)1(

1 ψϕϕψ ⇒←→

NzNyNxzyxzyx

N

nnnnnnnnn

rrr

,,,,,,,,,

,,

222111

21

K

rK

rr