Genady

Continuing exercise 4.21 above: a (continued). This property is generally true because product of projectors is not generally a projector: (P2P1)2 = P2P1P2P1 ≠ P2P1, generally. b. If S2 is a subspace of S1 then P2P1 = P2.

a. Initial state |0〉. Measure in {|+〉, |-〉}. Let's say, the resulting state is |+〉. Now measure in {|0〉, |1〉}. The resulting state can be |1〉. This effect cannot be achieved by a single measurement. Assume such measurement exists: |1〉 = (a|0〉 + b|1〉) (a*〈0| + b*〈1|) |0〉 = |a|2|0〉 + a*b|1〉. Thus, |a|2 = 0, a* = 0, |1〉 = 0.

The reference: a. Projector onto the subspace of agreeing measurements is |v1v2〉〈v1v2|+|v1⊥v2⊥〉〈v1⊥v2⊥|. |v1v2〉 = (cos θ1 |0〉 + sin θ1 |1〉)⊗(cos θ2 |0〉 + sin θ2 |1〉) When it acts on |ψ〉 only the terms |00〉 and |11〉 survive, i.e., 〈v1v2|ψ〉 = 1/√2 (cos θ1 cos θ2 + sin θ1 sin θ2) = 1/√2 cos (θ1 - θ2) Similarly, 〈v1⊥v2⊥|ψ〉 = 1/√2 (cos θ1 cos θ2 + sin θ1 sin θ2) = 1/√2 cos (θ1 - θ2). Thus, the projection results in the state 1/c (1/√2 cos (θ1 - θ2) |v1v2〉 + 1/√2 cos (θ1 - θ2) |v1⊥v2⊥〉) with the probability c2 = cos2 (θ1 - θ2). b. θ1 = θ2 c. θ1 - θ2= π/2 d. θ1 - θ2= π/4 e. For these pairs, the θ1 - θ2 is either 600 or 1200 with cos2 (θ1 - θ2) = 1/4.

The reference: |v〉〈v|-|v⊥〉〈v⊥| = cos2θ|0〉〈0| + sin2θ|1〉〈1| + cosθsinθ|0〉〈1| + cosθsinθ|1〉〈0| - (sin2θ|0〉〈0| + cos2θ|1〉〈1| - cosθsinθ|0〉〈1| - cosθsinθ|1〉〈0|) = cos2θ|0〉〈0| - cos2θ|1〉〈1| + sin2θ|0〉〈1| + sin2θ|1〉〈0| cos2θ sin2θ sin2θ -cos2θ

Let's assume that the two qubits are unentangled. Then their state can be factorized, v⊗w. If the first qubit is measured by projecting its state onto the subspace spanned by v, its result will be certain contrary to the assumption that there is no such measurement. Thus, they are entangled.

a. An unentangled state can be factorized by single-qubit states as v⊗w. A single-qubit measurement can be represented by an observable Q⊗I. The measurement with Q would change v into v'. The measurement with I would leave w unchanged. The result would be v'⊗w, which is unentangled. b. Yes, e.g., a measurement with Bell states would produce an entangled state. c. Yes, e.g., such a measuring of any Bell state.

They are asking for an observable that embodies this measurement. That is the Hermitian operator built out of the combination of the projectors. The other numbers are just decimal representations of the binary sequences, i.e. 1=001, 2=010, 3=011, ... The observable |0><0|+|7><7|-(|1><1|+|2><2|+...+|6><6|) has an eigenspace with eigenvalue 1 of states with equal bit values, and another eigenspace with eigenvalue -1 of states with unequal bit values.

(Finishing exercise 4.16 above) e. Let's make an upper triangular matrix T=UAU-1. If A is Hermitian, then T† = (U-1)†A†U† = UAU-1 = T Thus, T is upper triangular and T† is upper triangular, i.e., T is diagonal. It has n orthonormal eigenvectors (1,0,...0), ... (0,...0,1) with the same eigenvalues as A. Then the columns of U-1 are eigenvectors of A. They are orthogonal. Thus, the direct sum of their corresponding eigenspaces gives the whole space.

a. Rows of U† are conjugate transposed columns of U. Multiplying them by other columns results in 0, thus they are orthogonal. Multiplying them by the same column results in 1, thus they are normalized. b. (UOU-1)† = (U-1)†O†U†=UOU-1 c. The eigenvalue equation, (A-λI)v=0, constitutes n equations with n+1 unknowns. Hence it has at least one nontrivial solution. d. Use induction. Any 1x1 matrix is upper triangular. Assume it's true for any nxn matrix and take A to be a (n+1)x(n+1) matrix. Take the eigenvector v of A and choose any n orthonormal vectors orthogonal to v. Make matrix B, which has (λ,0,...0)T as its first column and the other chosen orthonormal vectors as its other columns. B is unitary. Make matrix C = B-1AB. C(1,0,...0)T = B-1Av = B-1λv= λB-1v = (λ,0,...0)T. Thus, (λ,0,...0)T is the C matrix' first column. The nxn matrix to the lower right of this column can be made upper triangular by the induction assumption. This completes the proof.

For the reference: a. B = (|0〉〈0|+π|1〉〈1|)⊗(|0〉〈0|+|1〉〈1|) b. Subspace decomposition of Q is V=S0⊕S1 and subspace decomposition of I is V=S. Subspace decomposition of Q⊗I is V⊗V=S0⊗S⊕S1⊕S. c. Subspace decomposition of Q is V=S0⊕S1 and subspace decomposition of I is V=S. Subspace decomposition of I⊗Q is V⊗V=S⊗S0⊕S⊕S1. It measures the second qubit.

For the reference: a. |01〉 = 1/√2 (|Φ+〉-|Φ-〉); 1/2 each |10〉 = 1/√2 (|Ψ+〉+|Ψ-〉); 1/2 each |11〉 = 1/√2 (|Ψ+〉-|Ψ-〉); 1/2 each b. = 1/√2 (a00(|Φ+〉+|Φ-〉)+a01(|Φ+〉-|Φ-〉)+a10(|Ψ+〉+|Ψ-〉)+a11(|Ψ+〉-|Ψ-〉)) = 1/√2 ((a00+a01)|Φ+〉+(a00-a01)|Φ-〉+(a10+a11)|Ψ+〉+(a10-a11)|Ψ-〉) Outcomes: |Φ+〉, |a00+a01|2/2; |Φ-〉, |a00-a01|2/2; |Ψ+〉, |a10+a11|2/2; |Ψ-〉, |a10-a11|2/2.

a. O = ∑i λiPi . For some i=j, Pj|ψ〉 = |φ〉. When the same measurement applied to |φ〉, Pj|φ〉=|φ〉, while for all i≠j, Pi|φ〉=0. b. Operators P but not O act on state vectors.

I am bad in this in any language. Perhaps, has something to do with my APD.

P1+P2-P3-P4

(|1><1|+|2><2|+|4><4|)+2(|3><3|+|5><5|+|6><6|)+3|7><7| The number of 1-bits is the eigenvalue.

|0><0|+|3><3|+|5><5|+|6><6|-(the other four: 1, 2, 4, 7)

Yes, I know how simple and inexpensive is notarizing in the States and how common notaries public are there (I lived in NYC 15 years.) Notarizing in the US Consulate is equally simple. Also, the Consul comes to Bonaire couple times a year to provide services to US citizens. I will avoid the Dutch notary and hope to never need them. I'm already avoiding them by NOT transferring my late father's real estate to my name. I asked what's involved and that was enough. The only limitation of not transferring is that I cannot sell it. But I don't plan to do so anyway. Otherwise, I can use it as I wish. BTW, I took the ferries from Cape Cod to Nantucket and to Martha's Vineyard several times. My wife used to live in Boston before eventually moving to NYC and marrying me.

|0><0|+|7><7|-(|1><1|+|2><2|+...+|6><6|)

On this exact point: (A|x〉)+ = (A|x〉)*T = (A*|x〉*)T = |x〉*TA*T= 〈x|A+

Yes, I think we do the same thing in different notations. I use the Dirac's notation because it seems to be the standard in this book and maybe in QC. The nested parenthesis might be easier to read if I give more spaces, e.g., (〈w|+〈w⊥|) (P (|v〉+|v⊥〉) ) = ( (〈w|+〈w⊥|) P ) (|v〉+|v⊥〉). Perhaps, I should've made the vectors |q> = |v〉+|v⊥〉 and |r> = |w〉+|w⊥〉, and the last line would be then <r| (P|q>) = (<r|P) |q>

They can and they do.

No, there is no ferry. I heard that there was one many years ago, but the sea between the islands is very rough, it took too long and was very unpleasant. It was discontinued eventually. OTOH, there are about dozen flights each way daily. It's like taking a bus. Many people go back and forth for work regularly. Takes about an hour one way*. Not terribly expensive. *The actual flight time is about 25 minutes.

Yes, I am. Thank you. Thank you for checking. The book does not give answers of the exercises, so I don't have a way to check myself.

Let's consider a direct sum decomposition of the space V = S ⊕ S⊥, where S and S⊥ are orthogonal, and let's consider a projection operator P of V onto S. Take two vectors (|v〉+|v⊥〉) and (|w〉+|w⊥〉) in V, where |v〉 and |w〉 are in S and |v⊥〉 and |w⊥〉 are in S⊥. By definition, P(|v〉+|v⊥〉)=|v〉. (〈w|+〈w⊥|)(P(|v〉+|v⊥〉)) = (〈w|+〈w⊥|)(|v〉) = 〈w|v〉. OTOH, by definition, P(|w〉+|w⊥〉)=|w〉. ((〈w|+〈w⊥|)P)(|v〉+|v⊥〉) = (〈w|)(|v〉+|v⊥〉) = 〈w|v〉. Thus, (〈w|+〈w⊥|)(P(|v〉+|v⊥〉)) = ((〈w|+〈w⊥|)P)(|v〉+|v⊥〉), i.e., P is its own adjoint.

Right. And I rather do this than pay to the pompous Dutch notary.