tensor product states.pdf
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8/18/2019 Tensor Product States.pdf
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Lecture 24:Tensor Product States
Phy851 Fall 2009
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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 "
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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
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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
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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"
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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(
! ! ! "=
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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
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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 :, ! ! ! ! "=
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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 !!=
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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 =
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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
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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
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‘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
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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 ,,),,( =
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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
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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
!=
!=
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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 ++=
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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
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