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Useful Tricks

Abstract

Some useful tricks that could prove useful in proofs

Contents:

Expectation Values Squared

Let AA be some operator and ∣ψ⟩\ket{\psi} be some state.

⟨ψ∣A∣ψ⟩2=⟨ψ∣⟨ψ∣A⊗A∣ψ⟩∣ψ⟩\bra{\psi} A \ket{\psi}^{2} = \bra{\psi}\bra{\psi} A \otimes A \ket{\psi}\ket{\psi}
(1)

Moving Operators on Entangled States

Let AA be some operator and ∣Φ+⟩\ket{\Phi^{+}} be a bipartite maximally entangled state.

(A⊗I)∣Φ+⟩=(I⊗At)∣Φ+⟩,(A \otimes \mathbb{I}) \ket{\Phi^{+}} = (\mathbb{I} \otimes A^{t}) \ket{\Phi^{+}},
(2)

where (⋅)t(\cdot)^{t} is the transpose operation.

Proof

Let AA be some operator and ∣Φ+⟩=1d∑i∣ii⟩\ket{\Phi^{+}} = \frac{1}{\sqrt{d}} \sum_{i} \ket{ii} be the bipartite maximally entangled state.

Consider the elements of the vector (A⊗I)(∣Φ+⟩)(A \otimes \mathbb{I}) \big( \ket{\Phi^{+}} \big) ,

⟨kl∣(A⊗I)(∣Φ+⟩)=1d∑i⟨kl∣(A⊗I)∣ii⟩,=1d∑i⟨k∣A∣i⟩⟨l∣∣i⟩,=1d∑i⟨k∣A∣i⟩δli,=1d⟨k∣A∣l⟩.\begin{align*} \bra{kl} (A \otimes \mathbb{I}) \big( \ket{\Phi^{+}} \big) &= \frac{1}{\sqrt{d}} \sum_{i} \bra{kl} (A \otimes \mathbb{I}) \ket{ii}, \\ &= \frac{1}{\sqrt{d}} \sum_{i} \bra{k} A \ket{i} \bra{l}\ket{i}, \\ &= \frac{1}{\sqrt{d}} \sum_{i} \bra{k} A \ket{i} \delta_{li}, \\ &= \frac{1}{\sqrt{d}} \bra{k} A \ket{l}. \end{align*}
(3)

Consider the elements of the vector (I⊗At)(∣Φ+⟩)(\mathbb{I} \otimes A^{t}) \big( \ket{\Phi^{+}} \big) ,

⟨kl∣(I⊗At)(∣Φ+⟩)=1d∑i⟨kl∣(I⊗At)∣ii⟩,=1d∑i⟨k∣∣i⟩⟨l∣At∣i⟩,=1d∑i⟨l∣At∣i⟩δki,=1d⟨l∣At∣k⟩,=1d⟨k∣A∣l⟩.\begin{align*} \bra{kl} (\mathbb{I} \otimes A^{t}) \big( \ket{\Phi^{+}} \big) &= \frac{1}{\sqrt{d}} \sum_{i} \bra{kl} (\mathbb{I} \otimes A^{t}) \ket{ii}, \\ &= \frac{1}{\sqrt{d}} \sum_{i} \bra{k}\ket{i} \bra{l} A^{t} \ket{i}, \\ &= \frac{1}{\sqrt{d}} \sum_{i} \bra{l} A^{t} \ket{i} \delta_{ki},\\ &= \frac{1}{\sqrt{d}} \bra{l} A^{t} \ket{k},\\ &= \frac{1}{\sqrt{d}} \bra{k} A \ket{l}. \end{align*}
(4)

If the elements of these two vector are the same, they must be the same vector. This completes the proof.

Alternative Bipartition of Entangled States

Let ∣Φ+⟩AB ∈ HA⊗HB\ket{\Phi^{+}}_{AB}~\in~\mathcal{H}_{A} \otimes \mathcal{H}_{B} be the maximally entangled stated between a space AA and BB. Let ∣Φ+⟩CD ∈ HC⊗HD\ket{\Phi^{+}}_{CD}~\in~\mathcal{H}_{C} \otimes \mathcal{H}_{D} be the maximally entangled stated between a space CC and DD

∣Φ+⟩AB⊗∣Φ+⟩CD=∣Φ+⟩(AC),(BD)\ket{\Phi^{+}}_{AB} \otimes \ket{\Phi^{+}}_{CD} = \ket{\Phi^{+}}_{(AC),(BD)}
(5)

where ∣Φ+⟩(AC),(BD)\ket{\Phi^{+}}_{(AC),(BD)} is the maximally entangled state between (AC)(AC) and (BD)(BD).

This can be generalised to more tensor products of the maximally entangled state, e.g

∣Φ+⟩AB⊗∣Φ+⟩CD⊗∣Φ+⟩EF=∣Φ+⟩(ACE),(BDF)\ket{\Phi^{+}}_{AB} \otimes \ket{\Phi^{+}}_{CD} \otimes \ket{\Phi^{+}}_{EF} = \ket{\Phi^{+}}_{(ACE),(BDF)}
(6)
Proof
∣Φ+⟩AB=12(∣00⟩AB+∣11⟩AB)∣Φ+⟩CD=12(∣00⟩CD+∣11⟩CD)\begin{align*} \ket{\Phi^{+}}_{AB} &= \frac{1}{\sqrt{2}} ( \ket{00}_{AB} + \ket{11}_{AB}) \\ \ket{\Phi^{+}}_{CD} &= \frac{1}{\sqrt{2}} ( \ket{00}_{CD} + \ket{11}_{CD}) \\ \end{align*}
(7)
∣Φ+⟩AB⊗∣Φ+⟩CD=12(∣00⟩AB+∣11⟩AB)⊗12(∣00⟩CD+∣11⟩CD)=12(∣0000⟩ABCD+∣0011⟩ABCD+∣1100⟩ABCD+∣1111⟩ABCD)\begin{align*} &\ket{\Phi^{+}}_{AB} \otimes \ket{\Phi^{+}}_{CD} \\ &= \frac{1}{\sqrt{2}} ( \ket{00}_{AB} + \ket{11}_{AB}) \otimes \frac{1}{\sqrt{2}} ( \ket{00}_{CD} + \ket{11}_{CD}) \\ &= \frac{1}{2} (\ket{0000}_{ABCD} + \ket{0011}_{ABCD} + \ket{1100}_{ABCD} + \ket{1111}_{ABCD}) \end{align*}
(8)

The labels on the BB and CC particles can then be swapped.

=12(∣0000⟩ACBD+∣0101⟩ACBD+∣1010⟩ACBD+∣1111⟩ACBD)\begin{align*} &= \frac{1}{2} (\ket{0000}_{ACBD} + \ket{0101}_{ACBD} + \ket{1010}_{ACBD} + \ket{1111}_{ACBD}) \\ \end{align*}
(9)

Rewriting the binary with

00→0,01→1,10→2,11→3,\begin{align*} 00 &\rightarrow 0, \\ 01 &\rightarrow 1, \\ 10 &\rightarrow 2, \\ 11 &\rightarrow 3, \\ \end{align*}
(10)

and grouping the spaces together gives

=12(∣00⟩(AC)(BD)+∣11⟩(AC)(BD)+∣22⟩(AC)(BD)+∣33⟩(AC)(BD))=∣Φ+⟩(AB)(CD),\begin{align*} &= \frac{1}{2} (\ket{00}_{(AC)(BD)} + \ket{11}_{(AC)(BD)} + \ket{22}_{(AC)(BD)} + \ket{33}_{(AC)(BD)}) \\ &= \ket{\Phi^{+}}_{(AB)(CD)}, \end{align*}
(11)

completing the proof.

Generating a Maximally Entangled Basis

Given a maximally entangled bipartite state, ∣Φ00+⟩\ket{\Phi^{+}_{00}}, one can generate a complete maximally entangled basis using the Heisenberg-Weyl operators.

In a space of local dimenions dd, the Heisenberg-Weyl operators are defined as

Wa,b=UaVb=∑k=0d−1Ωbk∣k+a⟩⟨k∣,W_{a,b} = U^{a}V^{b} = \sum_{k=0}^{d-1} \Omega^{bk} \ket{k+a}\bra{k},
(12)

where a,b ∈ {0,1,…,d−1}a,b~\in~\{0,1,\ldots,d-1 \} and

U=∑k=0d−1∣k+1⟩⟨k∣,  V=∑k=0d−1Ωk∣k⟩⟨k∣,  Ω=e2πid.U = \sum_{k=0}^{d-1} \ket{k+1}\bra{k}, ~ ~ V = \sum^{d-1}_{k=0} \Omega^{k} \ket{k}\bra{k}, ~ ~ \Omega = e^{\frac{2 \pi i}{d}}.
(13)

UU and VV are XX and ZZ Pauli operators generalized to higher dimension.

The set of states,

{∣Φab+⟩:a,b ∈ {0,1,…,d−1}},\{ \ket{\Phi_{ab}^{+}} : a,b~\in~\{0,1,\ldots,d-1 \} \},
(14)

where ∣Φab+⟩=(I⊗Wa,b)(∣Φ00+⟩)\ket{\Phi_{ab}^{+}} = (\mathbb{I} \otimes W_{a,b}) \big( \ket{\Phi^{+}_{00}} \big) then forms a complete maximally entangled basis.

Properties of Heisenberg-Weyl operators
(Wα,β)t=Ω−αβW−α,β,(Wα,β)†=ΩαβW−α,−β,Wα,βWγ,δ=ΩβγWα+γ,β+δ=Ωβγ−αδWγ,δWα,β.tr[Wa,b]={nif (a,b)=(0,0)0if otherwise\begin{split} & (W_{\alpha, \beta}) ^ {t} = \Omega^{-\alpha \beta}W_{-\alpha, \beta}, \\ & (W_{\alpha, \beta}) ^ {\dagger} = \Omega^{\alpha \beta}W_{-\alpha, -\beta}, \\ & W_{\alpha, \beta}W_{\gamma, \delta} = \Omega^{\beta\gamma}W_{\alpha + \gamma, \beta + \delta} = \Omega^{\beta\gamma - \alpha\delta}W_{\gamma, \delta}W_{\alpha, \beta}. \\ & \textrm{tr}\big[ W_{a,b} \big] = \begin{cases} n & \mbox{if } (a,b) = (0,0) \\ 0 & \mbox{if } \textrm{otherwise} \end{cases} \end{split}
(15)

For qubits, the Heisenberg-Weyl operators are

W0,0=I,  W0,1=σz,  W1,0=σx,  W1,1=−iσyW_{0,0} = \mathbb{I}, ~~ W_{0,1} = \sigma_{z}, ~~ W_{1,0} = \sigma_{x}, ~~ W_{1,1} = -i \sigma_{y}
(16)

Trace of Swap gate on two density operators

Let ρ,σ\rho, \sigma be density operators and SWAP\rm{SWAP} the swap gate.

tr[(ρ⊗σ)SWAP]=tr[ρσ]\textrm{tr} \big[ (\rho \otimes \sigma) \textrm{SWAP} \big] = \textrm{tr} \big[\rho \sigma \big]
(17)

This can be generalised further using the same proof technique as

tr[(ρ1⊗σ2⊗ω3)SWAP12SWAP13]=tr[ρσω],\textrm{tr} \big[ (\rho_1 \otimes \sigma_2 \otimes \omega_3) \textrm{SWAP}_{12} \textrm{SWAP}_{13} \big] = \textrm{tr}\big[\rho \sigma \omega \big],
(18)

where SWAPij\textrm{SWAP}_{ij} is a swap on the iith and jjth system.

Proof

Let

ρ=∑ijpij∣i⟩⟨j∣,σ=∑klqkl∣k⟩⟨l∣,\begin{align*} \rho = \sum_{ij} p_{ij} \ket{i} \bra{j}, \\ \sigma = \sum_{kl} q_{kl} \ket{k} \bra{l}, \end{align*}
(19)

be expansions of ρ \rho and σ \sigma in the computational basis, and note that

SWAP=∑nm∣n⟩⟨m∣⊗∣m⟩⟨n∣.\textrm{SWAP} = \sum_{nm} \ket{n}\bra{m} \otimes \ket{m}\bra{n}.
(20)

First, the left hand side gives

tr[(ρ⊗σ)SWAP]=tr[(∑ijklpijqkl∣i⟩⟨j∣⊗∣k⟩⟨l∣)∑nm∣n⟩⟨m∣⊗∣m⟩⟨n∣]=tr[∑ijkl∑nmpijqkl∣i⟩⟨m∣δjn⊗∣k⟩⟨n∣δml]=tr[∑inkmpinqkm∣i⟩⟨m∣⊗∣k⟩⟨n∣]=∑inkmpinqkm δimδkn=∑ikpikqki.\begin{align*} &\textrm{tr} \big[ (\rho \otimes \sigma) \textrm{SWAP} \big] \\ &= \textrm{tr} \bigg[ \bigg( \sum_{ijkl} p_{ij}q_{kl} \ket{i} \bra{j} \otimes \ket{k} \bra{l} \bigg) \sum_{nm} \ket{n}\bra{m} \otimes \ket{m}\bra{n} \bigg] \\ &= \textrm{tr} \bigg[ \sum_{ijkl} \sum_{nm} p_{ij}q_{kl} \ket{i} \bra{m} \delta_{jn} \otimes \ket{k} \bra{n} \delta_{ml} \bigg] \\ &= \textrm{tr} \bigg[ \sum_{inkm} p_{in}q_{km} \ket{i} \bra{m} \otimes \ket{k} \bra{n} \bigg] \\ &= \sum_{inkm} p_{in}q_{km} ~ \delta_{im} \delta_{kn} \\ &= \sum_{ik} p_{ik}q_{ki}. \end{align*}
(21)

The right hand side is then

tr[ρσ]=tr[(∑ijpij∣i⟩⟨j∣)(∑klqkl∣k⟩⟨l∣)]=tr[∑ijklpijqkl∣i⟩⟨l∣δjk]=tr[∑iklpikqkl∣i⟩⟨l∣]=∑iklpikqklδil=∑ikpikqki,\begin{align*} \textrm{tr} \big[\rho \sigma \big] &= \textrm{tr} \bigg[ \bigg( \sum_{ij} p_{ij} \ket{i} \bra{j} \bigg) \bigg( \sum_{kl} q_{kl} \ket{k} \bra{l} \bigg) \bigg] \\ &= \textrm{tr} \bigg[ \sum_{ijkl} p_{ij}q_{kl} \ket{i} \bra{l} \delta_{jk} \bigg] \\ &= \textrm{tr} \bigg[ \sum_{ikl} p_{ik}q_{kl} \ket{i} \bra{l} \bigg] \\ &= \sum_{ikl} p_{ik}q_{kl} \delta_{il} \\ &= \sum_{ik} p_{ik}q_{ki}, \end{align*}
(22)

completing the proof.

Eigenvalues of general 2×22 \times 2 Operator

The operators {I,X,Y,Z}\big\{ \mathbb{I}, X, Y, Z \big\}, which is the identity plus the Pauli-operators, form an operator basis for the space of 2×22 \times 2 matrices over the complex numbers.

Hence, if A∈M2,2(C)A \in \mathbb{M}_{2,2}(\mathbb{C}) then there exists an α∈R1\alpha \in \mathbb{R}^1 and n∈R3\mathbf{n} \in \mathbb{R}^3 such that

A=αI+n⋅σ=αI+nxX+nyY+nzZ,\begin{align*} A &= \alpha \mathbb{I} + \mathbf{n} \cdot \mathbf{\sigma} \\ &= \alpha \mathbb{I} + n_x X + n_y Y + n_z Z, \end{align*}
(23)

where

n=(nxnynz),   σ=(XYZ),\bm{n} = \begin{pmatrix} n_x \\ n_y \\ n_z \end{pmatrix}, ~ ~ ~ \bm{\sigma} = \begin{pmatrix} X \\ Y \\ Z \end{pmatrix},
(24)

as seen here

The eigenvalues of AA are then λ=α±∣n∣\lambda = \alpha \pm \vert \mathbf{n} \vert.

Proof

The fact that {I,X,Y,Z}\big\{ \mathbb{I}, X, Y, Z \big\} forms an operator basis for the space of 2×22 \times 2 complex matrices can be seen by comparing coefficients of a decomposition in this basis to a general 2×22 \times 2 matrix. This is not proved here, but taken as a given.

The result about the eigenvalues can then be seen by direct calculation. In this basis AA can written as

A=(α+nznx−inynx+inyα−nz).\begin{align*} A = \begin{pmatrix} \alpha + n_z & n_x - in_y \\ n_x + in_y & \alpha - n_z \end{pmatrix}. \end{align*}
(25)

One can then find the eigenvectors, λ\lambda, of AA from

Det(A−λI)=0.\mathrm{Det}\big( A - \lambda \mathbb{I}) = 0.
(26)

This can be easily solved for λ\lambda, giving

λ=α+nx2+ny2+nz2,\lambda = \alpha + \sqrt{n_x^2 + n_y^2 + n_z^2},
(27)

completing the proof.

Operator Inequalities and Supports

Let ΠA,ΠB\Pi_A, \Pi_B be hermitian projectors, such that ΠiΠi=Πi\Pi_i \Pi_i = \Pi_i and Πi†=Πi\Pi_i^\dagger = \Pi_i.

The following conditions holds:

ΠA≤ΠB   ⟺   Supp(A)⊆Supp(B).\Pi_A \leq \Pi_B ~ \iff ~ \mathrm{Supp}(A) \subseteq \mathrm{Supp}(B).
(28)
Proof

Firstly, we prove that ΠA≤ΠB   ⟹   Supp(A)⊆Supp(B)\Pi_A \leq \Pi_B ~ \implies ~ \mathrm{Supp}(A) \subseteq \mathrm{Supp}(B):

We have ΠB−ΠA≥0\Pi_B - \Pi_A \geq 0, meaning that ΠB−ΠA\Pi_B - \Pi_A is positive semi-definite. From here it holds that

⟨x,(ΠB−ΠA)x⟩≥0,\begin{align*} \langle x, (\Pi_B - \Pi_A) x \rangle \geq 0, \end{align*}
(29)

such that

⟨x,ΠBx⟩≥⟨x,ΠAx⟩,\begin{align*} \langle x, \Pi_B x \rangle \geq \langle x, \Pi_A x \rangle, \end{align*}
(30)

where ⟨⋅,⋅⟩\langle \cdot, \cdot \rangle is some inner product, and this holds for all xx. If x∈Ker(B)x \in \mathrm{Ker}(B), such that ⟨x,ΠBx⟩=0\langle x, \Pi_B x \rangle = 0, it follows that ⟨x,ΠAx⟩=0 \langle x, \Pi_A x \rangle = 0, as ⟨x,ΠAx⟩≥0\langle x, \Pi_A x \rangle \geq 0.

Therefore, if x∈Ker(B)x \in \mathrm{Ker}(B) then x∈Ker(A)x \in \mathrm{Ker}(A), and it can be concluded that

Ker(B)⊆Ker(A).\mathrm{Ker}(B) \subseteq \mathrm{Ker}(A).
(31)

As

Supp(A)=(Ker(A))⊥,\mathrm{Supp}(A) = \big( \mathrm{Ker}(A) \big)^{\perp},
(32)

meaning the support is the orthogonal complement of the kernal, it can be concluded that

Supp(A)⊆Supp(B).\mathrm{Supp}(A) \subseteq \mathrm{Supp}(B).
(33)

Now, we prove that Supp(A)⊆Supp(B)   ⟹   ΠA≤ΠB\mathrm{Supp}(A) \subseteq \mathrm{Supp}(B) ~ \implies ~ \Pi_A \leq \Pi_B:

If Supp(A)⊆Supp(B)\mathrm{Supp}(A) \subseteq \mathrm{Supp}(B) there exists some subspace MM such that

Supp(A)⊕M=Supp(B).\mathrm{Supp}(A) \oplus M = \mathrm{Supp}(B).
(34)

This implies that ΠA+ΠM=ΠB\Pi_{A} + \Pi_M = \Pi_B, where ΠM\Pi_M is the projector onto the subspace MM. This is defined as the projector onto ΠB−ΠA\Pi_B - \Pi_A, which, by assumption, is postive semi-definite, ΠM≥0\Pi_M \geq 0. Hence, ΠA≤ΠB\Pi_A \leq \Pi_B, completing the proof.

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