Seminar: Tensor Products and the Partial Trace

SEMINAR:TENSOR PRODUCTSANDTHE PARTIALTRACE

Georg Maringer

Monday 23rd April, 2018

Deﬁnition (dual bases): If {v1, ... , vm} and {w1, ... , wn} are bases of V and W then { f1,..., fm} and {g1,..., gn} denote the corresponding dual bases.

fk(vl) = δk,l, gs(wt) = δs,t (0.1)

Inner products using the tensor product The Riesz representation theorem states that for every element in the dual space of a Hilbert space there exists a unique element of the Hilbert space corresponding to the respective element of the dual space.

fi(v) = hvi, vi (0.2)

For a proof look at any book covering functional analysis.

1 1. Tensor products of Hilbert spaces

Deﬁnition: Let V =∼ Cm and W =∼ Cn be ﬁnite dimensional Hilbert spaces. The tensor product space V ⊗ W is the vector space consisting of all bilinear forms K on V∗ × W∗.

Let v ∈ V and w ∈ W. Then we deﬁne v ⊗ w ∈ V ⊗ W by

v ⊗ w( f , g) = f (v)g(w) ∀ f ∈ V∗, g ∈ W∗ . (1.1)

Elements of the form v ⊗ w are called product vectors. Consequence:

(µv1 + λv2) ⊗ w = µv1 ⊗ w + λv2 ⊗ w (1.2) and similarly in the second argument.

A general element in V ⊗ W is a superposition ∑i,j αi,jvi ⊗ wj with vi ∈ V and wj ∈ W and αi,j ∈ C and ! ∑ αi,jvi ⊗ wj ( f , g) = ∑ αi,j f (vi) f (wj) . (1.3) i,j i,j

The inner product on V ⊗ W is deﬁned for product vectors as

hv ⊗ w, v0 ⊗ w0i = hv, v0ihw, w0i (1.4) and this deﬁnition is linearly (in the second argument) respectively antilinearly (in the ﬁrst argument) extended to all of V ⊗ W.

m n Main message: If {ej}j=1 and { fk}k=1 are ONBs of V and W, respectively, then

{ej ⊗ fk}j=1,...,m (1.5) k=1,...,n is an ONS of V ⊗ W.

Let K be a bilinear mapping on V∗ ⊗ W∗.

K( f , g) = ∑ K( f (vi) fi, g(wj)gj) fi(v)gj(w) = (1.6) i,j

2 ∑ K( fi, gj) f (vi)g(wj) = ∑ K( fi, gj)[vi ⊗ wj]( f , g) (1.7) i,j i,j

Thus, K = ∑i,j K( fi, gj)[vi ⊗ wj] and {vi ⊗ wj : 1 ≤ i ≤ m, 1 ≤ j ≤ n} spans V ⊗ W.

Linear independence follows because

∑ bi,jvi ⊗ wj = 0 = (∑ bi,jvi ⊗ wj)( fk, gl) = ∑ bi,j fk(vi)gl(wj) = bk,l (1.8) i,j i,j i,j

Considering that

[v ⊗ w]i,j = fi(v)gj(w) = hvi, vihwj, wi (1.9)

Matrix isomorphism: There exists an isomorphism between V ⊗ W and the matrices

Matm×n via the correspondence

[K]i,j = K( fi, gj) (1.10)

Example: Let v ∈ V and w ∈ W. The matrix corresponding to v ⊗ w is given by (for bases assume the bases as before (dual and normal one))

[v ⊗ w]i,j = [v ⊗ w]( fi, gj) = fi(v)gj(w) = [v]i[w]j (1.11)

rank one matrix iff the corresponding tensor is a product vector in V ⊗ W.

What we want to have now is an inner product on V ⊗ W. This is established by the matrix correspondence given before. For matrices a possible inner product is the Hilbert-Schmidt inner product. Let B, C ∈ V ⊗ W

∗ hB, Ci = Tr([B] [C]) = ∑ Bi,jCj,i (1.12) i,j

0 0 0 0 0 0 hv ⊗ w, v ⊗ w i = ∑ hvi, vihwj, wihvi, v ihwj, w i = hv, v ihw, w i (1.13) i,j

3 Using this deﬁnition for the inner product it is easy to see that the system {vi ⊗ wj : 1 ≤ i ≤ m, 1 ≤ j ≤ n} is an orthonormal basis of V ⊗ W.

Composite systems Composite quantum systems can be described by the tensor product of the Hilbert spaces which describe the contributing systems.

Example: spin up |0i = (1 0)T and spin down |1i = (0 1)T. The rest of the states is the composition of the two. They span C2 = span{|0i, |1i}. As described earlier the tensor product space C2 ⊗ C2 = span{|0i ⊗ |0i, |0i ⊗ |1i, |1i ⊗ |0i, |1i ⊗ |1i}.

Bell ONB (maximally entangled state)

1 1 1 1 {√ (|00i + |11i), √ (|00i − |11i), √ (|01i + |10i), √ (|01i − |10i)} (1.14) 2 2 2 2 check orthogonality:

1 1 (h00| + h11|)(|00i − |11i) = (h00|00i − h00|11i + h11|00i − h11|11i) = 0 2 2 (1.15) check normalization:

1 (h00| + h11|)(|00i + |11i) = 1 (1.16) 2

Review: state: ρ∗ = ρ, ρ ≥ 0, tr(ρ) = 1. Pure state: ρ = |ΨihΨ| , kΨk = 1. Observable A = A∗.

Deﬁnition: Let ρ be a state and A = A∗ be an observable. Then expectation value of A in the state ρ is given by tr(Aρ).

tr(Aρ) = tr(A|ΨihΨ|) (1.17) d = ∑hj|A|ΨihΨ|ji (1.18) j=1 = ... (1.19)

4 Tensor product of matrices Operators on V ⊗ W: For A ∈ B(V) and B ∈ B(W), we deﬁne A ⊗ B ∈ B(V ⊗ W) by

(A ⊗ B)(v ⊗ w) = Av ⊗ Bw (1.20) where this deﬁnition is again linearly extended to all of V ⊗ W. We call an operator of the form A ⊗ B a product operator. Consequence:

(A1 ⊗ B1)(A2 ⊗ B2) = A1 A2 ⊗ B1B2 (1.21)

Remark: every operator X ∈ B(V ⊗ W) can be written as a linear combination of product operators.

X = ∑ x(i,j),(k,l)(|eii ⊗ | fji)(hek| ⊗ h fl|) (1.22) i,j,k,l

= ∑ xi,j,k,l|eiihek| ⊗ | fjih fl| . (1.23) i,j,k,l

Little examples: Let A and B be invertible matrices. Then

(A ⊗ B)−1 = A−1 ⊗ B−1 (1.24) just write down deﬁnitions.

More interesting is the adjoint:

(A ⊗ B)∗ = A∗ ⊗ B∗ (1.25)

Proof:

Let φ and ψ be arbitrary elements in V ⊗ W.

hφ, (A ⊗ B)ψi = h∑ αijvi ⊗ wj, (A ⊗ B)(∑ βklvk ⊗ wl)i (1.26) i,j k,l

∑ αijβklhvi ⊗ wj, Avk ⊗ Bwli = ∑ αijβklhvi, Avkihwj, Bwli (1.27) i,j,k,l i,j,k,l

5 ∗ ∗ ∗ ∗ = ∑ αijβklhA vi ⊗ B wj, vk ⊗ wli = h(A ⊗ B )(∑ αij(vi ⊗ wj)), ∑ βklvk ⊗ wli i,j,k,l i,j k,l (1.28) Consequently (A ⊗ B)∗ = A∗ ⊗ B∗.

Lemma:

tr(A ⊗ B) = tr(A)tr(B) . (1.29)

Proof:

tr(A ⊗ B) = ∑(hej| ⊗ h fk|)(A ⊗ B)(|eji ⊗ | fki) (1.30) j,k

= ∑hej|A|eji · h fk|B| fki (1.31) j,k = tr(A)tr(B) . (1.32)

Deﬁnition: local observable. An observable A ∈ B(H1 ⊗ H2) is called local on the ﬁrst system (i.e., on H1) if

A = O1 ⊗ I2 for some O1 ∈ B(V1) . (1.33)

Here I2 denotes the identity on V2.

Similarly, local on H2.

1 Example: σz ⊗ I . σz = |0ih0| − |1ih1| Compute expectation value in |Ψi = √ (|0i ⊗ 2 2 |1i + |1i ⊗ |0i).

By the rule for the product states of traces at the beginning

tr((σz ⊗ I2)|ψihψ|) = hψ|(σ ⊗ I)|ψi = (1.34) 1 = (h0| ⊗ h1| + h1| ⊗ h0|)(σ ⊗ I)(|0i ⊗ |1i + |1i ⊗ |0i) 2 z (1.35) 1 = (h0| ⊗ h1| + h1| ⊗ h0|)(|0i ⊗ |1i − |1i ⊗ |0i) (1.36) 2 1 = (h0|0ih1|1i − h1|1ih0|0i) = 0 (1.37) 2

6 A product state is deﬁned as

|ψi = |φ1i ⊗ φ2i. (1.38)

Next example 1 |ψi = √ (|0i ⊗ |0i + |1i ⊗ |1i) (1.39) 2 The question is now what the state of the ﬁrst respectively the second system is.

Motivation for the partial trace: ﬁnd mathematical object which reproduces all expectation values of local observables in a given state.

Deﬁnition Let ρ12 ∈ B(H1 ⊗ H2) an arbitrary operator. Then the unique operator ρ1 ∈ B(H1) satisfying

tr((O1 ⊗ I2)ρ12) = tr(O1ρ1) for all O1 ∈ B(H1) (1.40) is called the partial trace of ρ12.

Existence and uniquess: Riesz.

Notation: ρ1 = tr2ρ12. Alternative terminology: reduced density operator of ρ12.

tr2ρ12 = tr2(|φ1i ⊗ |φ2ihφ1| ⊗ hφ2|) (1.41)

= tr2(|φ1ihφ1| ⊗ |φ2ihφ2|) = |φ1ihφ1|tr(|φ2ihφ2|) (1.42)

= |φ1ihφ1| (1.43) alternatively, let O1 be arbitrary ∈ H1:

tr((O1 ⊗ I2)(|φ1i ⊗ |φ2ihφ1| ⊗ hφ2|)) = hφ1| ⊗ hφ2|(O1 ⊗ I2)|φ1i ⊗ |φ2i (1.44)

= hφ1| ⊗ hφ2)|(O1 ⊗ I2)|φ1i ⊗ |φ2i (1.45)

= hφ1| ⊗ hφ2|O1|φ1i ⊗ |φ2i (1.46)

= hφ1|O1φ1ihφ2|φ2i (1.47)

= hφ1|O1φ1i (1.48)

7 Example: ρ = |ΨihΨ| with |Ψi = √1 (|0i ⊗ |1i + |1i ⊗ |0i). Then ρ = I /2. 12 2 1 1

1 tr ρ = (|0i ⊗ |1i + |1i ⊗ |0i)(h0| ⊗ h1| + h1| ⊗ h0|) (1.49) 2 12 2 1 = (|0ih0| ⊗ |1ih1| + |1ih0| ⊗ |0ih1| + |0ih1| ⊗ |1ih0| + |1ih1| ⊗ |0ih0|) 2 (1.50)

Remark: the partial trace deﬁnes a map

tr2 : B(H1 ⊗ H2) → B(H1) . (1.51)

Properties:

tr2(A ⊗ B) = A · tr(B) . (1.52)

Proof: Let O1 ∈ H1 be arbitrary. Then

tr((O1 ⊗ I2)(A ⊗ B)) = tr (O1 A ⊗ B) (1.53)

= tr(O1 A)tr(B) (1.54)

= tr(O1(Atr(B))) . (1.55)

n Remark: This implies for example, that for any ONB { fj}j=1 on H2, we have

m tr2X = ∑(I1 ⊗ h fj|)X(I1 ⊗ | fji) . (1.56) j=1

Proof: write X as a sum of product operators, insert tr(Y) = ∑jh fj|Y| fji.

Examples: 1 |ψi = √ (h0| ⊗ h0| + h1| ⊗ h1|) (1.57) 2

8 1 |ψihψ| = (|00ih00| + |11ih00| + |00ih11| + |11ih11|) (1.58) 2 1 = (|0ih0| ⊗ |0ih0| + |1ih0| ⊗ |1ih0| + |0ih1| ⊗ |0ih1| + |1ih1| ⊗ |1ih1|) 2 (1.59)

  1 0 0 1       0 0 0 0 1 1   |ψihψ| = (|00hi00| + |00ih11| + |11ih00| + |11ih11|) =   (1.60) 2 2     0 0 0 0       1 0 0 1

  1 0 1 1 1   Tr2|ψihψ| = |0ih0| + |1ih1| =   (1.61) 2 2 2   0 1

  1 0 1 1 1   Tr1|ψihψ| = |0ih0| + |1ih1| =   (1.62) 2 2 2   0 1 fully mixed state (maximal entropy).

Next example

σz = |0ih0| − |1ih1| (1.63)

|ψi = |00i + |11i (1.64)

9 1 1 1 hψ|(σz ⊗ I)√ ψi = hψ|√ (σz|0i ⊗ I|0i + σz|1i ⊗ I|1i) = (1 − 1) = 0 (1.65) 2 2 2

Next we analyze for the product state

|ψi = |0i ⊗ |1i (1.66)

It is now shown that

hψ|(σz ⊗ I)|ψi = 1 (1.67)

hψ|(σz ⊗ I)|ψi = h0| ⊗ h1|(σz ⊗ I)(|0i ⊗ |1i) = h0| ⊗ h1|(σz|0i ⊗ |1i) (1.68)

hψ|(σz ⊗ I)|ψi = h0|σz|0ih1|1i = 1 (1.69)

Now the other way round (Identity operates on ﬁrst element and σz on the second element)

hψ|(I ⊗ σz)|ψi = −1 (1.70)

hψ|(I ⊗ σz)|ψi = h0| ⊗ h1|(I ⊗ σz)(|0i ⊗ |1i) = h0| ⊗ h1|(|0i ⊗ σz|1i) (1.71)

hψ|(I ⊗ σz)|ψi = h0|0ih1|σz|1i = −1 (1.72)

Access restricted to one system leads to ambiguity.

1 |ψi = √ (|00i + |11i) (1.73) 2

1 |φi = √ (|01i − |10i) (1.74) 2

1 Tr (|ψihψ|) = Tr (|00hi00| + |11ih00| + |11ih00| + |11ih11|) (1.75) 2 2 2

10 1 Tr (|ψihψ|) = Tr (|0ih0| ⊗ |0ih0| + |1ih0| ⊗ |1ih0| + |0ih1| ⊗ |0ih1| + |1ih1| ⊗ |1ih1|) 2 2 2 (1.76)

  1 0 1 1   Tr2(|ψihψ|) = (|0ih0| + |1ih1|) =   (1.77) 2 2   0 1

1 φi = √ (|01i − |10i) (1.78) 2

1 Tr (|φihφ|) = Tr (|01ih01| − |01ih10| − |10ih01| + |10ih10|) (1.79) 2 2 2

1 Tr (|φihφ|) = Tr (|0ih0| ⊗ |1ih1| − |0ih1| ⊗ |1ih0| − |1ih0| ⊗ |0ih| + |1ih1| ⊗ |0ih0|) 2 2 2 (1.80)

  1 0 1 1   Tr2(|φihφ|) = (|0ih0| + |1ih1|) =   (1.81) 2 2   0 1

One can see that the partial traces are equal. That means that the marginal distribu- tions are equal. This results in the fact that just by looking at the ﬁrst system one loses information (which is logical but mentioned here).

Example: We look at the operator given by

A = |ψihψ| (1.82)

hψ|A|ψi = 1, hφ|A|φi = 0 (1.83)

We see that the partial traces of |ψi and |φi are identical, but A ∈ B(H1 ⊗ H2) has distinct expectation values on these states. Therefore, A is not local.

11 Schmidt decomposition For any pure state ψ there exists a decomposition

r p |ψi = ∑ λi|eii ⊗ | fii (1.84) i=1 where λi > 0 for all i. These are called the Schmidt eigenvalues.

Proof: Let |ψi = ∑i,j αijvi ⊗ wj. This corresponds to the matrix A = [α]ij. Using the ∗ r ∗ singular value decomposition we obtain A = U1ΣV . Therefore A = ∑i=1 βiuivi .

∗ ∗ ∑(UΣV )ijvi ⊗ wj = ∑ UikΣkl(V )ljvi ⊗ wj (1.85) i,j i,j,k,l 2 ∗ = ∑ Uikσk (V )kjvi ⊗ wj (1.86) i,j,k,l 2 ∗ 2 ∗ = ∑ σk (V )kjvi ⊗ wj = ∑ σk (∑ Uikvi) ⊗ (∑(V )kjwj) k k i j (1.87) 2 = ∑ σk (ek ⊗ fk) (1.88) k

∗ The ek and the fk are ONBs because U and V are orthogonal matrices. Then

r p |ψi = |ψi = ∑ λi|eii ⊗ | fii (1.89) i=1 √ 2 Identify now σi with λi.

2 Consequence: ||ψ|| = 1 iff ∑i λi = 1 and q tr2|ψihψ| = tr2 ∑ λjλk|ejihek| ⊗ | fjih fk| (1.90) j,k

= ∑ λj|ejihej| (1.91) j

It follows that spec(tr2(|ψihψ|)) = {λ1,..., λr} and similarly for tr1.