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Lecture 24: Tensor Product States Phy851 Fall 2009 Basis sets for a particle in 3D • Clearly the Hilbert space of a particle in three dimensions is not the same as the Hilbert space for a particle in one-dimension • In one dimension, X and P are incompatible – If you specify the wave function in coordinatespace, 〈x|ψ〉, its momentum-space state is completely specified as well: 〈p|ψ〉 =∫dx〈p|x〉〈x|ψ〉 – You thus specify a state by assigning an amplitude to every possible position OR by assigning and amplitude to every possible momentum ψ ( x) = x ψ or ψ ( p) = p ψ • In three dimensions, X, Y, and Z, are compatible. – Thus, to specify a state, you must assign an € position in three amplitude to each possible dimensions. – This requires three quantum numbers ψ ( x, y , z ) = x, y , z ψ – So apparently, one basis set is: { x, y,z } ∃ x, y,z ∀ (x, y,z) ∈ R 3 Definition of Tensor product • Suppose you have a system with 10 possible states • Now you want to enlarge your system by adding 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 your system with a second system, which has ten states 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 Hilbert space 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 to specify which space the state belongs to ψ (1) ∈ H1 • Let the Hilbert space product of spaces ( 2) φ H12 ∈ H2 be the tensor- H1 and H2 . H 12 = H 1 ⊗ H 2 H12 is the ‘tensor product’ of 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’. ψ1 (1) → ← ψ2 (1) ⇒ ψ1 ψ 2 – BUT, Bras and kets in different Hilbert spaces do not ‘attach’ ψ1 (1) → ← ϕ1 ( 2) ⇒ ϕ1 ( 2) ψ1 (1) They just slide past each other Schmidt Basis • The easiest way to find a good basis for a tensor product space is to use tensor products of basis states from each sub-space If: H1 {|n2〉(2) } ; n2=1,2,…,N2 is a basis in H2 {|n1〉(1) } ; n1=1,2,…,N1 is a basis in It follows that: {|n1, n2〉(12) }; |n1, n2〉(12)=|n1〉(1)⊗|n2〉(2) is a basis in If System 1 is in state: ψ1 (1) N1 = ∑ an1 n1 H12. (1) n1 =1 and System 2 is in state: ψ 2 ( 2) N2 = ∑ bn2 n2 ( 2) n2 =1 Then the combined system is in state: ψ1,ψ 2 (12) = ψ1 (1) ⊗ ψ2 (2) N1 N2 = ∑ ∑ an1 bn 2 n1,n 2 (12) n1 =1n 2 =1 • Schmidt Decomposition Theorem: € All states in a tensor-product space can be expressed as a linear combination of tensor product states N N ψ 12 (12 ) 1 2 = ∑ ∑ cn1n2 n1, n2 n1 =1 n2 =1 (12 ) Entangled States • The essence of quantum `weirdness’ lies in the fact that there exist states in the tensorproduct space of physically distinct systems that are not tensor product states • A tensor-product state is of the form ψ (12 ) = ψ1 (1) ⊗ ψ2 ( 2) – Tensor-product states are called ‘factorizable’ • The most general state is ψ 12 (12 ) N1 N2 = ∑ ∑ cn1n2 n1, n2 (12 ) n1 =1 n2 =1 – This may or may-not be ‘factorizable’ • Non-factorizable states are called ‘entangled’ – For an `entangled state’, each subsystem has no independent objective reality Configuration Space • The state of a quantum system of N particles in 3 dimensions lives in ‘configuration space’ (We don’t know about spin yet) – There are three quantum numbers associated with each particle • It takes 3N quantum numbers to specify a state of the full system – Coordinate Basis: { x1, y1,z1, x 2, y 2 ,z2,..., x N , y N ,zN } or we could just write € r r r { r1, r2,..., rN – Wavefunction: } This form is the coordinate system independent representation r r r r r r € ψ (r1 , r2 ,..., rN ) = r1 , r2 ,..., rN ψ – To specify a state of N particles in d dimensions requires d·N quantum numbers So counting quantum numbers might be a good way to check if you are using a valid basis 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: A(1) = ∑ am am (1) am (1) B ( 2 ) = ∑ bm bm ( 2) bm ( 2) m m A(1) ⊗ B (2) = ∑ am bn am (1) am (1) (2) bn ⊗ bn (2) m,n ( = ∑ am bn am m,n € (1) ⊗ (12 ) C (12 ) = ∑ ambn am , bn bn (2) )( am , bn am (1) ⊗ bn (2) (12 ) m,n € • The action of C(12) on a tensor-product state: ψ 1 ,ψ 2 C (12 ) ψ 1 ,ψ 2 (12 ) (12 ) := ψ 1 (1) = ∑ ambn am , bn m,n ⊗ (12 ) ψ2 ( 2) am ψ 1 (1) bn ψ 2 ( 2) ) General form of Operators in Tensor-product spaces • The most general form of an operator in is: C (12 ) = ∑ cm ,n;m ',n ' m, n (12 ) m' , n' H12 (12 ) m,n cm ,n;m ',n ' := m, n (12 ) C (12 ) m′, n′ (12 ) – Here |m,n〉 may or may not be a tensor product state. The important thing is that it takes two quantum numbers to specify a basis state in H12 • A basis that is not formed from tensorproduct states is an ‘entangled-state’ basis • In the beginning, you should always start with a tensor-product basis as your ‘physical basis’ – Then all operators are well-defined – Just expand states and operators onto tensorproduct states – Then match up the bras and kets with their proper partners when taking inner products Upgrading Subspace Operators • Any operator in H1 can be upgraded to an H12 by taking the tensor product with the identity operator in H2 : operator in A(12 ) = A(1) ⊗ I ( 2 ) – If A1 is an observable in H1, then it is also an observable in H12 (since it remains Hermitian when upgraded). – The spectrum of A1 remains the same after upgrading Proof: Let A(1) am (1) = am am (1) Then: ( A(1) ⊗ I (2) am (1) ⊗ψ (2) ) ( (1) = A(1) am = am am ( = am am (1) € ⊗ψ (1) Note that |ψ2〉 is completely ) ( ⊗ I ⊗ arbitrary, but |a1〉⊗|ψ2〉 is an eigenstate of A(12) (2) ψ (2) ψ (2) ) (2) ) 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 ⊗ I2 )( I1 ⊗ B2 ) = A1I1 ⊗ I2 B2 = A1 ⊗ B2 € 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: [A, B]= AB − BA = (A1 ⊗ I 2 )(I1 ⊗ B2 )− (I1 ⊗ B2 )(A1 ⊗ I 2 ) = (A1 I1 )⊗ (I 2 B2 )− (I1 A1 )⊗ (B2 I 2 ) = A1 ⊗ B2 − A1 ⊗ B2 = 0 • 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〉. ‘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 A and B – Example: start with a system having 4 energy levels. Let it interact with a 2 level system. The Hilbert space of the combined system has 8 possible states. • Hilbert spaces are added when a system can have 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 Example #1 : Particle in Three Dimensions • Let H1 be the Hilbert space of functions in one dimension – The projector is: I1 = dx x x ∫ – So a basis is: {x } € • Then H3= (H1)3 would then be the Hilbert space of square integrable € functions in three dimensions. – Proof: 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 = x, y , z ≡ x ⊗ y ⊗ z € ψ ( x, y , z ) = x, y , z ψ ψ = ∫ dxdydz x, y, z ψ ( x, y, z ) • Note: H3 is also the Hilbert space of three particles in one-dimensional space Three-dimensional Operators • We can define the vector operators: r R = X xˆ + Y yˆ + Z zˆ r P = Px xˆ + Py yˆ + Pz zˆ r R x, y, z = (x xˆ + y yˆ + z zˆ ) x, y, z rr rr or R r = r r • Note that: X = X(1)⊗I(2) ⊗I(3) and Py = I(1)⊗P (2) ⊗I(3) so that [X, Py]=0. • With € R1 = X P1 = Px R2 = Y P2 = Py R3 = Z P3 = Pz • We can use: € [R ,R ] = [P ,P ] = 0 [R ,P ] = ihδ j k j k j k jk Example #2: Two particles in One Dimension • For two particles in one-dimensional space, the Hilbert space is (H1 )2. x1 , x2 = x1 (1) ⊗ x2 ( 2) ψ ( x1 , x2 ) = x1 , x2 ψ I= ∫ dx dx 1 2 x1, x 2 x1, x 2 = 1 X1 = X1 ⊗ I2 P2 = I1 ⊗ P2 € etc... [ X , X ] = [P ,P ] = 0 [ X ,P ] = ihδ j j € k k j k jk Hamiltonians • One particle in three dimensions: – Each component of momentum contributes additively to the Kinetic Energy 1 H= Px2 + Py2 + Pz2 + V ( X , Y , Z ) 2m v 1 r r = P ⋅ P + V ( R) 2m ( ) • Two particles in one dimension: P12 P22 H= + + V ( X1, X 2 ) 2m1 2m2 Conclusions • The ‘take home’ messages are: – The combined Hilbert space of two systems, dimensions d1 and d2, has dimension d1⊗2=d1·d2 – A physical basis set for the combined Hilbert space, H1⊗2 can be formed by taking all possible products of one basis state from space H1 with one basis state from H2. { n1 (1) } ⊇ H 1, n1,n 2 := n1 ∴ { n1,n 2 (1) { n2 ⊗ n2 (2) }⊇ H 2 (2) } ⊇ H 12 := H 1 ⊗ H 2 – In a tensor product space, a bra from one subspace can only attach to a ket from € the same (1subspace: ) (1) ψ1 ψ1 → ← ψ2 (1) → ← ϕ1 ( 2) ⇒ ψ1 ψ 2 ⇒ ϕ1 ( 2) ψ1 (1) – For N particles (spin 0) in d dimensions, d·N quantum numbers are required to specify in any basis r r a unit-vector r r1 , r2 , K rN n1x , n1 y , n1z , n2 x , n2 y , n2 z , K , nNx , nNy , nNz